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Calculus Volume 2










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Table of ContentsPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1Chapter 1: Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.1 Approximating Areas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.2 The Definite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271.3 The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . 471.4 Integration Formulas and the Net Change Theorem . . . . . . . . . . . . . . . . . . . . . . 641.5 Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 821.6 Integrals Involving Exponential and Logarithmic Functions . . . . . . . . . . . . . . . . . . 931.7 Integrals Resulting in Inverse Trigonometric Functions . . . . . . . . . . . . . . . . . . . . 106Chapter 2: Applications of Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1212.1 Areas between Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1222.2 Determining Volumes by Slicing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1342.3 Volumes of Revolution: Cylindrical Shells . . . . . . . . . . . . . . . . . . . . . . . . . . . 1542.4 Arc Length of a Curve and Surface Area . . . . . . . . . . . . . . . . . . . . . . . . . . . 1692.5 Physical Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1832.6 Moments and Centers of Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2012.7 Integrals, Exponential Functions, and Logarithms . . . . . . . . . . . . . . . . . . . . . . . 2192.8 Exponential Growth and Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2322.9 Calculus of the Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243Chapter 3: Techniques of Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2613.1 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2623.2 Trigonometric Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2733.3 Trigonometric Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2853.4 Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2983.5 Other Strategies for Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3113.6 Numerical Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3163.7 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330Chapter 4: Introduction to Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 3514.1 Basics of Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3524.2 Direction Fields and Numerical Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3654.3 Separable Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3814.4 The Logistic Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3934.5 First-order Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 408Chapter 5: Sequences and Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4275.1 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4275.2 Infinite Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4505.3 The Divergence and Integral Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4715.4 Comparison Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4855.5 Alternating Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4965.6 Ratio and Root Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 509Chapter 6: Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5316.1 Power Series and Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5326.2 Properties of Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5446.3 Taylor and Maclaurin Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5616.4 Working with Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 581Chapter 7: Parametric Equations and Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . 6057.1 Parametric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6067.2 Calculus of Parametric Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6257.3 Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6427.4 Area and Arc Length in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . 6627.5 Conic Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 671Appendix A: Table of Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 699Appendix B: Table of Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 705Appendix C: Review of Pre-Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 707Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 819




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PREFACE
Welcome to Calculus Volume 2, an OpenStax resource. This textbook was written to increase student access to high-quality learning materials, maintaining highest standards of academic rigor at little to no cost.
About OpenStax
OpenStax is a nonprofit based at Rice University, and it’s our mission to improve student access to education. Our first openly licensed college textbook was published in 2012, and our library has since scaled to over 20 books for college and AP courses used by hundreds of thousands of students. Our adaptive learning technology, designed to improve learning outcomes through personalized educational paths, is being piloted in college courses throughout the country. Through our partnerships with philanthropic foundations and our alliance with other educational resource organizations, OpenStax is breaking down the most common barriers to learning and empowering students and instructors to succeed.
About OpenStax's ResourcesCustomization
Calculus Volume 2 is licensed under a Creative Commons Attribution Non-Commercial ShareAlike (CC BY-NC-SA) license, which means that you can distribute, remix, and build upon the content, as long as you provide attribution to OpenStax and its content contributors.
Because our books are openly licensed, you are free to use the entire book or pick and choose the sections that are most relevant to the needs of your course. Feel free to remix the content by assigning your students certain chapters and sections in your syllabus, in the order that you prefer. You can even provide a direct link in your syllabus to the sections in the web view of your book.
Faculty also have the option of creating a customized version of their OpenStax book through the aerSelect platform. The custom version can be made available to students in low-cost print or digital form through their campus bookstore. Visit your book page on openstax.org for a link to your book on aerSelect.
Errata
All OpenStax textbooks undergo a rigorous review process. However, like any professional-grade textbook, errors sometimes occur. Since our books are web based, we can make updates periodically when deemed pedagogically necessary. If you have a correction to suggest, submit it through the link on your book page on openstax.org. Subject matter experts review all errata suggestions. OpenStax is committed to remaining transparent about all updates, so you will also find a list of past errata changes on your book page on openstax.org.
Format
You can access this textbook for free in web view or PDF through openstax.org, and for a low cost in print.
About Calculus Volume 2
Calculus is designed for the typical two- or three-semester general calculus course, incorporating innovative features to enhance student learning. The book guides students through the core concepts of calculus and helps them understand how those concepts apply to their lives and the world around them. Due to the comprehensive nature of the material, we are offering the book in three volumes for flexibility and efficiency. Volume 2 covers integration, differential equations, sequences and series, and parametric equations and polar coordinates.
Coverage and Scope
Our Calculus Volume 2 textbook adheres to the scope and sequence of most general calculus courses nationwide. We have worked to make calculus interesting and accessible to students while maintaining the mathematical rigor inherent in the subject. With this objective in mind, the content of the three volumes of Calculus have been developed and arranged to provide a logical progression from fundamental to more advanced concepts, building upon what students have already learned and emphasizing connections between topics and between theory and applications. The goal of each section is to enable students not just to recognize concepts, but work with them in ways that will be useful in later courses and future careers. The organization and pedagogical features were developed and vetted with feedback from mathematics educators dedicated to the project.
Volume 1


Preface 1




Chapter 1: Functions and Graphs
Chapter 2: Limits
Chapter 3: Derivatives
Chapter 4: Applications of Derivatives
Chapter 5: Integration
Chapter 6: Applications of Integration


Volume 2Chapter 1: Integration
Chapter 2: Applications of Integration
Chapter 3: Techniques of Integration
Chapter 4: Introduction to Differential Equations
Chapter 5: Sequences and Series
Chapter 6: Power Series
Chapter 7: Parametric Equations and Polar Coordinates


Volume 3Chapter 1: Parametric Equations and Polar Coordinates
Chapter 2: Vectors in Space
Chapter 3: Vector-Valued Functions
Chapter 4: Differentiation of Functions of Several Variables
Chapter 5: Multiple Integration
Chapter 6: Vector Calculus
Chapter 7: Second-Order Differential Equations


Pedagogical Foundation
Throughout Calculus Volume 2 you will find examples and exercises that present classical ideas and techniques as well asmodern applications and methods. Derivations and explanations are based on years of classroom experience on the partof long-time calculus professors, striving for a balance of clarity and rigor that has proven successful with their students.Motivational applications cover important topics in probability, biology, ecology, business, and economics, as well as areasof physics, chemistry, engineering, and computer science. Student Projects in each chapter give students opportunities toexplore interesting sidelights in pure and applied mathematics, from showing that the number e is irrational, to calculatingthe center of mass of the Grand Canyon Skywalk or the terminal speed of a skydiver. Chapter Opening Applicationspose problems that are solved later in the chapter, using the ideas covered in that chapter. Problems include the hydraulicforce against the Hoover Dam, and the comparison of the relative intensity of two earthquakes. Definitions, Rules, andTheorems are highlighted throughout the text, including over 60 Proofs of theorems.
Assessments That Reinforce Key Concepts
In-chapter Exampleswalk students through problems by posing a question, stepping out a solution, and then asking studentsto practice the skill with a “Check Your Learning” component. The book also includes assessments at the end of eachchapter so students can apply what they’ve learned through practice problems. Many exercises are marked with a [T] toindicate they are suitable for solution by technology, including calculators or Computer Algebra Systems (CAS). Answersfor selected exercises are available in the Answer Key at the back of the book. The book also includes assessments at theend of each chapter so students can apply what they’ve learned through practice problems.
Early or Late Transcendentals
Calculus Volume 2 is designed to accommodate both Early and Late Transcendental approaches to calculus. Exponentialand logarithmic functions are presented in Chapter 2. Integration of these functions is covered in Chapters 1 for instructorswho want to include them with other types of functions. These discussions, however, are in separate sections that can beskipped for instructors who prefer to wait until the integral definitions are given before teaching the calculus derivations ofexponentials and logarithms.
Comprehensive Art Program
Our art program is designed to enhance students’ understanding of concepts through clear and effective illustrations,


2 Preface


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diagrams, and photographs.


Additional ResourcesStudent and Instructor Resources
We’ve compiled additional resources for both students and instructors, including Getting Started Guides, an instructorsolution manual, and PowerPoint slides. Instructor resources require a verified instructor account, which can be requestedon your openstax.org log-in. Take advantage of these resources to supplement your OpenStax book.
Partner Resources
OpenStax Partners are our allies in the mission to make high-quality learning materials affordable and accessible to studentsand instructors everywhere. Their tools integrate seamlessly with our OpenStax titles at a low cost. To access the partnerresources for your text, visit your book page on openstax.org.
About The AuthorsSenior Contributing Authors
Gilbert Strang, Massachusetts Institute of TechnologyDr. Strang received his PhD from UCLA in 1959 and has been teaching mathematics at MIT ever since. His Calculus onlinetextbook is one of eleven that he has published and is the basis from which our final product has been derived and updatedfor today’s student. Strang is a decorated mathematician and past Rhodes Scholar at Oxford University.
Edwin “Jed” Herman, University of Wisconsin-Stevens PointDr. Herman earned a BS in Mathematics from Harvey Mudd College in 1985, an MA in Mathematics from UCLA in1987, and a PhD in Mathematics from the University of Oregon in 1997. He is currently a Professor at the University ofWisconsin-Stevens Point. He has more than 20 years of experience teaching college mathematics, is a student researchmentor, is experienced in course development/design, and is also an avid board game designer and player.


Preface 3




Contributing Authors
Catherine Abbott, Keuka CollegeNicoleta Virginia Bila, Fayetteville State UniversitySheri J. Boyd, Rollins CollegeJoyati Debnath, Winona State UniversityValeree Falduto, Palm Beach State CollegeJoseph Lakey, New Mexico State UniversityJulie Levandosky, Framingham State UniversityDavid McCune, William Jewell CollegeMichelle Merriweather, Bronxville High SchoolKirsten R. Messer, Colorado State University - PuebloAlfred K. Mulzet, Florida State College at JacksonvilleWilliam Radulovich (retired), Florida State College at JacksonvilleErica M. Rutter, Arizona State UniversityDavid Smith, University of the Virgin IslandsElaine A. Terry, Saint Joseph’s UniversityDavid Torain, Hampton University
Reviewers
Marwan A. Abu-Sawwa, Florida State College at JacksonvilleKenneth J. Bernard, Virginia State UniversityJohn Beyers, University of MarylandCharles Buehrle, Franklin & Marshall CollegeMatthew Cathey, Wofford CollegeMichael Cohen, Hofstra UniversityWilliam DeSalazar, Broward County School SystemMurray Eisenberg, University of Massachusetts AmherstKristyanna Erickson, Cecil CollegeTiernan Fogarty, Oregon Institute of TechnologyDavid French, Tidewater Community CollegeMarilyn Gloyer, Virginia Commonwealth UniversityShawna Haider, Salt Lake Community CollegeLance Hemlow, Raritan Valley Community CollegeJerry Jared, The Blue Ridge SchoolPeter Jipsen, Chapman UniversityDavid Johnson, Lehigh UniversityM.R. Khadivi, Jackson State UniversityRobert J. Krueger, Concordia UniversityTor A. Kwembe, Jackson State UniversityJean-Marie Magnier, Springfield Technical Community CollegeCheryl Chute Miller, SUNY PotsdamBagisa Mukherjee, Penn State University, Worthington Scranton CampusKasso Okoudjou, University of Maryland College ParkPeter Olszewski, Penn State Erie, The Behrend CollegeSteven Purtee, Valencia CollegeAlice Ramos, Bethel CollegeDoug Shaw, University of Northern IowaHussain Elalaoui-Talibi, Tuskegee UniversityJeffrey Taub, Maine Maritime AcademyWilliam Thistleton, SUNY Polytechnic InstituteA. David Trubatch, Montclair State UniversityCarmen Wright, Jackson State UniversityZhenbu Zhang, Jackson State University


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1 | INTEGRATION


Figure 1.1 Iceboating is a popular winter sport in parts of the northern United States and Europe. (credit: modification of workby Carter Brown, Flickr)


Chapter Outline
1.1 Approximating Areas
1.2 The Definite Integral
1.3 The Fundamental Theorem of Calculus
1.4 Integration Formulas and the Net Change Theorem
1.5 Substitution
1.6 Integrals Involving Exponential and Logarithmic Functions
1.7 Integrals Resulting in Inverse Trigonometric Functions


Introduction
Iceboats are a common sight on the lakes of Wisconsin and Minnesota on winter weekends. Iceboats are similar to sailboats,but they are fitted with runners, or “skates,” and are designed to run over the ice, rather than on water. Iceboats can movevery quickly, and many ice boating enthusiasts are drawn to the sport because of the speed. Top iceboat racers can attain


Chapter 1 | Integration 5




speeds up to five times the wind speed. If we know how fast an iceboat is moving, we can use integration to determine howfar it travels. We revisit this question later in the chapter (see Example 1.27).
Determining distance from velocity is just one of many applications of integration. In fact, integrals are used in a widevariety of mechanical and physical applications. In this chapter, we first introduce the theory behind integration and useintegrals to calculate areas. From there, we develop the Fundamental Theorem of Calculus, which relates differentiation andintegration. We then study some basic integration techniques and briefly examine some applications.
1.1 | Approximating Areas


Learning Objectives
1.1.1 Use sigma (summation) notation to calculate sums and powers of integers.
1.1.2 Use the sum of rectangular areas to approximate the area under a curve.
1.1.3 Use Riemann sums to approximate area.


Archimedes was fascinated with calculating the areas of various shapes—in other words, the amount of space enclosed bythe shape. He used a process that has come to be known as the method of exhaustion, which used smaller and smaller shapes,the areas of which could be calculated exactly, to fill an irregular region and thereby obtain closer and closer approximationsto the total area. In this process, an area bounded by curves is filled with rectangles, triangles, and shapes with exact areaformulas. These areas are then summed to approximate the area of the curved region.
In this section, we develop techniques to approximate the area between a curve, defined by a function f (x), and the x-axis
on a closed interval ⎡⎣a, b⎤⎦. Like Archimedes, we first approximate the area under the curve using shapes of known area
(namely, rectangles). By using smaller and smaller rectangles, we get closer and closer approximations to the area. Takinga limit allows us to calculate the exact area under the curve.
Let’s start by introducing some notation to make the calculations easier. We then consider the case when f (x) is continuous
and nonnegative. Later in the chapter, we relax some of these restrictions and develop techniques that apply in more generalcases.
Sigma (Summation) Notation
As mentioned, we will use shapes of known area to approximate the area of an irregular region bounded by curves. Thisprocess often requires adding up long strings of numbers. To make it easier to write down these lengthy sums, we look atsome new notation here, called sigma notation (also known as summation notation). The Greek capital letter Σ, sigma,
is used to express long sums of values in a compact form. For example, if we want to add all the integers from 1 to 20without sigma notation, we have to write


1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20.


We could probably skip writing a couple of terms and write
1 + 2 + 3 + 4 +⋯ + 19 + 20,


which is better, but still cumbersome. With sigma notation, we write this sum as

i = 1


20


i,


which is much more compact.
Typically, sigma notation is presented in the form



i = 1


n


ai


where ai describes the terms to be added, and the i is called the index. Each term is evaluated, then we sum all the values,
beginning with the value when i = 1 and ending with the value when i = n. For example, an expression like ∑


i = 2


7


si is


6 Chapter 1 | Integration


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1.1


interpreted as s2 + s3 + s4 + s5 + s6 + s7. Note that the index is used only to keep track of the terms to be added; it does
not factor into the calculation of the sum itself. The index is therefore called a dummy variable. We can use any letter welike for the index. Typically, mathematicians use i, j, k, m, and n for indices.
Let’s try a couple of examples of using sigma notation.
Example 1.1
Using Sigma Notation


a. Write in sigma notation and evaluate the sum of terms 3i for i = 1, 2, 3, 4, 5.
b. Write the sum in sigma notation:


1 + 1
4
+ 1


9
+ 1


16
+ 1


25
.


Solution
a. Write



i = 1


5


3i = 3 + 32 + 33 + 34 + 35


= 363.
b. The denominator of each term is a perfect square. Using sigma notation, this sum can be written as



i = 1


5
1
i2
.


Write in sigma notation and evaluate the sum of terms 2i for i = 3, 4, 5, 6.


The properties associated with the summation process are given in the following rule.
Rule: Properties of Sigma Notation
Let a1, a2 ,…, an and b1, b2 ,…, bn represent two sequences of terms and let c be a constant. The following
properties hold for all positive integers n and for integers m, with 1 ≤ m ≤ n.


1.
(1.1)



i = 1


n


c = nc


2.
(1.2)



i = 1


n


cai = c∑
i = 1


n


ai


3.
(1.3)



i = 1


n

⎝ai + bi



⎠ = ∑


i = 1


n


ai + ∑
i = 1


n


bi


4.
(1.4)



i = 1


n

⎝ai − bi



⎠ = ∑


i = 1


n


ai − ∑
i = 1


n


bi


Chapter 1 | Integration 7




5.
(1.5)



i = 1


n


ai = ∑
i = 1


m


ai + ∑
i = m + 1


n


ai


Proof
We prove properties 2. and 3. here, and leave proof of the other properties to the Exercises.
2. We have



i = 1


n


cai = ca1 + ca2 + ca3 +⋯+ can


= c(a1 + a2 + a3 +⋯+ an)


= c∑
i = 1


n


ai.


3. We have

i = 1


n

⎝ai + bi



⎠ = ⎛⎝a1 + b1



⎠+ ⎛⎝a2 + b2



⎠+ ⎛⎝a3 + b3



⎠+⋯+ ⎛⎝an + bn





= (a1 + a2 + a3 +⋯+ an) +

⎝b1 + b2 + b3 +⋯+ bn





= ∑
i = 1


n


ai + ∑
i = 1


n


bi.



A few more formulas for frequently found functions simplify the summation process further. These are shown in the nextrule, for sums and powers of integers, and we use them in the next set of examples.
Rule: Sums and Powers of Integers


1. The sum of n integers is given by

i = 1


n


i = 1 + 2 +⋯ + n = n(n + 1)
2


.


2. The sum of consecutive integers squared is given by

i = 1


n


i2 = 12 + 22 +⋯+ n2 = n(n + 1)(2n + 1)
6


.


3. The sum of consecutive integers cubed is given by

i = 1


n


i3 = 13 + 23 +⋯+ n3 = n
2 (n + 1)2


4
.


Example 1.2
Evaluation Using Sigma Notation
Write using sigma notation and evaluate:


a. The sum of the terms (i − 3)2 for i = 1, 2,…, 200.


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1.2


b. The sum of the terms ⎛⎝i3 − i2⎞⎠ for i = 1, 2, 3, 4, 5, 6.
Solution


a. Multiplying out (i − 3)2, we can break the expression into three terms.

i = 1


200


(i − 3)2 = ∑
i = 1


200

⎝i
2 − 6i + 9⎞⎠


= ∑
i = 1


200


i2 − ∑
i = 1


200


6i + ∑
i = 1


200


9


= ∑
i = 1


200


i2 − 6∑
i = 1


200


i + ∑
i = 1


200


9


= 200(200 + 1)(400 + 1)
6


− 6


200(200 + 1)


2

⎦+ 9(200)


= 2,686,700 − 120,600 + 1800
= 2,567,900


b. Use sigma notation property iv. and the rules for the sum of squared terms and the sum of cubed terms.

i = 1


6

⎝i
3 − i2⎞⎠ = ∑


i = 1


6


i3 − ∑
i = 1


6


i2


= 6
2 (6 + 1)2


4
− 6(6 + 1)



⎝2(6) + 1⎞⎠
6


= 1764
4


− 546
6


= 350


Find the sum of the values of 4 + 3i for i = 1, 2,…, 100.


Example 1.3
Finding the Sum of the Function Values
Find the sum of the values of f (x) = x3 over the integers 1, 2, 3,…, 10.
Solution
Using the formula, we have



i = 0


10


i3 = (10)
2 (10 + 1)2


4


= 100(121)
4


= 3025.


Chapter 1 | Integration 9




1.3 Evaluate the sum indicated by the notation ∑
k = 1


20


(2k + 1).


Approximating Area
Now that we have the necessary notation, we return to the problem at hand: approximating the area under a curve. Let f (x)
be a continuous, nonnegative function defined on the closed interval ⎡⎣a, b⎤⎦. We want to approximate the area A bounded by
f (x) above, the x-axis below, the line x = a on the left, and the line x = b on the right (Figure 1.2).


Figure 1.2 An area (shaded region) bounded by the curve
f (x) at top, the x-axis at bottom, the line x = a to the left, and
the line x = b at right.


How do we approximate the area under this curve? The approach is a geometric one. By dividing a region into many smallshapes that have known area formulas, we can sum these areas and obtain a reasonable estimate of the true area. We begin
by dividing the interval ⎡⎣a, b⎤⎦ into n subintervals of equal width, b − an . We do this by selecting equally spaced points
x0, x1, x2 ,…, xn with x0 = a, xn = b, and


xi − xi − 1 =
b − a
n


for i = 1, 2, 3,…, n.
We denote the width of each subinterval with the notation Δx, so Δx = b − an and


xi = x0 + iΔx


for i = 1, 2, 3,…, n. This notion of dividing an interval ⎡⎣a, b⎤⎦ into subintervals by selecting points from within the interval
is used quite often in approximating the area under a curve, so let’s define some relevant terminology.
Definition
A set of points P = {xi} for i = 0, 1, 2,…, n with a = x0 < x1 < x2 < ⋯ < xn = b, which divides the interval

⎣a, b⎤⎦ into subintervals of the form [x0, x1], [x1, x2],…, [xn − 1, xn] is called a partition of ⎡⎣a, b⎤⎦. If the
subintervals all have the same width, the set of points forms a regular partition of the interval ⎡⎣a, b⎤⎦.


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We can use this regular partition as the basis of a method for estimating the area under the curve. We next examine twomethods: the left-endpoint approximation and the right-endpoint approximation.
Rule: Left-Endpoint Approximation
On each subinterval [xi − 1, xi] (for i = 1, 2, 3,…, n), construct a rectangle with width Δx and height equal to
f (xi − 1), which is the function value at the left endpoint of the subinterval. Then the area of this rectangle is
f (xi − 1)Δx. Adding the areas of all these rectangles, we get an approximate value for A (Figure 1.3). We use the
notation Ln to denote that this is a left-endpoint approximation of A using n subintervals.


(1.6)A ≈ Ln = f (x0)Δx + f (x1)Δx +⋯+ f (xn − 1)Δx


= ∑
i = 1


n


f (xi − 1)Δx


Figure 1.3 In the left-endpoint approximation of area under acurve, the height of each rectangle is determined by the functionvalue at the left of each subinterval.


The second method for approximating area under a curve is the right-endpoint approximation. It is almost the same as theleft-endpoint approximation, but now the heights of the rectangles are determined by the function values at the right of eachsubinterval.
Rule: Right-Endpoint Approximation
Construct a rectangle on each subinterval [xi − 1, xi], only this time the height of the rectangle is determined by the
function value f (xi) at the right endpoint of the subinterval. Then, the area of each rectangle is f (xi)Δx and the
approximation for A is given by


(1.7)A ≈ Rn = f (x1)Δx + f (x2)Δx +⋯+ f (xn)Δx


= ∑
i = 1


n


f (xi)Δx.


The notation Rn indicates this is a right-endpoint approximation for A (Figure 1.4).


Chapter 1 | Integration 11




Figure 1.4 In the right-endpoint approximation of area undera curve, the height of each rectangle is determined by thefunction value at the right of each subinterval. Note that theright-endpoint approximation differs from the left-endpointapproximation in Figure 1.3.


The graphs in Figure 1.5 represent the curve f (x) = x2
2
. In graph (a) we divide the region represented by the interval


[0, 3] into six subintervals, each of width 0.5. Thus, Δx = 0.5. We then form six rectangles by drawing vertical lines
perpendicular to xi − 1, the left endpoint of each subinterval. We determine the height of each rectangle by calculating
f (xi − 1) for i = 1, 2, 3, 4, 5, 6. The intervals are ⎡⎣0, 0.5⎤⎦, ⎡⎣0.5, 1⎤⎦, ⎡⎣1, 1.5⎤⎦, ⎡⎣1.5, 2⎤⎦, ⎡⎣2, 2.5⎤⎦, ⎡⎣2.5, 3⎤⎦. We find the area
of each rectangle by multiplying the height by the width. Then, the sum of the rectangular areas approximates the areabetween f (x) and the x-axis. When the left endpoints are used to calculate height, we have a left-endpoint approximation.
Thus,


A ≈ L6 = ∑
i = 1


6


f (xi − 1)Δx = f (x0)Δx + f (x1)Δx + f (x2)Δx + f (x3)Δx + f (x4)Δx + f (x5)Δx


= f (0)0.5 + f (0.5)0.5 + f (1)0.5 + f (1.5)0.5 + f (2)0.5 + f (2.5)0.5


= (0)0.5 + (0.125)0.5 + (0.5)0.5 + (1.125)0.5 + (2)0.5 + (3.125)0.5
= 0 + 0.0625 + 0.25 + 0.5625 + 1 + 1.5625
= 3.4375.


Figure 1.5 Methods of approximating the area under a curve by using (a) the left endpointsand (b) the right endpoints.


In Figure 1.5(b), we draw vertical lines perpendicular to xi such that xi is the right endpoint of each subinterval, and
calculate f (xi) for i = 1, 2, 3, 4, 5, 6. We multiply each f (xi) by Δx to find the rectangular areas, and then add them.
This is a right-endpoint approximation of the area under f (x). Thus,


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A ≈ R6 = ∑
i = 1


6


f (xi)Δx = f (x1)Δx + f (x2)Δx + f (x3)Δx + f (x4)Δx + f (x5)Δx + f (x6)Δx


= f (0.5)0.5 + f (1)0.5 + f (1.5)0.5 + f (2)0.5 + f (2.5)0.5 + f (3)0.5


= (0.125)0.5 + (0.5)0.5 + (1.125)0.5 + (2)0.5 + (3.125)0.5 + (4.5)0.5
= 0.0625 + 0.25 + 0.5625 + 1 + 1.5625 + 2.25
= 5.6875.


Example 1.4
Approximating the Area Under a Curve
Use both left-endpoint and right-endpoint approximations to approximate the area under the curve of f (x) = x2
on the interval [0, 2]; use n = 4.
Solution
First, divide the interval [0, 2] into n equal subintervals. Using n = 4, Δx = (2 − 0)


4
= 0.5. This is the width of


each rectangle. The intervals ⎡⎣0, 0.5⎤⎦, ⎡⎣0.5, 1⎤⎦, ⎡⎣1, 1.5⎤⎦, ⎡⎣1.5, 2⎤⎦ are shown in Figure 1.6. Using a left-endpoint
approximation, the heights are f (0) = 0, f (0.5) = 0.25, f (1) = 1, f (1.5) = 2.25. Then,


L4 = f (x0)Δx + f (x1)Δx + f (x2)Δx + f (x3)Δx


= 0(0.5) + 0.25(0.5) + 1(0.5) + 2.25(0.5)
= 1.75.


Figure 1.6 The graph shows the left-endpoint approximation
of the area under f (x) = x2 from 0 to 2.


The right-endpoint approximation is shown in Figure 1.7. The intervals are the same, Δx = 0.5, but now use
the right endpoint to calculate the height of the rectangles. We have


R4 = f (x1)Δx + f (x2)Δx + f (x3)Δx + f (x4)Δx


= 0.25(0.5) + 1(0.5) + 2.25(0.5) + 4(0.5)
= 3.75.


Chapter 1 | Integration 13




1.4


Figure 1.7 The graph shows the right-endpoint approximation
of the area under f (x) = x2 from 0 to 2.


The left-endpoint approximation is 1.75; the right-endpoint approximation is 3.75.


Sketch left-endpoint and right-endpoint approximations for f (x) = 1x on [1, 2]; use n = 4.
Approximate the area using both methods.


Looking at Figure 1.5 and the graphs in Example 1.4, we can see that when we use a small number of intervals, neitherthe left-endpoint approximation nor the right-endpoint approximation is a particularly accurate estimate of the area underthe curve. However, it seems logical that if we increase the number of points in our partition, our estimate of A will improve.We will have more rectangles, but each rectangle will be thinner, so we will be able to fit the rectangles to the curve moreprecisely.
We can demonstrate the improved approximation obtained through smaller intervals with an example. Let’s explore the ideaof increasing n, first in a left-endpoint approximation with four rectangles, then eight rectangles, and finally 32 rectangles.Then, let’s do the same thing in a right-endpoint approximation, using the same sets of intervals, of the same curved region.
Figure 1.8 shows the area of the region under the curve f (x) = (x − 1)3 + 4 on the interval [0, 2] using a left-endpoint
approximation where n = 4. The width of each rectangle is


Δx = 2 − 0
4


= 1
2
.


The area is approximated by the summed areas of the rectangles, or
L4 = f (0)(0.5) + f (0.5)(0.5) + f (1)(0.5) + f (1.5)0.5


= 7.5.


Figure 1.8 With a left-endpoint approximation and dividingthe region from a to b into four equal intervals, the area underthe curve is approximately equal to the sum of the areas of therectangles.


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Figure 1.9 shows the same curve divided into eight subintervals. Comparing the graph with four rectangles in Figure 1.8with this graph with eight rectangles, we can see there appears to be less white space under the curve when n = 8. This
white space is area under the curve we are unable to include using our approximation. The area of the rectangles is


L8 = f (0)(0.25) + f (0.25)(0.25) + f (0.5)(0.25) + f (0.75)(0.25)


+ f (1)(0.25) + f (1.25)(0.25) + f (1.5)(0.25) + f (1.75)(0.25)


= 7.75.


Figure 1.9 The region under the curve is divided into n = 8
rectangular areas of equal width for a left-endpointapproximation.


The graph in Figure 1.10 shows the same function with 32 rectangles inscribed under the curve. There appears to be littlewhite space left. The area occupied by the rectangles is
L32 = f (0)(0.0625) + f (0.0625)(0.0625) + f (0.125)(0.0625) + ⋯ + f (1.9375)(0.0625)


= 7.9375.


Figure 1.10 Here, 32 rectangles are inscribed under the curvefor a left-endpoint approximation.


We can carry out a similar process for the right-endpoint approximation method. A right-endpoint approximation of thesame curve, using four rectangles (Figure 1.11), yields an area
R4 = f (0.5)(0.5) + f (1)(0.5) + f (1.5)(0.5) + f (2)(0.5)


= 8.5.


Chapter 1 | Integration 15




Figure 1.11 Now we divide the area under the curve into fourequal subintervals for a right-endpoint approximation.


Dividing the region over the interval [0, 2] into eight rectangles results in Δx = 2 − 0
8


= 0.25. The graph is shown in
Figure 1.12. The area is


R8 = f (0.25)(0.25) + f (0.5)(0.25) + f (0.75)(0.25) + f (1)(0.25)


+ f (1.25)(0.25) + f (1.5)(0.25) + f (1.75)(0.25) + f (2)(0.25)


= 8.25.


Figure 1.12 Here we use right-endpoint approximation for aregion divided into eight equal subintervals.


Last, the right-endpoint approximation with n = 32 is close to the actual area (Figure 1.13). The area is approximately
R32 = f (0.0625)(0.0625) + f (0.125)(0.0625) + f (0.1875)(0.0625) + ⋯ + f (2)(0.0625)


= 8.0625.


Figure 1.13 The region is divided into 32 equal subintervalsfor a right-endpoint approximation.


Based on these figures and calculations, it appears we are on the right track; the rectangles appear to approximate the areaunder the curve better as n gets larger. Furthermore, as n increases, both the left-endpoint and right-endpoint approximationsappear to approach an area of 8 square units. Table 1.1 shows a numerical comparison of the left- and right-endpoint


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methods. The idea that the approximations of the area under the curve get better and better as n gets larger and larger is veryimportant, and we now explore this idea in more detail.
Values of n Approximate Area Ln Approximate Area Rn
n = 4 7.5 8.5
n = 8 7.75 8.25
n = 32 7.94 8.06


Table 1.1 Converging Values of Left- and Right-Endpoint Approximationsas n Increases
Forming Riemann Sums
So far we have been using rectangles to approximate the area under a curve. The heights of these rectangles have beendetermined by evaluating the function at either the right or left endpoints of the subinterval [xi − 1, xi]. In reality, there is
no reason to restrict evaluation of the function to one of these two points only. We could evaluate the function at any point
ci in the subinterval [xi − 1, xi], and use f ⎛⎝xi* ⎞⎠ as the height of our rectangle. This gives us an estimate for the area of
the form


A ≈ ∑
i = 1


n


f ⎛⎝xi*

⎠Δx.


A sum of this form is called a Riemann sum, named for the 19th-century mathematician Bernhard Riemann, who developedthe idea.
Definition
Let f (x) be defined on a closed interval ⎡⎣a, b⎤⎦ and let P be a regular partition of ⎡⎣a, b⎤⎦. Let Δx be the width of each
subinterval [xi − 1, xi] and for each i, let xi* be any point in [xi − 1, xi]. A Riemann sum is defined for f (x) as



i = 1


n


f ⎛⎝xi*

⎠Δx.


Recall that with the left- and right-endpoint approximations, the estimates seem to get better and better as n get larger andlarger. The same thing happens with Riemann sums. Riemann sums give better approximations for larger values of n. Weare now ready to define the area under a curve in terms of Riemann sums.
Definition
Let f (x) be a continuous, nonnegative function on an interval ⎡⎣a, b⎤⎦, and let ∑


i = 1


n


f ⎛⎝xi*

⎠Δx be a Riemann sum for


f (x). Then, the area under the curve y = f (x) on ⎡⎣a, b⎤⎦ is given by
A = lim


n → ∞

i = 1


n


f ⎛⎝xi*

⎠Δx.


Chapter 1 | Integration 17




See a graphical demonstration (http://www.openstaxcollege.org/l/20_riemannsums) of theconstruction of a Riemann sum.


Some subtleties here are worth discussing. First, note that taking the limit of a sum is a little different from taking the limitof a function f (x) as x goes to infinity. Limits of sums are discussed in detail in the chapter on Sequences and Series;
however, for now we can assume that the computational techniques we used to compute limits of functions can also be usedto calculate limits of sums.
Second, we must consider what to do if the expression converges to different limits for different choices of ⎧



⎨xi*





⎬.


Fortunately, this does not happen. Although the proof is beyond the scope of this text, it can be shown that if f (x) is
continuous on the closed interval ⎡⎣a, b⎤⎦, then limn → ∞∑


i = 1


n


f ⎛⎝xi*

⎠Δx exists and is unique (in other words, it does not depend


on the choice of ⎧

⎨xi*





⎬).


We look at some examples shortly. But, before we do, let’s take a moment and talk about some specific choices for ⎧

⎨xi*





⎬.


Although any choice for ⎧

⎨xi*





⎬ gives us an estimate of the area under the curve, we don’t necessarily know whether that


estimate is too high (overestimate) or too low (underestimate). If it is important to know whether our estimate is high or
low, we can select our value for ⎧



⎨xi*





⎬ to guarantee one result or the other.


If we want an overestimate, for example, we can choose ⎧

⎨xi*





⎬ such that for i = 1, 2, 3,…, n, f ⎛⎝xi* ⎞⎠ ≥ f (x) for all


x ∈ [xi − 1, xi]. In other words, we choose ⎧⎩⎨xi* ⎫⎭⎬ so that for i = 1, 2, 3,…, n, f ⎛⎝xi* ⎞⎠ is the maximum function value on
the interval [xi − 1, xi]. If we select ⎧⎩⎨xi* ⎫⎭⎬ in this way, then the Riemann sum ∑


i = 1


n


f ⎛⎝xi*

⎠Δx is called an upper sum.


Similarly, if we want an underestimate, we can choose ⎧

⎨xi*





⎬ so that for i = 1, 2, 3,…, n, f ⎛⎝xi* ⎞⎠ is the minimum function


value on the interval [xi − 1, xi]. In this case, the associated Riemann sum is called a lower sum. Note that if f (x) is either
increasing or decreasing throughout the interval ⎡⎣a, b⎤⎦, then the maximum and minimum values of the function occur at the
endpoints of the subintervals, so the upper and lower sums are just the same as the left- and right-endpoint approximations.
Example 1.5
Finding Lower and Upper Sums
Find a lower sum for f (x) = 10 − x2 on [1, 2]; let n = 4 subintervals.
Solution
With n = 4 over the interval [1, 2], Δx = 1


4
. We can list the intervals as



⎣1, 1.25⎤⎦, ⎡⎣1.25, 1.5⎤⎦, ⎡⎣1.5, 1.75⎤⎦, ⎡⎣1.75, 2⎤⎦. Because the function is decreasing over the interval [1, 2], Figure
1.14 shows that a lower sum is obtained by using the right endpoints.


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1.5


Figure 1.14 The graph of f (x) = 10 − x2 is set up for a
right-endpoint approximation of the area bounded by the curveand the x-axis on [1, 2], and it shows a lower sum.


The Riemann sum is

k = 1


4

⎝10 − x


2⎞
⎠(0.25) = 0.25



⎣10 − (1.25)


2 + 10 − (1.5)2 + 10 − (1.75)2 + 10 − (2)2⎤⎦


= 0.25[8.4375 + 7.75 + 6.9375 + 6]
= 7.28.


The area of 7.28 is a lower sum and an underestimate.


a. Find an upper sum for f (x) = 10 − x2 on [1, 2]; let n = 4.
b. Sketch the approximation.


Example 1.6
Finding Lower and Upper Sums for f(x) = sinx
Find a lower sum for f (x) = sinx over the interval ⎡⎣a, b⎤⎦ = ⎡⎣0, π2⎤⎦; let n = 6.


Solution
Let’s first look at the graph in Figure 1.15 to get a better idea of the area of interest.


Chapter 1 | Integration 19




1.6


Figure 1.15 The graph of y = sinx is divided into six regions: Δx = π/2
6


= π
12


.


The intervals are ⎡⎣0, π12⎤⎦, ⎡⎣ π12, π6⎤⎦, ⎡⎣π6, π4⎤⎦, ⎡⎣π4, π3⎤⎦, ⎡⎣π3, 5π12⎤⎦, and ⎡⎣5π12, π2⎤⎦. Note that f (x) = sinx is
increasing on the interval ⎡⎣0, π2⎤⎦, so a left-endpoint approximation gives us the lower sum. A left-endpoint
approximation is the Riemann sum ∑


i = 0


5


sinxi


π
12

⎠. We have


A ≈ sin(0)⎛⎝
π
12

⎠+ sin




π
12




π
12

⎠+ sin




π
6




π
12

⎠+ sin




π
4




π
12

⎠+ sin




π
3




π
12

⎠+ sin





12




π
12



= 0.863.


Using the function f (x) = sinx over the interval ⎡⎣0, π2⎤⎦, find an upper sum; let n = 6.


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1.1 EXERCISES
1. State whether the given sums are equal or unequal.


a. ∑
i = 1


10


i and ∑
k = 1


10


k


b. ∑
i = 1


10


i and ∑
i = 6


15


(i − 5)


c. ∑
i = 1


10


i(i − 1) and ∑
j = 0


9

⎝ j + 1⎞⎠ j


d. ∑
i = 1


10


i(i − 1) and ∑
k = 1


10

⎝k


2 − k⎞⎠


In the following exercises, use the rules for sums of powersof integers to compute the sums.
2. ∑


i = 5


10


i


3. ∑
i = 5


10


i2


Suppose that ∑
i = 1


100


ai = 15 and ∑
i = 1


100


bi = −12. In the
following exercises, compute the sums.
4. ∑


i = 1


100

⎝ai + bi





5. ∑
i = 1


100

⎝ai − bi





6. ∑
i = 1


100

⎝3ai − 4bi





7. ∑
i = 1


100

⎝5ai + 4bi





In the following exercises, use summation properties andformulas to rewrite and evaluate the sums.
8. ∑


k = 1


20


100⎛⎝k
2 − 5k + 1⎞⎠


9. ∑
j = 1


50

⎝j


2 − 2 j⎞⎠


10. ∑
j = 11


20

⎝j


2 − 10 j⎞⎠


11. ∑
k = 1


25

⎣(2k)


2 − 100k⎤⎦


Let Ln denote the left-endpoint sum using n subintervals
and let Rn denote the corresponding right-endpoint sum.
In the following exercises, compute the indicated left andright sums for the given functions on the indicated interval.
12. L4 for f (x) = 1x − 1 on [2, 3]
13. R4 for g(x) = cos(πx) on [0, 1]
14. L6 for f (x) = 1x(x − 1) on ⎡⎣2, 5⎤⎦


15. R6 for f (x) = 1x(x − 1) on ⎡⎣2, 5⎤⎦


16. R4 for 1
x2 + 1


on [−2, 2]


17. L4 for 1
x2 + 1


on [−2, 2]


18. R4 for x2 − 2x + 1 on [0, 2]
19. L8 for x2 − 2x + 1 on [0, 2]
20. Compute the left and right Riemann sums—L4 and R4,respectively—for f (x) = (2 − |x|) on [−2, 2]. Compute
their average value and compare it with the area under thegraph of f.
21. Compute the left and right Riemann sums—L6 andR6, respectively—for f (x) = (3 − |3 − x|) on ⎡⎣0, 6⎤⎦.
Compute their average value and compare it with the areaunder the graph of f.
22. Compute the left and right Riemann sums—L4 and
R4, respectively—for f (x) = 4 − x2 on [−2, 2] and
compare their values.
23. Compute the left and right Riemann sums—L6 and
R6, respectively—for f (x) = 9 − (x − 3)2 on ⎡⎣0, 6⎤⎦ and
compare their values.
Express the following endpoint sums in sigma notation butdo not evaluate them.


Chapter 1 | Integration 21




24. L30 for f (x) = x2 on [1, 2]
25. L10 for f (x) = 4 − x2 on [−2, 2]
26. R20 for f (x) = sinx on [0, π]
27. R100 for lnx on [1, e]
In the following exercises, graph the function then use acalculator or a computer program to evaluate the followingleft and right endpoint sums. Is the area under the curvebetween the left and right endpoint sums?
28. [T] L100 and R100 for y = x2 − 3x + 1 on the interval
[−1, 1]


29. [T] L100 and R100 for y = x2 on the interval [0, 1]
30. [T] L50 and R50 for y = x + 1


x2 − 1
on the interval [2, 4]


31. [T] L100 and R100 for y = x3 on the interval [−1, 1]


32. [T] L50 and R50 for y = tan(x) on the interval ⎡⎣0, π4⎤⎦


33. [T] L100 and R100 for y = e2x on the interval [−1, 1]


34. Let tj denote the time that it took Tejay van Garterento ride the jth stage of the Tour de France in 2014. If there
were a total of 21 stages, interpret ∑


j = 1


21


t j.


35. Let r j denote the total rainfall in Portland on the jth
day of the year in 2009. Interpret ∑


j = 1


31


r j.


36. Let d j denote the hours of daylight and δ j denote the
increase in the hours of daylight from day j − 1 to day j
in Fargo, North Dakota, on the jth day of the year. Interpret
d1 + ∑


j = 2


365


δ j.


37. To help get in shape, Joe gets a new pair of running
shoes. If Joe runs 1 mi each day in week 1 and adds 1


10
mi


to his daily routine each week, what is the total mileage onJoe’s shoes after 25 weeks?
38. The following table gives approximate values of theaverage annual atmospheric rate of increase in carbondioxide (CO2) each decade since 1960, in parts per million(ppm). Estimate the total increase in atmospheric CO2between 1964 and 2013.


Decade Ppm/y
1964–1973 1.07
1974–1983 1.34
1984–1993 1.40
1994–2003 1.87
2004–2013 2.07


Table 1.2 Average AnnualAtmospheric CO2Increase,1964–2013 Source:http://www.esrl.noaa.gov/gmd/ccgg/trends/.


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39. The following table gives the approximate increase insea level in inches over 20 years starting in the given year.Estimate the net change in mean sea level from 1870 to2010.
Starting Year 20-Year Change
1870 0.3
1890 1.5
1910 0.2
1930 2.8
1950 0.7
1970 1.1
1990 1.5


Table 1.3 Approximate 20-Year SeaLevel Increases, 1870–1990 Source:http://link.springer.com/article/10.1007%2Fs10712-011-9119-1


40. The following table gives the approximate increase indollars in the average price of a gallon of gas per decadesince 1950. If the average price of a gallon of gas in 2010was $2.60, what was the average price of a gallon of gas in1950?
Starting Year 10-Year Change
1950 0.03
1960 0.05
1970 0.86
1980 −0.03
1990 0.29
2000 1.12


Table 1.4 Approximate 10-Year GasPrice Increases, 1950–2000 Source:http://epb.lbl.gov/homepages/Rick_Diamond/docs/lbnl55011-trends.pdf.


Chapter 1 | Integration 23




41. The following table gives the percent growth of theU.S. population beginning in July of the year indicated. Ifthe U.S. population was 281,421,906 in July 2000, estimatethe U.S. population in July 2010.
Year % Change/Year
2000 1.12
2001 0.99
2002 0.93
2003 0.86
2004 0.93
2005 0.93
2006 0.97
2007 0.96
2008 0.95
2009 0.88


Table 1.5 Annual PercentageGrowth of U.S. Population,2000–2009 Source:http://www.census.gov/popest/data.
(Hint: To obtain the population in July 2001, multiply thepopulation in July 2000 by 1.0112 to get 284,573,831.)
In the following exercises, estimate the areas under thecurves by computing the left Riemann sums, L8.
42.


43.


44.


45.


46. [T] Use a computer algebra system to compute theRiemann sum, LN, for N = 10, 30, 50 for
f (x) = 1 − x2 on [−1, 1].
47. [T] Use a computer algebra system to compute theRiemann sum, LN, for N = 10, 30, 50 for
f (x) = 1


1 + x2
on [−1, 1].


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48. [T] Use a computer algebra system to compute the
Riemann sum, LN, for N = 10, 30, 50 for f (x) = sin2 x
on [0, 2π]. Compare these estimates with π.
In the following exercises, use a calculator or a computerprogram to evaluate the endpoint sums RN and LN for
N = 1,10,100. How do these estimates compare with the
exact answers, which you can find via geometry?
49. [T] y = cos(πx) on the interval [0, 1]
50. [T] y = 3x + 2 on the interval ⎡⎣3, 5⎤⎦
In the following exercises, use a calculator or a computerprogram to evaluate the endpoint sums RN and LN for
N = 1,10,100.


51. [T] y = x4 − 5x2 + 4 on the interval [−2, 2],
which has an exact area of 32


15


52. [T] y = lnx on the interval [1, 2], which has an
exact area of 2ln(2) − 1
53. Explain why, if f (a) ≥ 0 and f is increasing on

⎣a, b⎤⎦, that the left endpoint estimate is a lower bound for
the area below the graph of f on ⎡⎣a, b⎤⎦.
54. Explain why, if f (b) ≥ 0 and f is decreasing on

⎣a, b⎤⎦, that the left endpoint estimate is an upper bound for
the area below the graph of f on ⎡⎣a, b⎤⎦.
55. Show that, in general,
RN − LN = (b − a) ×


f (b) − f (a)
N


.


56. Explain why, if f is increasing on ⎡⎣a, b⎤⎦, the error
between either LN or RN and the area A below the graph of
f is at most (b − a) f (b) − f (a)


N
.


57. For each of the three graphs:a. Obtain a lower bound L(A) for the area enclosed
by the curve by adding the areas of the squaresenclosed completely by the curve.b. Obtain an upper bound U(A) for the area by
adding to L(A) the areas B(A) of the squares
enclosed partially by the curve.


58. In the previous exercise, explain why L(A) gets no
smaller while U(A) gets no larger as the squares are
subdivided into four boxes of equal area.


Chapter 1 | Integration 25




59. A unit circle is made up of n wedges equivalent to theinner wedge in the figure. The base of the inner triangle
is 1 unit and its height is sin⎛⎝πn ⎞⎠. The base of the outer
triangle is B = cos⎛⎝πn ⎞⎠+ sin⎛⎝πn ⎞⎠tan⎛⎝πn ⎞⎠ and the height is
H = Bsin⎛⎝



n

⎠. Use this information to argue that the area


of a unit circle is equal to π.


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1.2 | The Definite Integral
Learning Objectives


1.2.1 State the definition of the definite integral.
1.2.2 Explain the terms integrand, limits of integration, and variable of integration.
1.2.3 Explain when a function is integrable.
1.2.4 Describe the relationship between the definite integral and net area.
1.2.5 Use geometry and the properties of definite integrals to evaluate them.
1.2.6 Calculate the average value of a function.


In the preceding section we defined the area under a curve in terms of Riemann sums:
A = lim


n → ∞

i = 1


n


f ⎛⎝xi*

⎠Δx.


However, this definition came with restrictions. We required f (x) to be continuous and nonnegative. Unfortunately, real-
world problems don’t always meet these restrictions. In this section, we look at how to apply the concept of the area underthe curve to a broader set of functions through the use of the definite integral.
Definition and Notation
The definite integral generalizes the concept of the area under a curve. We lift the requirements that f (x) be continuous
and nonnegative, and define the definite integral as follows.
Definition
If f (x) is a function defined on an interval ⎡⎣a, b⎤⎦, the definite integral of f from a to b is given by


(1.8)

a


b
f (x)dx = lim


n → ∞

i = 1


n


f ⎛⎝xi*

⎠Δx,


provided the limit exists. If this limit exists, the function f (x) is said to be integrable on ⎡⎣a, b⎤⎦, or is an integrable
function.


The integral symbol in the previous definition should look familiar. We have seen similar notation in the chapter onApplications of Derivatives (http://cnx.org/content/m53602/latest/) , where we used the indefinite integralsymbol (without the a and b above and below) to represent an antiderivative. Although the notation for indefinite integralsmay look similar to the notation for a definite integral, they are not the same. A definite integral is a number. An indefiniteintegral is a family of functions. Later in this chapter we examine how these concepts are related. However, close attentionshould always be paid to notation so we know whether we’re working with a definite integral or an indefinite integral.
Integral notation goes back to the late seventeenth century and is one of the contributions of Gottfried Wilhelm Leibniz, whois often considered to be the codiscoverer of calculus, along with Isaac Newton. The integration symbol ∫ is an elongated S,suggesting sigma or summation. On a definite integral, above and below the summation symbol are the boundaries of theinterval, ⎡⎣a, b⎤⎦. The numbers a and b are x-values and are called the limits of integration; specifically, a is the lower limit
and b is the upper limit. To clarify, we are using the word limit in two different ways in the context of the definite integral.First, we talk about the limit of a sum as n → ∞. Second, the boundaries of the region are called the limits of integration.
We call the function f (x) the integrand, and the dx indicates that f (x) is a function with respect to x, called the variable
of integration. Note that, like the index in a sum, the variable of integration is a dummy variable, and has no impact on thecomputation of the integral. We could use any variable we like as the variable of integration:



a


b
f (x)dx = ∫


a


b
f (t)dt = ∫


a


b
f (u)du


Chapter 1 | Integration 27




Previously, we discussed the fact that if f (x) is continuous on ⎡⎣a, b⎤⎦, then the limit limn → ∞∑
i = 1


n


f ⎛⎝xi*

⎠Δx exists and is


unique. This leads to the following theorem, which we state without proof.
Theorem 1.1: Continuous Functions Are Integrable
If f (x) is continuous on ⎡⎣a, b⎤⎦, then f is integrable on ⎡⎣a, b⎤⎦.


Functions that are not continuous on ⎡⎣a, b⎤⎦ may still be integrable, depending on the nature of the discontinuities. For
example, functions with a finite number of jump discontinuities on a closed interval are integrable.
It is also worth noting here that we have retained the use of a regular partition in the Riemann sums. This restriction is notstrictly necessary. Any partition can be used to form a Riemann sum. However, if a nonregular partition is used to definethe definite integral, it is not sufficient to take the limit as the number of subintervals goes to infinity. Instead, we must takethe limit as the width of the largest subinterval goes to zero. This introduces a little more complex notation in our limits andmakes the calculations more difficult without really gaining much additional insight, so we stick with regular partitions forthe Riemann sums.
Example 1.7
Evaluating an Integral Using the Definition
Use the definition of the definite integral to evaluate ∫


0


2
x2dx. Use a right-endpoint approximation to generate


the Riemann sum.
Solution
We first want to set up a Riemann sum. Based on the limits of integration, we have a = 0 and b = 2. For
i = 0, 1, 2,…, n, let P = {xi} be a regular partition of [0, 2]. Then


Δx = b − an =
2
n.


Since we are using a right-endpoint approximation to generate Riemann sums, for each i, we need to calculatethe function value at the right endpoint of the interval [xi − 1, xi]. The right endpoint of the interval is xi, and
since P is a regular partition,


xi = x0 + iΔx = 0 + i


2
n

⎦ =


2i
n .


Thus, the function value at the right endpoint of the interval is
f (xi) = xi


2 = ⎛⎝
2i
n



2
= 4i


2


n2
.


Then the Riemann sum takes the form

i = 1


n


f (xi)Δx = ∑
i = 1


n ⎛

4i2


n2


2
n = ∑


i = 1


n
8i2


n3
= 8


n3

i = 1


n


i2.


Using the summation formula for ∑
i = 1


n


i2, we have


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1.7



i = 1


n


f (xi)Δx =
8
n3

i = 1


n


i2


= 8
n3


n(n + 1)(2n + 1)


6



= 8
n3


2n3 + 3n2 + n


6



= 16n
3 + 24n2 + n


6n3


= 8
3
+ 4n +


1
6n2


.


Now, to calculate the definite integral, we need to take the limit as n → ∞. We get


0


2
x2dx = lim


n → ∞

i = 1


n


f (xi)Δx


= lim
n → ∞


8
3
+ 4n +


1
6n2



= lim
n → ∞




8
3

⎠+ limn → ∞




4
n

⎠+ limn → ∞





1
6n2



= 8
3
+ 0 + 0 = 8


3
.


Use the definition of the definite integral to evaluate ∫
0


3
(2x − 1)dx. Use a right-endpoint approximation


to generate the Riemann sum.


Evaluating Definite Integrals
Evaluating definite integrals this way can be quite tedious because of the complexity of the calculations. Later in this chapterwe develop techniques for evaluating definite integrals without taking limits of Riemann sums. However, for now, we canrely on the fact that definite integrals represent the area under the curve, and we can evaluate definite integrals by usinggeometric formulas to calculate that area. We do this to confirm that definite integrals do, indeed, represent areas, so we canthen discuss what to do in the case of a curve of a function dropping below the x-axis.
Example 1.8
Using Geometric Formulas to Calculate Definite Integrals
Use the formula for the area of a circle to evaluate ∫


3


6
9 − (x − 3)2dx.


Solution
The function describes a semicircle with radius 3. To find


Chapter 1 | Integration 29




1.8



3


6
9 − (x − 3)2dx,


we want to find the area under the curve over the interval ⎡⎣3, 6⎤⎦. The formula for the area of a circle is A = πr2.
The area of a semicircle is just one-half the area of a circle, or A = ⎛⎝12⎞⎠πr2. The shaded area in Figure 1.16
covers one-half of the semicircle, or A = ⎛⎝14⎞⎠πr2. Thus,



3


6
9 − (x − 3)2 = 1


4
π(3)2


= 9
4
π


≈ 7.069.


Figure 1.16 The value of the integral of the function f (x)
over the interval ⎡⎣3, 6⎤⎦ is the area of the shaded region.


Use the formula for the area of a trapezoid to evaluate ∫
2


4
(2x + 3)dx.


Area and the Definite Integral
When we defined the definite integral, we lifted the requirement that f (x) be nonnegative. But how do we interpret “the
area under the curve” when f (x) is negative?
Net Signed Area
Let us return to the Riemann sum. Consider, for example, the function f (x) = 2 − 2x2 (shown in Figure 1.17) on
the interval [0, 2]. Use n = 8 and choose ⎧



⎨xi* } as the left endpoint of each interval. Construct a rectangle on each


subinterval of height f ⎛⎝xi* ⎞⎠ and width Δx. When f ⎛⎝xi* ⎞⎠ is positive, the product f ⎛⎝xi* ⎞⎠Δx represents the area of the
rectangle, as before. When f ⎛⎝xi* ⎞⎠ is negative, however, the product f ⎛⎝xi* ⎞⎠Δx represents the negative of the area of the
rectangle. The Riemann sum then becomes



i = 1


8


f ⎛⎝xi*

⎠Δx = ⎛⎝Area of rectangles above the x-axis⎞⎠− ⎛⎝Area of rectangles below the x-axis⎞⎠


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Figure 1.17 For a function that is partly negative, theRiemann sum is the area of the rectangles above the x-axis lessthe area of the rectangles below the x-axis.


Taking the limit as n → ∞, the Riemann sum approaches the area between the curve above the x-axis and the x-axis, less
the area between the curve below the x-axis and the x-axis, as shown in Figure 1.18. Then,



0


2
f (x)dx = lim


n → ∞

i = 1


n


f (ci)Δx


= A1 − A2.


The quantity A1 − A2 is called the net signed area.


Figure 1.18 In the limit, the definite integral equals area A1less area A2, or the net signed area.


Notice that net signed area can be positive, negative, or zero. If the area above the x-axis is larger, the net signed area ispositive. If the area below the x-axis is larger, the net signed area is negative. If the areas above and below the x-axis areequal, the net signed area is zero.
Example 1.9
Finding the Net Signed Area


Chapter 1 | Integration 31




1.9


Find the net signed area between the curve of the function f (x) = 2x and the x-axis over the interval [−3, 3].
Solution
The function produces a straight line that forms two triangles: one from x = −3 to x = 0 and the other from
x = 0 to x = 3 (Figure 1.19). Using the geometric formula for the area of a triangle, A = 1


2
bh, the area of


triangle A1, above the axis, is
A1 =


1
2
3(6) = 9,


where 3 is the base and 2(3) = 6 is the height. The area of triangle A2, below the axis, is
A2 =


1
2
(3)(6) = 9,


where 3 is the base and 6 is the height. Thus, the net area is


−3


3
2xdx = A1 − A2 = 9 − 9 = 0.


Figure 1.19 The area above the curve and below the x-axisequals the area below the curve and above the x-axis.


Analysis
If A1 is the area above the x-axis and A2 is the area below the x-axis, then the net area is A1 − A2. Since the areas
of the two triangles are equal, the net area is zero.


Find the net signed area of f (x) = x − 2 over the interval ⎡⎣0, 6⎤⎦, illustrated in the following image.


Total Area
One application of the definite integral is finding displacement when given a velocity function. If v(t) represents the


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velocity of an object as a function of time, then the area under the curve tells us how far the object is from its originalposition. This is a very important application of the definite integral, and we examine it in more detail later in the chapter.For now, we’re just going to look at some basics to get a feel for how this works by studying constant velocities.
When velocity is a constant, the area under the curve is just velocity times time. This idea is already very familiar. If a cartravels away from its starting position in a straight line at a speed of 75 mph for 2 hours, then it is 150 mi away from itsoriginal position (Figure 1.20). Using integral notation, we have



0


2
75dt = 150.


Figure 1.20 The area under the curve v(t) = 75 tells us how far the car
is from its starting point at a given time.


In the context of displacement, net signed area allows us to take direction into account. If a car travels straight north at aspeed of 60 mph for 2 hours, it is 120 mi north of its starting position. If the car then turns around and travels south at aspeed of 40 mph for 3 hours, it will be back at it starting position (Figure 1.21). Again, using integral notation, we have


0


2
60dt + ∫


2


5
−40dt = 120 − 120


= 0.


In this case the displacement is zero.


Chapter 1 | Integration 33




Figure 1.21 The area above the axis and the area below the axisare equal, so the net signed area is zero.


Suppose we want to know how far the car travels overall, regardless of direction. In this case, we want to know the areabetween the curve and the x-axis, regardless of whether that area is above or below the axis. This is called the total area.
Graphically, it is easiest to think of calculating total area by adding the areas above the axis and the areas below the axis(rather than subtracting the areas below the axis, as we did with net signed area). To accomplish this mathematically, we usethe absolute value function. Thus, the total distance traveled by the car is



0


2
|60|dt + ∫


2


5
|−40|dt = ∫


0


2
60dt + ∫


2


5
40dt


= 120 + 120
= 240.


Bringing these ideas together formally, we state the following definitions.
Definition
Let f (x) be an integrable function defined on an interval ⎡⎣a, b⎤⎦. Let A1 represent the area between f (x) and the
x-axis that lies above the axis and let A2 represent the area between f (x) and the x-axis that lies below the axis. Then,
the net signed area between f (x) and the x-axis is given by



a


b
f (x)dx = A1 − A2.


The total area between f (x) and the x-axis is given by


a


b


| f (x)|dx = A1 + A2.


Example 1.10


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1.10


Finding the Total Area
Find the total area between f (x) = x − 2 and the x-axis over the interval ⎡⎣0, 6⎤⎦.
Solution
Calculate the x-intercept as (2, 0) (set y = 0, solve for x). To find the total area, take the area below the x-axis
over the subinterval [0, 2] and add it to the area above the x-axis on the subinterval ⎡⎣2, 6⎤⎦ (Figure 1.22).


Figure 1.22 The total area between the line and the x-axisover ⎡⎣0, 6⎤⎦ is A2 plus A1.


We have


0


6


|(x − 2)|dx = A2 + A1.


Then, using the formula for the area of a triangle, we obtain
A2 =


1
2
bh = 1


2
· 2 · 2 = 2


A1 =
1
2
bh = 1


2
· 4 · 4 = 8.


The total area, then, is
A1 + A2 = 8 + 2 = 10.


Find the total area between the function f (x) = 2x and the x-axis over the interval [−3, 3].


Properties of the Definite Integral
The properties of indefinite integrals apply to definite integrals as well. Definite integrals also have properties that relate tothe limits of integration. These properties, along with the rules of integration that we examine later in this chapter, help usmanipulate expressions to evaluate definite integrals.
Rule: Properties of the Definite Integral


1.
(1.9)∫


a


a
f (x)dx = 0


Chapter 1 | Integration 35




If the limits of integration are the same, the integral is just a line and contains no area.
2.


(1.10)

b


a
f (x)dx = −∫


a


b
f (x)dx


If the limits are reversed, then place a negative sign in front of the integral.
3.


(1.11)

a


b

⎣ f (x) + g(x)⎤⎦dx = ∫


a


b
f (x)dx + ∫


a


b
g(x)dx


The integral of a sum is the sum of the integrals.
4.


(1.12)

⌡a


b



⎣ f (x) − g(x)⎤⎦dx = ⌠


⌡a


b


f (x)dx − ∫
a


b
g(x)dx


The integral of a difference is the difference of the integrals.
5.


(1.13)

a


b
c f (x)dx = c∫


a


b
f (x)


for constant c. The integral of the product of a constant and a function is equal to the constant multiplied bythe integral of the function.
6.


(1.14)

a


b
f (x)dx = ∫


a


c
f (x)dx + ∫


c


b
f (x)dx


Although this formula normally applies when c is between a and b, the formula holds for all values of a, b, andc, provided f (x) is integrable on the largest interval.


Example 1.11
Using the Properties of the Definite Integral
Use the properties of the definite integral to express the definite integral of f (x) = −3x3 + 2x + 2 over the
interval [−2, 1] as the sum of three definite integrals.
Solution
Using integral notation, we have ∫


−2


1

⎝−3x


3 + 2x + 2⎞⎠dx. We apply properties 3. and 5. to get


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1.11


1.12



−2


1

⎝−3x


3 + 2x + 2⎞⎠dx = ∫
−2


1
−3x3dx + ∫


−2


1
2xdx + ∫


−2


1
2dx


= −3∫
−2


1
x3dx + 2∫


−2


1
xdx + ∫


−2


1
2dx.


Use the properties of the definite integral to express the definite integral of f (x) = 6x3 − 4x2 + 2x − 3
over the interval [1, 3] as the sum of four definite integrals.


Example 1.12
Using the Properties of the Definite Integral
If it is known that ∫


0


8
f (x)dx = 10 and ∫


0


5
f (x)dx = 5, find the value of ∫


5


8
f (x)dx.


Solution
By property 6.,



a


b
f (x)dx = ∫


a


c
f (x)dx + ∫


c


b
f (x)dx.


Thus,


0


8
f (x)dx = ∫


0


5
f (x)dx + ∫


5


8
f (x)dx


10 = 5 + ∫
5


8
f (x)dx


5 = ∫
5


8
f (x)dx.


If it is known that ∫
1


5
f (x)dx = −3 and ∫


2


5
f (x)dx = 4, find the value of ∫


1


2
f (x)dx.


Comparison Properties of Integrals
A picture can sometimes tell us more about a function than the results of computations. Comparing functions by their graphsas well as by their algebraic expressions can often give new insight into the process of integration. Intuitively, we might saythat if a function f (x) is above another function g(x), then the area between f (x) and the x-axis is greater than the area
between g(x) and the x-axis. This is true depending on the interval over which the comparison is made. The properties of
definite integrals are valid whether a < b, a = b, or a > b. The following properties, however, concern only the case
a ≤ b, and are used when we want to compare the sizes of integrals.


Chapter 1 | Integration 37




Theorem 1.2: Comparison Theorem
i. If f (x) ≥ 0 for a ≤ x ≤ b, then



a


b
f (x)dx ≥ 0.


ii. If f (x) ≥ g(x) for a ≤ x ≤ b, then

a


b
f (x)dx ≥ ∫


a


b
g(x)dx.


iii. If m and M are constants such that m ≤ f (x) ≤ M for a ≤ x ≤ b, then
m(b − a) ≤ ∫


a


b
f (x)dx


≤ M(b − a).


Example 1.13
Comparing Two Functions over a Given Interval
Compare f (x) = 1 + x2 and g(x) = 1 + x over the interval [0, 1].
Solution
Graphing these functions is necessary to understand how they compare over the interval [0, 1]. Initially, when
graphed on a graphing calculator, f (x) appears to be above g(x) everywhere. However, on the interval [0, 1],
the graphs appear to be on top of each other. We need to zoom in to see that, on the interval [0, 1], g(x) is above
f (x). The two functions intersect at x = 0 and x = 1 (Figure 1.23).


Figure 1.23 (a) The function f (x) appears above the function g(x)
except over the interval [0, 1] (b) Viewing the same graph with a greater
zoom shows this more clearly.


We can see from the graph that over the interval [0, 1], g(x) ≥ f (x). Comparing the integrals over the specified
interval [0, 1], we also see that ∫


0


1
g(x)dx ≥ ∫


0


1
f (x)dx (Figure 1.24). The thin, red-shaded area shows just


how much difference there is between these two integrals over the interval [0, 1].


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Figure 1.24 (a) The graph shows that over the interval
[0, 1], g(x) ≥ f (x), where equality holds only at the endpoints of the
interval. (b) Viewing the same graph with a greater zoom shows this moreclearly.


Average Value of a Function
We often need to find the average of a set of numbers, such as an average test grade. Suppose you received the followingtest scores in your algebra class: 89, 90, 56, 78, 100, and 69. Your semester grade is your average of test scores and youwant to know what grade to expect. We can find the average by adding all the scores and dividing by the number of scores.In this case, there are six test scores. Thus,


89 + 90 + 56 + 78 + 100 + 69
6


= 482
6


≈ 80.33.


Therefore, your average test grade is approximately 80.33, which translates to a B− at most schools.
Suppose, however, that we have a function v(t) that gives us the speed of an object at any time t, and we want to find the
object’s average speed. The function v(t) takes on an infinite number of values, so we can’t use the process just described.
Fortunately, we can use a definite integral to find the average value of a function such as this.
Let f (x) be continuous over the interval ⎡⎣a, b⎤⎦ and let ⎡⎣a, b⎤⎦ be divided into n subintervals of width Δx = (b − a)/n.
Choose a representative xi* in each subinterval and calculate f ⎛⎝xi* ⎞⎠ for i = 1, 2,…, n. In other words, consider each
f ⎛⎝xi*



⎠ as a sampling of the function over each subinterval. The average value of the function may then be approximated as


f ⎛⎝x1*

⎠+ f



⎝x2*



⎠+⋯+ f ⎛⎝xn*





n ,


which is basically the same expression used to calculate the average of discrete values.
But we know Δx = b − an , so n = b − aΔx , and we get


f ⎛⎝x1*

⎠+ f



⎝x2*



⎠+⋯+ f ⎛⎝xn*





n =
f ⎛⎝x1*



⎠+ f



⎝x2*



⎠+⋯+ f ⎛⎝xn*





(b − a)
Δx


.


Following through with the algebra, the numerator is a sum that is represented as ∑
i = 1


n


f ⎛⎝xi*

⎠, and we are dividing by a


fraction. To divide by a fraction, invert the denominator and multiply. Thus, an approximate value for the average value ofthe function is given by


Chapter 1 | Integration 39





i = 1


n


f ⎛⎝xi*



(b − a)
Δx


= ⎛⎝
Δx
b − a



⎠∑
i = 1


n


f ⎛⎝xi*



= ⎛⎝
1


b − a

⎠∑
i = 1


n


f ⎛⎝xi*

⎠Δx.


This is a Riemann sum. Then, to get the exact average value, take the limit as n goes to infinity. Thus, the average value ofa function is given by
1


b − a
lim


n → ∞

i = 1


n


f (xi)Δx =
1


b − a∫a
b
f (x)dx.


Definition
Let f (x) be continuous over the interval ⎡⎣a, b⎤⎦. Then, the average value of the function f (x) (or fave) on ⎡⎣a, b⎤⎦ is
given by


fave = 1b − a∫a
b
f (x)dx.


Example 1.14
Finding the Average Value of a Linear Function
Find the average value of f (x) = x + 1 over the interval ⎡⎣0, 5⎤⎦.
Solution
First, graph the function on the stated interval, as shown in Figure 1.25.


Figure 1.25 The graph shows the area under the function
f (x) = x + 1 over ⎡⎣0, 5⎤⎦.


The region is a trapezoid lying on its side, so we can use the area formula for a trapezoid A = 1
2
h(a + b), where


h represents height, and a and b represent the two parallel sides. Then,


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1.13



0


5
x + 1dx = 1


2
h(a + b)


= 1
2
· 5 · (1 + 6)


= 35
2
.


Thus the average value of the function is
1


5 − 0∫0
5
x + 1dx = 1


5
· 35
2


= 7
2
.


Find the average value of f (x) = 6 − 2x over the interval [0, 3].


Chapter 1 | Integration 41




1.2 EXERCISES
In the following exercises, express the limits as integrals.
60. lim


n → ∞

i = 1


n

⎝xi*



⎠Δx over [1, 3]


61. lim
n → ∞



i = 1


n

⎝5

⎝xi*




2 − 3⎛⎝xi*




3⎞
⎠Δx over [0, 2]


62. lim
n → ∞



i = 1


n


sin2 ⎛⎝2πxi*

⎠Δx over [0, 1]


63. lim
n → ∞



i = 1


n


cos2 ⎛⎝2πxi*

⎠Δx over [0, 1]


In the following exercises, given Ln or Rn as indicated,express their limits as n → ∞ as definite integrals,
identifying the correct intervals.
64. Ln = 1n∑


i = 1


n
i − 1
n


65. Rn = 1n∑
i = 1


n
i
n


66. Ln = 2n∑
i = 1


n

⎝1 + 2


i − 1
n



67. Rn = 3n∑
i = 1


n

⎝3 + 3


i
n



68. Ln = 2πn ∑
i = 1


n


2πi − 1n cos

⎝2π


i − 1
n



69. Rn = 1n∑
i = 1


n

⎝1 +


i
n

⎠log



⎝1 +


i
n



2⎞


In the following exercises, evaluate the integrals of thefunctions graphed using the formulas for areas of trianglesand circles, and subtracting the areas below the x-axis.
70.


71.


72.


73.


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74.


75.


In the following exercises, evaluate the integral using areaformulas.
76. ∫


0


3
(3 − x)dx


77. ∫
2


3
(3 − x)dx


78. ∫
−3


3
(3 − |x|)dx


79. ∫
0


6
(3 − |x − 3|)dx


80. ∫
−2


2
4 − x2dx


81. ∫
1


5
4 − (x − 3)2dx


82. ∫
0


12
36 − (x − 6)2dx


83. ∫
−2


3
(3 − |x|)dx


In the following exercises, use averages of values at the left(L) and right (R) endpoints to compute the integrals of thepiecewise linear functions with graphs that pass through thegiven list of points over the indicated intervals.
84. {(0, 0), (2, 1), (4, 3), (5, 0), (6, 0), (8, 3)} over
[0, 8]


85. {(0, 2), (1, 0), (3, 5), (5, 5), (6, 2), (8, 0)} over
[0, 8]


86. {(−4, −4), (−2, 0), (0, −2), (3, 3), (4, 3)} over
[−4, 4]


87. {(−4, 0), (−2, 2), (0, 0), (1, 2), (3, 2), (4, 0)}
over [−4, 4]


Suppose that ∫
0


4
f (x)dx = 5 and ∫


0


2
f (x)dx = −3, and



0


4
g(x)dx = −1 and ∫


0


2
g(x)dx = 2. In the following


exercises, compute the integrals.
88. ∫


0


4

⎝ f (x) + g(x)⎞⎠dx


89. ∫
2


4

⎝ f (x) + g(x)⎞⎠dx


90. ∫
0


2

⎝ f (x) − g(x)⎞⎠dx


91. ∫
2


4

⎝ f (x) − g(x)⎞⎠dx


92. ∫
0


2

⎝3 f (x) − 4g(x)⎞⎠dx


93. ∫
2


4

⎝4 f (x) − 3g(x)⎞⎠dx


In the following exercises, use the identity


−A


A
f (x)dx = ∫


−A


0
f (x)dx + ∫


0


A
f (x)dx to compute the


integrals.
94. ⌠
⌡−π


π
sin t
1 + t2


dt (Hint: sin(−t) = −sin(t))


Chapter 1 | Integration 43




95. ∫
− π


π
t


1 + cos t
dt


96. ∫
1


3
(2 − x)dx (Hint: Look at the graph of f.)


97. ∫
2


4
(x − 3)3dx (Hint: Look at the graph of f.)


In the following exercises, given that


0


1
xdx = 1


2
, ∫


0


1
x2dx = 1


3
, and ∫


0


1
x3dx = 1


4
,


compute the integrals.
98. ∫


0


1

⎝1 + x + x


2 + x3⎞⎠dx


99. ∫
0


1

⎝1 − x + x


2 − x3⎞⎠dx


100. ∫
0


1
(1 − x)2dx


101. ∫
0


1
(1 − 2x)3dx


102. ⌠
⌡0


1⎛
⎝6x −


4
3
x2⎞⎠dx


103. ∫
0


1

⎝7 − 5x


3⎞
⎠dx


In the following exercises, use the comparisontheorem.
104. Show that ∫


0


3

⎝x


2 − 6x + 9⎞⎠dx ≥ 0.


105. Show that ∫
−2


3
(x − 3)(x + 2)dx ≤ 0.


106. Show that ∫
0


1
1 + x3dx ≤ ∫


0


1
1 + x2dx.


107. Show that ∫
1


2
1 + xdx ≤ ∫


1


2
1 + x2dx.


108. Show that ∫
0


π/2
sin tdt ≥ π


4
. (Hint: sin t ≥ 2tπ over



⎣0,


π
2

⎦)


109. Show that ∫
−π/4


π/4
cos tdt ≥ π 2/4.


In the following exercises, find the average value fave of fbetween a and b, and find a point c, where f (c) = fave.
110. f (x) = x2, a = −1, b = 1
111. f (x) = x5, a = −1, b = 1
112. f (x) = 4 − x2, a = 0, b = 2
113. f (x) = (3 − |x|), a = −3, b = 3
114. f (x) = sinx, a = 0, b = 2π
115. f (x) = cosx, a = 0, b = 2π
In the following exercises, approximate the average valueusing Riemann sums L100 and R100. How does your answercompare with the exact given answer?
116. [T] y = ln(x) over the interval [1, 4]; the exact
solution is ln(256)


3
− 1.


117. [T] y = ex/2 over the interval [0, 1]; the exact
solution is 2( e − 1).
118. [T] y = tanx over the interval ⎡⎣0, π4⎤⎦; the exact
solution is 2ln(2)π .
119. [T] y = x + 1


4 − x2
over the interval [−1, 1]; the


exact solution is π
6
.


In the following exercises, compute the average value usingthe left Riemann sums LN for N = 1, 10, 100. How does
the accuracy compare with the given exact value?
120. [T] y = x2 − 4 over the interval [0, 2]; the exact
solution is −8


3
.


121. [T] y = xex2 over the interval [0, 2]; the exact
solution is 1


4

⎝e


4 − 1⎞⎠.


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122. [T] y = ⎛⎝12⎞⎠
x over the interval [0, 4]; the exact


solution is 15
64ln(2)


.


123. [T] y = xsin⎛⎝x2⎞⎠ over the interval [−π, 0]; the
exact solution is cos⎛⎝π2⎞⎠− 1



.


124. Suppose that A = ∫
0



sin2 tdt and


B = ∫
0



cos2 tdt. Show that A + B = 2π and A = B.


125. Suppose that A = ∫
−π/4


π/4
sec2 tdt = π and


B = ∫
−π/4


π/4
tan2 tdt. Show that A − B = π


2
.


126. Show that the average value of sin2 t over [0, 2π]
is equal to 1/2 Without further calculation, determine
whether the average value of sin2 t over [0, π] is also
equal to 1/2.
127. Show that the average value of cos2 t over [0, 2π]
is equal to 1/2. Without further calculation, determine
whether the average value of cos2 (t) over [0, π] is also
equal to 1/2.
128. Explain why the graphs of a quadratic function(parabola) p(x) and a linear function ℓ(x) can intersect
in at most two points. Suppose that p(a) = ℓ(a) and
p(b) = ℓ(b), and that ∫


a


b
p(t)dt > ∫


a


b
ℓ(t)dt. Explain


why ∫
c


d
p(t) > ∫


c


d
ℓ(t)dt whenever a ≤ c < d ≤ b.


129. Suppose that parabola p(x) = ax2 + bx + c opens
downward (a < 0) and has a vertex of y = −b


2a
> 0. For


which interval [A, B] is ∫
A


B

⎝ax


2 + bx + c⎞⎠dx as large as
possible?


130. Suppose ⎡⎣a, b⎤⎦ can be subdivided into subintervals
a = a0 < a1 < a2 < ⋯ < aN = b such that either
f ≥ 0 over [ai − 1, ai] or f ≤ 0 over [ai − 1, ai]. Set
Ai = ∫


ai − 1


ai
f (t)dt.


a. Explain why ∫
a


b
f (t)dt = A1 + A2 +⋯+ AN.


b. Then, explain why |∫ab f (t)dt| ≤ ∫ab| f (t)|dt.
131. Suppose f and g are continuous functions such that

c


d
f (t)dt ≤ ∫


c


d
g(t)dt for every subinterval ⎡⎣c, d⎤⎦ of



⎣a, b⎤⎦. Explain why f (x) ≤ g(x) for all values of x.
132. Suppose the average value of f over ⎡⎣a, b⎤⎦ is 1 and
the average value of f over ⎡⎣b, c⎤⎦ is 1 where a < c < b.
Show that the average value of f over [a, c] is also 1.
133. Suppose that ⎡⎣a, b⎤⎦ can be partitioned. taking
a = a0 < a1 < ⋯ < aN = b such that the average value
of f over each subinterval [ai − 1, ai] = 1 is equal to 1 for
each i = 1,…, N. Explain why the average value of f over

⎣a, b⎤⎦ is also equal to 1.
134. Suppose that for each i such that 1 ≤ i ≤ N one has

i − 1


i
f (t)dt = i. Show that ∫


0


N
f (t)dt = N(N + 1)


2
.


135. Suppose that for each i such that 1 ≤ i ≤ N one
has ∫


i − 1


i
f (t)dt = i2. Show that



0


N
f (t)dt = N(N + 1)(2N + 1)


6
.


136. [T] Compute the left and right Riemann sums L10
and R10 and their average L10 + R102 for f (t) = t2 over
[0, 1]. Given that ∫


0


1
t2dt = 0.33



, to how many


decimal places is L10 + R10
2


accurate?


Chapter 1 | Integration 45




137. [T] Compute the left and right Riemann sums, L10
and R10, and their average L10 + R102 for f (t) = ⎛⎝4 − t2⎞⎠
over [1, 2]. Given that ∫


1


2

⎝4 − t


2⎞
⎠dt = 1.66



, to how


many decimal places is L10 + R10
2


accurate?


138. If ∫
1


5
1 + t4dt = 41.7133..., what is



1


5
1 + u4du?


139. Estimate ∫
0


1
tdt using the left and right endpoint


sums, each with a single rectangle. How does the averageof these left and right endpoint sums compare with the
actual value ∫


0


1
tdt?


140. Estimate ∫
0


1
tdt by comparison with the area of a


single rectangle with height equal to the value of t at the
midpoint t = 1


2
. How does this midpoint estimate compare


with the actual value ∫
0


1
tdt?


141. From the graph of sin(2πx) shown:
a. Explain why ∫


0


1
sin(2πt)dt = 0.


b. Explain why, in general, ∫
a


a + 1
sin(2πt)dt = 0 for


any value of a.


142. If f is 1-periodic ⎛⎝ f (t + 1) = f (t)⎞⎠, odd, and
integrable over [0, 1], is it always true that


0


1
f (t)dt = 0?


143. If f is 1-periodic and ∫
0


1
f (t)dt = A, is it


necessarily true that ∫
a


1 + a
f (t)dt = A for all A?


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1.3 | The Fundamental Theorem of Calculus
Learning Objectives


1.3.1 Describe the meaning of the Mean Value Theorem for Integrals.
1.3.2 State the meaning of the Fundamental Theorem of Calculus, Part 1.
1.3.3 Use the Fundamental Theorem of Calculus, Part 1, to evaluate derivatives of integrals.
1.3.4 State the meaning of the Fundamental Theorem of Calculus, Part 2.
1.3.5 Use the Fundamental Theorem of Calculus, Part 2, to evaluate definite integrals.
1.3.6 Explain the relationship between differentiation and integration.


In the previous two sections, we looked at the definite integral and its relationship to the area under the curve of a function.Unfortunately, so far, the only tools we have available to calculate the value of a definite integral are geometric areaformulas and limits of Riemann sums, and both approaches are extremely cumbersome. In this section we look at somemore powerful and useful techniques for evaluating definite integrals.
These new techniques rely on the relationship between differentiation and integration. This relationship was discovered andexplored by both Sir Isaac Newton and Gottfried Wilhelm Leibniz (among others) during the late 1600s and early 1700s,and it is codified in what we now call the Fundamental Theorem of Calculus, which has two parts that we examine in thissection. Its very name indicates how central this theorem is to the entire development of calculus.


Isaac Newton’s contributions to mathematics and physics changed the way we look at the world. The relationshipshe discovered, codified as Newton’s laws and the law of universal gravitation, are still taught as foundationalmaterial in physics today, and his calculus has spawned entire fields of mathematics. To learn more, read a briefbiography (http://www.openstaxcollege.org/l/20_newtonbio) of Newton with multimedia clips.
Before we get to this crucial theorem, however, let’s examine another important theorem, the Mean Value Theorem forIntegrals, which is needed to prove the Fundamental Theorem of Calculus.
The Mean Value Theorem for Integrals
The Mean Value Theorem for Integrals states that a continuous function on a closed interval takes on its average value atthe same point in that interval. The theorem guarantees that if f (x) is continuous, a point c exists in an interval ⎡⎣a, b⎤⎦ such
that the value of the function at c is equal to the average value of f (x) over ⎡⎣a, b⎤⎦. We state this theorem mathematically
with the help of the formula for the average value of a function that we presented at the end of the preceding section.
Theorem 1.3: The Mean Value Theorem for Integrals
If f (x) is continuous over an interval ⎡⎣a, b⎤⎦, then there is at least one point c ∈ ⎡⎣a, b⎤⎦ such that


(1.15)
f (c) = 1


b − a∫a
b
f (x)dx.


This formula can also be stated as

a


b
f (x)dx = f (c)(b − a).


Proof
Since f (x) is continuous on ⎡⎣a, b⎤⎦, by the extreme value theorem (seeMaxima and Minima (http://cnx.org/content/
m53611/latest/) ), it assumes minimum and maximum values—m and M, respectively—on ⎡⎣a, b⎤⎦. Then, for all x in

⎣a, b⎤⎦, we have m ≤ f (x) ≤ M. Therefore, by the comparison theorem (see The Definite Integral), we have


Chapter 1 | Integration 47




m(b − a) ≤ ∫
a


b
f (x)dx ≤ M(b − a).


Dividing by b − a gives us
m ≤ 1


b − a∫a
b
f (x)dx ≤ M.


Since 1
b − a∫a


b
f (x)dx is a number between m and M, and since f (x) is continuous and assumes the values m and M


over ⎡⎣a, b⎤⎦, by the Intermediate Value Theorem (see Continuity (http://cnx.org/content/m53489/latest/) ), there is
a number c over ⎡⎣a, b⎤⎦ such that


f (c) = 1
b − a∫a


b
f (x)dx,


and the proof is complete.

Example 1.15
Finding the Average Value of a Function
Find the average value of the function f (x) = 8 − 2x over the interval [0, 4] and find c such that f (c) equals
the average value of the function over [0, 4].
Solution
The formula states the mean value of f (x) is given by


1
4 − 0∫0


4
(8 − 2x)dx.


We can see in Figure 1.26 that the function represents a straight line and forms a right triangle bounded by the
x- and y-axes. The area of the triangle is A = 1


2
(base)⎛⎝height⎞⎠. We have
A = 1


2
(4)(8) = 16.


The average value is found by multiplying the area by 1/(4 − 0). Thus, the average value of the function is
1
4
(16) = 4.


Set the average value equal to f (c) and solve for c.
8 − 2c = 4


c = 2


At c = 2, f (2) = 4.


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1.14


Figure 1.26 By the Mean Value Theorem, the continuousfunction f (x) takes on its average value at c at least once over
a closed interval.


Find the average value of the function f (x) = x
2
over the interval ⎡⎣0, 6⎤⎦ and find c such that f (c)


equals the average value of the function over [0, 6].


Example 1.16
Finding the Point Where a Function Takes on Its Average Value
Given ∫


0


3
x2dx = 9, find c such that f (c) equals the average value of f (x) = x2 over [0, 3].


Solution
We are looking for the value of c such that


f (c) = 1
3 − 0∫0


3
x2dx = 1


3
(9) = 3.


Replacing f (c) with c2, we have
c2 = 3
c = ± 3.


Since − 3 is outside the interval, take only the positive value. Thus, c = 3 (Figure 1.27).


Chapter 1 | Integration 49




1.15


Figure 1.27 Over the interval [0, 3], the function
f (x) = x2 takes on its average value at c = 3.


Given ∫
0


3

⎝2x


2 − 1⎞⎠dx = 15, find c such that f (c) equals the average value of f (x) = 2x2 − 1 over
[0, 3].


Fundamental Theorem of Calculus Part 1: Integrals andAntiderivatives
As mentioned earlier, the Fundamental Theorem of Calculus is an extremely powerful theorem that establishes therelationship between differentiation and integration, and gives us a way to evaluate definite integrals without using Riemannsums or calculating areas. The theorem is comprised of two parts, the first of which, the Fundamental Theorem ofCalculus, Part 1, is stated here. Part 1 establishes the relationship between differentiation and integration.
Theorem 1.4: Fundamental Theorem of Calculus, Part 1
If f (x) is continuous over an interval ⎡⎣a, b⎤⎦, and the function F(x) is defined by


(1.16)
F(x) = ∫


a


x
f (t)dt,


then F′ (x) = f (x) over ⎡⎣a, b⎤⎦.


Before we delve into the proof, a couple of subtleties are worth mentioning here. First, a comment on the notation. Note thatwe have defined a function, F(x), as the definite integral of another function, f (t), from the point a to the point x. At
first glance, this is confusing, because we have said several times that a definite integral is a number, and here it looks likeit’s a function. The key here is to notice that for any particular value of x, the definite integral is a number. So the function
F(x) returns a number (the value of the definite integral) for each value of x.


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Second, it is worth commenting on some of the key implications of this theorem. There is a reason it is called theFundamental Theorem of Calculus. Not only does it establish a relationship between integration and differentiation, butalso it guarantees that any integrable function has an antiderivative. Specifically, it guarantees that any continuous functionhas an antiderivative.
Proof
Applying the definition of the derivative, we have


F′ (x) = lim
h → 0


F(x + h) − F(x)
h


= lim
h → 0


1
h





⎢∫


a


x + h
f (t)dt − ∫


a


x
f (t)dt







= lim
h → 0


1
h





⎢∫


a


x + h
f (t)dt + ∫


x


a
f (t)dt







= lim
h → 0


1
h∫x


x + h
f (t)dt.


Looking carefully at this last expression, we see 1
h∫x


x + h
f (t)dt is just the average value of the function f (x) over the


interval ⎡⎣x, x + h⎤⎦. Therefore, by The Mean Value Theorem for Integrals, there is some number c in ⎡⎣x, x + h⎤⎦ such
that


1
h∫x


x + h
f (x)dx = f (c).


In addition, since c is between x and h, c approaches x as h approaches zero. Also, since f (x) is continuous, we have
lim
h → 0


f (c) = lim
c → x


f (c) = f (x). Putting all these pieces together, we have


F′ (x) = lim
h → 0


1
h∫x


x + h
f (x)dx


= lim
h → 0


f (c)


= f (x),


and the proof is complete.

Example 1.17
Finding a Derivative with the Fundamental Theorem of Calculus
Use the Fundamental Theorem of Calculus, Part 1 to find the derivative of


g(x) = ⌠
⌡1


x
1


t3 + 1
dt.


Solution
According to the Fundamental Theorem of Calculus, the derivative is given by


g′ (x) = 1
x3 + 1


.


Chapter 1 | Integration 51




1.16


1.17


Use the Fundamental Theorem of Calculus, Part 1 to find the derivative of g(r) = ∫
0


r
x2 + 4dx.


Example 1.18
Using the Fundamental Theorem and the Chain Rule to Calculate Derivatives
Let F(x) = ∫


1


x
sin tdt. Find F′ (x).


Solution
Letting u(x) = x, we have F(x) = ∫


1


u(x)
sin tdt. Thus, by the Fundamental Theorem of Calculus and the chain


rule,
F′ (x) = sin⎛⎝u(x)⎞⎠du


dx


= sin(u(x)) · ⎛⎝
1
2
x−1/2⎞⎠


= sin x
2 x


.


Let F(x) = ∫
1


x3


cos tdt. Find F′ (x).


Example 1.19
Using the Fundamental Theorem of Calculus with Two Variable Limits ofIntegration
Let F(x) = ∫


x


2x
t3dt. Find F′ (x).


Solution
We have F(x) = ∫


x


2x
t3dt. Both limits of integration are variable, so we need to split this into two integrals. We


get
F(x) = ∫


x


2x
t3dt


= ∫
x


0
t3dt + ∫


0


2x
t3dt


= −∫
0


x
t3dt + ∫


0


2x
t3dt.


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1.18


Differentiating the first term, we obtain
d
dx

⎣−∫0


x
t3dt

⎦ = −x


3.


Differentiating the second term, we first let u(x) = 2x. Then,
d
dx





⎢∫


0


2x
t3dt



⎥ = d


dx





⎢∫


0


u(x)
t3dt





= (u(x))3 du
dx


= (2x)3 · 2


= 16x3.


Thus,


F′ (x) = d
dx

⎣−∫0


x
t3dt

⎦+


d
dx





⎢∫


0


2x
t3dt





= −x3 + 16x3


= 15x3.


Let F(x) = ∫
x


x2


cos tdt. Find F′ (x).


Fundamental Theorem of Calculus, Part 2: The Evaluation Theorem
The Fundamental Theorem of Calculus, Part 2, is perhaps the most important theorem in calculus. After tireless effortsby mathematicians for approximately 500 years, new techniques emerged that provided scientists with the necessary toolsto explain many phenomena. Using calculus, astronomers could finally determine distances in space and map planetaryorbits. Everyday financial problems such as calculating marginal costs or predicting total profit could now be handled withsimplicity and accuracy. Engineers could calculate the bending strength of materials or the three-dimensional motion ofobjects. Our view of the world was forever changed with calculus.
After finding approximate areas by adding the areas of n rectangles, the application of this theorem is straightforward bycomparison. It almost seems too simple that the area of an entire curved region can be calculated by just evaluating anantiderivative at the first and last endpoints of an interval.
Theorem 1.5: The Fundamental Theorem of Calculus, Part 2
If f is continuous over the interval ⎡⎣a, b⎤⎦ and F(x) is any antiderivative of f (x), then


(1.17)

a


b
f (x)dx = F(b) − F(a).


We often see the notation F(x)|ab to denote the expression F(b) − F(a). We use this vertical bar and associated limits a
and b to indicate that we should evaluate the function F(x) at the upper limit (in this case, b), and subtract the value of the
function F(x) evaluated at the lower limit (in this case, a).
The Fundamental Theorem of Calculus, Part 2 (also known as the evaluation theorem) states that if we can find an


Chapter 1 | Integration 53




antiderivative for the integrand, then we can evaluate the definite integral by evaluating the antiderivative at the endpointsof the interval and subtracting.
Proof
Let P = {xi}, i = 0, 1,…, n be a regular partition of ⎡⎣a, b⎤⎦. Then, we can write


F(b) − F(a) = F(xn) − F(x0)


= ⎡⎣F(xn) − F(xn − 1)

⎦+ ⎡⎣F(xn − 1) − F(xn − 2)



⎦+… + ⎡⎣F(x1) − F(x0)





= ∑
i = 1


n

⎣F(xi) − F(xi − 1)



⎦.


Now, we know F is an antiderivative of f over ⎡⎣a, b⎤⎦, so by the Mean Value Theorem (see The Mean Value Theorem
(http://cnx.org/content/m53612/latest/) ) for i = 0, 1,…, n we can find ci in [xi − 1, xi] such that


F(xi) − F(xi − 1) = F′ (ci

⎠(xi − xi − 1) = f (ci)Δx.


Then, substituting into the previous equation, we have
F(b) − F(a) = ∑


i = 1


n


f (ci)Δx.


Taking the limit of both sides as n → ∞, we obtain


F(b) − F(a) = lim
n → ∞



i = 1


n


f (ci)Δx


= ∫
a


b
f (x)dx.



Example 1.20
Evaluating an Integral with the Fundamental Theorem of Calculus
Use The Fundamental Theorem of Calculus, Part 2 to evaluate



−2


2

⎝t
2 − 4⎞⎠dt.


Solution
Recall the power rule for Antiderivatives (http://cnx.org/content/m53621/latest/) :


If y = xn, ∫ xndx = x
n + 1


n + 1
+ C.


Use this rule to find the antiderivative of the function and then apply the theorem. We have


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−2


2

⎝t
2 − 4⎞⎠dt =


t3


3
− 4t|−2


2


=


(2)3


3
− 4(2)

⎦−


(−2)3


3
− 4(−2)





= ⎛⎝
8
3
− 8⎞⎠−



⎝−


8
3
+ 8⎞⎠


= 8
3
− 8 + 8


3
− 8


= 16
3


− 16


= − 32
3
.


Analysis
Notice that we did not include the “+ C” term when we wrote the antiderivative. The reason is that, accordingto the Fundamental Theorem of Calculus, Part 2, any antiderivative works. So, for convenience, we chose theantiderivative with C = 0. If we had chosen another antiderivative, the constant term would have canceled out.
This always happens when evaluating a definite integral.
The region of the area we just calculated is depicted in Figure 1.28. Note that the region between the curveand the x-axis is all below the x-axis. Area is always positive, but a definite integral can still produce a negativenumber (a net signed area). For example, if this were a profit function, a negative number indicates the companyis operating at a loss over the given interval.


Figure 1.28 The evaluation of a definite integral can producea negative value, even though area is always positive.


Example 1.21
Evaluating a Definite Integral Using the Fundamental Theorem of Calculus, Part 2
Evaluate the following integral using the Fundamental Theorem of Calculus, Part 2:



1


9
x − 1


x dx.


Solution


Chapter 1 | Integration 55




1.19


First, eliminate the radical by rewriting the integral using rational exponents. Then, separate the numerator termsby writing each one over the denominator:

⌡1


9
x − 1
x1/2


dx = ⌠
⌡1


9⎛


x
x1/2


− 1
x1/2

⎠dx.


Use the properties of exponents to simplify:

⌡1


9⎛


x
x1/2


− 1
x1/2

⎠dx = ∫1


9

⎝x


1/2 − x−1/2⎞⎠dx.


Now, integrate using the power rule:



1


9

⎝x


1/2 − x−1/2⎞⎠dx =



⎜x


3/2


3
2


− x
1/2


1
2





⎟|1
9


=



⎢(9)


3/2


3
2


− (9)
1/2


1
2





⎥−



⎢(1)


3/2


3
2


− (1)
1/2


1
2







= ⎡⎣
2
3
(27) − 2(3)⎤⎦−




2
3
(1) − 2(1)⎤⎦


= 18 − 6 − 2
3
+ 2


= 40
3
.


See Figure 1.29.


Figure 1.29 The area under the curve from x = 1 to x = 9
can be calculated by evaluating a definite integral.


Use The Fundamental Theorem of Calculus, Part 2 to evaluate ∫
1


2
x−4dx.


Example 1.22
A Roller-Skating Race
James and Kathy are racing on roller skates. They race along a long, straight track, and whoever has gone the


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1.20


farthest after 5 sec wins a prize. If James can skate at a velocity of f (t) = 5 + 2t ft/sec and Kathy can skate at a
velocity of g(t) = 10 + cos⎛⎝π2 t⎞⎠ ft/sec, who is going to win the race?


Solution
We need to integrate both functions over the interval ⎡⎣0, 5⎤⎦ and see which value is bigger. For James, we want to
calculate



0


5
(5 + 2t)dt.


Using the power rule, we have


0


5
(5 + 2t)dt = ⎛⎝5t + t


2⎞
⎠|0
5


= (25 + 25) = 50.


Thus, James has skated 50 ft after 5 sec. Turning now to Kathy, we want to calculate


0


5
10 + cos⎛⎝


π
2
t⎞⎠dt.


We know sin t is an antiderivative of cos t, so it is reasonable to expect that an antiderivative of cos⎛⎝π2 t⎞⎠ would
involve sin⎛⎝π2 t⎞⎠. However, when we differentiate sin⎛⎝π2 t⎞⎠, we get π2cos⎛⎝π2 t⎞⎠ as a result of the chain rule, so we
have to account for this additional coefficient when we integrate. We obtain



0


5
10 + cos⎛⎝


π
2
t⎞⎠dt =



⎝10t +


2
πsin


π
2
t⎞⎠

⎠|0
5


= ⎛⎝50 +
2
π

⎠−

⎝0 −


2
πsin0





≈ 50.6.


Kathy has skated approximately 50.6 ft after 5 sec. Kathy wins, but not by much!


Suppose James and Kathy have a rematch, but this time the official stops the contest after only 3 sec.Does this change the outcome?


Chapter 1 | Integration 57




A Parachutist in Free Fall


Figure 1.30 Skydivers can adjust the velocity of their dive by changing the position of their body during thefree fall. (credit: Jeremy T. Lock)


Julie is an avid skydiver. She has more than 300 jumps under her belt and has mastered the art of making adjustmentsto her body position in the air to control how fast she falls. If she arches her back and points her belly toward theground, she reaches a terminal velocity of approximately 120 mph (176 ft/sec). If, instead, she orients her body withher head straight down, she falls faster, reaching a terminal velocity of 150 mph (220 ft/sec).
Since Julie will be moving (falling) in a downward direction, we assume the downward direction is positive to simplifyour calculations. Julie executes her jumps from an altitude of 12,500 ft. After she exits the aircraft, she immediatelystarts falling at a velocity given by v(t) = 32t. She continues to accelerate according to this velocity function until she
reaches terminal velocity. After she reaches terminal velocity, her speed remains constant until she pulls her ripcordand slows down to land.
On her first jump of the day, Julie orients herself in the slower “belly down” position (terminal velocity is 176 ft/sec).Using this information, answer the following questions.


1. How long after she exits the aircraft does Julie reach terminal velocity?
2. Based on your answer to question 1, set up an expression involving one or more integrals that represents thedistance Julie falls after 30 sec.
3. If Julie pulls her ripcord at an altitude of 3000 ft, how long does she spend in a free fall?
4. Julie pulls her ripcord at 3000 ft. It takes 5 sec for her parachute to open completely and for her to slow down,during which time she falls another 400 ft. After her canopy is fully open, her speed is reduced to 16 ft/sec.Find the total time Julie spends in the air, from the time she leaves the airplane until the time her feet touch theground.


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On Julie’s second jump of the day, she decides she wants to fall a little faster and orients herself in the “headdown” position. Her terminal velocity in this position is 220 ft/sec. Answer these questions based on thisvelocity:
5. How long does it take Julie to reach terminal velocity in this case?
6. Before pulling her ripcord, Julie reorients her body in the “belly down” position so she is not moving quite asfast when her parachute opens. If she begins this maneuver at an altitude of 4000 ft, how long does she spendin a free fall before beginning the reorientation?Some jumpers wear “ wingsuits” (see Figure 1.31). These suits have fabric panels between the arms and legsand allow the wearer to glide around in a free fall, much like a flying squirrel. (Indeed, the suits are sometimescalled “flying squirrel suits.”) When wearing these suits, terminal velocity can be reduced to about 30 mph (44ft/sec), allowing the wearers a much longer time in the air. Wingsuit flyers still use parachutes to land; althoughthe vertical velocities are within the margin of safety, horizontal velocities can exceed 70 mph, much too fastto land safely.


Figure 1.31 The fabric panels on the arms and legs of a wingsuit work to reduce the vertical velocity of askydiver’s fall. (credit: Richard Schneider)


Answer the following question based on the velocity in a wingsuit.
7. If Julie dons a wingsuit before her third jump of the day, and she pulls her ripcord at an altitude of 3000 ft, howlong does she get to spend gliding around in the air?


Chapter 1 | Integration 59




1.3 EXERCISES
144. Consider two athletes running at variable speeds
v1 (t) and v2 (t). The runners start and finish a race at
exactly the same time. Explain why the two runners mustbe going the same speed at some point.
145. Two mountain climbers start their climb at basecamp, taking two different routes, one steeper than theother, and arrive at the peak at exactly the same time. Is itnecessarily true that, at some point, both climbers increasedin altitude at the same rate?
146. To get on a certain toll road a driver has to take acard that lists the mile entrance point. The card also has atimestamp. When going to pay the toll at the exit, the driveris surprised to receive a speeding ticket along with the toll.Explain how this can happen.
147. Set F(x) = ∫


1


x
(1 − t)dt. Find F′ (2) and the


average value of F ′ over [1, 2].
In the following exercises, use the Fundamental Theoremof Calculus, Part 1, to find each derivative.
148. d


dx∫1
x
e−t


2
dt


149. d
dx∫1


x
ecos t dt


150. d
dx∫3


x
9 − y2dy


151. d
dx

⌡4


x
ds


16 − s2


152. d
dx∫x


2x
tdt


153. d
dx∫0


x
tdt


154. d
dx∫0


sinx
1 − t2dt


155. d
dx∫cosx


1
1 − t2dt


156. d
dx

⌡1


x
t2


1 + t4
dt


157. d
dx

⌡1


x2
t


1 + t
dt


158. d
dx∫0


lnx
et dt


159. d
dx∫1


e2


lnu2du


160. The graph of y = ∫
0


x
f (t)dt, where f is a piecewise


constant function, is shown here.


a. Over which intervals is f positive? Over whichintervals is it negative? Over which intervals, ifany, is it equal to zero?b. What are the maximum and minimum values of f?c. What is the average value of f?
161. The graph of y = ∫


0


x
f (t)dt, where f is a piecewise


constant function, is shown here.


a. Over which intervals is f positive? Over whichintervals is it negative? Over which intervals, ifany, is it equal to zero?b. What are the maximum and minimum values of f?c. What is the average value of f?


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162. The graph of y = ∫
0


x
ℓ(t)dt, where ℓ is a piecewise


linear function, is shown here.


a. Over which intervals is ℓ positive? Over whichintervals is it negative? Over which, if any, is itzero?b. Over which intervals is ℓ increasing? Over which isit decreasing? Over which, if any, is it constant?c. What is the average value of ℓ?
163. The graph of y = ∫


0


x
ℓ(t)dt, where ℓ is a piecewise


linear function, is shown here.


a. Over which intervals is ℓ positive? Over whichintervals is it negative? Over which, if any, is itzero?b. Over which intervals is ℓ increasing? Over whichis it decreasing? Over which intervals, if any, is itconstant?c. What is the average value of ℓ?
In the following exercises, use a calculator to estimate thearea under the curve by computing T10, the average ofthe left- and right-endpoint Riemann sums using N = 10
rectangles. Then, using the Fundamental Theorem ofCalculus, Part 2, determine the exact area.
164. [T] y = x2 over [0, 4]
165. [T] y = x3 + 6x2 + x − 5 over [−4, 2]
166. [T] y = x3 over ⎡⎣0, 6⎤⎦
167. [T] y = x + x2 over [1, 9]
168. [T] ∫ (cosx − sinx)dx over [0, π]


169. [T] ⌠

4
x2
dx over [1, 4]


In the following exercises, evaluate each definite integralusing the Fundamental Theorem of Calculus, Part 2.
170. ∫


−1


2

⎝x


2 − 3x⎞⎠dx


171. ∫
−2


3

⎝x


2 + 3x − 5⎞⎠dx


172. ∫
−2


3
(t + 2)(t − 3)dt


173. ∫
2


3

⎝t
2 − 9⎞⎠



⎝4 − t


2⎞
⎠dt


174. ∫
1


2
x9dx


175. ∫
0


1
x99dx


176. ∫
4


8

⎝4t


5/2 − 3t3/2⎞⎠dt


177. ⌠
⌡1/4


4 ⎛
⎝x


2 − 1
x2

⎠dx


178. ⌠
⌡1


2
2
x3
dx


179. ⌠
⌡1


4
1


2 x
dx


180. ⌠
⌡1


4
2 − t
t2


dt


181. ⌠
⌡1


16
dt
t1/4


182. ∫
0



cosθdθ


183. ∫
0


π/2
sinθdθ


Chapter 1 | Integration 61




184. ∫
0


π/4
sec2 θdθ


185. ∫
0


π/4
secθ tanθ


186. ∫
π/3


π/4
cscθcotθdθ


187. ∫
π/4


π/2
csc2 θdθ


188. ⌠
⌡1


2⎛

1
t2


− 1
t3

⎠dt


189. ⌠
⌡−2


−1⎛

1
t2


− 1
t3

⎠dt


In the following exercises, use the evaluation theorem toexpress the integral as a function F(x).
190. ∫


a


x
t2dt


191. ∫
1


x
et dt


192. ∫
0


x
cos tdt


193. ∫
−x


x
sin tdt


In the following exercises, identify the roots of theintegrand to remove absolute values, then evaluate usingthe Fundamental Theorem of Calculus, Part 2.
194. ∫


−2


3
|x|dx


195. ∫
−2


4


|t2 − 2t − 3|dt


196. ∫
0


π
|cos t|dt


197. ∫
−π/2


π/2


|sin t|dt


198. Suppose that the number of hours of daylight ona given day in Seattle is modeled by the function
−3.75cos⎛⎝


πt
6

⎠+ 12.25, with t given in months and


t = 0 corresponding to the winter solstice.
a. What is the average number of daylight hours in ayear?b. At which times t1 and t2, where


0 ≤ t1 < t2 < 12, do the number of daylight
hours equal the average number?c. Write an integral that expresses the total number ofdaylight hours in Seattle between t1 and t2.


d. Compute the mean hours of daylight in Seattlebetween t1 and t2, where 0 ≤ t1 < t2 < 12,
and then between t2 and t1, and show that the
average of the two is equal to the average daylength.


199. Suppose the rate of gasoline consumption in theUnited States can be modeled by a sinusoidal function of
the form ⎛⎝11.21 − cos⎛⎝πt6 ⎞⎠⎞⎠× 109 gal/mo.


a. What is the average monthly consumption, and forwhich values of t is the rate at time t equal to theaverage rate?b. What is the number of gallons of gasolineconsumed in the United States in a year?c. Write an integral that expresses the averagemonthly U.S. gas consumption during the part ofthe year between the beginning of April (t = 3)
and the end of September ⎛⎝t = 9).


200. Explain why, if f is continuous over ⎡⎣a, b⎤⎦, there
is at least one point c ∈ ⎡⎣a, b⎤⎦ such that
f (c) = 1


b − a∫a
b
f (t)dt.


201. Explain why, if f is continuous over ⎡⎣a, b⎤⎦ and is not
equal to a constant, there is at least one point M ∈ ⎡⎣a, b⎤⎦
such that f (M) = 1


b − a∫a
b
f (t)dt and at least one point


m ∈ ⎡⎣a, b⎤⎦ such that f (m) < 1
b − a∫a


b
f (t)dt.


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202. Kepler’s first law states that the planets move inelliptical orbits with the Sun at one focus. The closest pointof a planetary orbit to the Sun is called the perihelion (forEarth, it currently occurs around January 3) and the farthestpoint is called the aphelion (for Earth, it currently occursaround July 4). Kepler’s second law states that planetssweep out equal areas of their elliptical orbits in equaltimes. Thus, the two arcs indicated in the following figureare swept out in equal times. At what time of year is Earthmoving fastest in its orbit? When is it moving slowest?


203. A point on an ellipse with major axis length 2aand minor axis length 2b has the coordinates
(acosθ, bsinθ), 0 ≤ θ ≤ 2π.


a. Show that the distance from this point to the focusat (−c, 0) is d(θ) = a + ccosθ, where
c = a2 − b2.


b. Use these coordinates to show that the average
distance d– from a point on the ellipse to the focus
at (−c, 0), with respect to angle θ, is a.


204. As implied earlier, according to Kepler’s laws,Earth’s orbit is an ellipse with the Sun at one focus. Theperihelion for Earth’s orbit around the Sun is 147,098,290km and the aphelion is 152,098,232 km.a. By placing the major axis along the x-axis, find theaverage distance from Earth to the Sun.b. The classic definition of an astronomical unit (AU)is the distance from Earth to the Sun, and its valuewas computed as the average of the perihelion andaphelion distances. Is this definition justified?
205. The force of gravitational attraction between the Sun
and a planet is F(θ) = GmM


r2 (θ)
, where m is the mass of the


planet, M is the mass of the Sun, G is a universal constant,and r(θ) is the distance between the Sun and the planet
when the planet is at an angle θ with the major axis of itsorbit. Assuming that M, m, and the ellipse parameters aand b (half-lengths of the major and minor axes) are given,set up—but do not evaluate—an integral that expresses interms of G, m, M, a, b the average gravitational force
between the Sun and the planet.


206. The displacement from rest of a mass attached toa spring satisfies the simple harmonic motion equation
x(t) = Acos⎛⎝ωt − ϕ⎞⎠, where ϕ is a phase constant, ω is
the angular frequency, and A is the amplitude. Find theaverage velocity, the average speed (magnitude ofvelocity), the average displacement, and the averagedistance from rest (magnitude of displacement) of themass.


Chapter 1 | Integration 63




1.21


1.4 | Integration Formulas and the Net Change Theorem
Learning Objectives


1.4.1 Apply the basic integration formulas.
1.4.2 Explain the significance of the net change theorem.
1.4.3 Use the net change theorem to solve applied problems.
1.4.4 Apply the integrals of odd and even functions.


In this section, we use some basic integration formulas studied previously to solve some key applied problems. It isimportant to note that these formulas are presented in terms of indefinite integrals. Although definite and indefinite integralsare closely related, there are some key differences to keep in mind. A definite integral is either a number (when the limitsof integration are constants) or a single function (when one or both of the limits of integration are variables). An indefiniteintegral represents a family of functions, all of which differ by a constant. As you become more familiar with integration,you will get a feel for when to use definite integrals and when to use indefinite integrals. You will naturally select the correctapproach for a given problem without thinking too much about it. However, until these concepts are cemented in your mind,think carefully about whether you need a definite integral or an indefinite integral and make sure you are using the propernotation based on your choice.
Basic Integration Formulas
Recall the integration formulas given in the table in Antiderivatives (http://cnx.org/content/m53621/latest/#fs-id1165043092431) and the rule on properties of definite integrals. Let’s look at a few examples of how to apply theserules.
Example 1.23
Integrating a Function Using the Power Rule
Use the power rule to integrate the function ∫


1


4
t(1 + t)dt.


Solution
The first step is to rewrite the function and simplify it so we can apply the power rule:



1


4
t(1 + t)dt = ∫


1


4
t1/2(1 + t)dt


= ∫
1


4

⎝t
1/2 + t3/2⎞⎠dt.


Now apply the power rule:


1


4

⎝t
1/2 + t3/2⎞⎠dt =




2
3
t3/2 + 2


5
t5/2⎞⎠|1


4


= ⎡⎣
2
3
(4)3/2 + 2


5
(4)5/2⎤⎦−




2
3
(1)3/2 + 2


5
(1)5/2⎤⎦


= 256
15


.


Find the definite integral of f (x) = x2 − 3x over the interval [1, 3].


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The Net Change Theorem
The net change theorem considers the integral of a rate of change. It says that when a quantity changes, the new valueequals the initial value plus the integral of the rate of change of that quantity. The formula can be expressed in two ways.The second is more familiar; it is simply the definite integral.
Theorem 1.6: Net Change Theorem
The new value of a changing quantity equals the initial value plus the integral of the rate of change:


(1.18)
F(b) = F(a) + ∫


a


b
F '(x)dx


or



a


b
F '(x)dx = F(b) − F(a).


Subtracting F(a) from both sides of the first equation yields the second equation. Since they are equivalent formulas, which
one we use depends on the application.
The significance of the net change theorem lies in the results. Net change can be applied to area, distance, and volume, toname only a few applications. Net change accounts for negative quantities automatically without having to write more thanone integral. To illustrate, let’s apply the net change theorem to a velocity function in which the result is displacement.
We looked at a simple example of this in The Definite Integral. Suppose a car is moving due north (the positive direction)at 40 mph between 2 p.m. and 4 p.m., then the car moves south at 30 mph between 4 p.m. and 5 p.m. We can graph thismotion as shown in Figure 1.32.


Figure 1.32 The graph shows speed versus time for the givenmotion of a car.


Just as we did before, we can use definite integrals to calculate the net displacement as well as the total distance traveled.The net displacement is given by


2


5
v(t)dt = ∫


2


4
40dt +⌠


⌡4


5


−30dt


= 80 − 30
= 50.


Thus, at 5 p.m. the car is 50 mi north of its starting position. The total distance traveled is given by


Chapter 1 | Integration 65





2


5


|v(t)|dt = ∫
2


4
40dt +⌠


⌡4


5


30dt


= 80 + 30
= 110.


Therefore, between 2 p.m. and 5 p.m., the car traveled a total of 110 mi.
To summarize, net displacement may include both positive and negative values. In other words, the velocity functionaccounts for both forward distance and backward distance. To find net displacement, integrate the velocity function overthe interval. Total distance traveled, on the other hand, is always positive. To find the total distance traveled by an object,regardless of direction, we need to integrate the absolute value of the velocity function.
Example 1.24
Finding Net Displacement
Given a velocity function v(t) = 3t − 5 (in meters per second) for a particle in motion from time t = 0 to time
t = 3, find the net displacement of the particle.
Solution
Applying the net change theorem, we have



0


3
(3t − 5)dt = 3t


2


2
− 5t|0


3


=


3(3)2


2
− 5(3)

⎦− 0


= 27
2


− 15


= 27
2


− 30
2


= − 3
2
.


The net displacement is −3
2
m (Figure 1.33).


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Figure 1.33 The graph shows velocity versus time for aparticle moving with a linear velocity function.


Example 1.25
Finding the Total Distance Traveled
Use Example 1.24 to find the total distance traveled by a particle according to the velocity function
v(t) = 3t − 5 m/sec over a time interval [0, 3].
Solution
The total distance traveled includes both the positive and the negative values. Therefore, we must integrate theabsolute value of the velocity function to find the total distance traveled.
To continue with the example, use two integrals to find the total distance. First, find the t-intercept of the function,since that is where the division of the interval occurs. Set the equation equal to zero and solve for t. Thus,


3t − 5 = 0
3t = 5


t = 5
3
.


The two subintervals are ⎡⎣0, 53⎤⎦ and ⎡⎣53, 3⎤⎦. To find the total distance traveled, integrate the absolute value of
the function. Since the function is negative over the interval ⎡⎣0, 53⎤⎦, we have |v(t)| = −v(t) over that interval.
Over ⎡⎣53, 3⎤⎦, the function is positive, so |v(t)| = v(t). Thus, we have


Chapter 1 | Integration 67




1.22



0


3


|v(t)|dt = ⌠
⌡0


5/3


−v(t)dt + ∫
5/3


3
v(t)dt


= ∫
0


5/3
5 − 3tdt + ∫


5/3


3
3t − 5dt


=

⎝5t −


3t2
2

⎠|0
5/3


+


3t2
2


− 5t

⎠|5/3
3


=

⎣5


5
3

⎠−


3(5/3)2


2

⎦− 0 +




27
2


− 15⎤⎦−


3(5/3)2


2
− 25


3



= 25
3


− 25
6


+ 27
2


− 15 − 25
6


+ 25
3


= 41
6
.


So, the total distance traveled is 14
6
m.


Find the net displacement and total distance traveled in meters given the velocity function
f (t) = 1


2
et − 2 over the interval [0, 2].


Applying the Net Change Theorem
The net change theorem can be applied to the flow and consumption of fluids, as shown in Example 1.26.
Example 1.26
How Many Gallons of Gasoline Are Consumed?
If the motor on a motorboat is started at t = 0 and the boat consumes gasoline at the rate of 5 − t3 gal/hr, how
much gasoline is used in the first 2 hours?
Solution
Express the problem as a definite integral, integrate, and evaluate using the Fundamental Theorem of Calculus.The limits of integration are the endpoints of the interval [0, 2]. We have



0


2

⎝5 − t


3⎞
⎠dt =



⎝5t −


t4
4

⎠|0
2


=

⎣5(2) −


(2)4


4

⎦− 0


= 10 − 16
4


= 6.


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Thus, the motorboat uses 6 gal of gas in 2 hours.


Example 1.27
Chapter Opener: Iceboats


Figure 1.34 (credit: modification of work by Carter Brown,Flickr)


As we saw at the beginning of the chapter, top iceboat racers (Figure 1.1) can attain speeds of up to five times thewind speed. Andrew is an intermediate iceboater, though, so he attains speeds equal to only twice the wind speed.Suppose Andrew takes his iceboat out one morning when a light 5-mph breeze has been blowing all morning. AsAndrew gets his iceboat set up, though, the wind begins to pick up. During his first half hour of iceboating, thewind speed increases according to the function v(t) = 20t + 5. For the second half hour of Andrew’s outing, the
wind remains steady at 15 mph. In other words, the wind speed is given by


v(t) =




20t + 5 for 0 ≤ t ≤ 1


2


15 for 1
2
≤ t ≤ 1.


Recalling that Andrew’s iceboat travels at twice the wind speed, and assuming he moves in a straight line awayfrom his starting point, how far is Andrew from his starting point after 1 hour?
Solution
To figure out how far Andrew has traveled, we need to integrate his velocity, which is twice the wind speed. Then
Distance = ∫


0


1
2v(t)dt.


Substituting the expressions we were given for v(t), we get


Chapter 1 | Integration 69




1.23



0


1
2v(t)dt = ⌠


⌡0


1/2


2v(t)dt + ∫
1/2


1
2v(t)dt


= ⌠
⌡0


1/2


2(20t + 5)dt + ∫
1/3


1
2(15)dt


= ⌠
⌡0


1/2


(40t + 10)dt + ∫
1/2


1
30dt


= ⎡⎣20t
2 + 10t⎤⎦|0


1/2 + [30t]|1/2
1


= ⎛⎝
20
4


+ 5⎞⎠− 0 + (30 − 15)


= 25.


Andrew is 25 mi from his starting point after 1 hour.


Suppose that, instead of remaining steady during the second half hour of Andrew’s outing, the windstarts to die down according to the function v(t) = −10t + 15. In other words, the wind speed is given by


v(t) =




20t + 5 for 0 ≤ t ≤ 1


2


−10t + 15 for 1
2
≤ t ≤ 1.


Under these conditions, how far from his starting point is Andrew after 1 hour?


Integrating Even and Odd Functions
We saw in Functions and Graphs (http://cnx.org/content/m53472/latest/) that an even function is a function inwhich f (−x) = f (x) for all x in the domain—that is, the graph of the curve is unchanged when x is replaced with −x. The
graphs of even functions are symmetric about the y-axis. An odd function is one in which f (−x) = − f (x) for all x in the
domain, and the graph of the function is symmetric about the origin.
Integrals of even functions, when the limits of integration are from −a to a, involve two equal areas, because they aresymmetric about the y-axis. Integrals of odd functions, when the limits of integration are similarly [−a, a], evaluate to
zero because the areas above and below the x-axis are equal.
Rule: Integrals of Even and Odd Functions
For continuous even functions such that f (−x) = f (x),



⌡−a


a


f (x)dx = 2∫
0


a
f (x)dx.


For continuous odd functions such that f (−x) = − f (x),


−a


a
f (x)dx = 0.


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Example 1.28
Integrating an Even Function
Integrate the even function ∫


−2


2

⎝3x


8 − 2⎞⎠dx and verify that the integration formula for even functions holds.


Solution
The symmetry appears in the graphs in Figure 1.35. Graph (a) shows the region below the curve and above thex-axis. We have to zoom in to this graph by a huge amount to see the region. Graph (b) shows the region abovethe curve and below the x-axis. The signed area of this region is negative. Both views illustrate the symmetryabout the y-axis of an even function. We have



−2


2

⎝3x


8 − 2⎞⎠dx =


x9


3
− 2x

⎠|−2
2


=



⎢ (2)


9


3
− 2(2)





⎥ −



⎢ (−2)


9


3
− 2(−2)







= ⎛⎝
512
3


− 4⎞⎠−

⎝−


512
3


+ 4⎞⎠


= 1000
3


.


To verify the integration formula for even functions, we can calculate the integral from 0 to 2 and double it, thencheck to make sure we get the same answer.


0


2

⎝3x


8 − 2⎞⎠dx =


x9


3
− 2x

⎠|0
2


= 512
3


− 4


= 500
3


Since 2 · 500
3


= 1000
3


, we have verified the formula for even functions in this particular example.


Figure 1.35 Graph (a) shows the positive area between the curve and the x-axis, whereas graph (b) shows the negative areabetween the curve and the x-axis. Both views show the symmetry about the y-axis.


Chapter 1 | Integration 71




1.24


Example 1.29
Integrating an Odd Function
Evaluate the definite integral of the odd function −5sinx over the interval [−π, π].
Solution
The graph is shown in Figure 1.36. We can see the symmetry about the origin by the positive area above thex-axis over [−π, 0], and the negative area below the x-axis over [0, π]. We have



−π


π
−5sinxdx = −5(−cosx)|−π


π


= 5cosx|−π
π


= [5cosπ] − ⎡⎣5cos(−π)⎤⎦
= −5 − (−5)
= 0.


Figure 1.36 The graph shows areas between a curve and thex-axis for an odd function.


Integrate the function ∫
−2


2
x4dx.


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1.4 EXERCISES
Use basic integration formulas to compute the followingantiderivatives.
207. ∫ ⎛⎝ x − 1x⎞⎠dx


208. ⌠


⎝e


2x − 1
2
ex/2⎞⎠dx


209. ⌠

dx
2x


210. ⌠

x − 1
x2


dx


211. ∫
0


π
(sinx − cosx)dx


212. ∫
0


π/2
(x − sinx)dx


213. Write an integral that expresses the increase in theperimeter P(s) of a square when its side length s increases
from 2 units to 4 units and evaluate the integral.
214. Write an integral that quantifies the change in the
area A(s) = s2 of a square when the side length doubles
from S units to 2S units and evaluate the integral.
215. A regular N-gon (an N-sided polygon with sides thathave equal length s, such as a pentagon or hexagon) hasperimeter Ns. Write an integral that expresses the increasein perimeter of a regular N-gon when the length of each sideincreases from 1 unit to 2 units and evaluate the integral.
216. The area of a regular pentagon with side length
a > 0 is pa2 with p = 1


4
5 + 5 + 2 5. The Pentagon in


Washington, DC, has inner sides of length 360 ft and outersides of length 920 ft. Write an integral to express the areaof the roof of the Pentagon according to these dimensionsand evaluate this area.
217. A dodecahedron is a Platonic solid with a surface thatconsists of 12 pentagons, each of equal area. By how muchdoes the surface area of a dodecahedron increase as the sidelength of each pentagon doubles from 1 unit to 2 units?
218. An icosahedron is a Platonic solid with a surface thatconsists of 20 equilateral triangles. By how much does thesurface area of an icosahedron increase as the side length ofeach triangle doubles from a unit to 2a units?
219. Write an integral that quantifies the change in thearea of the surface of a cube when its side length doublesfrom s unit to 2s units and evaluate the integral.


220. Write an integral that quantifies the increase in thevolume of a cube when the side length doubles from s unitto 2s units and evaluate the integral.
221. Write an integral that quantifies the increase in thesurface area of a sphere as its radius doubles from R unit to2R units and evaluate the integral.
222. Write an integral that quantifies the increase in thevolume of a sphere as its radius doubles from R unit to 2Runits and evaluate the integral.
223. Suppose that a particle moves along a straight linewith velocity v(t) = 4 − 2t, where 0 ≤ t ≤ 2 (in meters
per second). Find the displacement at time t and the totaldistance traveled up to t = 2.
224. Suppose that a particle moves along a straight line
with velocity defined by v(t) = t2 − 3t − 18, where
0 ≤ t ≤ 6 (in meters per second). Find the displacement at
time t and the total distance traveled up to t = 6.
225. Suppose that a particle moves along a straight linewith velocity defined by v(t) = |2t − 6|, where
0 ≤ t ≤ 6 (in meters per second). Find the displacement at
time t and the total distance traveled up to t = 6.
226. Suppose that a particle moves along a straight linewith acceleration defined by a(t) = t − 3, where
0 ≤ t ≤ 6 (in meters per second). Find the velocity and
displacement at time t and the total distance traveled up to
t = 6 if v(0) = 3 and d(0) = 0.
227. A ball is thrown upward from a height of 1.5 m atan initial speed of 40 m/sec. Acceleration resulting fromgravity is −9.8 m/sec2. Neglecting air resistance, solve forthe velocity v(t) and the height h(t) of the ball t seconds
after it is thrown and before it returns to the ground.
228. A ball is thrown upward from a height of 3 m atan initial speed of 60 m/sec. Acceleration resulting fromgravity is −9.8 m/sec2. Neglecting air resistance, solve forthe velocity v(t) and the height h(t) of the ball t seconds
after it is thrown and before it returns to the ground.
229. The area A(t) of a circular shape is growing at a
constant rate. If the area increases from 4π units to 9π unitsbetween times t = 2 and t = 3, find the net change in the
radius during that time.


Chapter 1 | Integration 73




230. A spherical balloon is being inflated at a constantrate. If the volume of the balloon changes from 36π in.3 to288π in.3 between time t = 30 and t = 60 seconds, find
the net change in the radius of the balloon during that time.
231. Water flows into a conical tank with cross-sectional
area πx2 at height x and volume πx3


3
up to height x. If


water flows into the tank at a rate of 1 m3/min, find theheight of water in the tank after 5 min. Find the change inheight between 5 min and 10 min.
232. A horizontal cylindrical tank has cross-sectional area
A(x) = 4⎛⎝6x − x


2⎞
⎠m


2 at height x meters above the bottom
when x ≤ 3.


a. The volume V between heights a and b is

a


b
A(x)dx. Find the volume at heights between 2


m and 3 m.b. Suppose that oil is being pumped into the tankat a rate of 50 L/min. Using the chain rule,
dx
dt


= dx
dV


dV
dt


, at how many meters per minute is
the height of oil in the tank changing, expressed interms of x, when the height is at x meters?c. How long does it take to fill the tank to 3 m startingfrom a fill level of 2 m?


233. The following table lists the electrical power ingigawatts—the rate at which energy is consumed—used ina certain city for different hours of the day, in a typical24-hour period, with hour 1 corresponding to midnight to 1a.m.
Hour Power Hour Power
1 28 13 48
2 25 14 49
3 24 15 49
4 23 16 50
5 24 17 50
6 27 18 50
7 29 19 46
8 32 20 43
9 34 21 42
10 39 22 40
11 42 23 37
12 46 24 34


Find the total amount of power in gigawatt-hours (gW-h)consumed by the city in a typical 24-hour period.


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234. The average residential electrical power use (inhundreds of watts) per hour is given in the following table.
Hour Power Hour Power
1 8 13 12
2 6 14 13
3 5 15 14
4 4 16 15
5 5 17 17
6 6 18 19
7 7 19 18
8 8 20 17
9 9 21 16
10 10 22 16
11 10 23 13
12 11 24 11


a. Compute the average total energy used in a day inkilowatt-hours (kWh).b. If a ton of coal generates 1842 kWh, how long doesit take for an average residence to burn a ton ofcoal?c. Explain why the data might fit a plot of the form
p(t) = 11.5 − 7.5sin⎛⎝


πt
12

⎠.


235. The data in the following table are used to estimatethe average power output produced by Peter Sagan for eachof the last 18 sec of Stage 1 of the 2012 Tour de France.
Second Watts Second Watts
1 600 10 1200
2 500 11 1170
3 575 12 1125
4 1050 13 1100
5 925 14 1075
6 950 15 1000
7 1050 16 950
8 950 17 900
9 1100 18 780


Table 1.6 Average Power Output Source:sportsexercisengineering.com
Estimate the net energy used in kilojoules (kJ), noting that1W = 1 J/s, and the average power output by Sagan duringthis time interval.


Chapter 1 | Integration 75




236. The data in the following table are used to estimatethe average power output produced by Peter Sagan for each15-min interval of Stage 1 of the 2012 Tour de France.
Minutes Watts Minutes Watts
15 200 165 170
30 180 180 220
45 190 195 140
60 230 210 225
75 240 225 170
90 210 240 210
105 210 255 200
120 220 270 220
135 210 285 250
150 150 300 400


Table 1.7 Average Power Output Source:sportsexercisengineering.com
Estimate the net energy used in kilojoules, noting that 1W= 1 J/s.


237. The distribution of incomes as of 2012 in the UnitedStates in $5000 increments is given in the following table.The kth row denotes the percentage of households withincomes between $5000xk and 5000xk + 4999. The row
k = 40 contains all households with income between
$200,000 and $250,000 and k = 41 accounts for all
households with income exceeding $250,000.


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0 3.5 21 1.5
1 4.1 22 1.4
2 5.9 23 1.3
3 5.7 24 1.3
4 5.9 25 1.1
5 5.4 26 1.0
6 5.5 27 0.75
7 5.1 28 0.8
8 4.8 29 1.0
9 4.1 30 0.6
10 4.3 31 0.6
11 3.5 32 0.5
12 3.7 33 0.5
13 3.2 34 0.4
14 3.0 35 0.3
15 2.8 36 0.3
16 2.5 37 0.3
17 2.2 38 0.2
18 2.2 39 1.8


Table 1.8 IncomeDistributions Source:http://www.census.gov/prod/2013pubs/p60-245.pdf


19 1.8 40 2.3
20 2.1 41


Table 1.8 IncomeDistributions Source:http://www.census.gov/prod/2013pubs/p60-245.pdf
a. Estimate the percentage of U.S. households in 2012with incomes less than $55,000.b. What percentage of households had incomesexceeding $85,000?c. Plot the data and try to fit its shape to that of a


graph of the form a(x + c)e−b(x + e) for suitable
a, b, c.


238. Newton’s law of gravity states that the gravitationalforce exerted by an object of mass M and one of massm with centers that are separated by a distance r is
F = GmM


r2
, with G an empirical constant


G = 6.67x10−11 m3 /⎛⎝kg · s
2⎞
⎠. The work done by a


variable force over an interval ⎡⎣a, b⎤⎦ is defined as
W = ∫


a


b
F(x)dx. If Earth has mass 5.97219 × 1024 and


radius 6371 km, compute the amount of work to elevatea polar weather satellite of mass 1400 kg to its orbitingaltitude of 850 km above Earth.
239. For a given motor vehicle, the maximum achievabledeceleration from braking is approximately 7 m/sec2 on dryconcrete. On wet asphalt, it is approximately 2.5 m/sec2.Given that 1 mph corresponds to 0.447 m/sec, find the totaldistance that a car travels in meters on dry concrete after thebrakes are applied until it comes to a complete stop if theinitial velocity is 67 mph (30 m/sec) or if the initial brakingvelocity is 56 mph (25 m/sec). Find the correspondingdistances if the surface is slippery wet asphalt.
240. John is a 25-year old man who weighs 160 lb. Heburns 500 − 50t calories/hr while riding his bike for t
hours. If an oatmeal cookie has 55 cal and John eats 4tcookies during the tth hour, how many net calories has helost after 3 hours riding his bike?
241. Sandra is a 25-year old woman who weighs 120lb. She burns 300 − 50t cal/hr while walking on her
treadmill. Her caloric intake from drinking Gatorade is 100tcalories during the tth hour. What is her net decrease incalories after walking for 3 hours?


Chapter 1 | Integration 77




242. A motor vehicle has a maximum efficiency of 33mpg at a cruising speed of 40 mph. The efficiency drops ata rate of 0.1 mpg/mph between 40 mph and 50 mph, and ata rate of 0.4 mpg/mph between 50 mph and 80 mph. Whatis the efficiency in miles per gallon if the car is cruising at50 mph? What is the efficiency in miles per gallon if the caris cruising at 80 mph? If gasoline costs $3.50/gal, what isthe cost of fuel to drive 50 mi at 40 mph, at 50 mph, and at80 mph?
243. Although some engines are more efficient at givena horsepower than others, on average, fuel efficiencydecreases with horsepower at a rate of 1/25 mpg/
horsepower. If a typical 50-horsepower engine has anaverage fuel efficiency of 32 mpg, what is the average fuelefficiency of an engine with the following horsepower: 150,300, 450?
244. [T] The following table lists the 2013 schedule offederal income tax versus taxable income.


Taxable IncomeRange The Tax Is…
… Of theAmountOver


$0–$8925 10% $0


$8925–$36,250 $892.50 +15% $8925


$36,250–$87,850 $4,991.25 +25% $36,250


$87,850–$183,250 $17,891.25+ 28% $87,850


$183,250–$398,350 $44,603.25+ 33% $183,250


$398,350–$400,000 $115,586.25+ 35% $398,350


> $400,000 $116,163.75+ 39.6% $400,000


Table 1.9 Federal Income Tax Versus TaxableIncome Source: http://www.irs.gov/pub/irs-prior/i1040tt--2013.pdf.
Suppose that Steve just received a $10,000 raise. Howmuch of this raise is left after federal taxes if Steve’s salarybefore receiving the raise was $40,000? If it was $90,000?If it was $385,000?


245. [T] The following table provides hypothetical dataregarding the level of service for a certain highway.
HighwaySpeed Range(mph)


Vehicles perHour perLane
DensityRange(vehicles/mi)


> 60 < 600 < 10
60–57 600–1000 10–20
57–54 1000–1500 20–30
54–46 1500–1900 30–45
46–30 1900–2100 45–70
<30 Unstable 70–200


Table 1.10
a. Plot vehicles per hour per lane on the x-axis andhighway speed on the y-axis.b. Compute the average decrease in speed (in milesper hour) per unit increase in congestion (vehiclesper hour per lane) as the latter increases from 600 to1000, from 1000 to 1500, and from 1500 to 2100.Does the decrease in miles per hour depend linearlyon the increase in vehicles per hour per lane?c. Plot minutes per mile (60 times the reciprocal ofmiles per hour) as a function of vehicles per hourper lane. Is this function linear?


For the next two exercises use the data in the followingtable, which displays bald eagle populations from 1963 to2000 in the continental United States.


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Year Population of Breeding Pairs ofBald Eagles
1963 487
1974 791
1981 1188
1986 1875
1992 3749
1996 5094
2000 6471


Table 1.11 Population of Breeding Bald EaglePairs Source: http://www.fws.gov/Midwest/eagle/population/chtofprs.html.
246. [T] The graph below plots the quadratic
p(t) = 6.48t2 − 80.3 1t + 585.69 against the data in
preceding table, normalized so that t = 0 corresponds to
1963. Estimate the average number of bald eagles per yearpresent for the 37 years by computing the average value ofp over [0, 37].


247. [T] The graph below plots the cubic
p(t) = 0.07t3 + 2.42t2 − 25.63t + 521.23 against the
data in the preceding table, normalized so that t = 0
corresponds to 1963. Estimate the average number of baldeagles per year present for the 37 years by computing theaverage value of p over [0, 37].


248. [T] Suppose you go on a road trip and record yourspeed at every half hour, as compiled in the followingtable. The best quadratic fit to the data is
q(t) = 5x2 − 11x + 49, shown in the accompanying
graph. Integrate q to estimate the total distance driven overthe 3 hours.


Time (hr) Speed (mph)
0 (start) 50
1 40
2 50
3 60


As a car accelerates, it does not accelerate at a constantrate; rather, the acceleration is variable. For the followingexercises, use the following table, which contains theacceleration measured at every second as a driver mergesonto a freeway.


Chapter 1 | Integration 79




Time (sec) Acceleration (mph/sec)
1 11.2
2 10.6
3 8.1
4 5.4
5 0


249. [T] The accompanying graph plots the best quadratic
fit, a(t) = −0.70t2 + 1.44t + 10.44, to the data from the
preceding table. Compute the average value of a(t) to
estimate the average acceleration between t = 0 and
t = 5.


250. [T] Using your acceleration equation from theprevious exercise, find the corresponding velocityequation. Assuming the final velocity is 0 mph, find thevelocity at time t = 0.
251. [T] Using your velocity equation from the previousexercise, find the corresponding distance equation,assuming your initial distance is 0 mi. How far did youtravel while you accelerated your car? (Hint: You will needto convert time units.)


252. [T] The number of hamburgers sold at a restaurantthroughout the day is given in the following table, with theaccompanying graph plotting the best cubic fit to the data,
b(t) = 0.12t3 − 2.13t3 + 12.13t + 3.91, with t = 0
corresponding to 9 a.m. and t = 12 corresponding to 9
p.m. Compute the average value of b(t) to estimate the
average number of hamburgers sold per hour.


Hours Past Midnight No. of Burgers Sold
9 3
12 28
15 20
18 30
21 45


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253. [T] An athlete runs by a motion detector, whichrecords her speed, as displayed in the following table. Thebest linear fit to this data, ℓ(t) = −0.068t + 5.14, is
shown in the accompanying graph. Use the average valueof ℓ(t) between t = 0 and t = 40 to estimate the
runner’s average speed.


Minutes Speed (m/sec)
0 5
10 4.8
20 3.6
30 3.0
40 2.5


Chapter 1 | Integration 81




1.5 | Substitution
Learning Objectives


1.5.1 Use substitution to evaluate indefinite integrals.
1.5.2 Use substitution to evaluate definite integrals.


The Fundamental Theorem of Calculus gave us a method to evaluate integrals without using Riemann sums. The drawbackof this method, though, is that we must be able to find an antiderivative, and this is not always easy. In this section weexamine a technique, called integration by substitution, to help us find antiderivatives. Specifically, this method helps usfind antiderivatives when the integrand is the result of a chain-rule derivative.
At first, the approach to the substitution procedure may not appear very obvious. However, it is primarily a visual task—thatis, the integrand shows you what to do; it is a matter of recognizing the form of the function. So, what are we supposed to
see? We are looking for an integrand of the form f ⎡⎣g(x)⎤⎦g′ (x)dx. For example, in the integral ⌠




⎝x


2 − 3⎞⎠
3
2xdx, we have


f (x) = x3, g(x) = x2 − 3, and g '(x) = 2x. Then,
f ⎡⎣g(x)⎤⎦g′ (x) = ⎛⎝x


2 − 3⎞⎠
3
(2x),


and we see that our integrand is in the correct form.
The method is called substitution because we substitute part of the integrand with the variable u and part of the integrandwith du. It is also referred to as change of variables because we are changing variables to obtain an expression that is easierto work with for applying the integration rules.
Theorem 1.7: Substitution with Indefinite Integrals
Let u = g(x), , where g′ (x) is continuous over an interval, let f (x) be continuous over the corresponding range of
g, and let F(x) be an antiderivative of f (x). Then,


(1.19)∫ f ⎡⎣g(x)⎤⎦g′ (x)dx = ∫ f (u)du
= F(u) + C
= F⎛⎝g(x)⎞⎠+ C.


Proof
Let f, g, u, and F be as specified in the theorem. Then


d
dx


F(g(x)) = F′ (g(x)⎞⎠g′ (x)


= f ⎡⎣g(x)⎤⎦g′ (x).


Integrating both sides with respect to x, we see that
∫ f ⎡⎣g(x)⎤⎦g′ (x)dx = F⎛⎝g(x)⎞⎠+ C.


If we now substitute u = g(x), and du = g '(x)dx, we get
∫ f ⎡⎣g(x)⎤⎦g′ (x)dx = ∫ f (u)du


= F(u) + C
= F⎛⎝g(x)⎞⎠+ C.




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Returning to the problem we looked at originally, we let u = x2 − 3 and then du = 2xdx. Rewrite the integral in terms of
u:





⎝x


2 − 3⎞⎠⎫⎭ ⎬


u


3


(2xdx)

du


= ∫ u3du.


Using the power rule for integrals, we have


u3du = u


4


4
+ C.


Substitute the original expression for x back into the solution:
u4
4


+ C =

⎝x


2 − 3⎞⎠
4


4
+ C.


We can generalize the procedure in the following Problem-Solving Strategy.
Problem-Solving Strategy: Integration by Substitution


1. Look carefully at the integrand and select an expression g(x) within the integrand to set equal to u. Let’s select
g(x). such that g′ (x) is also part of the integrand.


2. Substitute u = g(x) and du = g′ (x)dx. into the integral.
3. We should now be able to evaluate the integral with respect to u. If the integral can’t be evaluated we need togo back and select a different expression to use as u.
4. Evaluate the integral in terms of u.
5. Write the result in terms of x and the expression g(x).


Example 1.30
Using Substitution to Find an Antiderivative
Use substitution to find the antiderivative of ⌠



6x⎛⎝3x


2 + 4⎞⎠
4
dx.


Solution
The first step is to choose an expression for u. We choose u = 3x2 + 4. because then du = 6xdx., and we
already have du in the integrand. Write the integral in terms of u:




6x⎛⎝3x


2 + 4⎞⎠
4
dx = ∫ u4du.


Remember that du is the derivative of the expression chosen for u, regardless of what is inside the integrand. Nowwe can evaluate the integral with respect to u:


Chapter 1 | Integration 83




1.25


∫ u4du = u
5


5
+ C


=

⎝3x


2 + 4⎞⎠
5


5
+ C.


Analysis
We can check our answer by taking the derivative of the result of integration. We should obtain the integrand.
Picking a value for C of 1, we let y = 1


5

⎝3x


2 + 4⎞⎠
5
+ 1. We have


y = 1
5

⎝3x


2 + 4⎞⎠
5
+ 1,


so
y′ = ⎛⎝


1
5

⎠5

⎝3x


2 + 4⎞⎠
4
6x


= 6x⎛⎝3x
2 + 4⎞⎠


4
.


This is exactly the expression we started with inside the integrand.


Use substitution to find the antiderivative of ⌠

3x2 ⎛⎝x


3 − 3⎞⎠
2
dx.


Sometimes we need to adjust the constants in our integral if they don’t match up exactly with the expressions we aresubstituting.
Example 1.31
Using Substitution with Alteration
Use substitution to find the antiderivative of ∫ z z2 − 5dz.


Solution
Rewrite the integral as ⌠



z⎛⎝z


2 − 5⎞⎠
1/2


dz. Let u = z2 − 5 and du = 2z dz. Now we have a problem because
du = 2z dz and the original expression has only z dz. We have to alter our expression for du or the integral in
u will be twice as large as it should be. If we multiply both sides of the du equation by 1


2
. we can solve this


problem. Thus,
u = z2 − 5


du = 2z dz
1
2
du = 1


2
(2z)dz = z dz.


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1.26


Write the integral in terms of u, but pull the 1
2
outside the integration symbol:




z⎛⎝z


2 − 5⎞⎠
1/2


dz = 1
2∫ u


1/2du.


Integrate the expression in u:
1
2∫ u


1/2du = ⎛⎝
1
2


u3/2
3
2


+ C


= ⎛⎝
1
2




2
3

⎠u


3/2 + C


= 1
3
u3/2 + C


= 1
3

⎝z


2 − 5⎞⎠
3/2


+ C.


Use substitution to find the antiderivative of ⌠

x2 ⎛⎝x


3 + 5⎞⎠
9
dx.


Example 1.32
Using Substitution with Integrals of Trigonometric Functions
Use substitution to evaluate the integral ⌠



sin t
cos3 t


dt.


Solution
We know the derivative of cos t is −sin t, so we set u = cos t. Then du = −sin tdt. Substituting into the
integral, we have





sin t
cos3 t


dt = −⌠

du
u3


.


Evaluating the integral, we get
−⌠

du
u3


= −∫ u−3du


= −⎛⎝−
1
2

⎠u


−2 + C.


Putting the answer back in terms of t, we get



sin t
cos3 t


dt = 1
2u2


+ C


= 1
2cos2 t


+ C.


Chapter 1 | Integration 85




1.27


1.28


Use substitution to evaluate the integral ⌠

cos t
sin2 t


dt.


Sometimes we need to manipulate an integral in ways that are more complicated than just multiplying or dividing by aconstant. We need to eliminate all the expressions within the integrand that are in terms of the original variable. When weare done, u should be the only variable in the integrand. In some cases, this means solving for the original variable in termsof u. This technique should become clear in the next example.
Example 1.33
Finding an Antiderivative Using u-Substitution
Use substitution to find the antiderivative of ∫ x


x − 1
dx.


Solution
If we let u = x − 1, then du = dx. But this does not account for the x in the numerator of the integrand. We
need to express x in terms of u. If u = x − 1, then x = u + 1. Now we can rewrite the integral in terms of u:


∫ x
x − 1


dx = ∫ u + 1u du


= ∫ u + 1udu


= ∫ ⎛⎝u1/2 + u−1/2⎞⎠du.


Then we integrate in the usual way, replace u with the original expression, and factor and simplify the result.Thus,
∫ ⎛⎝u1/2 + u−1/2⎞⎠du = 23u


3/2 + 2u1/2 + C


= 2
3
(x − 1)3/2 + 2(x − 1)1/2 + C


= (x − 1)1/2 ⎡⎣
2
3
(x − 1) + 2⎤⎦+ C


= (x − 1)1/2 ⎛⎝
2
3
x − 2


3
+ 6


3



= (x − 1)1/2 ⎛⎝
2
3
x + 4


3



= 2
3
(x − 1)1/2 (x + 2) + C.


Use substitution to evaluate the indefinite integral ∫ cos3 t sin t dt.


Substitution for Definite Integrals
Substitution can be used with definite integrals, too. However, using substitution to evaluate a definite integral requires achange to the limits of integration. If we change variables in the integrand, the limits of integration change as well.


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Theorem 1.8: Substitution with Definite Integrals
Let u = g(x) and let g′ be continuous over an interval ⎡⎣a, b⎤⎦, and let f be continuous over the range of u = g(x).
Then,



⌡a


b


f ⎛⎝g(x)⎞⎠g′ (x)dx = ∫
g(a)


g(b)


f (u)du.


Although we will not formally prove this theorem, we justify it with some calculations here. From the substitution rule forindefinite integrals, if F(x) is an antiderivative of f (x), we have
∫ f ⎛⎝g(x)⎞⎠g′ (x)dx = F⎛⎝g(x)⎞⎠+ C.


Then
(1.20)



a


b
f ⎡⎣g(x)⎤⎦g′ (x)dx = F⎛⎝g(x)⎞⎠|x = ax = b


= F⎛⎝g(b)⎞⎠− F⎛⎝g(a)⎞⎠


= F(u)|u = g(a)
u = g(b)


= ∫
g(a)


g(b)


f (u)du,


and we have the desired result.
Example 1.34
Using Substitution to Evaluate a Definite Integral


Use substitution to evaluate ⌠
⌡0


1


x2 ⎛⎝1 + 2x
3⎞


5
dx.


Solution
Let u = 1 + 2x3, so du = 6x2dx. Since the original function includes one factor of x2 and du = 6x2dx,
multiply both sides of the du equation by 1/6. Then,


du = 6x2dx
1
6
du = x2dx.


To adjust the limits of integration, note that when x = 0, u = 1 + 2(0) = 1, and when
x = 1, u = 1 + 2(1) = 3. Then



⌡0


1


x2 ⎛⎝1 + 2x
3⎞


5
dx = 1


6∫1
3
u5du.


Evaluating this expression, we get


Chapter 1 | Integration 87




1.29


1.30


1
6∫1


3
u5du = ⎛⎝


1
6




u6


6

⎠|1
3


= 1
36

⎣(3)


6 − (1)6⎤⎦


= 182
9


.


Use substitution to evaluate the definite integral ⌠
⌡−1


0


y⎛⎝2y
2 − 3⎞⎠


5
dy.


Example 1.35
Using Substitution with an Exponential Function
Use substitution to evaluate ∫


0


1
xe4x


2 + 3dx.


Solution
Let u = 4x3 + 3. Then, du = 8xdx. To adjust the limits of integration, we note that when x = 0, u = 3, and
when x = 1, u = 7. So our substitution gives



0


1
xe4x


2 + 3dx = 1
8∫3


7
eudu


= 1
8
eu |3


7


= e
7 − e3
8


≈ 134.568.


Use substitution to evaluate ⌠
⌡0


1
x2 cos⎛⎝


π
2
x3⎞⎠dx.


Substitution may be only one of the techniques needed to evaluate a definite integral. All of the properties and rules ofintegration apply independently, and trigonometric functions may need to be rewritten using a trigonometric identity beforewe can apply substitution. Also, we have the option of replacing the original expression for u after we find the antiderivative,which means that we do not have to change the limits of integration. These two approaches are shown in Example 1.36.
Example 1.36


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Using Substitution to Evaluate a Trigonometric Integral
Use substitution to evaluate ∫


0


π/2
cos2 θ dθ.


Solution
Let us first use a trigonometric identity to rewrite the integral. The trig identity cos2 θ = 1 + cos2θ


2
allows us


to rewrite the integral as


0


π/2
cos2 θdθ = ⌠


⌡0


π/2
1 + cos2θ


2
dθ.


Then,

⌡0


π/2⎛

1 + cos2θ


2

⎠dθ =



⌡0


π/2⎛

1
2
+ 1


2
cos2θ⎞⎠dθ


= 1
2∫0


π/2
dθ + ∫


0


π/2
cos2θdθ.


We can evaluate the first integral as it is, but we need to make a substitution to evaluate the second integral. Let
u = 2θ. Then, du = 2dθ, or 1


2
du = dθ. Also, when θ = 0, u = 0, and when θ = π/2, u = π. Expressing


the second integral in terms of u, we have


1
2

⌡0


π/2


dθ + 1
2∫0


π/2
cos2θdθ = 1


2

⌡0


π/2


dθ + 1
2


1
2

⎠∫


0


π
cosudu


= θ
2|θ = 0


θ = π/2
+ 1


4
sinu|u = 0


u = θ


= ⎛⎝
π
4
− 0⎞⎠+ (0 − 0) =


π
4
.


Chapter 1 | Integration 89




1.5 EXERCISES
254. Why is u-substitution referred to as change ofvariable?
255. 2. If f = g ∘h, when reversing the chain rule,
d
dx


(g ∘h)(x) = g′ (h(x)⎞⎠h′ (x), should you take u = g(x)
or u = h(x)?
In the following exercises, verify each identity usingdifferentiation. Then, using the indicated u-substitution,
identify f such that the integral takes the form ∫ f (u)du.
256.
∫ x x + 1dx = 215(x + 1)


3/2 (3x − 2) + C; u = x + 1


257.



x2


x − 1
dx(x > 1) = 2


15
x − 1⎛⎝3x


2 + 4x + 8⎞⎠+ C; u = x − 1


258.


x 4x2 + 9dx = 1


12

⎝4x


2 + 9⎞⎠
3/2


+ C; u = 4x2 + 9


259. ⌠


x


4x2 + 9
dx = 1


4
4x2 + 9 + C; u = 4x2 + 9


260. ⌠


x
(4x2 + 9)2


dx = − 1
8(4x2 + 9)


; u = 4x2 + 9


In the following exercises, find the antiderivative using theindicated substitution.
261. ∫ (x + 1)4dx; u = x + 1


262. ∫ (x − 1)5dx; u = x − 1


263. ∫ (2x − 3)−7dx; u = 2x − 3


264. ∫ (3x − 2)−11dx; u = 3x − 2


265. ⌠


x


x2 + 1
dx; u = x2 + 1


266. ⌠


x


1 − x2
dx; u = 1 − x2


267. ⌠

(x − 1)⎛⎝x


2 − 2x⎞⎠
3
dx; u = x2 − 2x


268. ⌠


⎝x


2 − 2x⎞⎠

⎝x


3 − 3x2⎞⎠
2
dx; u = x3 = 3x2


269. ∫ cos3 θdθ; u = sinθ (Hint: cos2 θ = 1 − sin2 θ)


270. ∫ sin3 θdθ; u = cosθ (Hint: sin2 θ = 1 − cos2 θ)


In the following exercises, use a suitable change ofvariables to determine the indefinite integral.
271. ∫ x(1 − x)99dx


272. ⌠

t⎛⎝1 − t


2⎞


10
dt


273. ∫ (11x − 7)−3dx


274. ∫ (7x − 11)4dx


275. ∫ cos3 θsinθdθ


276. ∫ sin7 θcosθdθ


277. ∫ cos2 (πt)sin(πt)dt


278. ∫ sin2 xcos3 xdx (Hint: sin2 x + cos2 x = 1)


279. ∫ t sin⎛⎝t2⎞⎠cos⎛⎝t2⎞⎠dt


280. ∫ t2cos2 ⎛⎝t3⎞⎠sin⎛⎝t3⎞⎠dt


281. ⌠

⎮ x


2



⎝x


3 − 3⎞⎠
2
dx


282. ⌠


x3


1 − x2
dx


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283. ⌠

⎮ y


5



⎝1 − y


3⎞


3/2
dy


284. ∫ cosθ(1 − cosθ)99 sinθdθ


285. ∫ ⎛⎝1 − cos3 θ⎞⎠10 cos2 θsinθdθ


286. ⌠

(cosθ − 1)⎛⎝cos


2 θ − 2cosθ⎞⎠
3
sinθdθ


287. ⌠


⎝sin


2 θ − 2sinθ⎞⎠

⎝sin


3 θ − 3sin2 θ⎞⎠
3
cosθdθ


In the following exercises, use a calculator to estimate thearea under the curve using left Riemann sums with 50terms, then use substitution to solve for the exact answer.
288. [T] y = 3(1 − x)2 over [0, 2]


289. [T] y = x⎛⎝1 − x2⎞⎠3 over [−1, 2]
290. [T] y = sinx(1 − cosx)2 over [0, π]
291. [T] y = x



⎝x


2 + 1⎞⎠
2
over [−1, 1]


In the following exercises, use a change of variables toevaluate the definite integral.
292. ∫


0


1
x 1 − x2dx


293. ⌠
⌡0


1
x


1 + x2
dx


294. ⌠
⌡0


2
t


5 + t2
dt


295. ⌠
⌡0


1
t


1 + t3
dt


296. ∫
0


π/4
sec2 θ tanθdθ


297. ⌠
⌡0


π/4
sinθ
cos4 θ




In the following exercises, evaluate the indefinite integral
∫ f (x)dx with constant C = 0 using u-substitution.
Then, graph the function and the antiderivative over theindicated interval. If possible, estimate a value of C thatwould need to be added to the antiderivative to make it
equal to the definite integral F(x) = ∫


a


x
f (t)dt, with a the


left endpoint of the given interval.
298. [T] ∫ (2x + 1)ex2 + x − 6dx over [−3, 2]


299. [T] ∫ cos⎛⎝ln(2x)⎞⎠x dx on [0, 2]


300. [T] ⌠


3x2 + 2x + 1


x3 + x2 + x + 4
dx over [−1, 2]


301. [T] ⌠


sinx
cos3 x


dx over ⎡⎣−π3, π3⎤⎦


302. [T] ∫ (x + 2)e−x2 − 4x + 3dx over ⎡⎣−5, 1⎤⎦


303. [T] ∫ 3x2 2x3 + 1dx over [0, 1]


304. If h(a) = h(b) in ∫
a


b
g '⎛⎝h(x)⎞⎠h(x)dx, what can you


say about the value of the integral?
305. Is the substitution u = 1 − x2 in the definite integral

⌡0


2
x


1 − x2
dx okay? If not, why not?


In the following exercises, use a change of variables toshow that each definite integral is equal to zero.
306. ∫


0


π
cos2 (2θ)sin(2θ)dθ


307. ∫
0


π
tcos⎛⎝t


2⎞
⎠sin

⎝t
2⎞
⎠dt


308. ∫
0


1
(1 − 2t)dt


Chapter 1 | Integration 91




309. ⌠






0


1


1 − 2t

⎝1 +



⎝t −


1
2



2⎞


dt


310. ⌠
⌡0


π


sin



⎜⎛⎝t −


π
2



3⎞



⎟cos⎛⎝t −


π
2

⎠dt


311. ∫
0


2
(1 − t)cos(πt)dt


312. ∫
π/4


3π/4
sin2 tcos tdt


313. Show that the average value of f (x) over an interval

⎣a, b⎤⎦ is the same as the average value of f (cx) over the
interval ⎡⎣ac , bc⎤⎦ for c > 0.
314. Find the area under the graph of f (t) = t



⎝1 + t


2⎞


a


between t = 0 and t = x where a > 0 and a ≠ 1 is
fixed, and evaluate the limit as x → ∞.
315. Find the area under the graph of g(t) = t



⎝1 − t


2⎞


a


between t = 0 and t = x, where 0 < x < 1 and a > 0
is fixed. Evaluate the limit as x → 1.
316. The area of a semicircle of radius 1 can be expressed
as ∫


−1


1
1 − x2dx. Use the substitution x = cos t to


express the area of a semicircle as the integral of atrigonometric function. You do not need to compute theintegral.
317. The area of the top half of an ellipse with a majoraxis that is the x-axis from x = −1 to a and with a minor
axis that is the y-axis from y = −b to b can be written
as ⌠
⌡−a


a


b 1 − x
2


a2
dx. Use the substitution x = acos t to


express this area in terms of an integral of a trigonometricfunction. You do not need to compute the integral.


318. [T] The following graph is of a function of the form
f (t) = asin(nt) + bsin(mt). Estimate the coefficients a
and b, and the frequency parameters n and m. Use these
estimates to approximate ∫


0


π
f (t)dt.


319. [T] The following graph is of a function of the form
f (x) = acos(nt) + bcos(mt). Estimate the coefficients a
and b and the frequency parameters n and m. Use these
estimates to approximate ∫


0


π
f (t)dt.


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1.31


1.6 | Integrals Involving Exponential and Logarithmic
Functions


Learning Objectives
1.6.1 Integrate functions involving exponential functions.
1.6.2 Integrate functions involving logarithmic functions.


Exponential and logarithmic functions are used to model population growth, cell growth, and financial growth, as well asdepreciation, radioactive decay, and resource consumption, to name only a few applications. In this section, we exploreintegration involving exponential and logarithmic functions.
Integrals of Exponential Functions
The exponential function is perhaps the most efficient function in terms of the operations of calculus. The exponentialfunction, y = ex, is its own derivative and its own integral.


Rule: Integrals of Exponential Functions
Exponential functions can be integrated using the following formulas.


(1.21)∫ ex dx = ex + C
∫ ax dx = a


x


lna
+ C


Example 1.37
Finding an Antiderivative of an Exponential Function
Find the antiderivative of the exponential function e−x.
Solution
Use substitution, setting u = −x, and then du = −1dx. Multiply the du equation by −1, so you now have
−du = dx. Then,


∫ e−xdx = −∫ eudu


= −eu + C
= −e−x + C.


Find the antiderivative of the function using substitution: x2 e−2x3.


A common mistake when dealing with exponential expressions is treating the exponent on e the same way we treatexponents in polynomial expressions. We cannot use the power rule for the exponent on e. This can be especially confusingwhen we have both exponentials and polynomials in the same expression, as in the previous checkpoint. In these cases, weshould always double-check to make sure we’re using the right rules for the functions we’re integrating.


Chapter 1 | Integration 93




1.32


Example 1.38
Square Root of an Exponential Function
Find the antiderivative of the exponential function ex 1 + ex.
Solution
First rewrite the problem using a rational exponent:


∫ ex 1 + exdx = ∫ ex (1 + ex)1/2dx.


Using substitution, choose u = 1 + ex.u = 1 + ex. Then, du = ex dx. We have (Figure 1.37)
∫ ex (1 + ex)1/2dx = ∫ u1/2du.


Then


u1/2du = u


3/2


3/2
+ C = 2


3
u3/2 + C = 2


3
(1 + ex)3/2 + C.


Figure 1.37 The graph shows an exponential function timesthe square root of an exponential function.


Find the antiderivative of ex (3ex − 2)2.


Example 1.39
Using Substitution with an Exponential Function
Use substitution to evaluate the indefinite integral ∫ 3x2 e2x3dx.


Solution
Here we choose to let u equal the expression in the exponent on e. Let u = 2x3 and du = 6x2dx.. Again, du
is off by a constant multiplier; the original function contains a factor of 3x2, not 6x2. Multiply both sides of the
equation by 1


2
so that the integrand in u equals the integrand in x. Thus,


∫ 3x2 e2x
3
dx = 1


2∫ e
udu.


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1.33


Integrate the expression in u and then substitute the original expression in x back into the u integral:
1
2∫ e


udu = 1
2
eu + C = 1


2
e2x


3
+ C.


Evaluate the indefinite integral ∫ 2x3 ex4dx.


As mentioned at the beginning of this section, exponential functions are used in many real-life applications. The number e isoften associated with compounded or accelerating growth, as we have seen in earlier sections about the derivative. Althoughthe derivative represents a rate of change or a growth rate, the integral represents the total change or the total growth. Let’slook at an example in which integration of an exponential function solves a common business application.
A price–demand function tells us the relationship between the quantity of a product demanded and the price of the product.In general, price decreases as quantity demanded increases. The marginal price–demand function is the derivative of theprice–demand function and it tells us how fast the price changes at a given level of production. These functions are used inbusiness to determine the price–elasticity of demand, and to help companies determine whether changing production levelswould be profitable.
Example 1.40
Finding a Price–Demand Equation
Find the price–demand equation for a particular brand of toothpaste at a supermarket chain when the demand is50 tubes per week at $2.35 per tube, given that the marginal price—demand function, p′ (x), for x number of
tubes per week, is given as


p '(x) = −0.015e−0.01x.


If the supermarket chain sells 100 tubes per week, what price should it set?
Solution
To find the price–demand equation, integrate the marginal price–demand function. First find the antiderivative,then look at the particulars. Thus,


p(x) = ∫ −0.015e−0.01x dx


= −0.015∫ e−0.01xdx.


Using substitution, let u = −0.01x and du = −0.01dx. Then, divide both sides of the du equation by −0.01.
This gives


−0.015
−0.01 ∫ e


udu = 1.5∫ eudu


= 1.5eu + C


= 1.5e−0.01x + C.


The next step is to solve for C. We know that when the price is $2.35 per tube, the demand is 50 tubes per week.This means


Chapter 1 | Integration 95




p(50) = 1.5e−0.01(50) + C


= 2.35.


Now, just solve for C:
C = 2.35 − 1.5e−0.5


= 2.35 − 0.91
= 1.44.


Thus,
p(x) = 1.5e−0.01x + 1.44.


If the supermarket sells 100 tubes of toothpaste per week, the price would be
p(100) = 1.5e−0.01(100) + 1.44 = 1.5e−1 + 1.44 ≈ 1.99.


The supermarket should charge $1.99 per tube if it is selling 100 tubes per week.


Example 1.41
Evaluating a Definite Integral Involving an Exponential Function
Evaluate the definite integral ∫


1


2
e1 − x dx.


Solution
Again, substitution is the method to use. Let u = 1 − x, so du = −1dx or −du = dx. Then
∫ e1 − x dx = −∫ eudu. Next, change the limits of integration. Using the equation u = 1 − x, we have


u = 1 − (1) = 0
u = 1 − (2) = −1.


The integral then becomes


1


2
e1 − x dx = −∫


0


−1
eudu


= ∫
−1


0
eudu


= eu|−10


= e0 − ⎛⎝e
−1⎞


= −e−1 + 1.


See Figure 1.38.


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1.34


1.35


Figure 1.38 The indicated area can be calculated byevaluating a definite integral using substitution.


Evaluate ∫
0


2
e2x dx.


Example 1.42
Growth of Bacteria in a Culture
Suppose the rate of growth of bacteria in a Petri dish is given by q(t) = 3t, where t is given in hours and q(t)
is given in thousands of bacteria per hour. If a culture starts with 10,000 bacteria, find a function Q(t) that gives
the number of bacteria in the Petri dish at any time t. How many bacteria are in the dish after 2 hours?
Solution
We have


Q(t) = ⌠

3t dt = 3


t


ln3
+ C.


Then, at t = 0 we have Q(0) = 10 = 1
ln3


+ C, so C ≈ 9.090 and we get
Q(t) = 3


t


ln3
+ 9.090.


At time t = 2, we have
Q(2) = 3


2


ln3
+ 9.090


= 17.282.


After 2 hours, there are 17,282 bacteria in the dish.


From Example 1.42, suppose the bacteria grow at a rate of q(t) = 2t. Assume the culture still starts
with 10,000 bacteria. Find Q(t). How many bacteria are in the dish after 3 hours?


Chapter 1 | Integration 97




1.36


Example 1.43
Fruit Fly Population Growth
Suppose a population of fruit flies increases at a rate of g(t) = 2e0.02t, in flies per day. If the initial population
of fruit flies is 100 flies, how many flies are in the population after 10 days?
Solution
Let G(t) represent the number of flies in the population at time t. Applying the net change theorem, we have


G(10) = G(0) + ∫
0


10
2e0.02t dt


= 100 + ⎡⎣
2


0.02
e0.02t⎤⎦|0


10


= 100 + ⎡⎣100e
0.02t⎤
⎦|0
10


= 100 + 100e0.2 − 100
≈ 122.


There are 122 flies in the population after 10 days.


Suppose the rate of growth of the fly population is given by g(t) = e0.01t, and the initial fly population
is 100 flies. How many flies are in the population after 15 days?


Example 1.44
Evaluating a Definite Integral Using Substitution
Evaluate the definite integral using substitution: ⌠


⌡1


2
e1/x


x2
dx.


Solution
This problem requires some rewriting to simplify applying the properties. First, rewrite the exponent on e as apower of x, then bring the x2 in the denominator up to the numerator using a negative exponent. We have



⌡1


2
e1/x


x2
dx = ∫


1


2
ex


−1
x−2dx.


Let u = x−1, the exponent on e. Then
du = −x−2dx


−du = x−2dx.


Bringing the negative sign outside the integral sign, the problem now reads


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1.37


−∫ eudu.


Next, change the limits of integration:
u = (1)−1 = 1


u = (2)−1 = 1
2
.


Notice that now the limits begin with the larger number, meaning we must multiply by −1 and interchange thelimits. Thus,


−∫
1


1/2
eudu = ∫


1/2


1
eudu


= eu |1/2
1


= e − e1/2


= e − e.


Evaluate the definite integral using substitution: ⌠
⌡1


2
1
x3
e4x


−2
dx.


Integrals Involving Logarithmic Functions
Integrating functions of the form f (x) = x−1 result in the absolute value of the natural log function, as shown in the
following rule. Integral formulas for other logarithmic functions, such as f (x) = lnx and f (x) = loga x, are also included
in the rule.
Rule: Integration Formulas Involving Logarithmic Functions
The following formulas can be used to evaluate integrals involving logarithmic functions.


(1.22)∫ x−1dx = ln|x| + C
∫ lnx dx = x lnx − x + C = x(lnx − 1) + C


∫ logax dx = xlna(lnx − 1) + C


Example 1.45
Finding an Antiderivative Involving ln x
Find the antiderivative of the function 3


x − 10
.


Chapter 1 | Integration 99




1.38


Solution
First factor the 3 outside the integral symbol. Then use the u−1 rule. Thus,





3
x − 10


dx = 3⌠


1
x − 10


dx


= 3∫ duu
= 3ln|u| + C
= 3ln|x − 10| + C, x ≠ 10.


See Figure 1.39.


Figure 1.39 The domain of this function is x ≠ 10.


Find the antiderivative of 1
x + 2


.


Example 1.46
Finding an Antiderivative of a Rational Function
Find the antiderivative of 2x3 + 3x


x4 + 3x2
.


Solution
This can be rewritten as ⌠




⎝2x


3 + 3x⎞⎠

⎝x


4 + 3x2⎞⎠
−1


dx. Use substitution. Let u = x4 + 3x2, then
du = 4x3 + 6x. Alter du by factoring out the 2. Thus,


du = ⎛⎝4x
3 + 6x⎞⎠dx


= 2⎛⎝2x
3 + 3x⎞⎠dx


1
2
du = ⎛⎝2x


3 + 3x⎞⎠dx.


Rewrite the integrand in u:


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1.39





⎝2x


3 + 3x⎞⎠

⎝x


4 + 3x2⎞⎠
−1


dx = 1
2∫ u


−1du.


Then we have
1
2∫ u


−1du = 1
2
ln|u| + C


= 1
2
ln|x4 + 3x2| + C.


Example 1.47
Finding an Antiderivative of a Logarithmic Function
Find the antiderivative of the log function log2 x.
Solution
Follow the format in the formula listed in the rule on integration formulas involving logarithmic functions. Basedon this format, we have


∫ log2 xdx = xln2(lnx − 1) + C.


Find the antiderivative of log3 x.


Example 1.48 is a definite integral of a trigonometric function. With trigonometric functions, we often have to apply atrigonometric property or an identity before we can move forward. Finding the right form of the integrand is usually the keyto a smooth integration.
Example 1.48
Evaluating a Definite Integral
Find the definite integral of ⌠


⌡0


π/2
sinx


1 + cosx
dx.


Solution
We need substitution to evaluate this problem. Let u = 1 + cosx, , so du = −sinx dx. Rewrite the integral in
terms of u, changing the limits of integration as well. Thus,


u = 1 + cos(0) = 2
u = 1 + cos⎛⎝


π
2

⎠ = 1.


Chapter 1 | Integration 101




Then

⌡0


π/2
sinx


1 + cosx
= −∫


2


1
u−1du


= ∫
1


2
u−1du


= ln|u||1
2


= [ln2 − ln1]
= ln2.


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1.6 EXERCISES
In the following exercises, compute each indefiniteintegral.
320. ∫ e2xdx


321. ∫ e−3xdx


322. ∫ 2xdx


323. ∫ 3−xdx


324. ⌠

1
2x


dx


325. ∫ 2xdx


326. ⌠

1
x2
dx


327. ∫ 1xdx
In the following exercises, find each indefinite integral byusing appropriate substitutions.
328. ∫ lnxx dx


329. ⌠


dx
x(lnx)2


330. ⌠

dx
x lnx


(x > 1)


331. ⌠


dx
x lnx ln(lnx)


332. ∫ tanθ dθ


333. ∫ cosx − xsinxxcosx dx


334. ⌠

ln(sinx)
tanx dx


335. ∫ ln(cosx)tanxdx


336. ∫ xe−x2dx


337. ∫ x2e−x3dx


338. ∫ esinx cosxdx


339. ∫ etanxsec2 xdx


340. ∫ elnx dxx


341. ⌠

e
ln(1 − t)


1 − t
dt


In the following exercises, verify by differentiation that
∫ lnx dx = x(lnx − 1) + C, then use appropriate
changes of variables to compute the integral.
342. ∫ lnxdx (Hint: ⌠⌡lnxdx = 12∫ xln⎛⎝x2⎞⎠dx)


343. ∫ x2ln2 x dx


344. ⌠

lnx
x2


dx (Hint: Set u = 1x .)


345. ∫ lnxx dx (Hint: Set u = x.)
346. Write an integral to express the area under the graph
of y = 1t from t = 1 to ex and evaluate the integral.
347. Write an integral to express the area under the graph
of y = et between t = 0 and t = lnx, and evaluate the
integral.
In the following exercises, use appropriate substitutionsto express the trigonometric integrals in terms ofcompositions with logarithms.
348. ∫ tan(2x)dx


349. ⌠

sin(3x) − cos(3x)
sin(3x) + cos(3x)


dx


350. ⌠

xsin⎛⎝x


2⎞


cos⎛⎝x
2⎞


dx


351. ∫ xcsc⎛⎝x2⎞⎠dx


Chapter 1 | Integration 103




352. ∫ ln(cosx)tanx dx


353. ∫ ln(cscx)cotxdx


354. ⌠

ex − e−x


ex + e−x
dx


In the following exercises, evaluate the definite integral.


355. ⌠
⌡1


2
1 + 2x + x2


3x + 3x2 + x3
dx


356. ∫
0


π/4
tanx dx


357. ⌠
⌡0


π/3
sinx − cosx
sinx + cosx


dx


358. ∫
π/6


π/2
cscxdx


359. ∫
π/4


π/3
cotxdx


In the following exercises, integrate using the indicatedsubstitution.
360. ∫ xx − 100dx; u = x − 100


361. ⌠

y − 1
y + 1


dy; u = y + 1


362. ⌠

1 − x2


3x − x3
dx; u = 3x − x3


363. ⌠

sinx + cosx
sinx − cosx


dx; u = sinx − cosx


364. ∫ e2x 1 − e2xdx; u = e2x


365. ⌠

ln(x) 1 − (lnx)


2


x dx; u = lnx


In the following exercises, does the right-endpointapproximation overestimate or underestimate the exactarea? Calculate the right endpoint estimate R50 and solvefor the exact area.
366. [T] y = ex over [0, 1]
367. [T] y = e−x over [0, 1]


368. [T] y = ln(x) over [1, 2]
369. [T] y = x + 1


x2 + 2x + 6
over [0, 1]


370. [T] y = 2x over [−1, 0]
371. [T] y = −2−x over [0, 1]
In the following exercises, f (x) ≥ 0 for a ≤ x ≤ b. Find
the area under the graph of f (x) between the given values
a and b by integrating.
372. f (x) = log10 (x)x ; a = 10, b = 100


373. f (x) = log2 (x)x ; a = 32, b = 64
374. f (x) = 2−x; a = 1, b = 2
375. f (x) = 2−x; a = 3, b = 4
376. Find the area under the graph of the function
f (x) = xe−x


2 between x = 0 and x = 5.
377. Compute the integral of f (x) = xe−x2 and find the
smallest value of N such that the area under the graph
f (x) = xe−x


2 between x = N and x = N + 10 is, at
most, 0.01.
378. Find the limit, as N tends to infinity, of the area under
the graph of f (x) = xe−x2 between x = 0 and x = 5.


379. Show that ∫
a


b
dt
t = ∫1/b


1/a
dt
t when 0 < a ≤ b.


380. Suppose that f (x) > 0 for all x and that f and g are
differentiable. Use the identity f g = eg ln f and the chain
rule to find the derivative of f g.
381. Use the previous exercise to find the antiderivative of
h(x) = xx (1 + lnx) and evaluate ∫


2


3
xx (1 + lnx)dx.


382. Show that if c > 0, then the integral of 1/x from
ac to bc (0 < a < b) is the same as the integral of 1/x
from a to b.
The following exercises are intended to derive thefundamental properties of the natural log starting from the


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definition ln(x) = ∫
1


x
dt
t , using properties of the definite


integral and making no further assumptions.
383. Use the identity ln(x) = ∫


1


x
dt
t to derive the identity


ln⎛⎝
1
x

⎠ = −lnx.


384. Use a change of variable in the integral ∫
1


xy
1
t dt to


show that lnxy = lnx + lny for x, y > 0.
385. Use the identity lnx = ∫


1


x
dt
x to show that ln(x)


is an increasing function of x on [0, ∞), and use the
previous exercises to show that the range of ln(x) is
(−∞, ∞). Without any further assumptions, conclude that
ln(x) has an inverse function defined on (−∞, ∞).
386. Pretend, for the moment, that we do not know that
ex is the inverse function of ln(x), but keep in mind
that ln(x) has an inverse function defined on (−∞, ∞).
Call it E. Use the identity lnxy = lnx + lny to deduce that
E(a + b) = E(a)E(b) for any real numbers a, b.
387. Pretend, for the moment, that we do not know that
ex is the inverse function of lnx, but keep in mind that
lnx has an inverse function defined on (−∞, ∞). Call it
E. Show that E '(t) = E(t).
388. The sine integral, defined as S(x) = ∫


0


x
sin t
t dt is


an important quantity in engineering. Although it does nothave a simple closed formula, it is possible to estimateits behavior for large x. Show that for
k ≥ 1, |S(2πk) − S⎛⎝2π(k + 1)⎞⎠| ≤ 1k(2k + 1)π .


(Hint: sin(t + π) = −sin t)


389. [T] The normal distribution in probability is given
by p(x) = 1


σ 2π
e
−(x − µ)2 /2σ2


, where σ is the standard
deviation and μ is the average. The standard normaldistribution in probability, ps, corresponds to
µ = 0 and σ = 1. Compute the left endpoint estimates
R10 and R100 of ⌠⌡−1


1
1


e−x
2/2


dx.


390. [T] Compute the right endpoint estimates
R50 and R100 of ⌠⌡−3


5
1


2 2π
e−


(x − 1)2 /8.


Chapter 1 | Integration 105




1.7 | Integrals Resulting in Inverse Trigonometric
Functions


Learning Objectives
1.7.1 Integrate functions resulting in inverse trigonometric functions


In this section we focus on integrals that result in inverse trigonometric functions. We have worked with these functionsbefore. Recall from Functions and Graphs (http://cnx.org/content/m53472/latest/) that trigonometric functionsare not one-to-one unless the domains are restricted. When working with inverses of trigonometric functions, we alwaysneed to be careful to take these restrictions into account. Also in Derivatives (http://cnx.org/content/m53494/latest/), we developed formulas for derivatives of inverse trigonometric functions. The formulas developed there give rise directlyto integration formulas involving inverse trigonometric functions.
Integrals that Result in Inverse Sine Functions
Let us begin this last section of the chapter with the three formulas. Along with these formulas, we use substitution toevaluate the integrals. We prove the formula for the inverse sine integral.
Rule: Integration Formulas Resulting in Inverse Trigonometric Functions
The following integration formulas yield inverse trigonometric functions:


1.
(1.23)⌠



du


a2 − u2
= sin−1 ua + C


2.
(1.24)⌠



du


a2 + u2
= 1a tan


−1 u
a + C


3.
(1.25)⌠



du


u u2 − a2
= 1asec


−1 u
a + C


Proof
Let y = sin−1 xa. Then asiny = x. Now let’s use implicit differentiation. We obtain


d
dx



⎝asiny⎞⎠ = ddx


(x)


acosy
dy
dx


= 1


dy
dx


= 1acosy.


For −π
2
≤ y ≤ π


2
, cosy ≥ 0. Thus, applying the Pythagorean identity sin2 y + cos2 y = 1, we have


cosy = 1 = sin2 y. This gives


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1.40


1
acosy =


1
a 1 − sin2 y


= 1
a2 − a2 sin2 y


= 1
a2 − x2


.


Then for −a ≤ x ≤ a, we have



1
a2 − u2


du = sin−1⎛⎝ua

⎠+ C.



Example 1.49
Evaluating a Definite Integral Using Inverse Trigonometric Functions


Evaluate the definite integral ⌠
⌡0


1
dx


1 − x2
.


Solution
We can go directly to the formula for the antiderivative in the rule on integration formulas resulting in inversetrigonometric functions, and then evaluate the definite integral. We have



⌡0


1
dx


1 − x2
= sin−1 x|0


1


= sin−11 − sin−10
= π


2
− 0


= π
2
.


Find the antiderivative of ⌠


dx


1 − 16x2
.


Example 1.50
Finding an Antiderivative Involving an Inverse Trigonometric Function
Evaluate the integral ⌠



dx


4 − 9x2
.


Solution


Chapter 1 | Integration 107




1.41


Substitute u = 3x. Then du = 3dx and we have



dx


4 − 9x2
= 1


3



du


4 − u2
.


Applying the formula with a = 2, we obtain



dx


4 − 9x2
= 1


3



du


4 − u2


= 1
3
sin−1 ⎛⎝


u
2

⎠+ C


= 1
3
sin−1 ⎛⎝


3x
2

⎠+ C.


Find the indefinite integral using an inverse trigonometric function and substitution for ⌠


dx


9 − x2
.


Example 1.51
Evaluating a Definite Integral


Evaluate the definite integral ⌠
⌡0


3/2
du


1 − u2
.


Solution
The format of the problem matches the inverse sine formula. Thus,



⌡0


3/2
du


1 − u2
= sin−1u|0


3/2


=

⎣sin


−1 ⎛


3
2



⎦−

⎣sin


−1 (0)⎤⎦


= π
3
.


Integrals Resulting in Other Inverse Trigonometric Functions
There are six inverse trigonometric functions. However, only three integration formulas are noted in the rule on integrationformulas resulting in inverse trigonometric functions because the remaining three are negative versions of the ones we use.The only difference is whether the integrand is positive or negative. Rather than memorizing three more formulas, if theintegrand is negative, simply factor out −1 and evaluate the integral using one of the formulas already provided. To closethis section, we examine one more formula: the integral resulting in the inverse tangent function.


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1.42


1.43


Example 1.52
Finding an Antiderivative Involving the Inverse Tangent Function
Find an antiderivative of ⌠



1


1 + 4x2
dx.


Solution
Comparing this problem with the formulas stated in the rule on integration formulas resulting in inverse
trigonometric functions, the integrand looks similar to the formula for tan−1u + C. So we use substitution,
letting u = 2x, then du = 2dx and 1/2du = dx. Then, we have


1
2



1
1 + u2


du = 1
2
tan−1u + C = 1


2
tan−1 (2x) + C.


Use substitution to find the antiderivative of ⌠


dx
25 + 4x2


.


Example 1.53
Applying the Integration Formulas
Find the antiderivative of ⌠



1


9 + x2
dx.


Solution
Apply the formula with a = 3. Then,





dx
9 + x2


= 1
3
tan−1 ⎛⎝


x
3

⎠+ C.


Find the antiderivative of ⌠


dx
16 + x2


.


Example 1.54
Evaluating a Definite Integral


Evaluate the definite integral ⌠


3/3


3
dx


1 + x2
.


Chapter 1 | Integration 109




1.44


Solution
Use the formula for the inverse tangent. We have





3/3


3
dx


1 + x2
= tan−1 x|


3/3


3


= ⎡⎣tan
−1 ⎛
⎝ 3



⎦−

⎣tan


−1 ⎛


3
3





= π
6
.


Evaluate the definite integral ⌠
⌡0


2
dx


4 + x2
.


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1.7 EXERCISES
In the following exercises, evaluate each integral in termsof an inverse trigonometric function.


391. ⌠
⌡0


3/2
dx


1 − x2


392. ⌠
⌡−1/2


1/2
dx


1 − x2


393. ⌠


3


1
dx


1 + x2


394. ⌠

1/ 3


3
dx


1 + x2


395. ⌠
⌡1


2
dx


|x| x2 − 1


396. ⌠
⌡1


2/ 3
dx


|x| x2 − 1


In the following exercises, find each indefinite integral,using appropriate substitutions.
397. ⌠


dx


9 − x2


398. ⌠


dx


1 − 16x2


399. ⌠


dx
9 + x2


400. ⌠


dx
25 + 16x2


401. ⌠


dx


|x| x2 − 9


402. ⌠


dx


|x| 4x2 − 16


403. Explain the relationship
−cos−1 t + C = ⌠



dt


1 − t2
= sin−1 t + C. Is it true, in


general, that cos−1 t = −sin−1 t?
404. Explain the relationship
sec−1 t + C = ⌠



dt


|t| t2 − 1
= −csc−1 t + C. Is it true, in


general, that sec−1 t = −csc−1 t?
405. Explain what is wrong with the following integral:

⌡1


2
dt


1 − t2
.


406. Explain what is wrong with the following integral:

⌡−1


1
dt


|t| t2 − 1
.


In the following exercises, solve for the antiderivative ∫ f
of f with C = 0, then use a calculator to graph f and
the antiderivative over the given interval ⎡⎣a, b⎤⎦. Identify a
value of C such that adding C to the antiderivative recovers
the definite integral F(x) = ∫


a


x
f (t)dt.


407. [T] ⌠


1
9 − x2


dx over [−3, 3]


408. [T] ⌠


9
9 + x2


dx over ⎡⎣−6, 6⎤⎦


409. [T] ⌠


cosx
4 + sin2 x


dx over ⎡⎣−6, 6⎤⎦


410. [T] ⌠


ex


1 + e2x
dx over ⎡⎣−6, 6⎤⎦


In the following exercises, compute the antiderivative usingappropriate substitutions.
411. ⌠

sin−1 tdt


1 − t2


412. ⌠


dt


sin−1 t 1 − t2


Chapter 1 | Integration 111




413. ⌠

tan−1 (2t)
1 + 4t2


dt


414. ⌠

ttan−1 ⎛⎝t


2⎞


1 + t4
dt


415. ⌠

sec−1 ⎛⎝


t
2



|t| t2 − 4
dt


416. ⌠

tsec−1 ⎛⎝t


2⎞


t2 t4 − 1
dt


In the following exercises, use a calculator to graph the
antiderivative ∫ f with C = 0 over the given interval

⎣a, b⎤⎦. Approximate a value of C, if possible, such that
adding C to the antiderivative gives the same value as the
definite integral F(x) = ∫


a


x
f (t)dt.


417. [T] ⌠


1
x x2 − 4


dx over ⎡⎣2, 6⎤⎦


418. [T] ⌠


1
(2x + 2) x


dx over ⎡⎣0, 6⎤⎦


419. [T] ⌠

(sinx + xcosx)
1 + x2 sin2 x


dx over ⎡⎣−6, 6⎤⎦


420. [T] ⌠


2e−2x


1 − e−4x
dx over [0, 2]


421. [T] ⌠


1
x + xln2 x


over [0, 2]


422. [T] ⌠

sin−1 x
1 − x2


over [−1, 1]


In the following exercises, compute each integral usingappropriate substitutions.
423. ⌠


ex


1 − e2t
dt


424. ⌠


et


1 + e2t
dt


425. ⌠


dt


t 1 − ln2 t


426. ⌠


dt
t⎛⎝1 + ln


2 t⎞⎠


427. ⌠

cos−1 (2t)


1 − 4t2
dt


428. ⌠

et cos−1 ⎛⎝e


t⎞


1 − e2t
dt


In the following exercises, compute each definite integral.


429. ⌠
⌡0


1/2
tan⎛⎝sin


−1 t⎞⎠


1 − t2
dt


430. ⌠
⌡1/4


1/2
tan⎛⎝cos


−1 t⎞⎠


1 − t2
dt


431. ⌠
⌡0


1/2
sin⎛⎝tan


−1 t⎞⎠


1 + t2
dt


432. ⌠
⌡0


1/2
cos⎛⎝tan


−1 t⎞⎠


1 + t2
dt


433. For A > 0, compute I(A) = ⌠
⌡−A


A
dt


1 + t2
and


evaluate lim
a → ∞


I(A), the area under the graph of 1
1 + t2


on [−∞, ∞].


434. For 1 < B < ∞, compute I(B) = ⌠
⌡1


B
dt


t t2 − 1
and


evaluate lim
B → ∞


I(B), the area under the graph of
1


t t2 − 1
over [1, ∞).


435. Use the substitution u = 2 cotx and the identity
1 + cot2 x = csc2 x to evaluate ⌠



dx


1 + cos2 x
. (Hint:


Multiply the top and bottom of the integrand by csc2 x.)


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436. [T] Approximate the points at which the graphs of
f (x) = 2x2 − 1 and g(x) = ⎛⎝1 + 4x2⎞⎠−3/2 intersect, and
approximate the area between their graphs accurate to threedecimal places.
437. 47. [T] Approximate the points at which the graphs
of f (x) = x2 − 1 and f (x) = x2 − 1 intersect, and
approximate the area between their graphs accurate to threedecimal places.
438. Use the following graph to prove that

⌡0


x
1 − t2dt = 1


2
x 1 − x2 + 1


2
sin−1 x.


Chapter 1 | Integration 113




average value of a function
change of variables
definite integral
fundamental theorem of calculus
fundamental theorem of calculus, part 1
fundamental theorem of calculus, part 2
integrable function
integrand
integration by substitution
left-endpoint approximation
limits of integration
lower sum
mean value theorem for integrals
net change theorem
net signed area
partition
regular partition
riemann sum


right-endpoint approximation
sigma notation
total area
upper sum
variable of integration


CHAPTER 1 REVIEW
KEY TERMS


(or fave) the average value of a function on an interval can be found by calculating thedefinite integral of the function and dividing that value by the length of the interval
the substitution of a variable, such as u, for an expression in the integrand


a primary operation of calculus; the area between the curve and the x-axis over a given interval is adefinite integral
the theorem, central to the entire development of calculus, that establishes therelationship between differentiation and integration


uses a definite integral to define an antiderivative of a function
(also, evaluation theorem) we can evaluate a definite integral byevaluating the antiderivative of the integrand at the endpoints of the interval and subtracting


a function is integrable if the limit defining the integral exists; in other words, if the limit of theRiemann sums as n goes to infinity exists
the function to the right of the integration symbol; the integrand includes the function being integrated


a technique for integration that allows integration of functions that are the result of achain-rule derivative
an approximation of the area under a curve computed by using the left endpoint of eachsubinterval to calculate the height of the vertical sides of each rectangle


these values appear near the top and bottom of the integral sign and define the interval over whichthe function should be integrated
a sum obtained by using the minimum value of f (x) on each subinterval


guarantees that a point c exists such that f (c) is equal to the average value of the
function


if we know the rate of change of a quantity, the net change theorem says the future quantity isequal to the initial quantity plus the integral of the rate of change of the quantity
the area between a function and the x-axis such that the area below the x-axis is subtracted from the areaabove the x-axis; the result is the same as the definite integral of the function


a set of points that divides an interval into subintervals
a partition in which the subintervals all have the same width


an estimate of the area under the curve of the form A ≈ ∑
i = 1


n


f (xi* )Δx


the right-endpoint approximation is an approximation of the area of the rectanglesunder a curve using the right endpoint of each subinterval to construct the vertical sides of each rectangle
(also, summation notation) the Greek letter sigma (Σ) indicates addition of the values; the values of theindex above and below the sigma indicate where to begin the summation and where to end it


total area between a function and the x-axis is calculated by adding the area above the x-axis and the areabelow the x-axis; the result is the same as the definite integral of the absolute value of the function
a sum obtained by using the maximum value of f (x) on each subinterval


indicates which variable you are integrating with respect to; if it is x, then the function in theintegrand is followed by dx


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KEY EQUATIONS
• Properties of Sigma Notation



i = 1


n


c = nc



i = 1


n


cai = c∑
i = 1


n


ai



i = 1


n

⎝ai + bi



⎠ = ∑


i = 1


n


ai + ∑
i = 1


n


bi



i = 1


n

⎝ai − bi



⎠ = ∑


i = 1


n


ai − ∑
i = 1


n


bi



i = 1


n


ai = ∑
i = 1


m


ai + ∑
i = m + 1


n


ai


• Sums and Powers of Integers

i = 1


n


i = 1 + 2 +⋯ + n = n(n + 1)
2



i = 1


n


i2 = 12 + 22 +⋯+ n2 = n(n + 1)(2n + 1)
6



i = 0


n


i3 = 13 + 23 +⋯+ n3 = n
2 (n + 1)2


4


• Left-Endpoint Approximation
A ≈ Ln = f (x0)Δx + f (x1)Δx +⋯+ f (xn − 1)Δx = ∑


i = 1


n


f (xi − 1)Δx


• Right-Endpoint Approximation
A ≈ Rn = f (x1)Δx + f (x2)Δx +⋯+ f (xn)Δx = ∑


i = 1


n


f (xi)Δx


• Definite Integral

a


b
f (x)dx = lim


n → ∞

i = 1


n


f ⎛⎝xi*

⎠Δx


• Properties of the Definite Integral

a


a
f (x)dx = 0



b


a
f (x)dx = −∫


a


b
f (x)dx



a


b

⎣ f (x) + g(x)⎤⎦dx = ∫


a


b
f (x)dx + ∫


a


b
g(x)dx



⌡a


b



⎣ f (x) − g(x)⎤⎦dx = ⌠


⌡a


b


f (x)dx − ∫
a


b
g(x)dx



a


b
c f (x)dx = c∫


a


b
f (x) for constant c



a


b
f (x)dx = ∫


a


c
f (x)dx + ∫


c


b
f (x)dx


• Mean Value Theorem for Integrals


Chapter 1 | Integration 115




If f (x) is continuous over an interval ⎡⎣a, b⎤⎦, then there is at least one point c ∈ ⎡⎣a, b⎤⎦ such that
f (c) = 1


b − a∫a
b
f (x)dx.


• Fundamental Theorem of Calculus Part 1
If f (x) is continuous over an interval ⎡⎣a, b⎤⎦, and the function F(x) is defined by F(x) = ∫


a


x
f (t)dt, then


F′ (x) = f (x).


• Fundamental Theorem of Calculus Part 2
If f is continuous over the interval ⎡⎣a, b⎤⎦ and F(x) is any antiderivative of f (x), then ∫


a


b
f (x)dx = F(b) − F(a).


• Net Change Theorem
F(b) = F(a) + ∫


a


b
F '(x)dx or ∫


a


b
F '(x)dx = F(b) − F(a)


• Substitution with Indefinite Integrals
∫ f ⎡⎣g(x)⎤⎦g′ (x)dx = ∫ f (u)du = F(u) + C = F⎛⎝g(x)⎞⎠+ C


• Substitution with Definite Integrals

⌡a


b


f ⎛⎝g(x)⎞⎠g '(x)dx = ∫
g(a)


g(b)


f (u)du


• Integrals of Exponential Functions
∫ ex dx = ex + C




ax dx = a


x


lna
+ C


• Integration Formulas Involving Logarithmic Functions
∫ x−1dx = ln|x| + C


∫ lnx dx = x lnx − x + C = x(lnx − 1) + C


∫ loga x dx = xlna(lnx − 1) + C


• Integrals That Produce Inverse Trigonometric Functions



du


a2 − u2
= sin−1 ⎛⎝ua



⎠+ C





du
a2 + u2


= 1a tan
−1 ⎛

u
a

⎠+ C





du


u u2 − a2
= 1asec


−1 ⎛

u
a

⎠+ C


KEY CONCEPTS
1.1 Approximating Areas


• The use of sigma (summation) notation of the form ∑
i = 1


n


ai is useful for expressing long sums of values in compact
form.


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• For a continuous function defined over an interval ⎡⎣a, b⎤⎦, the process of dividing the interval into n equal parts,
extending a rectangle to the graph of the function, calculating the areas of the series of rectangles, and then summingthe areas yields an approximation of the area of that region.


• The width of each rectangle is Δx = b − an .
• Riemann sums are expressions of the form ∑


i = 1


n


f ⎛⎝xi*

⎠Δx, and can be used to estimate the area under the curve


y = f (x). Left- and right-endpoint approximations are special kinds of Riemann sums where the values of ⎧

⎨xi*







are chosen to be the left or right endpoints of the subintervals, respectively.
• Riemann sums allow for much flexibility in choosing the set of points ⎧



⎨xi*





⎬ at which the function is evaluated,


often with an eye to obtaining a lower sum or an upper sum.
1.2 The Definite Integral


• The definite integral can be used to calculate net signed area, which is the area above the x-axis less the area belowthe x-axis. Net signed area can be positive, negative, or zero.
• The component parts of the definite integral are the integrand, the variable of integration, and the limits ofintegration.
• Continuous functions on a closed interval are integrable. Functions that are not continuous may still be integrable,depending on the nature of the discontinuities.
• The properties of definite integrals can be used to evaluate integrals.
• The area under the curve of many functions can be calculated using geometric formulas.
• The average value of a function can be calculated using definite integrals.


1.3 The Fundamental Theorem of Calculus
• The Mean Value Theorem for Integrals states that for a continuous function over a closed interval, there is a value csuch that f (c) equals the average value of the function. See The Mean Value Theorem for Integrals.
• The Fundamental Theorem of Calculus, Part 1 shows the relationship between the derivative and the integral. SeeFundamental Theorem of Calculus, Part 1.
• The Fundamental Theorem of Calculus, Part 2 is a formula for evaluating a definite integral in terms of anantiderivative of its integrand. The total area under a curve can be found using this formula. See TheFundamental Theorem of Calculus, Part 2.


1.4 Integration Formulas and the Net Change Theorem
• The net change theorem states that when a quantity changes, the final value equals the initial value plus the integralof the rate of change. Net change can be a positive number, a negative number, or zero.
• The area under an even function over a symmetric interval can be calculated by doubling the area over the positivex-axis. For an odd function, the integral over a symmetric interval equals zero, because half the area is negative.


1.5 Substitution
• Substitution is a technique that simplifies the integration of functions that are the result of a chain-rule derivative.The term ‘substitution’ refers to changing variables or substituting the variable u and du for appropriate expressionsin the integrand.
• When using substitution for a definite integral, we also have to change the limits of integration.


Chapter 1 | Integration 117




1.6 Integrals Involving Exponential and Logarithmic Functions
• Exponential and logarithmic functions arise in many real-world applications, especially those involving growth anddecay.
• Substitution is often used to evaluate integrals involving exponential functions or logarithms.


1.7 Integrals Resulting in Inverse Trigonometric Functions
• Formulas for derivatives of inverse trigonometric functions developed in Derivatives of Exponential andLogarithmic Functions (http://cnx.org/content/m53584/latest/) lead directly to integration formulasinvolving inverse trigonometric functions.
• Use the formulas listed in the rule on integration formulas resulting in inverse trigonometric functions to match upthe correct format and make alterations as necessary to solve the problem.
• Substitution is often required to put the integrand in the correct form.


CHAPTER 1 REVIEW EXERCISES
True or False. Justify your answer with a proof or acounterexample. Assume all functions f and g are
continuous over their domains.
439. If f (x) > 0, f ′ (x) > 0 for all x, then the right-
hand rule underestimates the integral ∫


a


b
f (x). Use a graph


to justify your answer.


440. ∫
a


b
f (x)2dx = ∫


a


b
f (x)dx∫


a


b
f (x)dx


441. If f (x) ≤ g(x) for all x ∈ ⎡⎣a, b⎤⎦, then

a


b
f (x) ≤ ∫


a


b
g(x).


442. All continuous functions have an antiderivative.
Evaluate the Riemann sums L4 and R4 for the following
functions over the specified interval. Compare your answerwith the exact answer, when possible, or use a calculator todetermine the answer.
443. y = 3x2 − 2x + 1 over [−1, 1]


444. y = ln⎛⎝x2 + 1⎞⎠ over [0, e]


445. y = x2 sinx over [0, π]


446. y = x + 1x over [1, 4]
Evaluate the following integrals.


447. ∫
−1


1

⎝x


3 − 2x2 + 4x⎞⎠dx


448. ⌠
⌡0


4
3t


1 + 6t2
dt


449. ∫
π/3


π/2
2sec(2θ)tan(2θ)dθ


450. ∫
0


π/4
ecos


2 xsinxcosdx


Find the antiderivative.
451. ⌠


dx
(x + 4)3


452. ∫ x ln⎛⎝x2⎞⎠dx


453. ⌠


4x2


1 − x6
dx


454. ⌠


e2x


1 + e4x
dx


Find the derivative.
455. d


dt

⌡0


t
sinx
1 + x2


dx


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456. d
dx∫1


x3


4 − t2dt


457. d
dx∫1


ln(x)

⎝4t + e


t⎞
⎠dt


458. d
dx∫0


cosx
et


2
dt


The following problems consider the historic average costper gigabyte of RAM on a computer.
Year 5-Year Change ($)
1980 0
1985 −5,468,750
1990 −755,495
1995 −73,005
2000 −29,768
2005 −918
2010 −177


459. If the average cost per gigabyte of RAM in 2010 is$12, find the average cost per gigabyte of RAM in 1980.
460. The average cost per gigabyte of RAM can beapproximated by the function
C(t) = 8, 500, 000(0.65)t, where t is measured in years
since 1980, and C is cost in US$. Find the average cost per
gigabyte of RAM for 1980 to 2010.
461. Find the average cost of 1GB RAM for 2005 to2010.
462. The velocity of a bullet from a rifle can be
approximated by v(t) = 6400t2 − 6505t + 2686, where
t is seconds after the shot and v is the velocity measured
in feet per second. This equation only models the velocityfor the first half-second after the shot: 0 ≤ t ≤ 0.5. What
is the total distance the bullet travels in 0.5 sec?


463. What is the average velocity of the bullet for the firsthalf-second?


Chapter 1 | Integration 119




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2 | APPLICATIONS OFINTEGRATION


Figure 2.1 Hoover Dam is one of the United States’ iconic landmarks, and provides irrigation and hydroelectric power formillions of people in the southwest United States. (credit: modification of work by Lynn Betts, Wikimedia)


Chapter Outline
2.1 Areas between Curves
2.2 Determining Volumes by Slicing
2.3 Volumes of Revolution: Cylindrical Shells
2.4 Arc Length of a Curve and Surface Area
2.5 Physical Applications
2.6 Moments and Centers of Mass
2.7 Integrals, Exponential Functions, and Logarithms
2.8 Exponential Growth and Decay
2.9 Calculus of the Hyperbolic Functions


Chapter 2 | Applications of Integration 121




Introduction
The Hoover Dam is an engineering marvel. When Lake Mead, the reservoir behind the dam, is full, the dam withstands agreat deal of force. However, water levels in the lake vary considerably as a result of droughts and varying water demands.Later in this chapter, we use definite integrals to calculate the force exerted on the dam when the reservoir is full and weexamine how changing water levels affect that force (see Example 2.28).
Hydrostatic force is only one of the many applications of definite integrals we explore in this chapter. From geometricapplications such as surface area and volume, to physical applications such as mass and work, to growth and decay models,definite integrals are a powerful tool to help us understand and model the world around us.
2.1 | Areas between Curves


Learning Objectives
2.1.1 Determine the area of a region between two curves by integrating with respect to theindependent variable.
2.1.2 Find the area of a compound region.
2.1.3 Determine the area of a region between two curves by integrating with respect to thedependent variable.


In Introduction to Integration, we developed the concept of the definite integral to calculate the area below a curve ona given interval. In this section, we expand that idea to calculate the area of more complex regions. We start by finding thearea between two curves that are functions of x, beginning with the simple case in which one function value is always
greater than the other. We then look at cases when the graphs of the functions cross. Last, we consider how to calculate thearea between two curves that are functions of y.
Area of a Region between Two Curves
Let f (x) and g(x) be continuous functions over an interval ⎡⎣a, b⎤⎦ such that f (x) ≥ g(x) on ⎡⎣a, b⎤⎦. We want to find the
area between the graphs of the functions, as shown in the following figure.


Figure 2.2 The area between the graphs of two functions,
f (x) and g(x), on the interval [a, b].


As we did before, we are going to partition the interval on the x-axis and approximate the area between the graphs
of the functions with rectangles. So, for i = 0, 1, 2,…, n, let P = {xi} be a regular partition of ⎡⎣a, b⎤⎦. Then, for
i = 1, 2,…, n, choose a point xi* ∈ [xi − 1, xi], and on each interval [xi − 1, xi] construct a rectangle that extends
vertically from g(xi* ) to f (xi* ). Figure 2.3(a) shows the rectangles when xi* is selected to be the left endpoint of the
interval and n = 10. Figure 2.3(b) shows a representative rectangle in detail.


Use this calculator (http://www.openstaxcollege.org/l/20_CurveCalc) to learn more about the areasbetween two curves.


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Figure 2.3 (a)We can approximate the area between thegraphs of two functions, f (x) and g(x), with rectangles. (b)
The area of a typical rectangle goes from one curve to the other.


The height of each individual rectangle is f (xi* ) − g(xi* ) and the width of each rectangle is Δx. Adding the areas of all
the rectangles, we see that the area between the curves is approximated by


A ≈ ∑
i = 1


n

⎣ f (xi* ) − g(xi* )



⎦Δx.


This is a Riemann sum, so we take the limit as n → ∞ and we get
A = lim


n → ∞

i = 1


n

⎣ f (xi* ) − g(xi* )



⎦Δx = ∫


a


b

⎣ f (x) − g(x)⎤⎦dx.


These findings are summarized in the following theorem.
Theorem 2.1: Finding the Area between Two Curves
Let f (x) and g(x) be continuous functions such that f (x) ≥ g(x) over an interval ⎡⎣a, b⎤⎦. Let R denote the region
bounded above by the graph of f (x), below by the graph of g(x), and on the left and right by the lines x = a and
x = b, respectively. Then, the area of R is given by


(2.1)
A = ∫


a


b

⎣ f (x) − g(x)⎤⎦dx.


We apply this theorem in the following example.
Example 2.1
Finding the Area of a Region between Two Curves 1
If R is the region bounded above by the graph of the function f (x) = x + 4 and below by the graph of the
function g(x) = 3 − x


2
over the interval [1, 4], find the area of region R.


Solution
The region is depicted in the following figure.


Chapter 2 | Applications of Integration 123




2.1


Figure 2.4 A region between two curves is shown where onecurve is always greater than the other.


We have
A = ∫


a


b

⎣ f (x) − g(x)⎤⎦dx


= ∫
1


4

⎣(x + 4) −



⎝3 −


x
2



⎦dx = ∫


1


4


3x
2


+ 1⎤⎦dx


=


3x2
4


+ x

⎦ |1


4


= ⎛⎝16 −
7
4

⎠ =


57
4
.


The area of the region is 57
4


units2.


If R is the region bounded by the graphs of the functions f (x) = x
2
+ 5 and g(x) = x + 1


2
over the


interval ⎡⎣1, 5⎤⎦, find the area of region R.


In Example 2.1, we defined the interval of interest as part of the problem statement. Quite often, though, we want to defineour interval of interest based on where the graphs of the two functions intersect. This is illustrated in the following example.
Example 2.2
Finding the Area of a Region between Two Curves 2
If R is the region bounded above by the graph of the function f (x) = 9 − (x/2)2 and below by the graph of the
function g(x) = 6 − x, find the area of region R.


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2.2


Solution
The region is depicted in the following figure.


Figure 2.5 This graph shows the region below the graph of
f (x) and above the graph of g(x).


We first need to compute where the graphs of the functions intersect. Setting f (x) = g(x), we get
f (x) = g(x)


9 − ⎛⎝
x
2



2
= 6 − x


9 − x
2


4
= 6 − x


36 − x2 = 24 − 4x


x2 − 4x − 12 = 0
(x − 6)(x + 2) = 0.


The graphs of the functions intersect when x = 6 or x = −2, so we want to integrate from −2 to 6. Since
f (x) ≥ g(x) for −2 ≤ x ≤ 6, we obtain


A = ∫
a


b

⎣ f (x) − g(x)⎤⎦dx


= ∫
−2


6 ⎡



⎢9 − ⎛⎝


x
2



2
− (6 − x)





⎥dx = ∫


−2


6 ⎡
⎣3 −


x2
4


+ x

⎦dx


=

⎣3x −


x3


12
+ x


2


2

⎦ |−2


6


= 64
3
.


The area of the region is 64/3 units2.


If R is the region bounded above by the graph of the function f (x) = x and below by the graph of the
function g(x) = x4, find the area of region R.


Chapter 2 | Applications of Integration 125




Areas of Compound Regions
So far, we have required f (x) ≥ g(x) over the entire interval of interest, but what if we want to look at regions bounded by
the graphs of functions that cross one another? In that case, we modify the process we just developed by using the absolutevalue function.
Theorem 2.2: Finding the Area of a Region between Curves That Cross
Let f (x) and g(x) be continuous functions over an interval ⎡⎣a, b⎤⎦. Let R denote the region between the graphs of
f (x) and g(x), and be bounded on the left and right by the lines x = a and x = b, respectively. Then, the area of
R is given by


A = ∫
a


b


| f (x) − g(x)|dx.


In practice, applying this theorem requires us to break up the interval ⎡⎣a, b⎤⎦ and evaluate several integrals, depending on
which of the function values is greater over a given part of the interval. We study this process in the following example.
Example 2.3
Finding the Area of a Region Bounded by Functions That Cross
If R is the region between the graphs of the functions f (x) = sin x and g(x) = cos x over the interval [0, π],
find the area of region R.
Solution
The region is depicted in the following figure.


Figure 2.6 The region between two curves can be broken intotwo sub-regions.


The graphs of the functions intersect at x = π/4. For x ∈ [0, π/4], cos x ≥ sin x, so
| f (x) − g(x)| = |sin x − cos x| = cos x − sin x.


On the other hand, for x ∈ [π/4, π], sin x ≥ cos x, so


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2.3


| f (x) − g(x)| = |sin x − cos x| = sin x − cos x.


Then
A = ∫


a


b


| f (x) − g(x)|dx


= ∫
0


π


|sin x − cos x|dx = ∫
0


π/4
(cos x − sin x)dx + ∫


π/4


π
(sin x − cos x)dx


= [sin x + cos x] |0
π/4 + [−cos x − sin x] |π/4


π


= ( 2 − 1) + ⎛⎝1 + 2

⎠ = 2 2.


The area of the region is 2 2 units2.


If R is the region between the graphs of the functions f (x) = sin x and g(x) = cos x over the interval
[π/2, 2π], find the area of region R.


Example 2.4
Finding the Area of a Complex Region
Consider the region depicted in Figure 2.7. Find the area of R.


Figure 2.7 Two integrals are required to calculate the area ofthis region.
Solution
As with Example 2.3, we need to divide the interval into two pieces. The graphs of the functions intersect at
x = 1 (set f (x) = g(x) and solve for x), so we evaluate two separate integrals: one over the interval [0, 1] and
one over the interval [1, 2].
Over the interval [0, 1], the region is bounded above by f (x) = x2 and below by the x-axis, so we have


A1 = ∫
0


1
x2dx = x


3


3 |0
1


= 1
3
.


Over the interval [1, 2], the region is bounded above by g(x) = 2 − x and below by the x-axis, so we have


Chapter 2 | Applications of Integration 127




2.4


A2 = ∫
1


2
(2 − x)dx =



⎣2x −


x2
2

⎦ |1


2


= 1
2
.


Adding these areas together, we obtain
A = A1 + A2 =


1
3
+ 1


2
= 5


6
.


The area of the region is 5/6 units2.


Consider the region depicted in the following figure. Find the area of R.


Regions Defined with Respect to y
In Example 2.4, we had to evaluate two separate integrals to calculate the area of the region. However, there is anotherapproach that requires only one integral. What if we treat the curves as functions of y, instead of as functions of x?
Review Figure 2.7. Note that the left graph, shown in red, is represented by the function y = f (x) = x2. We could just
as easily solve this for x and represent the curve by the function x = v(y) = y. (Note that x = − y is also a valid
representation of the function y = f (x) = x2 as a function of y. However, based on the graph, it is clear we are interested
in the positive square root.) Similarly, the right graph is represented by the function y = g(x) = 2 − x, but could just as
easily be represented by the function x = u(y) = 2 − y. When the graphs are represented as functions of y, we see the
region is bounded on the left by the graph of one function and on the right by the graph of the other function. Therefore, ifwe integrate with respect to y, we need to evaluate one integral only. Let’s develop a formula for this type of integration.
Let u(y) and v(y) be continuous functions over an interval ⎡⎣c, d⎤⎦ such that u(y) ≥ v(y) for all y ∈ ⎡⎣c, d⎤⎦. We want to
find the area between the graphs of the functions, as shown in the following figure.


Figure 2.8 We can find the area between the graphs of twofunctions, u(y) and v(y).


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This time, we are going to partition the interval on the y-axis and use horizontal rectangles to approximate the area between
the functions. So, for i = 0, 1, 2,…, n, let Q = {yi} be a regular partition of ⎡⎣c, d⎤⎦. Then, for i = 1, 2,…, n, choose
a point yi* ∈ [yi − 1, yi], then over each interval [yi − 1, yi] construct a rectangle that extends horizontally from v⎛⎝yi* ⎞⎠
to u⎛⎝yi* ⎞⎠. Figure 2.9(a) shows the rectangles when yi* is selected to be the lower endpoint of the interval and n = 10.
Figure 2.9(b) shows a representative rectangle in detail.


Figure 2.9 (a) Approximating the area between the graphs oftwo functions, u(y) and v(y), with rectangles. (b) The area of
a typical rectangle.


The height of each individual rectangle is Δy and the width of each rectangle is u⎛⎝yi* ⎞⎠− v⎛⎝yi* ⎞⎠. Therefore, the area
between the curves is approximately


A ≈ ∑
i = 1


n

⎣u

⎝yi*



⎠− v



⎝yi*





⎦Δy.


This is a Riemann sum, so we take the limit as n → ∞, obtaining
A = lim


n → ∞

i = 1


n

⎣u

⎝yi*



⎠− v



⎝yi*





⎦Δy = ∫


c


d

⎣u(y) − v(y)⎤⎦dy.


These findings are summarized in the following theorem.
Theorem 2.3: Finding the Area between Two Curves, Integrating along the y-axis
Let u(y) and v(y) be continuous functions such that u(y) ≥ v(y) for all y ∈ ⎡⎣c, d⎤⎦. Let R denote the region bounded
on the right by the graph of u(y), on the left by the graph of v(y), and above and below by the lines y = d and
y = c, respectively. Then, the area of R is given by


(2.2)
A = ∫


c


d

⎣u(y) − v(y)⎤⎦dy.


Example 2.5
Integrating with Respect to y


Chapter 2 | Applications of Integration 129




2.5


Let’s revisit Example 2.4, only this time let’s integrate with respect to y. Let R be the region depicted in
Figure 2.10. Find the area of R by integrating with respect to y.


Figure 2.10 The area of region R can be calculated using
one integral only when the curves are treated as functions of y.


Solution
We must first express the graphs as functions of y. As we saw at the beginning of this section, the curve on
the left can be represented by the function x = v(y) = y, and the curve on the right can be represented by the
function x = u(y) = 2 − y.
Now we have to determine the limits of integration. The region is bounded below by the x-axis, so the lower limitof integration is y = 0. The upper limit of integration is determined by the point where the two graphs intersect,
which is the point (1, 1), so the upper limit of integration is y = 1. Thus, we have ⎡⎣c, d⎤⎦ = [0, 1].
Calculating the area of the region, we get


A = ∫
c


d

⎣u(y) − v(y)⎤⎦dy


= ∫
0


1



⎝2 − y⎞⎠− y⎤⎦dy =





⎢2y −


y2


2
− 2


3
y3/2



⎥ |0


1


= 5
6
.


The area of the region is 5/6 units2.


Let’s revisit the checkpoint associated with Example 2.4, only this time, let’s integrate with respect to
y. Let be the region depicted in the following figure. Find the area of R by integrating with respect to y.


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2.1 EXERCISES
For the following exercises, determine the area of theregion between the two curves in the given figure byintegrating over the x-axis.
1. y = x2 − 3 and y = 1


2. y = x2 and y = 3x + 4


For the following exercises, split the region between thetwo curves into two smaller regions, then determine thearea by integrating over the x-axis. Note that you will
have two integrals to solve.
3. y = x3 and y = x2 + x


4. y = cos θ and y = 0.5, for 0 ≤ θ ≤ π


For the following exercises, determine the area of theregion between the two curves by integrating over the
y-axis.


5. x = y2 and x = 9


6. y = x and x = y2


For the following exercises, graph the equations and shadethe area of the region between the curves. Determine itsarea by integrating over the x-axis.
7. y = x2 and y = −x2 + 18x
8. y = 1x , y = 1x2, and x = 3


9. y = cos x and y = cos2 x on x = [−π, π]
10. y = ex, y = e2x − 1, and x = 0
11. y = ex, y = e−x, x = −1 and x = 1


Chapter 2 | Applications of Integration 131




12. y = e, y = ex, and y = e−x
13. y = |x| and y = x2
For the following exercises, graph the equations and shadethe area of the region between the curves. If necessary,break the region into sub-regions to determine its entirearea.
14. y = sin(πx), y = 2x, and x > 0
15. y = 12 − x, y = x, and y = 1
16. y = sin x and y = cos x over x = [−π, π]
17. y = x3 and y = x2 − 2x over x = [−1, 1]
18. y = x2 + 9 and y = 10 + 2x over x = [−1, 3]
19. y = x3 + 3x and y = 4x
For the following exercises, graph the equations and shadethe area of the region between the curves. Determine itsarea by integrating over the y-axis.
20. x = y3 and x = 3y − 2
21. x = 2y and x = y3 − y
22. x = −3 + y2 and x = y − y2
23. y2 = x and x = y + 2
24. x = |y| and 2x = −y2 + 2
25. x = sin y, x = cos(2y), y = π/2, and y = −π/2
For the following exercises, graph the equations and shadethe area of the region between the curves. Determine itsarea by integrating over the x-axis or y-axis, whicheverseems more convenient.
26. x = y4 and x = y5
27. y = xex, y = ex, x = 0, and x = 1
28. y = x6 and y = x4
29. x = y3 + 2y2 + 1 and x = −y2 + 1
30. y = |x| and y = x2 − 1


31. y = 4 − 3x and y = 1x
32. y = sin x, x = −π/6, x = π/6, and y = cos3 x
33. y = x2 − 3x + 2 and y = x3 − 2x2 − x + 2
34. y = 2 cos3 (3x), y = −1, x = π


4
, and x = − π


4


35. y + y3 = x and 2y = x
36. y = 1 − x2 and y = x2 − 1
37. y = cos−1 x, y = sin−1 x, x = −1, and x = 1
For the following exercises, find the exact area of theregion bounded by the given equations if possible. If youare unable to determine the intersection points analytically,use a calculator to approximate the intersection points withthree decimal places and determine the approximate area ofthe region.
38. [T] x = ey and y = x − 2
39. [T] y = x2 and y = 1 − x2
40. [T] y = 3x2 + 8x + 9 and 3y = x + 24
41. [T] x = 4 − y2 and y2 = 1 + x2
42. [T] x2 = y3 and x = 3y
43. [T]
y = sin3 x + 2, y = tan x, x = −1.5, and x = 1.5


44. [T] y = 1 − x2 and y2 = x2


45. [T] y = 1 − x2 and y = x2 + 2x + 1
46. [T] x = 4 − y2 and x = 1 + 3y + y2
47. [T] y = cos x, y = ex, x = −π, and x = 0


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48. The largest triangle with a base on the x-axis that
fits inside the upper half of the unit circle y2 + x2 = 1
is given by y = 1 + x and y = 1 − x. See the following
figure. What is the area inside the semicircle but outside thetriangle?


49. A factory selling cell phones has a marginal cost
function C(x) = 0.01x2 − 3x + 229, where x represents
the number of cell phones, and a marginal revenue functiongiven by R(x) = 429 − 2x. Find the area between the
graphs of these curves and x = 0. What does this area
represent?
50. An amusement park has a marginal cost function
C(x) = 1000e−x + 5, where x represents the number
of tickets sold, and a marginal revenue function given by
R(x) = 60 − 0.1x. Find the total profit generated when
selling 550 tickets. Use a calculator to determine
intersection points, if necessary, to two decimal places.
51. The tortoise versus the hare: The speed of the hareis given by the sinusoidal function H(t) = 1 − cos⎛⎝(πt)/2⎞⎠
whereas the speed of the tortoise is
T(t) = (1/2)tan−1 (t/4), where t is time measured in
hours and the speed is measured in miles per hour. Find thearea between the curves from time t = 0 to the first time
after one hour when the tortoise and hare are traveling atthe same speed. What does it represent? Use a calculator todetermine the intersection points, if necessary, accurate tothree decimal places.
52. The tortoise versus the hare: The speed of the hareis given by the sinusoidal function
H(t) = (1/2) − (1/2)cos(2πt) whereas the speed of the
tortoise is T(t) = t, where t is time measured in hours
and speed is measured in kilometers per hour. If the race isover in 1 hour, who won the race and by how much? Use a
calculator to determine the intersection points, if necessary,accurate to three decimal places.
For the following exercises, find the area between thecurves by integrating with respect to x and then with
respect to y. Is one method easier than the other? Do you


obtain the same answer?
53. y = x2 + 2x + 1 and y = −x2 − 3x + 4
54. y = x4 and x = y5
55. x = y2 − 2 and x = 2y
For the following exercises, solve using calculus, thencheck your answer with geometry.
56. Determine the equations for the sides of the squarethat touches the unit circle on all four sides, as seen in thefollowing figure. Find the area between the perimeter ofthis square and the unit circle. Is there another way to solvethis without using calculus?


57. Find the area between the perimeter of the unit circleand the triangle created from y = 2x + 1, y = 1 − 2x and
y = − 3


5
, as seen in the following figure. Is there a way


to solve this without using calculus?


Chapter 2 | Applications of Integration 133




2.2 | Determining Volumes by Slicing
Learning Objectives


2.2.1 Determine the volume of a solid by integrating a cross-section (the slicing method).
2.2.2 Find the volume of a solid of revolution using the disk method.
2.2.3 Find the volume of a solid of revolution with a cavity using the washer method.


In the preceding section, we used definite integrals to find the area between two curves. In this section, we use definiteintegrals to find volumes of three-dimensional solids. We consider three approaches—slicing, disks, and washers—forfinding these volumes, depending on the characteristics of the solid.
Volume and the Slicing Method
Just as area is the numerical measure of a two-dimensional region, volume is the numerical measure of a three-dimensionalsolid. Most of us have computed volumes of solids by using basic geometric formulas. The volume of a rectangular solid,for example, can be computed by multiplying length, width, and height: V = lwh. The formulas for the volume of a sphere

⎝V =


4
3
πr3⎞⎠, a cone ⎛⎝V = 13πr2h⎞⎠, and a pyramid ⎛⎝V = 13Ah⎞⎠ have also been introduced. Although some of these


formulas were derived using geometry alone, all these formulas can be obtained by using integration.
We can also calculate the volume of a cylinder. Although most of us think of a cylinder as having a circular base, such asa soup can or a metal rod, in mathematics the word cylinder has a more general meaning. To discuss cylinders in this moregeneral context, we first need to define some vocabulary.
We define the cross-section of a solid to be the intersection of a plane with the solid. A cylinder is defined as any solidthat can be generated by translating a plane region along a line perpendicular to the region, called the axis of the cylinder.Thus, all cross-sections perpendicular to the axis of a cylinder are identical. The solid shown in Figure 2.11 is an exampleof a cylinder with a noncircular base. To calculate the volume of a cylinder, then, we simply multiply the area of the cross-
section by the height of the cylinder: V = A · h. In the case of a right circular cylinder (soup can), this becomes V = πr2h.


Figure 2.11 Each cross-section of a particular cylinder is identical to the others.


If a solid does not have a constant cross-section (and it is not one of the other basic solids), we may not have a formula forits volume. In this case, we can use a definite integral to calculate the volume of the solid. We do this by slicing the solidinto pieces, estimating the volume of each slice, and then adding those estimated volumes together. The slices should all beparallel to one another, and when we put all the slices together, we should get the whole solid. Consider, for example, thesolid S shown in Figure 2.12, extending along the x-axis.


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Figure 2.12 A solid with a varying cross-section.


We want to divide S into slices perpendicular to the x-axis. As we see later in the chapter, there may be times when we
want to slice the solid in some other direction—say, with slices perpendicular to the y-axis. The decision of which way toslice the solid is very important. If we make the wrong choice, the computations can get quite messy. Later in the chapter,we examine some of these situations in detail and look at how to decide which way to slice the solid. For the purposes ofthis section, however, we use slices perpendicular to the x-axis.
Because the cross-sectional area is not constant, we let A(x) represent the area of the cross-section at point x. Now let
P = ⎧⎩⎨x0, x1…, Xn





⎬ be a regular partition of ⎡⎣a, b⎤⎦, and for i = 1, 2,…n, let Si represent the slice of S stretching from


xi − 1 to xi. The following figure shows the sliced solid with n = 3.


Figure 2.13 The solid S has been divided into three slices
perpendicular to the x-axis.


Finally, for i = 1, 2,…n, let xi* be an arbitrary point in [xi − 1, xi]. Then the volume of slice Si can be estimated by
V ⎛⎝Si



⎠ ≈ A⎛⎝xi*



⎠Δx. Adding these approximations together, we see the volume of the entire solid S can be approximated by


V(S) ≈ ∑
i = 1


n


A⎛⎝xi*

⎠Δx.


By now, we can recognize this as a Riemann sum, and our next step is to take the limit as n → ∞. Then we have
V(S) = lim


n → ∞

i = 1


n


A⎛⎝xi*

⎠Δx = ∫


a


b


A(x)dx.


Chapter 2 | Applications of Integration 135




The technique we have just described is called the slicing method. To apply it, we use the following strategy.
Problem-Solving Strategy: Finding Volumes by the Slicing Method


1. Examine the solid and determine the shape of a cross-section of the solid. It is often helpful to draw a pictureif one is not provided.
2. Determine a formula for the area of the cross-section.
3. Integrate the area formula over the appropriate interval to get the volume.


Recall that in this section, we assume the slices are perpendicular to the x-axis. Therefore, the area formula is in terms of
x and the limits of integration lie on the x-axis. However, the problem-solving strategy shown here is valid regardless of
how we choose to slice the solid.
Example 2.6
Deriving the Formula for the Volume of a Pyramid
We know from geometry that the formula for the volume of a pyramid is V = 1


3
Ah. If the pyramid has a square


base, this becomes V = 1
3
a2h, where a denotes the length of one side of the base. We are going to use the


slicing method to derive this formula.
Solution
We want to apply the slicing method to a pyramid with a square base. To set up the integral, consider the pyramidshown in Figure 2.14, oriented along the x-axis.


Figure 2.14 (a) A pyramid with a square base is oriented along the x-axis. (b) A two-dimensional view of thepyramid is seen from the side.


We first want to determine the shape of a cross-section of the pyramid. We are know the base is a square, so thecross-sections are squares as well (step 1). Now we want to determine a formula for the area of one of these cross-sectional squares. Looking at Figure 2.14(b), and using a proportion, since these are similar triangles, we have
s
a =


x
h
or s = ax


h
.


Therefore, the area of one of the cross-sectional squares is


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2.6


A(x) = s2 = ⎛⎝
ax
h



2

⎝step 2⎞⎠.


Then we find the volume of the pyramid by integrating from 0 to h (step 3):
V = ∫


0


h


A(x)dx


= ∫
0


h


ax
h



2
dx = a


2


h2

0


h


x2dx


=


a2


h2


1
3
x3⎞⎠

⎦ |0


h


= 1
3
a2h.


This is the formula we were looking for.


Use the slicing method to derive the formula V = 1
3
πr2h for the volume of a circular cone.


Solids of Revolution
If a region in a plane is revolved around a line in that plane, the resulting solid is called a solid of revolution, as shown inthe following figure.


Chapter 2 | Applications of Integration 137




Figure 2.15 (a) This is the region that is revolved around the x-axis.(b) As the region begins to revolve around the axis, it sweeps out asolid of revolution. (c) This is the solid that results when therevolution is complete.


Solids of revolution are common in mechanical applications, such as machine parts produced by a lathe. We spend the restof this section looking at solids of this type. The next example uses the slicing method to calculate the volume of a solid ofrevolution.
Use an online integral calculator (http://www.openstaxcollege.org/l/20_IntCalc2) to learn more.


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Example 2.7
Using the Slicing Method to find the Volume of a Solid of Revolution
Use the slicing method to find the volume of the solid of revolution bounded by the graphs of
f (x) = x2 − 4x + 5, x = 1, and x = 4, and rotated about the x-axis.
Solution
Using the problem-solving strategy, we first sketch the graph of the quadratic function over the interval [1, 4] as
shown in the following figure.


Figure 2.16 A region used to produce a solid of revolution.


Next, revolve the region around the x-axis, as shown in the following figure.


Figure 2.17 Two views, (a) and (b), of the solid of revolution produced by revolving the regionin Figure 2.16 about the x-axis.


Chapter 2 | Applications of Integration 139




2.7


Since the solid was formed by revolving the region around the x-axis, the cross-sections are circles (step 1).
The area of the cross-section, then, is the area of a circle, and the radius of the circle is given by f (x). Use the
formula for the area of the circle:


A(x) = πr2 = π⎡⎣ f (x)⎤⎦2 = π⎛⎝x
2 − 4x + 5⎞⎠


2
(step 2).


The volume, then, is (step 3)
V = ∫


a


h


A(x)dx


= ∫
1


4
π⎛⎝x


2 − 4x + 5⎞⎠
2
dx = π∫


1


4

⎝x


4 − 8x3 + 26x2 − 40x + 25⎞⎠dx


= π


x5


5
− 2x4 + 26x


3


3
− 20x2 + 25x



⎠|1
4


= 78
5
π.


The volume is 78π/5.


Use the method of slicing to find the volume of the solid of revolution formed by revolving the regionbetween the graph of the function f (x) = 1/x and the x-axis over the interval [1, 2] around the x-axis. See
the following figure.


The Disk Method
When we use the slicing method with solids of revolution, it is often called the disk method because, for solids ofrevolution, the slices used to over approximate the volume of the solid are disks. To see this, consider the solid of revolution
generated by revolving the region between the graph of the function f (x) = (x − 1)2 + 1 and the x-axis over the interval
[−1, 3] around the x-axis. The graph of the function and a representative disk are shown in Figure 2.18(a) and (b). The
region of revolution and the resulting solid are shown in Figure 2.18(c) and (d).


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Figure 2.18 (a) A thin rectangle for approximating the area under a curve. (b) A representative disk formed byrevolving the rectangle about the x-axis. (c) The region under the curve is revolved about the x-axis, resulting in
(d) the solid of revolution.


We already used the formal Riemann sum development of the volume formula when we developed the slicing method. Weknow that
V = ∫


a


b
A(x)dx.


The only difference with the disk method is that we know the formula for the cross-sectional area ahead of time; it is thearea of a circle. This gives the following rule.
Rule: The Disk Method
Let f (x) be continuous and nonnegative. Define R as the region bounded above by the graph of f (x), below by the


Chapter 2 | Applications of Integration 141




x-axis, on the left by the line x = a, and on the right by the line x = b. Then, the volume of the solid of revolution
formed by revolving R around the x-axis is given by


(2.3)
V = ∫


a


b
π⎡⎣ f (x)⎤⎦2dx.


The volume of the solid we have been studying (Figure 2.18) is given by
V = ∫


a


b
π⎡⎣ f (x)⎤⎦2dx


= ∫
−1


3
π⎡⎣(x − 1)


2 + 1⎤⎦
2
dx = π∫


−1


3

⎣(x − 1)


4 + 2(x − 1)2 + 1⎤⎦
2
dx


= π⎡⎣
1
5
(x − 1)5 + 2


3
(x − 1)3 + x⎤⎦ |−1


3
= π⎡⎣


32
5


+ 16
3


+ 3⎞⎠−

⎝−


32
5


− 16
3


− 1⎞⎠

⎦ =


412π
15


units3.


Let’s look at some examples.
Example 2.8
Using the Disk Method to Find the Volume of a Solid of Revolution 1
Use the disk method to find the volume of the solid of revolution generated by rotating the region between thegraph of f (x) = x and the x-axis over the interval [1, 4] around the x-axis.
Solution
The graphs of the function and the solid of revolution are shown in the following figure.


Figure 2.19 (a) The function f (x) = x over the interval [1, 4]. (b) The solid of revolution
obtained by revolving the region under the graph of f (x) about the x-axis.


We have


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2.8


V = ∫
a


b
π⎡⎣ f (x)⎤⎦2dx


= ∫
1


4
π[ x]2dx = π∫


1


4
x dx


= π
2
x2|1


4
= 15π


2
.


The volume is (15π)/2 units3.


Use the disk method to find the volume of the solid of revolution generated by rotating the region
between the graph of f (x) = 4 − x and the x-axis over the interval [0, 4] around the x-axis.


So far, our examples have all concerned regions revolved around the x-axis, but we can generate a solid of revolution by
revolving a plane region around any horizontal or vertical line. In the next example, we look at a solid of revolution that hasbeen generated by revolving a region around the y-axis. The mechanics of the disk method are nearly the same as when
the x-axis is the axis of revolution, but we express the function in terms of y and we integrate with respect to y as well.
This is summarized in the following rule.
Rule: The Disk Method for Solids of Revolution around the y-axis
Let g(y) be continuous and nonnegative. Define Q as the region bounded on the right by the graph of g(y), on the
left by the y-axis, below by the line y = c, and above by the line y = d. Then, the volume of the solid of revolution
formed by revolving Q around the y-axis is given by


(2.4)
V = ∫


c


d
π⎡⎣g(y)⎤⎦2dy.


The next example shows how this rule works in practice.
Example 2.9
Using the Disk Method to Find the Volume of a Solid of Revolution 2
Let R be the region bounded by the graph of g(y) = 4 − y and the y-axis over the y-axis interval [0, 4].
Use the disk method to find the volume of the solid of revolution generated by rotating R around the y-axis.
Solution
Figure 2.20 shows the function and a representative disk that can be used to estimate the volume. Notice thatsince we are revolving the function around the y-axis, the disks are horizontal, rather than vertical.


Chapter 2 | Applications of Integration 143




Figure 2.20 (a) Shown is a thin rectangle between the curve of the function g(y) = 4 − y
and the y-axis. (b) The rectangle forms a representative disk after revolution around the y-axis.


The region to be revolved and the full solid of revolution are depicted in the following figure.


Figure 2.21 (a) The region to the left of the function g(y) = 4 − y over the y-axis interval
[0, 4]. (b) The solid of revolution formed by revolving the region about the y-axis.


To find the volume, we integrate with respect to y. We obtain
V = ∫


c


d
π⎡⎣g(y)⎤⎦2dy


= ∫
0


4
π⎡⎣ 4 − y




2dy = π∫


0


4

⎝4 − y⎞⎠dy


= π



⎢4y −


y2


2





⎥ |0


4


= 8π.


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2.9


The volume is 8π units3.


Use the disk method to find the volume of the solid of revolution generated by rotating the regionbetween the graph of g(y) = y and the y-axis over the interval [1, 4] around the y-axis.


The Washer Method
Some solids of revolution have cavities in the middle; they are not solid all the way to the axis of revolution. Sometimes,this is just a result of the way the region of revolution is shaped with respect to the axis of revolution. In other cases, cavitiesarise when the region of revolution is defined as the region between the graphs of two functions. A third way this can happenis when an axis of revolution other than the x-axis or y-axis is selected.
When the solid of revolution has a cavity in the middle, the slices used to approximate the volume are not disks, but washers(disks with holes in the center). For example, consider the region bounded above by the graph of the function f (x) = x
and below by the graph of the function g(x) = 1 over the interval [1, 4]. When this region is revolved around the x-axis,
the result is a solid with a cavity in the middle, and the slices are washers. The graph of the function and a representativewasher are shown in Figure 2.22(a) and (b). The region of revolution and the resulting solid are shown in Figure 2.22(c)and (d).


Chapter 2 | Applications of Integration 145




Figure 2.22 (a) A thin rectangle in the region between two curves. (b) Arepresentative disk formed by revolving the rectangle about the x-axis. (c) The region
between the curves over the given interval. (d) The resulting solid of revolution.


The cross-sectional area, then, is the area of the outer circle less the area of the inner circle. In this case,
A(x) = π( x)2 − π(1)2 = π(x − 1).


Then the volume of the solid is
V = ∫


a


b
A(x)dx


= ∫
1


4
π(x − 1)dx = π




x2
2


− x

⎦ |1


4


= 9
2
π units3.


Generalizing this process gives the washer method.
Rule: The Washer Method
Suppose f (x) and g(x) are continuous, nonnegative functions such that f (x) ≥ g(x) over ⎡⎣a, b⎤⎦. Let R denote the
region bounded above by the graph of f (x), below by the graph of g(x), on the left by the line x = a, and on


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2.10


the right by the line x = b. Then, the volume of the solid of revolution formed by revolving R around the x-axis is
given by


(2.5)
V = ∫


a


b
π⎡⎣

⎝ f (x)⎞⎠2 − ⎛⎝g(x)⎞⎠2⎤⎦dx.


Example 2.10
Using the Washer Method
Find the volume of a solid of revolution formed by revolving the region bounded above by the graph of f (x) = x
and below by the graph of g(x) = 1/x over the interval [1, 4] around the x-axis.
Solution
The graphs of the functions and the solid of revolution are shown in the following figure.


Figure 2.23 (a) The region between the graphs of the functions f (x) = x and
g(x) = 1/x over the interval [1, 4]. (b) Revolving the region about the x-axis generates
a solid of revolution with a cavity in the middle.


We have
V = ∫


a


b
π⎡⎣

⎝ f (x)⎞⎠2 − ⎛⎝g(x)⎞⎠2⎤⎦dx


= π∫
1


4⎡
⎣x


2 − ⎛⎝
1
x



2⎤
⎦dx = π




x3


3
+ 1x

⎦ |1


4


= 81π
4


units3.


Find the volume of a solid of revolution formed by revolving the region bounded by the graphs of
f (x) = x and g(x) = 1/x over the interval [1, 3] around the x-axis.


As with the disk method, we can also apply the washer method to solids of revolution that result from revolving a regionaround the y-axis. In this case, the following rule applies.


Chapter 2 | Applications of Integration 147




Rule: The Washer Method for Solids of Revolution around the y-axis
Suppose u(y) and v(y) are continuous, nonnegative functions such that v(y) ≤ u(y) for y ∈ ⎡⎣c, d⎤⎦. Let Q denote
the region bounded on the right by the graph of u(y), on the left by the graph of v(y), below by the line y = c,
and above by the line y = d. Then, the volume of the solid of revolution formed by revolving Q around the y-axis
is given by


V = ∫
c


d
π⎡⎣

⎝u(y)⎞⎠2 − ⎛⎝v(y)⎞⎠2⎤⎦dy.


Rather than looking at an example of the washer method with the y-axis as the axis of revolution, we now consider an
example in which the axis of revolution is a line other than one of the two coordinate axes. The same general methodapplies, but you may have to visualize just how to describe the cross-sectional area of the volume.
Example 2.11
The Washer Method with a Different Axis of Revolution
Find the volume of a solid of revolution formed by revolving the region bounded above by f (x) = 4 − x and
below by the x-axis over the interval [0, 4] around the line y = −2.
Solution
The graph of the region and the solid of revolution are shown in the following figure.


Figure 2.24 (a) The region between the graph of the function f (x) = 4 − x and the x-axis
over the interval [0, 4]. (b) Revolving the region about the line y = −2 generates a solid of
revolution with a cylindrical hole through its middle.


We can’t apply the volume formula to this problem directly because the axis of revolution is not one of the


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2.11


coordinate axes. However, we still know that the area of the cross-section is the area of the outer circle less thearea of the inner circle. Looking at the graph of the function, we see the radius of the outer circle is given by
f (x) + 2, which simplifies to


f (x) + 2 = (4 − x) + 2 = 6 − x.


The radius of the inner circle is g(x) = 2. Therefore, we have
V = ∫


0


4
π⎡⎣(6 − x)


2 − (2)2⎤⎦dx


= π∫
0


4

⎝x


2 − 12x + 32⎞⎠dx = π


x3


3
− 6x2 + 32x



⎦ |0


4


= 160π
3


units3.


Find the volume of a solid of revolution formed by revolving the region bounded above by the graph of
f (x) = x + 2 and below by the x-axis over the interval [0, 3] around the line y = −1.


Chapter 2 | Applications of Integration 149




2.2 EXERCISES
58. Derive the formula for the volume of a sphere usingthe slicing method.
59. Use the slicing method to derive the formula for thevolume of a cone.
60. Use the slicing method to derive the formula for thevolume of a tetrahedron with side length a.
61. Use the disk method to derive the formula for thevolume of a trapezoidal cylinder.
62. Explain when you would use the disk method versusthe washer method. When are they interchangeable?
For the following exercises, draw a typical slice and findthe volume using the slicing method for the given volume.
63. A pyramid with height 6 units and square base of side2 units, as pictured here.


64. A pyramid with height 4 units and a rectangular basewith length 2 units and width 3 units, as pictured here.


65. A tetrahedron with a base side of 4 units, as seen here.


66. A pyramid with height 5 units, and an isoscelestriangular base with lengths of 6 units and 8 units, as seenhere.


67. A cone of radius r and height h has a smaller cone of
radius r/2 and height h/2 removed from the top, as seen
here. The resulting solid is called a frustum.


For the following exercises, draw an outline of the solid andfind the volume using the slicing method.
68. The base is a circle of radius a. The slices
perpendicular to the base are squares.
69. The base is a triangle with vertices (0, 0), (1, 0),
and (0, 1). Slices perpendicular to the xy-plane are
semicircles.
70. The base is the region under the parabola y = 1 − x2
in the first quadrant. Slices perpendicular to the xy-planeare squares.
71. The base is the region under the parabola y = 1 − x2
and above the x-axis. Slices perpendicular to the y-axis
are squares.


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72. The base is the region enclosed by y = x2 and
y = 9. Slices perpendicular to the x-axis are right isosceles
triangles.
73. The base is the area between y = x and y = x2.
Slices perpendicular to the x-axis are semicircles.
For the following exercises, draw the region bounded bythe curves. Then, use the disk method to find the volumewhen the region is rotated around the x-axis.
74. x + y = 8, x = 0, and y = 0
75. y = 2x2, x = 0, x = 4, and y = 0
76. y = ex + 1, x = 0, x = 1, and y = 0
77. y = x4, x = 0, and y = 1
78. y = x, x = 0, x = 4, and y = 0
79. y = sin x, y = cos x, and x = 0
80. y = 1x , x = 2, and y = 3
81. x2 − y2 = 9 and x + y = 9, y = 0 and x = 0
For the following exercises, draw the region bounded bythe curves. Then, find the volume when the region isrotated around the y-axis.
82. y = 4 − 1


2
x, x = 0, and y = 0


83. y = 2x3, x = 0, x = 1, and y = 0
84. y = 3x2, x = 0, and y = 3
85. y = 4 − x2, y = 0, and x = 0
86. y = 1


x + 1
, x = 0, and x = 3


87. x = sec(y) and y = π
4
, y = 0 and x = 0


88. y = 1
x + 1


, x = 0, and x = 2


89. y = 4 − x, y = x, and x = 0
For the following exercises, draw the region bounded bythe curves. Then, find the volume when the region isrotated around the x-axis.


90. y = x + 2, y = x + 6, x = 0, and x = 5
91. y = x2 and y = x + 2
92. x2 = y3 and x3 = y2
93. y = 4 − x2 and y = 2 − x
94. [T] y = cos x, y = e−x, x = 0, and x = 1.2927
95. y = x and y = x2
96. y = sin x, y = 5 sin x, x = 0 and x = π
97. y = 1 + x2 and y = 4 − x2
For the following exercises, draw the region bounded bythe curves. Then, use the washer method to find the volumewhen the region is revolved around the y-axis.
98. y = x, x = 4, and y = 0
99. y = x + 2, y = 2x − 1, and x = 0
100. y = x3 and y = x3


101. x = e2y, x = y2, y = 0, and y = ln(2)
102. x = 9 − y2, x = e−y, y = 0, and y = 3


Chapter 2 | Applications of Integration 151




103. Yogurt containers can be shaped like frustums.
Rotate the line y = 1mx around the y-axis to find the
volume between y = a and y = b.


104. Rotate the ellipse ⎛⎝x2 /a2⎞⎠+ ⎛⎝y2 /b2⎞⎠ = 1 around the
x-axis to approximate the volume of a football, as seenhere.


105. Rotate the ellipse ⎛⎝x2 /a2⎞⎠+ ⎛⎝y2 /b2⎞⎠ = 1 around the
y-axis to approximate the volume of a football.


106. A better approximation of the volume of a footballis given by the solid that comes from rotating y = sin x
around the x-axis from x = 0 to x = π. What is the
volume of this football approximation, as seen here?


107. What is the volume of the Bundt cake that comesfrom rotating y = sin x around the y-axis from x = 0 to
x = π ?


For the following exercises, find the volume of the soliddescribed.


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108. The base is the region between y = x and y = x2.
Slices perpendicular to the x-axis are semicircles.
109. The base is the region enclosed by the generic ellipse

⎝x


2 /a2⎞⎠+

⎝y


2 /b2⎞⎠ = 1. Slices perpendicular to the x-axis
are semicircles.
110. Bore a hole of radius a down the axis of a right cone
and through the base of radius b, as seen here.


111. Find the volume common to two spheres of radius r
with centers that are 2h apart, as shown here.


112. Find the volume of a spherical cap of height h and
radius r where h < r, as seen here.


113. Find the volume of a sphere of radius R with a cap
of height h removed from the top, as seen here.


Chapter 2 | Applications of Integration 153




2.3 | Volumes of Revolution: Cylindrical Shells
Learning Objectives


2.3.1 Calculate the volume of a solid of revolution by using the method of cylindrical shells.
2.3.2 Compare the different methods for calculating a volume of revolution.


In this section, we examine the method of cylindrical shells, the final method for finding the volume of a solid of revolution.We can use this method on the same kinds of solids as the disk method or the washer method; however, with the disk andwasher methods, we integrate along the coordinate axis parallel to the axis of revolution. With the method of cylindricalshells, we integrate along the coordinate axis perpendicular to the axis of revolution. The ability to choose which variableof integration we want to use can be a significant advantage with more complicated functions. Also, the specific geometryof the solid sometimes makes the method of using cylindrical shells more appealing than using the washer method. In thelast part of this section, we review all the methods for finding volume that we have studied and lay out some guidelines tohelp you determine which method to use in a given situation.
The Method of Cylindrical Shells
Again, we are working with a solid of revolution. As before, we define a region R, bounded above by the graph of a
function y = f (x), below by the x-axis, and on the left and right by the lines x = a and x = b, respectively, as shown
in Figure 2.25(a). We then revolve this region around the y-axis, as shown in Figure 2.25(b). Note that this is differentfrom what we have done before. Previously, regions defined in terms of functions of x were revolved around the x-axis
or a line parallel to it.


Figure 2.25 (a) A region bounded by the graph of a function of x. (b) The solid of revolution formed when the
region is revolved around the y-axis.


As we have done many times before, partition the interval ⎡⎣a, b⎤⎦ using a regular partition, P = {x0, x1 ,…, xn} and,
for i = 1, 2,…, n, choose a point xi* ∈ [xi − 1, xi]. Then, construct a rectangle over the interval [xi − 1, xi] of height
f (xi* ) and width Δx. A representative rectangle is shown in Figure 2.26(a). When that rectangle is revolved around the


y-axis, instead of a disk or a washer, we get a cylindrical shell, as shown in the following figure.


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Figure 2.26 (a) A representative rectangle. (b) When this rectangle is revolved around the y-axis, the result is a cylindrical
shell. (c) When we put all the shells together, we get an approximation of the original solid.


To calculate the volume of this shell, consider Figure 2.27.


Figure 2.27 Calculating the volume of the shell.


The shell is a cylinder, so its volume is the cross-sectional area multiplied by the height of the cylinder. The cross-sectionsare annuli (ring-shaped regions—essentially, circles with a hole in the center), with outer radius xi and inner radius xi − 1.
Thus, the cross-sectional area is πxi2 − πxi − 12 . The height of the cylinder is f (xi* ). Then the volume of the shell is


Vshell = f (xi* )(πxi
2 − πxi − 1


2 )


= π f (xi* )

⎝xi


2 − xi − 1
2 ⎞


= π f (xi* )(xi + xi − 1)(xi − xi − 1)


= 2π f (xi* )


xi + xi − 1


2

⎠(xi − xi − 1).


Note that xi − xi − 1 = Δx, so we have


Chapter 2 | Applications of Integration 155




Vshell = 2π f (xi* )


xi + xi − 1


2

⎠Δx.


Furthermore, xi + xi − 1
2


is both the midpoint of the interval [xi − 1, xi] and the average radius of the shell, and we can
approximate this by xi* . We then have


Vshell ≈ 2π f (xi* )xi* Δx.


Another way to think of this is to think of making a vertical cut in the shell and then opening it up to form a flat plate(Figure 2.28).


Figure 2.28 (a) Make a vertical cut in a representative shell. (b) Open the shell up to form a flat plate.


In reality, the outer radius of the shell is greater than the inner radius, and hence the back edge of the plate would be slightlylonger than the front edge of the plate. However, we can approximate the flattened shell by a flat plate of height f (xi* ),
width 2πxi* , and thickness Δx (Figure 2.28). The volume of the shell, then, is approximately the volume of the flat
plate. Multiplying the height, width, and depth of the plate, we get


Vshell ≈ f (xi* )

⎝2πxi*



⎠Δx,


which is the same formula we had before.
To calculate the volume of the entire solid, we then add the volumes of all the shells and obtain


V ≈ ∑
i = 1


n

⎝2πxi* f (xi* )Δx



⎠.


Here we have another Riemann sum, this time for the function 2πx f (x). Taking the limit as n → ∞ gives us
V = lim


n → ∞

i = 1


n

⎝2πxi* f (xi* )Δx



⎠ = ∫


a


b

⎝2πx f (x)⎞⎠dx.


This leads to the following rule for the method of cylindrical shells.
Rule: The Method of Cylindrical Shells
Let f (x) be continuous and nonnegative. Define R as the region bounded above by the graph of f (x), below by the
x-axis, on the left by the line x = a, and on the right by the line x = b. Then the volume of the solid of revolution


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2.12


formed by revolving R around the y-axis is given by
(2.6)


V = ∫
a


b

⎝2πx f (x)⎞⎠dx.


Now let’s consider an example.
Example 2.12
The Method of Cylindrical Shells 1
Define R as the region bounded above by the graph of f (x) = 1/x and below by the x-axis over the interval
[1, 3]. Find the volume of the solid of revolution formed by revolving R around the y-axis.
Solution
First we must graph the region R and the associated solid of revolution, as shown in the following figure.


Figure 2.29 (a) The region R under the graph of f (x) = 1/x over the
interval [1, 3]. (b) The solid of revolution generated by revolving R about
the y-axis.


Then the volume of the solid is given by
V = ∫


a


b

⎝2πx f (x)⎞⎠dx


= ∫
1


3

⎝2πx


1
x



⎠dx


= ∫
1


3
2π dx = 2πx|1


3 = 4π units3 .


Define R as the region bounded above by the graph of f (x) = x2 and below by the x-axis over the
interval [1, 2]. Find the volume of the solid of revolution formed by revolving R around the y-axis.


Chapter 2 | Applications of Integration 157




2.13


Example 2.13
The Method of Cylindrical Shells 2
Define R as the region bounded above by the graph of f (x) = 2x − x2 and below by the x-axis over the interval
[0, 2]. Find the volume of the solid of revolution formed by revolving R around the y-axis.
Solution
First graph the region R and the associated solid of revolution, as shown in the following figure.


Figure 2.30 (a) The region R under the graph of f (x) = 2x − x2 over
the interval [0, 2]. (b) The volume of revolution obtained by revolving
R about the y-axis.


Then the volume of the solid is given by
V = ∫


a


b

⎝2πx f (x)⎞⎠dx


= ∫
0


2

⎝2πx



⎝2x − x


2⎞


⎠dx = 2π∫


0


2

⎝2x


2 − x3⎞⎠dx


= 2π


2x3
3


− x
4


4

⎦ |0


2


= 8π
3


units3 .


Define R as the region bounded above by the graph of f (x) = 3x − x2 and below by the x-axis over
the interval [0, 2]. Find the volume of the solid of revolution formed by revolving R around the y-axis.


As with the disk method and the washer method, we can use the method of cylindrical shells with solids of revolution,revolved around the x-axis, when we want to integrate with respect to y. The analogous rule for this type of solid is given
here.
Rule: The Method of Cylindrical Shells for Solids of Revolution around the x-axis
Let g(y) be continuous and nonnegative. Define Q as the region bounded on the right by the graph of g(y), on
the left by the y-axis, below by the line y = c, and above by the line y = d. Then, the volume of the solid of


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2.14


revolution formed by revolving Q around the x-axis is given by
V = ∫


c


d

⎝2πyg(y)⎞⎠dy.


Example 2.14
The Method of Cylindrical Shells for a Solid Revolved around the x-axis
Define Q as the region bounded on the right by the graph of g(y) = 2 y and on the left by the y-axis for
y ∈ [0, 4]. Find the volume of the solid of revolution formed by revolving Q around the x-axis.
Solution
First, we need to graph the region Q and the associated solid of revolution, as shown in the following figure.


Figure 2.31 (a) The region Q to the left of the function g(y) over the interval
[0, 4]. (b) The solid of revolution generated by revolving Q around the x-axis.


Label the shaded region Q. Then the volume of the solid is given by
V = ∫


c


d

⎝2πyg(y)⎞⎠dy


= ∫
0


4

⎝2πy⎛⎝2 y⎞⎠⎞⎠dy = 4π∫


0


4
y3/2dy


= 4π



⎢2y


5/2


5





⎥ |0


4


= 256π
5


units3 .


Define Q as the region bounded on the right by the graph of g(y) = 3/y and on the left by the y-axis
for y ∈ [1, 3]. Find the volume of the solid of revolution formed by revolving Q around the x-axis.


Chapter 2 | Applications of Integration 159




For the next example, we look at a solid of revolution for which the graph of a function is revolved around a line other thanone of the two coordinate axes. To set this up, we need to revisit the development of the method of cylindrical shells. Recallthat we found the volume of one of the shells to be given by
Vshell = f (xi* )(πxi


2 − πxi − 1
2 )


= π f (xi* )

⎝xi


2 − xi − 1
2 ⎞


= π f (xi* )(xi + xi − 1)(xi − xi − 1)


= 2π f (xi* )


xi + xi − 1


2

⎠(xi − xi − 1).


This was based on a shell with an outer radius of xi and an inner radius of xi − 1. If, however, we rotate the region around
a line other than the y-axis, we have a different outer and inner radius. Suppose, for example, that we rotate the region
around the line x = −k, where k is some positive constant. Then, the outer radius of the shell is xi + k and the inner
radius of the shell is xi − 1 + k. Substituting these terms into the expression for volume, we see that when a plane region is
rotated around the line x = −k, the volume of a shell is given by


Vshell = 2π f (xi* )



⎝xi + k



⎠+ ⎛⎝xi − 1 + k





2





⎝xi + k



⎠− ⎛⎝xi − 1 + k







= 2π f (xi* )




xi + xi − 2


2

⎠+ k

⎠Δx.


As before, we notice that xi + xi − 1
2


is the midpoint of the interval [xi − 1, xi] and can be approximated by xi* . Then,
the approximate volume of the shell is


Vshell ≈ 2π

⎝xi* + k



⎠ f (xi* )Δx.


The remainder of the development proceeds as before, and we see that
V = ∫


a


b

⎝2π(x + k) f (x)⎞⎠dx.


We could also rotate the region around other horizontal or vertical lines, such as a vertical line in the right half plane. Ineach case, the volume formula must be adjusted accordingly. Specifically, the x-term in the integral must be replaced with
an expression representing the radius of a shell. To see how this works, consider the following example.
Example 2.15
A Region of Revolution Revolved around a Line
Define R as the region bounded above by the graph of f (x) = x and below by the x-axis over the interval
[1, 2]. Find the volume of the solid of revolution formed by revolving R around the line x = −1.
Solution
First, graph the region R and the associated solid of revolution, as shown in the following figure.


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2.15


Figure 2.32 (a) The region R between the graph of f (x) and the x-axis over the interval [1, 2]. (b) The
solid of revolution generated by revolving R around the line x = −1.


Note that the radius of a shell is given by x + 1. Then the volume of the solid is given by
V = ∫


1


2

⎝2π(x + 1) f (x)⎞⎠dx


= ∫
1


2
(2π(x + 1)x)dx = 2π∫


1


2

⎝x


2 + x⎞⎠dx


= 2π


x3


3
+ x


2


2

⎦ |1


2


= 23π
3


units3 .


Define R as the region bounded above by the graph of f (x) = x2 and below by the x-axis over the
interval [0, 1]. Find the volume of the solid of revolution formed by revolving R around the line x = −2.


For our final example in this section, let’s look at the volume of a solid of revolution for which the region of revolution isbounded by the graphs of two functions.
Example 2.16
A Region of Revolution Bounded by the Graphs of Two Functions
Define R as the region bounded above by the graph of the function f (x) = x and below by the graph of the
function g(x) = 1/x over the interval [1, 4]. Find the volume of the solid of revolution generated by revolving
R around the y-axis.


Chapter 2 | Applications of Integration 161




2.16


Solution
First, graph the region R and the associated solid of revolution, as shown in the following figure.


Figure 2.33 (a) The region R between the graph of f (x) and the graph of g(x) over the interval [1, 4]. (b)
The solid of revolution generated by revolving R around the y-axis.


Note that the axis of revolution is the y-axis, so the radius of a shell is given simply by x. We don’t need to
make any adjustments to the x-term of our integrand. The height of a shell, though, is given by f (x) − g(x), so
in this case we need to adjust the f (x) term of the integrand. Then the volume of the solid is given by


V = ∫
1


4

⎝2πx⎛⎝ f (x) − g(x)⎞⎠⎞⎠dx


= ∫
1


4

⎝2πx

⎝ x −


1
x



⎠dx = 2π∫


1


4

⎝x


3/2 − 1⎞⎠dx


= 2π


2x5/2
5


− x

⎦ |1


4


= 94π
5


units3.


Define R as the region bounded above by the graph of f (x) = x and below by the graph of g(x) = x2
over the interval [0, 1]. Find the volume of the solid of revolution formed by revolving R around the y-axis.


Which Method Should We Use?
We have studied several methods for finding the volume of a solid of revolution, but how do we know which method to use?It often comes down to a choice of which integral is easiest to evaluate. Figure 2.34 describes the different approachesfor solids of revolution around the x-axis. It’s up to you to develop the analogous table for solids of revolution around the
y-axis.


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Figure 2.34


Let’s take a look at a couple of additional problems and decide on the best approach to take for solving them.
Example 2.17
Selecting the Best Method
For each of the following problems, select the best method to find the volume of a solid of revolution generatedby revolving the given region around the x-axis, and set up the integral to find the volume (do not evaluate the
integral).


a. The region bounded by the graphs of y = x, y = 2 − x, and the x-axis.
b. The region bounded by the graphs of y = 4x − x2 and the x-axis.


Solution
a. First, sketch the region and the solid of revolution as shown.


Chapter 2 | Applications of Integration 163




Figure 2.35 (a) The region R bounded by two lines and the x-axis. (b) The solid of
revolution generated by revolving R about the x-axis.


Looking at the region, if we want to integrate with respect to x, we would have to break the integral
into two pieces, because we have different functions bounding the region over [0, 1] and [1, 2]. In this
case, using the disk method, we would have


V = ∫
0


1

⎝πx


2⎞
⎠dx + ∫


1


2

⎝π(2 − x)


2⎞
⎠dx.


If we used the shell method instead, we would use functions of y to represent the curves, producing
V = ∫


0


1

⎝2πy⎡⎣⎛⎝2 − y⎞⎠− y⎤⎦⎞⎠dy


= ∫
0


1

⎝2πy⎡⎣2 − 2y⎤⎦⎞⎠dy.


Neither of these integrals is particularly onerous, but since the shell method requires only one integral,and the integrand requires less simplification, we should probably go with the shell method in this case.
b. First, sketch the region and the solid of revolution as shown.


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2.17


Figure 2.36 (a) The region R between the curve and the x-axis. (b) The solid of
revolution generated by revolving R about the x-axis.


Looking at the region, it would be problematic to define a horizontal rectangle; the region is bounded onthe left and right by the same function. Therefore, we can dismiss the method of shells. The solid has nocavity in the middle, so we can use the method of disks. Then
V = ∫


0


4
π⎛⎝4x − x


2⎞


2
dx.


Select the best method to find the volume of a solid of revolution generated by revolving the givenregion around the x-axis, and set up the integral to find the volume (do not evaluate the integral): the region
bounded by the graphs of y = 2 − x2 and y = x2.


Chapter 2 | Applications of Integration 165




2.3 EXERCISES
For the following exercise, find the volume generated whenthe region between the two curves is rotated around thegiven axis. Use both the shell method and the washermethod. Use technology to graph the functions and draw atypical slice by hand.
114. [T] Over the curve of y = 3x, x = 0, and y = 3
rotated around the y-axis.
115. [T] Under the curve of y = 3x, x = 0, and x = 3
rotated around the y-axis.
116. [T] Over the curve of y = 3x, x = 0, and y = 3
rotated around the x-axis.
117. [T] Under the curve of y = 3x, x = 0, and x = 3
rotated around the x-axis.
118. [T] Under the curve of y = 2x3, x = 0, and x = 2
rotated around the y-axis.
119. [T] Under the curve of y = 2x3, x = 0, and x = 2
rotated around the x-axis.
For the following exercises, use shells to find the volumesof the given solids. Note that the rotated regions lie betweenthe curve and the x-axis and are rotated around the
y-axis.


120. y = 1 − x2, x = 0, and x = 1
121. y = 5x3, x = 0, and x = 1
122. y = 1x , x = 1, and x = 100
123. y = 1 − x2, x = 0, and x = 1
124. y = 1


1 + x2
, x = 0, and x = 3


125. y = sinx2, x = 0, and x = π
126. y = 1


1 − x2
, x = 0, and x = 1


2


127. y = x, x = 0, and x = 1


128. y = ⎛⎝1 + x2⎞⎠3, x = 0, and x = 1


129. y = 5x3 − 2x4, x = 0, and x = 2
For the following exercises, use shells to find the volumegenerated by rotating the regions between the given curveand y = 0 around the x-axis.
130. y = 1 − x2, x = 0, and x = 1
131. y = x2, x = 0, and x = 2
132. y = ex, x = 0, and x = 1
133. y = ln(x), x = 1, and x = e
134. x = 1


1 + y2
, y = 1, and y = 4


135. x = 1 + y2y , y = 0, and y = 2
136. x = cos y, y = 0, and y = π
137. x = y3 − 4y2, x = −1, and x = 2
138. x = yey , x = −1, and x = 2
139. x = cos yey, x = 0, and x = π
For the following exercises, find the volume generatedwhen the region between the curves is rotated around thegiven axis.
140. y = 3 − x, y = 0, x = 0, and x = 2 rotated around
the y-axis.
141. y = x3, y = 0, and y = 8 rotated around the
y-axis.


142. y = x2, y = x, rotated around the y-axis.
143. y = x, x = 0, and x = 1 rotated around the line
x = 2.


144. y = 1
4 − x


, x = 1, and x = 2 rotated around the
line x = 4.
145. y = x and y = x2 rotated around the y-axis.
146. y = x and y = x2 rotated around the line x = 2.


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147. x = y3, y = 1x , x = 1, and y = 2 rotated around
the x-axis.
148. x = y2 and y = x rotated around the line y = 2.
149. [T] Left of x = sin(πy), right of y = x, around
the y-axis.
For the following exercises, use technology to graph theregion. Determine which method you think would beeasiest to use to calculate the volume generated when thefunction is rotated around the specified axis. Then, use yourchosen method to find the volume.
150. [T] y = x2 and y = 4x rotated around the y-axis.
151. [T] y = cos(πx), y = sin(πx), x = 1


4
, and x = 5


4
rotated around the y-axis.
152. [T] y = x2 − 2x, x = 2, and x = 4 rotated around
the y-axis.
153. [T] y = x2 − 2x, x = 2, and x = 4 rotated around
the x-axis.
154. [T] y = 3x3 − 2, y = x, and x = 2 rotated around
the x-axis.
155. [T] y = 3x3 − 2, y = x, and x = 2 rotated around
the y-axis.
156. [T] x = sin⎛⎝πy2⎞⎠ and x = 2y rotated around the
x-axis.


157. [T] x = y2, x = y2 − 2y + 1, and x = 2 rotated
around the y-axis.
For the following exercises, use the method of shells toapproximate the volumes of some common objects, whichare pictured in accompanying figures.
158. Use the method of shells to find the volume of asphere of radius r.


159. Use the method of shells to find the volume of a conewith radius r and height h.


160. Use the method of shells to find the volume of an
ellipse ⎛⎝x2/a2⎞⎠+ ⎛⎝y2/b2⎞⎠ = 1 rotated around the x-axis.


161. Use the method of shells to find the volume of acylinder with radius r and height h.


162. Use the method of shells to find the volume of the
donut created when the circle x2 + y2 = 4 is rotated
around the line x = 4.


163. Consider the region enclosed by the graphs of
y = f (x), y = 1 + f (x), x = 0, y = 0, and x = a > 0.
What is the volume of the solid generated when this regionis rotated around the y-axis? Assume that the function is
defined over the interval [0, a].


Chapter 2 | Applications of Integration 167




164. Consider the function y = f (x), which decreases
from f (0) = b to f (1) = 0. Set up the integrals for
determining the volume, using both the shell method andthe disk method, of the solid generated when this region,with x = 0 and y = 0, is rotated around the y-axis.
Prove that both methods approximate the same volume.Which method is easier to apply? (Hint: Since f (x) is one-
to-one, there exists an inverse f −1(y).)


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2.4 | Arc Length of a Curve and Surface Area
Learning Objectives


2.4.1 Determine the length of a curve, y = f (x), between two points.
2.4.2 Determine the length of a curve, x = g(y), between two points.
2.4.3 Find the surface area of a solid of revolution.


In this section, we use definite integrals to find the arc length of a curve. We can think of arc length as the distance youwould travel if you were walking along the path of the curve. Many real-world applications involve arc length. If a rocketis launched along a parabolic path, we might want to know how far the rocket travels. Or, if a curve on a map represents aroad, we might want to know how far we have to drive to reach our destination.
We begin by calculating the arc length of curves defined as functions of x, then we examine the same process for curves
defined as functions of y. (The process is identical, with the roles of x and y reversed.) The techniques we use to find arc
length can be extended to find the surface area of a surface of revolution, and we close the section with an examination ofthis concept.
Arc Length of the Curve y = f(x)
In previous applications of integration, we required the function f (x) to be integrable, or at most continuous. However,
for calculating arc length we have a more stringent requirement for f (x). Here, we require f (x) to be differentiable, and
furthermore we require its derivative, f ′(x), to be continuous. Functions like this, which have continuous derivatives, are
called smooth. (This property comes up again in later chapters.)
Let f (x) be a smooth function defined over ⎡⎣a, b⎤⎦. We want to calculate the length of the curve from the point ⎛⎝a, f (a)⎞⎠
to the point ⎛⎝b, f (b)⎞⎠. We start by using line segments to approximate the length of the curve. For i = 0, 1, 2,…, n,
let P = {xi} be a regular partition of ⎡⎣a, b⎤⎦. Then, for i = 1, 2,…, n, construct a line segment from the point

⎝xi − 1, f (xi − 1)



⎠ to the point ⎛⎝xi, f (xi)⎞⎠. Although it might seem logical to use either horizontal or vertical line segments,


we want our line segments to approximate the curve as closely as possible. Figure 2.37 depicts this construct for n = 5.


Figure 2.37 We can approximate the length of a curve byadding line segments.


To help us find the length of each line segment, we look at the change in vertical distance as well as the change in horizontaldistance over each interval. Because we have used a regular partition, the change in horizontal distance over each interval isgiven by Δx. The change in vertical distance varies from interval to interval, though, so we use Δyi = f (xi) − f (xi − 1)
to represent the change in vertical distance over the interval [xi − 1, xi], as shown in Figure 2.38. Note that some (or all)
Δyi may be negative.


Chapter 2 | Applications of Integration 169




Figure 2.38 A representative line segment approximates thecurve over the interval [xi − 1, xi].


By the Pythagorean theorem, the length of the line segment is (Δx)2 + ⎛⎝Δyi⎞⎠2. We can also write this as
Δx 1 + ⎛⎝⎛⎝Δyi



⎠/(Δx)⎞⎠2. Now, by the Mean Value Theorem, there is a point xi* ∈ [xi − 1, xi] such that


f ′(xi* ) =

⎝Δyi



⎠/(Δx). Then the length of the line segment is given by Δx 1 + ⎡⎣ f ′(xi* )⎤⎦2. Adding up the lengths of all


the line segments, we get
Arc Length ≈ ∑


i = 1


n


1 + ⎡⎣ f ′(xi* )


2 Δx.


This is a Riemann sum. Taking the limit as n → ∞, we have
Arc Length = lim


n → ∞

i = 1


n


1 + ⎡⎣ f ′(xi* )


2 Δx = ∫


a


b
1 + ⎡⎣ f ′(x)⎤⎦2 dx.


We summarize these findings in the following theorem.
Theorem 2.4: Arc Length for y = f(x)
Let f (x) be a smooth function over the interval ⎡⎣a, b⎤⎦. Then the arc length of the portion of the graph of f (x) from
the point ⎛⎝a, f (a)⎞⎠ to the point ⎛⎝b, f (b)⎞⎠ is given by


(2.7)
Arc Length = ∫


a


b
1 + ⎡⎣ f ′(x)⎤⎦2 dx.


Note that we are integrating an expression involving f ′(x), so we need to be sure f ′(x) is integrable. This is why we
require f (x) to be smooth. The following example shows how to apply the theorem.
Example 2.18
Calculating the Arc Length of a Function of x
Let f (x) = 2x3/2. Calculate the arc length of the graph of f (x) over the interval [0, 1]. Round the answer to
three decimal places.


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2.18


2.19


Solution
We have f ′(x) = 3x1/2, so ⎡⎣ f ′(x)⎤⎦2 = 9x. Then, the arc length is


Arc Length = ∫
a


b
1 + ⎡⎣ f ′(x)⎤⎦2 dx


= ∫
0


1
1 + 9x dx.


Substitute u = 1 + 9x. Then, du = 9 dx. When x = 0, then u = 1, and when x = 1, then u = 10. Thus,
Arc Length = ∫


0


1
1 + 9x dx


= 1
9


0


1
1 + 9x9dx = 1


9


1


10
u du


= 1
9
· 2
3
u3/2|1


10
= 2


27

⎣10 10 − 1



⎦ ≈ 2.268 units.


Let f (x) = (4/3)x3/2. Calculate the arc length of the graph of f (x) over the interval [0, 1]. Round the
answer to three decimal places.


Although it is nice to have a formula for calculating arc length, this particular theorem can generate expressions that aredifficult to integrate. We study some techniques for integration in Introduction to Techniques of Integration. In somecases, we may have to use a computer or calculator to approximate the value of the integral.
Example 2.19
Using a Computer or Calculator to Determine the Arc Length of a Function of x
Let f (x) = x2. Calculate the arc length of the graph of f (x) over the interval [1, 3].
Solution
We have f ′(x) = 2x, so ⎡⎣ f ′(x)⎤⎦2 = 4x2. Then the arc length is given by


Arc Length = ∫
a


b
1 + ⎡⎣ f ′(x)⎤⎦2 dx = ∫


1


3
1 + 4x2 dx.


Using a computer to approximate the value of this integral, we get


1


3
1 + 4x2 dx ≈ 8.26815.


Let f (x) = sin x. Calculate the arc length of the graph of f (x) over the interval [0, π]. Use a
computer or calculator to approximate the value of the integral.


Chapter 2 | Applications of Integration 171




Arc Length of the Curve x = g(y)
We have just seen how to approximate the length of a curve with line segments. If we want to find the arc length of thegraph of a function of y, we can repeat the same process, except we partition the y-axis instead of the x-axis. Figure
2.39 shows a representative line segment.


Figure 2.39 A representative line segment over the interval
[yi − 1, yi].


Then the length of the line segment is ⎛⎝Δy⎞⎠2 + ⎛⎝Δxi⎞⎠2, which can also be written as Δy 1 + ⎛⎝⎛⎝Δxi⎞⎠/⎛⎝Δy⎞⎠⎞⎠2. If we now
follow the same development we did earlier, we get a formula for arc length of a function x = g(y).


Theorem 2.5: Arc Length for x = g(y)
Let g(y) be a smooth function over an interval ⎡⎣c, d⎤⎦. Then, the arc length of the graph of g(y) from the point

⎝c, g(c)⎞⎠ to the point ⎛⎝d, g(d)⎞⎠ is given by


(2.8)
Arc Length = ∫


c


d
1 + ⎡⎣g′(y)⎤⎦2 dy.


Example 2.20
Calculating the Arc Length of a Function of y
Let g(y) = 3y3. Calculate the arc length of the graph of g(y) over the interval [1, 2].
Solution
We have g′(y) = 9y2, so ⎡⎣g′(y)⎤⎦2 = 81y4. Then the arc length is


Arc Length = ∫
c


d
1 + ⎡⎣g′(y)⎤⎦2 dy = ∫


1


2
1 + 81y4 dy.


Using a computer to approximate the value of this integral, we obtain


1


2
1 + 81y4 dy ≈ 21.0277.


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2.20 Let g(y) = 1/y. Calculate the arc length of the graph of g(y) over the interval [1, 4]. Use a computer
or calculator to approximate the value of the integral.


Area of a Surface of Revolution
The concepts we used to find the arc length of a curve can be extended to find the surface area of a surface of revolution.Surface area is the total area of the outer layer of an object. For objects such as cubes or bricks, the surface area of theobject is the sum of the areas of all of its faces. For curved surfaces, the situation is a little more complex. Let f (x) be a
nonnegative smooth function over the interval ⎡⎣a, b⎤⎦. We wish to find the surface area of the surface of revolution created
by revolving the graph of y = f (x) around the x-axis as shown in the following figure.


Figure 2.40 (a) A curve representing the function f (x). (b) The surface of revolution
formed by revolving the graph of f (x) around the x-axis.


As we have done many times before, we are going to partition the interval ⎡⎣a, b⎤⎦ and approximate the surface area by
calculating the surface area of simpler shapes. We start by using line segments to approximate the curve, as we did earlierin this section. For i = 0, 1, 2,…, n, let P = {xi} be a regular partition of ⎡⎣a, b⎤⎦. Then, for i = 1, 2,…, n, construct a
line segment from the point ⎛⎝xi − 1, f (xi − 1)⎞⎠ to the point ⎛⎝xi, f (xi)⎞⎠. Now, revolve these line segments around the x-axis
to generate an approximation of the surface of revolution as shown in the following figure.


Figure 2.41 (a) Approximating f (x) with line segments. (b) The surface of revolution
formed by revolving the line segments around the x-axis.


Notice that when each line segment is revolved around the axis, it produces a band. These bands are actually pieces of cones


Chapter 2 | Applications of Integration 173




(think of an ice cream cone with the pointy end cut off). A piece of a cone like this is called a frustum of a cone.
To find the surface area of the band, we need to find the lateral surface area, S, of the frustum (the area of just the slanted
outside surface of the frustum, not including the areas of the top or bottom faces). Let r1 and r2 be the radii of the wide
end and the narrow end of the frustum, respectively, and let l be the slant height of the frustum as shown in the following
figure.


Figure 2.42 A frustum of a cone can approximate a small partof surface area.


We know the lateral surface area of a cone is given by
Lateral Surface Area = πrs,


where r is the radius of the base of the cone and s is the slant height (see the following figure).


Figure 2.43 The lateral surface area of the cone is given by
πrs.


Since a frustum can be thought of as a piece of a cone, the lateral surface area of the frustum is given by the lateral surfacearea of the whole cone less the lateral surface area of the smaller cone (the pointy tip) that was cut off (see the followingfigure).


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Figure 2.44 Calculating the lateral surface area of a frustumof a cone.


The cross-sections of the small cone and the large cone are similar triangles, so we see that
r2
r1


= s − ls .


Solving for s, we get
r2
r1


= s − ls


r2 s = r1 (s − l)


r2 s = r1 s − r1 l
r1 l = r1 s − r2 s


r1 l = (r1 − r2)s


r1 l
r1 − r2


= s.


Then the lateral surface area (SA) of the frustum is
S = (Lateral SA of large cone) − (Lateral SA of small cone)


= πr1 s − πr2 (s − l)


= πr1



r1 l
r1 − r2



⎠− πr2





r1 l
r1 − r2


− l



=
πr1


2 l
r1 − r2



πr1 r2 l
r1 − r2


+ πr2 l


=
πr1


2 l
r1 − r2



πr1 r2 l
r1 − r2


+
πr2 l(r1 − r2)


r1 − r2


=
πr1


2 l
r1 − r2



πr1 r2 l
r1 − r2


+
πr1 r2 l
r1 − r2



πr2


2 l
r1 − r2


=
π⎛⎝r1


2 − r2
2⎞
⎠l


r1 − r2
=


π(r1 − r2)(r1 + r2)l
r1 − r2


= π(r1 + r2)l.


Let’s now use this formula to calculate the surface area of each of the bands formed by revolving the line segments aroundthe x-axis. A representative band is shown in the following figure.


Chapter 2 | Applications of Integration 175




Figure 2.45 A representative band used for determiningsurface area.


Note that the slant height of this frustum is just the length of the line segment used to generate it. So, applying the surfacearea formula, we have
S = π(r1 + r2)l


= π⎛⎝ f (xi − 1) + f (xi)

⎠ Δx2 + ⎛⎝Δyi




2


= π⎛⎝ f (xi − 1) + f (xi)

⎠Δx 1 +




Δyi
Δx



2


.


Now, as we did in the development of the arc length formula, we apply the Mean Value Theorem to select xi* ∈ [xi − 1, xi]
such that f ′(xi* ) = ⎛⎝Δyi⎞⎠/Δx. This gives us


S = π⎛⎝ f (xi − 1) + f (xi)

⎠Δx 1 + ⎛⎝ f ′(xi* )




2.


Furthermore, since f (x) is continuous, by the Intermediate Value Theorem, there is a point xi* * ∈ [xi − 1, xi] such that
f (xi


* * ) = (1/2)⎡⎣ f (xi − 1) + f (xi)

⎦, so we get


S = 2π f (xi
* * )Δx 1 + ⎛⎝ f ′(xi* )




2.


Then the approximate surface area of the whole surface of revolution is given by
Surface Area ≈ ∑


i = 1


n


2π f (xi
* * )Δx 1 + ⎛⎝ f ′(xi* )




2.


This almost looks like a Riemann sum, except we have functions evaluated at two different points, xi* and xi* * , over
the interval [xi − 1, xi]. Although we do not examine the details here, it turns out that because f (x) is smooth, if we let
n → ∞, the limit works the same as a Riemann sum even with the two different evaluation points. This makes sense
intuitively. Both xi* and xi* * are in the interval [xi − 1, xi], so it makes sense that as n → ∞, both xi* and xi* *
approach x. Those of you who are interested in the details should consult an advanced calculus text.
Taking the limit as n → ∞, we get


Surface Area = lim
n → ∞



i = 1


n


2π f (xi
* * )Δx 1 + ⎛⎝ f ′(xi* )




2 = ∫


a


b⎛
⎝2π f (x) 1 +



⎝ f ′(x)⎞⎠2



⎠dx.


As with arc length, we can conduct a similar development for functions of y to get a formula for the surface area of surfaces
of revolution about the y-axis. These findings are summarized in the following theorem.


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Theorem 2.6: Surface Area of a Surface of Revolution
Let f (x) be a nonnegative smooth function over the interval ⎡⎣a, b⎤⎦. Then, the surface area of the surface of revolution
formed by revolving the graph of f (x) around the x-axis is given by


(2.9)
Surface Area = ∫


a


b⎛
⎝2π f (x) 1 +



⎝ f ′(x)⎞⎠2



⎠dx.


Similarly, let g(y) be a nonnegative smooth function over the interval ⎡⎣c, d⎤⎦. Then, the surface area of the surface of
revolution formed by revolving the graph of g(y) around the y-axis is given by


Surface Area = ∫
c


d⎛
⎝2πg(y) 1 +



⎝g′(y)⎞⎠2



⎠dy.


Example 2.21
Calculating the Surface Area of a Surface of Revolution 1
Let f (x) = x over the interval [1, 4]. Find the surface area of the surface generated by revolving the graph of
f (x) around the x-axis. Round the answer to three decimal places.
Solution
The graph of f (x) and the surface of rotation are shown in the following figure.


Figure 2.46 (a) The graph of f (x). (b) The surface of revolution.


We have f (x) = x. Then, f ′(x) = 1/(2 x) and ⎛⎝ f ′(x)⎞⎠2 = 1/(4x). Then,


Chapter 2 | Applications of Integration 177




2.21


Surface Area = ∫
a


b⎛
⎝2π f (x) 1 +



⎝ f ′(x)⎞⎠2



⎠dx


= ∫
1


4⎛
⎝2π x 1 +


1
4x

⎠dx


= ∫
1


4⎛
⎝2π x +


1
4

⎠dx.


Let u = x + 1/4. Then, du = dx. When x = 1, u = 5/4, and when x = 4, u = 17/4. This gives us


0


1⎛
⎝2π x +


1
4

⎠dx = ∫5/4


17/4
2π u du


= 2π⎡⎣
2
3
u3/2⎤⎦ |5/4


17/4
= π


6

⎣17 17 − 5 5



⎦ ≈ 30.846.


Let f (x) = 1 − x over the interval [0, 1/2]. Find the surface area of the surface generated by
revolving the graph of f (x) around the x-axis. Round the answer to three decimal places.


Example 2.22
Calculating the Surface Area of a Surface of Revolution 2
Let f (x) = y = 3x3 . Consider the portion of the curve where 0 ≤ y ≤ 2. Find the surface area of the surface
generated by revolving the graph of f (x) around the y-axis.
Solution
Notice that we are revolving the curve around the y-axis, and the interval is in terms of y, so we want to
rewrite the function as a function of y. We get x = g(y) = (1/3)y3. The graph of g(y) and the surface of rotation
are shown in the following figure.


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2.22


Figure 2.47 (a) The graph of g(y). (b) The surface of revolution.


We have g(y) = (1/3)y3, so g′(y) = y2 and ⎛⎝g′(y)⎞⎠2 = y4. Then
Surface Area = ∫


c


d⎛
⎝2πg(y) 1 +



⎝g′(y)⎞⎠2



⎠dy


= ∫
0


2⎛
⎝2π


1
3
y3⎞⎠ 1 + y


4⎞
⎠dy


= 2π
3


0


2⎛
⎝y


3 1 + y4⎞⎠dy.


Let u = y4 + 1. Then du = 4y3dy. When y = 0, u = 1, and when y = 2, u = 17. Then

3


0


2⎛
⎝y


3 1 + y4⎞⎠dy =

3


1


17
1
4


udu


= π
6


2
3
u3/2⎤⎦ |1


17
= π


9

⎣(17)


3/2 − 1⎤⎦ ≈ 24.118.


Let g(y) = 9 − y2 over the interval y ∈ [0, 2]. Find the surface area of the surface generated by
revolving the graph of g(y) around the y-axis.


Chapter 2 | Applications of Integration 179




2.4 EXERCISES
For the following exercises, find the length of the functionsover the given interval.
165. y = 5x from x = 0 to x = 2
166. y = − 1


2
x + 25 from x = 1 to x = 4


167. x = 4y from y = −1 to y = 1
168. Pick an arbitrary linear function x = g(y) over any
interval of your choice (y1, y2). Determine the length of
the function and then prove the length is correct by usinggeometry.
169. Find the surface area of the volume generated whenthe curve y = x revolves around the x-axis from (1, 1)
to (4, 2), as seen here.


170. Find the surface area of the volume generated when
the curve y = x2 revolves around the y-axis from (1, 1)
to (3, 9).


For the following exercises, find the lengths of thefunctions of x over the given interval. If you cannot


evaluate the integral exactly, use technology toapproximate it.
171. y = x3/2 from (0, 0) to (1, 1)
172. y = x2/3 from (1, 1) to (8, 4)


173. y = 1
3

⎝x


2 + 2⎞⎠
3/2 from x = 0 to x = 1


174. y = 1
3

⎝x


2 − 2⎞⎠
3/2 from x = 2 to x = 4


175. [T] y = ex on x = 0 to x = 1
176. y = x3


3
+ 1


4x
from x = 1 to x = 3


177. y = x4
4


+ 1
8x2


from x = 1 to x = 2


178. y = 2x3/2
3


− x
1/2


2
from x = 1 to x = 4


179. y = 1
27

⎝9x


2 + 6⎞⎠
3/2 from x = 0 to x = 2


180. [T] y = sin x on x = 0 to x = π
For the following exercises, find the lengths of thefunctions of y over the given interval. If you cannot
evaluate the integral exactly, use technology toapproximate it.
181. y = 5 − 3x


4
from y = 0 to y = 4


182. x = 1
2

⎝e
y + e


−y⎞
⎠ from y = −1 to y = 1


183. x = 5y3/2 from y = 0 to y = 1
184. [T] x = y2 from y = 0 to y = 1
185. x = y from y = 0 to y = 1


186. x = 2
3

⎝y


2 + 1⎞⎠
3/2 from y = 1 to y = 3


187. [T] x = tan y from y = 0 to y = 3
4


188. [T] x = cos2 y from y = − π
2
to y = π


2


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189. [T] x = 4y from y = 0 to y = 2
190. [T] x = ln(y) on y = 1e to y = e
For the following exercises, find the surface area of thevolume generated when the following curves revolvearound the x-axis. If you cannot evaluate the integral
exactly, use your calculator to approximate it.
191. y = x from x = 2 to x = 6
192. y = x3 from x = 0 to x = 1
193. y = 7x from x = −1 to x = 1
194. [T] y = 1


x2
from x = 1 to x = 3


195. y = 4 − x2 from x = 0 to x = 2
196. y = 4 − x2 from x = −1 to x = 1
197. y = 5x from x = 1 to x = 5
198. [T] y = tan x from x = − π


4
to x = π


4


For the following exercises, find the surface area of thevolume generated when the following curves revolvearound the y-axis. If you cannot evaluate the integral
exactly, use your calculator to approximate it.
199. y = x2 from x = 0 to x = 2
200. y = 1


2
x2 + 1


2
from x = 0 to x = 1


201. y = x + 1 from x = 0 to x = 3
202. [T] y = 1x from x = 12 to x = 1
203. y = x3 from x = 1 to x = 27
204. [T] y = 3x4 from x = 0 to x = 1
205. [T] y = 1x from x = 1 to x = 3
206. [T] y = cos x from x = 0 to x = π


2


207. The base of a lamp is constructed by revolving a
quarter circle y = 2x − x2 around the y-axis from
x = 1 to x = 2, as seen here. Create an integral for the
surface area of this curve and compute it.


208. A light bulb is a sphere with radius 1/2 in. with the
bottom sliced off to fit exactly onto a cylinder of radius
1/4 in. and length 1/3 in., as seen here. The sphere is
cut off at the bottom to fit exactly onto the cylinder, sothe radius of the cut is 1/4 in. Find the surface area (not
including the top or bottom of the cylinder).


209. [T] A lampshade is constructed by rotating y = 1/x
around the x-axis from y = 1 to y = 2, as seen here.
Determine how much material you would need to constructthis lampshade—that is, the surface area—accurate to fourdecimal places.


210. [T] An anchor drags behind a boat according to
the function y = 24e−x/2 − 24, where y represents the
depth beneath the boat and x is the horizontal distance of
the anchor from the back of the boat. If the anchor is 23 ft
below the boat, how much rope do you have to pull to reachthe anchor? Round your answer to three decimal places.


Chapter 2 | Applications of Integration 181




211. [T] You are building a bridge that will span 10
ft. You intend to add decorative rope in the shape of
y = 5|sin⎛⎝(xπ)/5⎞⎠|, where x is the distance in feet from
one end of the bridge. Find out how much rope you need tobuy, rounded to the nearest foot.
For the following exercises, find the exact arc length for thefollowing problems over the given interval.
212. y = ln(sin x) from x = π/4 to x = (3π)/4. (Hint:
Recall trigonometric identities.)
213. Draw graphs of y = x2, y = x6, and y = x10.
For y = xn, as n increases, formulate a prediction on
the arc length from (0, 0) to (1, 1). Now, compute the
lengths of these three functions and determine whether yourprediction is correct.
214. Compare the lengths of the parabola x = y2 and the
line x = by from (0, 0) to ⎛⎝b2, b⎞⎠ as b increases. What
do you notice?
215. Solve for the length of x = y2 from
(0, 0) to (1, 1). Show that x = (1/2)y2 from (0, 0) to
(2, 2) is twice as long. Graph both functions and explain
why this is so.
216. [T] Which is longer between (1, 1) and (2, 1/2):
the hyperbola y = 1/x or the graph of x + 2y = 3?
217. Explain why the surface area is infinite when
y = 1/x is rotated around the x-axis for 1 ≤ x < ∞,
but the volume is finite.


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2.5 | Physical Applications
Learning Objectives


2.5.1 Determine the mass of a one-dimensional object from its linear density function.
2.5.2 Determine the mass of a two-dimensional circular object from its radial density function.
2.5.3 Calculate the work done by a variable force acting along a line.
2.5.4 Calculate the work done in pumping a liquid from one height to another.
2.5.5 Find the hydrostatic force against a submerged vertical plate.


In this section, we examine some physical applications of integration. Let’s begin with a look at calculating mass from adensity function. We then turn our attention to work, and close the section with a study of hydrostatic force.
Mass and Density
We can use integration to develop a formula for calculating mass based on a density function. First we consider a thin rodor wire. Orient the rod so it aligns with the x-axis, with the left end of the rod at x = a and the right end of the rod at
x = b (Figure 2.48). Note that although we depict the rod with some thickness in the figures, for mathematical purposes
we assume the rod is thin enough to be treated as a one-dimensional object.


Figure 2.48 We can calculate the mass of a thin rod orientedalong the x-axis by integrating its density function.


If the rod has constant density ρ, given in terms of mass per unit length, then the mass of the rod is just the product of the
density and the length of the rod: (b − a)ρ. If the density of the rod is not constant, however, the problem becomes a little
more challenging. When the density of the rod varies from point to point, we use a linear density function, ρ(x), to denote
the density of the rod at any point, x. Let ρ(x) be an integrable linear density function. Now, for i = 0, 1, 2,…, n let
P = {xi} be a regular partition of the interval ⎡⎣a, b⎤⎦, and for i = 1, 2,…, n choose an arbitrary point xi* ∈ [xi − 1, xi].
Figure 2.49 shows a representative segment of the rod.


Figure 2.49 A representative segment of the rod.


The mass mi of the segment of the rod from xi − 1 to xi is approximated by
mi ≈ ρ(xi* )(xi − xi − 1) = ρ(xi* )Δx.


Adding the masses of all the segments gives us an approximation for the mass of the entire rod:


Chapter 2 | Applications of Integration 183




2.23


m = ∑
i = 1


n


mi ≈ ∑
i = 1


n


ρ(xi* )Δx.


This is a Riemann sum. Taking the limit as n → ∞, we get an expression for the exact mass of the rod:
m = lim


n → ∞

i = 1


n


ρ(xi* )Δx = ∫a
b
ρ(x)dx.


We state this result in the following theorem.
Theorem 2.7: Mass–Density Formula of a One-Dimensional Object
Given a thin rod oriented along the x-axis over the interval ⎡⎣a, b⎤⎦, let ρ(x) denote a linear density function giving
the density of the rod at a point x in the interval. Then the mass of the rod is given by


(2.10)
m = ∫


a


b
ρ(x)dx.


We apply this theorem in the next example.
Example 2.23
Calculating Mass from Linear Density
Consider a thin rod oriented on the x-axis over the interval [π/2, π]. If the density of the rod is given by
ρ(x) = sin x, what is the mass of the rod?
Solution
Applying Equation 2.10 directly, we have


m = ∫
a


b
ρ(x)dx = ∫


π/2


π
sin x dx = −cos x|π/2


π = 1.


Consider a thin rod oriented on the x-axis over the interval [1, 3]. If the density of the rod is given by
ρ(x) = 2x2 + 3, what is the mass of the rod?


We now extend this concept to find the mass of a two-dimensional disk of radius r. As with the rod we looked at in
the one-dimensional case, here we assume the disk is thin enough that, for mathematical purposes, we can treat it as atwo-dimensional object. We assume the density is given in terms of mass per unit area (called area density), and furtherassume the density varies only along the disk’s radius (called radial density). We orient the disk in the xy-plane, with
the center at the origin. Then, the density of the disk can be treated as a function of x, denoted ρ(x). We assume
ρ(x) is integrable. Because density is a function of x, we partition the interval from [0, r] along the x-axis. For
i = 0, 1, 2,…, n, let P = {xi} be a regular partition of the interval [0, r], and for i = 1, 2,…, n, choose an arbitrary
point xi* ∈ [xi − 1, xi]. Now, use the partition to break up the disk into thin (two-dimensional) washers. A disk and a
representative washer are depicted in the following figure.


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Figure 2.50 (a) A thin disk in the xy-plane. (b) A representative washer.


We now approximate the density and area of the washer to calculate an approximate mass, mi. Note that the area of the
washer is given by


Ai = π(xi)
2 − π(xi − 1)


2


= π⎡⎣xi
2 − xi − 1


2 ⎤


= π(xi + xi − 1)(xi − xi − 1)


= π(xi + xi − 1)Δx.


You may recall that we had an expression similar to this when we were computing volumes by shells. As we did there, weuse xi* ≈ (xi + xi − 1)/2 to approximate the average radius of the washer. We obtain
Ai = π(xi + xi − 1)Δx ≈ 2πxi* Δx.


Using ρ(xi* ) to approximate the density of the washer, we approximate the mass of the washer by
mi ≈ 2πxi* ρ(xi* )Δx.


Adding up the masses of the washers, we see the mass m of the entire disk is approximated by
m = ∑


i = 1


n


mi ≈ ∑
i = 1


n


2πxi* ρ(xi* )Δx.


We again recognize this as a Riemann sum, and take the limit as n → ∞. This gives us
m = lim


n → ∞

i = 1


n


2πxi* ρ(xi* )Δx = ∫
0


r
2πxρ(x)dx.


We summarize these findings in the following theorem.
Theorem 2.8: Mass–Density Formula of a Circular Object
Let ρ(x) be an integrable function representing the radial density of a disk of radius r. Then the mass of the disk is
given by


(2.11)
m = ∫


0


r
2πxρ(x)dx.


Chapter 2 | Applications of Integration 185




2.24


Example 2.24
Calculating Mass from Radial Density
Let ρ(x) = x represent the radial density of a disk. Calculate the mass of a disk of radius 4.
Solution
Applying the formula, we find


m = ∫
0


r
2πxρ(x)dx


= ∫
0


4
2πx xdx = 2π∫


0


4
x3/2dx


= 2π2
5
x5/2|0


4
= 4π


5
[32] = 128π


5
.


Let ρ(x) = 3x + 2 represent the radial density of a disk. Calculate the mass of a disk of radius 2.


Work Done by a Force
We now consider work. In physics, work is related to force, which is often intuitively defined as a push or pull on an object.When a force moves an object, we say the force does work on the object. In other words, work can be thought of as theamount of energy it takes to move an object. According to physics, when we have a constant force, work can be expressedas the product of force and distance.
In the English system, the unit of force is the pound and the unit of distance is the foot, so work is given in foot-pounds. Inthe metric system, kilograms and meters are used. One newton is the force needed to accelerate 1 kilogram of mass at the
rate of 1 m/sec2. Thus, the most common unit of work is the newton-meter. This same unit is also called the joule. Both
are defined as kilograms times meters squared over seconds squared ⎛⎝kg ·m2/s2⎞⎠.
When we have a constant force, things are pretty easy. It is rare, however, for a force to be constant. The work done tocompress (or elongate) a spring, for example, varies depending on how far the spring has already been compressed (orstretched). We look at springs in more detail later in this section.
Suppose we have a variable force F(x) that moves an object in a positive direction along the x-axis from point a to point
b. To calculate the work done, we partition the interval ⎡⎣a, b⎤⎦ and estimate the work done over each subinterval. So, for
i = 0, 1, 2,…, n, let P = {xi} be a regular partition of the interval ⎡⎣a, b⎤⎦, and for i = 1, 2,…, n, choose an arbitrary
point xi* ∈ [xi − 1, xi]. To calculate the work done to move an object from point xi − 1 to point xi, we assume the
force is roughly constant over the interval, and use F(xi* ) to approximate the force. The work done over the interval
[xi − 1, xi], then, is given by


Wi ≈ F(xi* )(xi − xi − 1) = F(xi* )Δx.


Therefore, the work done over the interval ⎡⎣a, b⎤⎦ is approximately
W = ∑


i = 1


n


Wi ≈ ∑
i = 1


n


F(xi* )Δx.


Taking the limit of this expression as n → ∞ gives us the exact value for work:


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W = lim
n → ∞



i = 1


n


F(xi* )Δx = ∫a
b
F(x)dx.


Thus, we can define work as follows.
Definition
If a variable force F(x) moves an object in a positive direction along the x-axis from point a to point b, then the work
done on the object is


(2.12)
W = ∫


a


b
F(x)dx.


Note that if F is constant, the integral evaluates to F · (b − a) = F · d, which is the formula we stated at the beginning of
this section.
Now let’s look at the specific example of the work done to compress or elongate a spring. Consider a block attached to ahorizontal spring. The block moves back and forth as the spring stretches and compresses. Although in the real world wewould have to account for the force of friction between the block and the surface on which it is resting, we ignore frictionhere and assume the block is resting on a frictionless surface. When the spring is at its natural length (at rest), the system issaid to be at equilibrium. In this state, the spring is neither elongated nor compressed, and in this equilibrium position theblock does not move until some force is introduced. We orient the system such that x = 0 corresponds to the equilibrium
position (see the following figure).


Figure 2.51 A block attached to a horizontal spring atequilibrium, compressed, and elongated.


According to Hooke’s law, the force required to compress or stretch a spring from an equilibrium position is given by
F(x) = kx, for some constant k. The value of k depends on the physical characteristics of the spring. The constant k
is called the spring constant and is always positive. We can use this information to calculate the work done to compress orelongate a spring, as shown in the following example.
Example 2.25


Chapter 2 | Applications of Integration 187




2.25


The Work Required to Stretch or Compress a Spring
Suppose it takes a force of 10 N (in the negative direction) to compress a spring 0.2 m from the equilibrium
position. How much work is done to stretch the spring 0.5 m from the equilibrium position?
Solution
First find the spring constant, k. When x = −0.2, we know F(x) = −10, so


F(x) = kx
−10 = k(−0.2)


k = 50


and F(x) = 50x. Then, to calculate work, we integrate the force function, obtaining
W = ∫


a


b
F(x)dx = ∫


0


0.5
50x dx = 25x2|0


0.5
= 6.25.


The work done to stretch the spring is 6.25 J.


Suppose it takes a force of 8 lb to stretch a spring 6 in. from the equilibrium position. How much work
is done to stretch the spring 1 ft from the equilibrium position?


Work Done in Pumping
Consider the work done to pump water (or some other liquid) out of a tank. Pumping problems are a little more complicatedthan spring problems because many of the calculations depend on the shape and size of the tank. In addition, instead ofbeing concerned about the work done to move a single mass, we are looking at the work done to move a volume of water,and it takes more work to move the water from the bottom of the tank than it does to move the water from the top of thetank.
We examine the process in the context of a cylindrical tank, then look at a couple of examples using tanks of differentshapes. Assume a cylindrical tank of radius 4 m and height 10 m is filled to a depth of 8 m. How much work does it take
to pump all the water over the top edge of the tank?
The first thing we need to do is define a frame of reference. We let x represent the vertical distance below the top of the
tank. That is, we orient the x-axis vertically, with the origin at the top of the tank and the downward direction being positive
(see the following figure).


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Figure 2.52 How much work is needed to empty a tankpartially filled with water?


Using this coordinate system, the water extends from x = 2 to x = 10. Therefore, we partition the interval [2, 10] and
look at the work required to lift each individual “layer” of water. So, for i = 0, 1, 2,…, n, let P = {xi} be a regular
partition of the interval [2, 10], and for i = 1, 2,…, n, choose an arbitrary point xi* ∈ [xi − 1, xi]. Figure 2.53
shows a representative layer.


Figure 2.53 A representative layer of water.


In pumping problems, the force required to lift the water to the top of the tank is the force required to overcome gravity, soit is equal to the weight of the water. Given that the weight-density of water is 9800 N/m3, or 62.4 lb/ft3, calculating the
volume of each layer gives us the weight. In this case, we have


V = π(4)2Δx = 16πΔx.


Then, the force needed to lift each layer is
F = 9800 · 16πΔx = 156,800πΔx.


Note that this step becomes a little more difficult if we have a noncylindrical tank. We look at a noncylindrical tank in thenext example.
We also need to know the distance the water must be lifted. Based on our choice of coordinate systems, we can use xi* as
an approximation of the distance the layer must be lifted. Then the work to lift the ith layer of water Wi is approximately


Wi ≈ 156,800πxi* Δx.


Adding the work for each layer, we see the approximate work to empty the tank is given by


Chapter 2 | Applications of Integration 189




W = ∑
i = 1


n


Wi ≈ ∑
i = 1


n


156,800πxi* Δx.


This is a Riemann sum, so taking the limit as n → ∞, we get
W = lim


n → ∞

i = 1


n


156,800πxi* Δx


= 156,800π∫
2


10
xdx


= 156,800π


x2
2

⎦ |2


10


= 7,526,400π ≈ 23,644,883.


The work required to empty the tank is approximately 23,650,000 J.
For pumping problems, the calculations vary depending on the shape of the tank or container. The following problem-solving strategy lays out a step-by-step process for solving pumping problems.
Problem-Solving Strategy: Solving Pumping Problems


1. Sketch a picture of the tank and select an appropriate frame of reference.
2. Calculate the volume of a representative layer of water.
3. Multiply the volume by the weight-density of water to get the force.
4. Calculate the distance the layer of water must be lifted.
5. Multiply the force and distance to get an estimate of the work needed to lift the layer of water.
6. Sum the work required to lift all the layers. This expression is an estimate of the work required to pump outthe desired amount of water, and it is in the form of a Riemann sum.
7. Take the limit as n → ∞ and evaluate the resulting integral to get the exact work required to pump out the


desired amount of water.


We now apply this problem-solving strategy in an example with a noncylindrical tank.
Example 2.26
A Pumping Problem with a Noncylindrical Tank
Assume a tank in the shape of an inverted cone, with height 12 ft and base radius 4 ft. The tank is full to start
with, and water is pumped over the upper edge of the tank until the height of the water remaining in the tank is 4
ft. How much work is required to pump out that amount of water?
Solution
The tank is depicted in Figure 2.54. As we did in the example with the cylindrical tank, we orient the x-axis
vertically, with the origin at the top of the tank and the downward direction being positive (step 1).


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Figure 2.54 A water tank in the shape of an inverted cone.


The tank starts out full and ends with 4 ft of water left, so, based on our chosen frame of reference, we need
to partition the interval [0, 8]. Then, for i = 0, 1, 2,…, n, let P = {xi} be a regular partition of the interval
[0, 8], and for i = 1, 2,…, n, choose an arbitrary point xi* ∈ [xi − 1, xi]. We can approximate the volume
of a layer by using a disk, then use similar triangles to find the radius of the disk (see the following figure).


Figure 2.55 Using similar triangles to express the radius of a disk of water.


Chapter 2 | Applications of Integration 191




2.26


From properties of similar triangles, we have
ri


12 − xi*
= 4


12
= 1


3


3ri = 12 − xi*


ri =
12 − xi*


3


= 4 −
xi*
3


.


Then the volume of the disk is
Vi = π

⎝4 −


xi*
3



2


Δx (step 2).


The weight-density of water is 62.4 lb/ft3, so the force needed to lift each layer is approximately
Fi ≈ 62.4π



⎝4 −


xi*
3



2


Δx (step 3).


Based on the diagram, the distance the water must be lifted is approximately xi* feet (step 4), so the approximate
work needed to lift the layer is


Wi ≈ 62.4πxi*

⎝4 −


xi*
3



2


Δx (step 5).


Summing the work required to lift all the layers, we get an approximate value of the total work:
W = ∑


i = 1


n


Wi ≈ ∑
i = 1


n


62.4πxi*

⎝4 −


xi*
3



2


Δx (step 6).


Taking the limit as n → ∞, we obtain
W = lim


n → ∞

i = 1


n


62.4πxi*

⎝4 −


xi*
3



2


Δx


= ∫
0


8
62.4πx⎛⎝4 −


x
3



2
dx


= 62.4π∫
0


8
x

⎝16 −


8x
3


+ x
2


9

⎠dx = 62.4π∫0


8⎛
⎝16x −


8x2
3


+ x
3


9

⎠dx


= 62.4π

⎣8x


2 − 8x
3


9
+ x


4


36

⎦ |0


8


= 10,649.6π ≈ 33,456.7.


It takes approximately 33,450 ft-lb of work to empty the tank to the desired level.


A tank is in the shape of an inverted cone, with height 10 ft and base radius 6 ft. The tank is filled to a
depth of 8 ft to start with, and water is pumped over the upper edge of the tank until 3 ft of water remain in thetank. How much work is required to pump out that amount of water?


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Hydrostatic Force and Pressure
In this last section, we look at the force and pressure exerted on an object submerged in a liquid. In the English system, forceis measured in pounds. In the metric system, it is measured in newtons. Pressure is force per unit area, so in the Englishsystem we have pounds per square foot (or, perhaps more commonly, pounds per square inch, denoted psi). In the metricsystem we have newtons per square meter, also called pascals.
Let’s begin with the simple case of a plate of area A submerged horizontally in water at a depth s (Figure 2.56). Then, the
force exerted on the plate is simply the weight of the water above it, which is given by F = ρAs, where ρ is the weight
density of water (weight per unit volume). To find the hydrostatic pressure—that is, the pressure exerted by water on asubmerged object—we divide the force by the area. So the pressure is p = F/A = ρs.


Figure 2.56 A plate submerged horizontally in water.


By Pascal’s principle, the pressure at a given depth is the same in all directions, so it does not matter if the plate is submergedhorizontally or vertically. So, as long as we know the depth, we know the pressure. We can apply Pascal’s principle to findthe force exerted on surfaces, such as dams, that are oriented vertically. We cannot apply the formula F = ρAs directly,
because the depth varies from point to point on a vertically oriented surface. So, as we have done many times before, weform a partition, a Riemann sum, and, ultimately, a definite integral to calculate the force.
Suppose a thin plate is submerged in water. We choose our frame of reference such that the x-axis is oriented vertically, withthe downward direction being positive, and point x = 0 corresponding to a logical reference point. Let s(x) denote the
depth at point x. Note we often let x = 0 correspond to the surface of the water. In this case, depth at any point is simply
given by s(x) = x. However, in some cases we may want to select a different reference point for x = 0, so we proceed
with the development in the more general case. Last, let w(x) denote the width of the plate at the point x.
Assume the top edge of the plate is at point x = a and the bottom edge of the plate is at point x = b. Then, for
i = 0, 1, 2,…, n, let P = {xi} be a regular partition of the interval ⎡⎣a, b⎤⎦, and for i = 1, 2,…, n, choose an arbitrary
point xi* ∈ [xi − 1, xi]. The partition divides the plate into several thin, rectangular strips (see the following figure).


Chapter 2 | Applications of Integration 193




Figure 2.57 A thin plate submerged vertically in water.


Let’s now estimate the force on a representative strip. If the strip is thin enough, we can treat it as if it is at a constant depth,
s(xi* ). We then have


Fi = ρAs = ρ

⎣w(xi* )Δx



⎦s(xi* ).


Adding the forces, we get an estimate for the force on the plate:
F ≈ ∑


i = 1


n


Fi = ∑
i = 1


n


ρ⎡⎣w(xi* )Δx

⎦s(xi* ).


This is a Riemann sum, so taking the limit gives us the exact force. We obtain
(2.13)


F = lim
n → ∞



i = 1


n


ρ⎡⎣w(xi* )Δx

⎦s(xi* ) = ∫a


b
ρw(x)s(x)dx.


Evaluating this integral gives us the force on the plate. We summarize this in the following problem-solving strategy.
Problem-Solving Strategy: Finding Hydrostatic Force


1. Sketch a picture and select an appropriate frame of reference. (Note that if we select a frame of reference otherthan the one used earlier, we may have to adjust Equation 2.13 accordingly.)
2. Determine the depth and width functions, s(x) and w(x).
3. Determine the weight-density of whatever liquid with which you are working. The weight-density of water is


62.4 lb/ft3, or 9800 N/m3.
4. Use the equation to calculate the total force.


Example 2.27
Finding Hydrostatic Force
A water trough 15 ft long has ends shaped like inverted isosceles triangles, with base 8 ft and height 3 ft. Find theforce on one end of the trough if the trough is full of water.
Solution


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2.27


Figure 2.58 shows the trough and a more detailed view of one end.


Figure 2.58 (a) A water trough with a triangular cross-section. (b)Dimensions of one end of the water trough.


Select a frame of reference with the x-axis oriented vertically and the downward direction being positive. Select
the top of the trough as the point corresponding to x = 0 (step 1). The depth function, then, is s(x) = x. Using
similar triangles, we see that w(x) = 8 − (8/3)x (step 2). Now, the weight density of water is 62.4 lb/ft3 (step
3), so applying Equation 2.13, we obtain


F = ∫
a


b
ρw(x)s(x)dx


= ∫
0


3
62.4⎛⎝8 −


8
3
x⎞⎠x dx = 62.4∫0


3

⎝8x −


8
3
x2⎞⎠dx


= 62.4⎡⎣4x
2 − 8


9
x3⎤⎦ |0


3
= 748.8.


The water exerts a force of 748.8 lb on the end of the trough (step 4).


A water trough 12 m long has ends shaped like inverted isosceles triangles, with base 6 m and height 4m. Find the force on one end of the trough if the trough is full of water.


Chapter 2 | Applications of Integration 195




Example 2.28
Chapter Opener: Finding Hydrostatic Force
We now return our attention to the Hoover Dam, mentioned at the beginning of this chapter. The actual dam isarched, rather than flat, but we are going to make some simplifying assumptions to help us with the calculations.Assume the face of the Hoover Dam is shaped like an isosceles trapezoid with lower base 750 ft, upper base
1250 ft, and height 750 ft (see the following figure).


When the reservoir is full, Lake Mead’s maximum depth is about 530 ft, and the surface of the lake is about 10 ftbelow the top of the dam (see the following figure).


Figure 2.59 A simplified model of the Hoover Dam withassumed dimensions.
a. Find the force on the face of the dam when the reservoir is full.
b. The southwest United States has been experiencing a drought, and the surface of Lake Mead is about 125ft below where it would be if the reservoir were full. What is the force on the face of the dam under thesecircumstances?


Solution
a. We begin by establishing a frame of reference. As usual, we choose to orient the x-axis vertically, with


the downward direction being positive. This time, however, we are going to let x = 0 represent the top
of the dam, rather than the surface of the water. When the reservoir is full, the surface of the water is 10
ft below the top of the dam, so s(x) = x − 10 (see the following figure).


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Figure 2.60 We first choose a frame of reference.


To find the width function, we again turn to similar triangles as shown in the figure below.


Figure 2.61 We use similar triangles to determine a functionfor the width of the dam. (a) Assumed dimensions of the dam;(b) highlighting the similar triangles.


From the figure, we see that w(x) = 750 + 2r. Using properties of similar triangles, we get
r = 250 − (1/3)x. Thus,


w(x) = 1250 − 2
3
x (step 2).


Using a weight-density of 62.4 lb/ft3 (step 3) and applying Equation 2.13, we get


Chapter 2 | Applications of Integration 197




2.28


F = ∫
a


b
ρw(x)s(x)dx


= ∫
10


540
62.4⎛⎝1250 −


2
3
x⎞⎠(x − 10)dx = 62.4∫10


540
−2
3

⎣x


2 − 1885x + 18750⎤⎦dx


= −62.4⎛⎝
2
3




x3


3
− 1885x


2


2
+ 18750x



⎦ |10


540


≈ 8,832,245,000 lb = 4,416,122.5 t.


Note the change from pounds to tons (2000 lb = 1 ton) (step 4). This changes our depth function, s(x), and our
limits of integration. We have s(x) = x − 135. The lower limit of integration is 135. The upper limit remains
540. Evaluating the integral, we get


F = ∫
a


b
ρw(x)s(x)dx


= ∫
135


540
62.4⎛⎝1250 −


2
3
x⎞⎠(x − 135)dx


= −62.4⎛⎝
2
3

⎠∫135


540
(x − 1875)(x − 135)dx = −62.4⎛⎝


2
3

⎠∫135


540

⎝x


2 − 2010x + 253125⎞⎠dx


= −62.4⎛⎝
2
3




x3


3
− 1005x2 + 253125x



⎦ |135


540


≈ 5,015,230,000 lb = 2,507,615 t.


When the reservoir is at its average level, the surface of the water is about 50 ft below where it would beif the reservoir were full. What is the force on the face of the dam under these circumstances?


To learn more about Hoover Dam, see this article (http://www.openstaxcollege.org/l/20_HooverDam)published by the History Channel.


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2.5 EXERCISES
For the following exercises, find the work done.
218. Find the work done when a constant force F = 12
lb moves a chair from x = 0.9 to x = 1.1 ft.
219. How much work is done when a person lifts a 50 lb
box of comics onto a truck that is 3 ft off the ground?
220. What is the work done lifting a 20 kg child from the
floor to a height of 2 m? (Note that 1 kg equates to 9.8
N)
221. Find the work done when you push a box alongthe floor 2 m, when you apply a constant force of
F = 100 N.


222. Compute the work done for a force F = 12/x2 N
from x = 1 to x = 2 m.
223. What is the work done moving a particle from x = 0
to x = 1 m if the force acting on it is F = 3x2 N?
For the following exercises, find the mass of the one-dimensional object.
224. A wire that is 2 ft long (starting at x = 0) and has
a density function of ρ(x) = x2 + 2x lb/ft
225. A car antenna that is 3 ft long (starting at x = 0)
and has a density function of ρ(x) = 3x + 2 lb/ft
226. A metal rod that is 8 in. long (starting at x = 0) and
has a density function of ρ(x) = e1/2x lb/in.
227. A pencil that is 4 in. long (starting at x = 2) and
has a density function of ρ(x) = 5/x oz/in.
228. A ruler that is 12 in. long (starting at x = 5) and
has a density function of ρ(x) = ln(x) + (1/2)x2 oz/in.
For the following exercises, find the mass of the two-dimensional object that is centered at the origin.
229. An oversized hockey puck of radius 2 in. with
density function ρ(x) = x3 − 2x + 5
230. A frisbee of radius 6 in. with density function
ρ(x) = e−x


231. A plate of radius 10 in. with density function
ρ(x) = 1 + cos(πx)


232. A jar lid of radius 3 in. with density function
ρ(x) = ln(x + 1)


233. A disk of radius 5 cm with density function
ρ(x) = 3x


234. A 12 -in. spring is stretched to 15 in. by a force of
75 lb. What is the spring constant?
235. A spring has a natural length of 10 cm. It takes 2
J to stretch the spring to 15 cm. How much work would it
take to stretch the spring from 15 cm to 20 cm?
236. A 1 -m spring requires 10 J to stretch the spring to
1.1 m. How much work would it take to stretch the spring
from 1 m to 1.2 m?
237. A spring requires 5 J to stretch the spring from 8
cm to 12 cm, and an additional 4 J to stretch the spring
from 12 cm to 14 cm. What is the natural length of the
spring?
238. A shock absorber is compressed 1 in. by a weight of1 t. What is the spring constant?
239. A force of F = 20x − x3 N stretches a nonlinear
spring by x meters. What work is required to stretch the
spring from x = 0 to x = 2 m?
240. Find the work done by winding up a hanging cable oflength 100 ft and weight-density 5 lb/ft.
241. For the cable in the preceding exercise, how muchwork is done to lift the cable 50 ft?
242. For the cable in the preceding exercise, how muchadditional work is done by hanging a 200 lb weight at the
end of the cable?
243. [T] A pyramid of height 500 ft has a square base
800 ft by 800 ft. Find the area A at height h. If the
rock used to build the pyramid weighs approximately
w = 100 lb/ft3, how much work did it take to lift all the
rock?


Chapter 2 | Applications of Integration 199




244. [T] For the pyramid in the preceding exercise,assume there were 1000 workers each working 10 hours
a day, 5 days a week, 50 weeks a year. If the workers, on
average, lifted 10 100 lb rocks 2 ft/hr, how long did it take
to build the pyramid?
245. [T] The force of gravity on a mass m is
F = −⎛⎝(GMm)/x


2⎞
⎠ newtons. For a rocket of mass


m = 1000 kg, compute the work to lift the rocket from
x = 6400 to x = 6500 km. (Note:
G = 6 × 10−17 N m2 /kg2 and M = 6 × 1024 kg.)
246. [T] For the rocket in the preceding exercise, find thework to lift the rocket from x = 6400 to x = ∞.
247. [T] A rectangular dam is 40 ft high and 60 ft wide.
Compute the total force F on the dam when


a. the surface of the water is at the top of the dam andb. the surface of the water is halfway down the dam.
248. [T] Find the work required to pump all the water outof a cylinder that has a circular base of radius 5 ft and
height 200 ft. Use the fact that the density of water is 62
lb/ft3.
249. [T] Find the work required to pump all the water outof the cylinder in the preceding exercise if the cylinder isonly half full.
250. [T] How much work is required to pump out aswimming pool if the area of the base is 800 ft2, the water
is 4 ft deep, and the top is 1 ft above the water level?
Assume that the density of water is 62 lb/ft3.
251. A cylinder of depth H and cross-sectional area A
stands full of water at density ρ. Compute the work to
pump all the water to the top.
252. For the cylinder in the preceding exercise, computethe work to pump all the water to the top if the cylinder isonly half full.
253. A cone-shaped tank has a cross-sectional area that
increases with its depth: A = ⎛⎝πr2h2⎞⎠/H3. Show that the
work to empty it is half the work for a cylinder with thesame height and base.


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2.6 | Moments and Centers of Mass
Learning Objectives


2.6.1 Find the center of mass of objects distributed along a line.
2.6.2 Locate the center of mass of a thin plate.
2.6.3 Use symmetry to help locate the centroid of a thin plate.
2.6.4 Apply the theorem of Pappus for volume.


In this section, we consider centers of mass (also called centroids, under certain conditions) and moments. The basic ideaof the center of mass is the notion of a balancing point. Many of us have seen performers who spin plates on the ends ofsticks. The performers try to keep several of them spinning without allowing any of them to drop. If we look at a single plate(without spinning it), there is a sweet spot on the plate where it balances perfectly on the stick. If we put the stick anywhereother than that sweet spot, the plate does not balance and it falls to the ground. (That is why performers spin the plates; thespin helps keep the plates from falling even if the stick is not exactly in the right place.) Mathematically, that sweet spot iscalled the center of mass of the plate.
In this section, we first examine these concepts in a one-dimensional context, then expand our development to considercenters of mass of two-dimensional regions and symmetry. Last, we use centroids to find the volume of certain solids byapplying the theorem of Pappus.
Center of Mass and Moments
Let’s begin by looking at the center of mass in a one-dimensional context. Consider a long, thin wire or rod of negligiblemass resting on a fulcrum, as shown in Figure 2.62(a). Now suppose we place objects having masses m1 and m2 at
distances d1 and d2 from the fulcrum, respectively, as shown in Figure 2.62(b).


Figure 2.62 (a) A thin rod rests on a fulcrum. (b) Masses areplaced on the rod.


The most common real-life example of a system like this is a playground seesaw, or teeter-totter, with children of differentweights sitting at different distances from the center. On a seesaw, if one child sits at each end, the heavier child sinksdown and the lighter child is lifted into the air. If the heavier child slides in toward the center, though, the seesaw balances.Applying this concept to the masses on the rod, we note that the masses balance each other if and only if m1d1 = m2d2.
In the seesaw example, we balanced the system by moving the masses (children) with respect to the fulcrum. However,we are really interested in systems in which the masses are not allowed to move, and instead we balance the system bymoving the fulcrum. Suppose we have two point masses, m1 and m2, located on a number line at points x1 and x2,
respectively (Figure 2.63). The center of mass, x– , is the point where the fulcrum should be placed to make the system
balance.


Chapter 2 | Applications of Integration 201




Figure 2.63 The center of mass x– is the balance point of
the system.


Thus, we have
m1 |x1 − x– | = m2 |x2 − x– |
m1

⎝ x
– − x1



⎠ = m2



⎝x2 − x


– ⎞


m1 x
– − m1 x1 = m2 x2 − m2 x




x– (m1 + m2) = m1 x1 + m2 x2


x– =
m1 x1 + m2 x2


m1 + m2
.


The expression in the numerator, m1 x1 + m2 x2, is called the first moment of the system with respect to the origin. If the
context is clear, we often drop the word first and just refer to this expression as the moment of the system. The expressionin the denominator, m1 + m2, is the total mass of the system. Thus, the center of mass of the system is the point at which
the total mass of the system could be concentrated without changing the moment.
This idea is not limited just to two point masses. In general, if n masses, m1, m2 ,…, mn, are placed on a number line at
points x1, x2 ,…, xn, respectively, then the center of mass of the system is given by


x– =

i = 1


n


mixi



i = 1


n


mi


.


Theorem 2.9: Center of Mass of Objects on a Line
Let m1, m2 ,…, mn be point masses placed on a number line at points x1, x2 ,…, xn, respectively, and let
m = ∑


i = 1


n


mi denote the total mass of the system. Then, the moment of the system with respect to the origin is given
by


(2.14)
M = ∑


i = 1


n


mi xi


and the center of mass of the system is given by
(2.15)x– = Mm .


We apply this theorem in the following example.
Example 2.29
Finding the Center of Mass of Objects along a Line
Suppose four point masses are placed on a number line as follows:


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2.29


m1 = 30 kg, placed at x1 = −2m m2 = 5 kg, placed at x2 = 3m


m3 = 10 kg, placed at x3 = 6m m4 = 15 kg, placed at x4 = −3m.


Find the moment of the system with respect to the origin and find the center of mass of the system.
Solution
First, we need to calculate the moment of the system:


M = ∑
i = 1


4


mixi


= −60 + 15 + 60 − 45 = −30.


Now, to find the center of mass, we need the total mass of the system:
m = ∑


i = 1


4


mi


= 30 + 5 + 10 + 15 = 60 kg.


Then we have
x– = Mm =


−30
60


= − 1
2
.


The center of mass is located 1/2 m to the left of the origin.


Suppose four point masses are placed on a number line as follows:
m1 = 12 kg, placed at x1 = −4m m2 = 12 kg, placed at x2 = 4m


m3 = 30 kg, placed at x3 = 2m m4 = 6 kg, placed at x4 = −6m.


Find the moment of the system with respect to the origin and find the center of mass of the system.
We can generalize this concept to find the center of mass of a system of point masses in a plane. Let m1 be a point
mass located at point (x1, y1) in the plane. Then the moment Mx of the mass with respect to the x-axis is given by
Mx = m1 y1. Similarly, the moment My with respect to the y-axis is given by My = m1 x1. Notice that the x-coordinate
of the point is used to calculate the moment with respect to the y-axis, and vice versa. The reason is that the x-coordinategives the distance from the point mass to the y-axis, and the y-coordinate gives the distance to the x-axis (see the followingfigure).


Figure 2.64 Point mass m1 is located at point (x1, y1) in
the plane.


If we have several point masses in the xy-plane, we can use the moments with respect to the x- and y-axes to calculate the


Chapter 2 | Applications of Integration 203




x- and y-coordinates of the center of mass of the system.
Theorem 2.10: Center of Mass of Objects in a Plane
Let m1, m2 ,…, mn be point masses located in the xy-plane at points (x1, y1), (x2, y2),…, (xn, yn), respectively,
and let m = ∑


i = 1


n


mi denote the total mass of the system. Then the moments Mx and My of the system with respect
to the x- and y-axes, respectively, are given by


(2.16)
Mx = ∑


i = 1


n


mi yi and My = ∑
i = 1


n


mi xi.


Also, the coordinates of the center of mass ⎛⎝ x– , y– ⎞⎠ of the system are
(2.17)


x– =
My
m and y


– = Mxm .


The next example demonstrates how to apply this theorem.
Example 2.30
Finding the Center of Mass of Objects in a Plane
Suppose three point masses are placed in the xy-plane as follows (assume coordinates are given in meters):


m1 = 2 kg, placed at (−1, 3),


m2 = 6 kg, placed at (1, 1),


m3 = 4 kg, placed at (2, −2).


Find the center of mass of the system.
Solution
First we calculate the total mass of the system:


m = ∑
i = 1


3


mi = 2 + 6 + 4 = 12 kg.


Next we find the moments with respect to the x- and y-axes:
My = ∑


i = 1


3


mixi = −2 + 6 + 8 = 12,


Mx = ∑
i = 1


3


miyi = 6 + 6 − 8 = 4.


Then we have
x– =


My
m =


12
12


= 1 and y– = Mxm =
4
12


= 1
3
.


The center of mass of the system is (1, 1/3), in meters.


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2.30 Suppose three point masses are placed on a number line as follows (assume coordinates are given inmeters):
m1 = 5 kg, placed at (−2, −3),


m2 = 3 kg, placed at (2, 3),


m3 = 2 kg, placed at (−3, −2).


Find the center of mass of the system.


Center of Mass of Thin Plates
So far we have looked at systems of point masses on a line and in a plane. Now, instead of having the mass of a systemconcentrated at discrete points, we want to look at systems in which the mass of the system is distributed continuouslyacross a thin sheet of material. For our purposes, we assume the sheet is thin enough that it can be treated as if it is two-dimensional. Such a sheet is called a lamina. Next we develop techniques to find the center of mass of a lamina. In thissection, we also assume the density of the lamina is constant.
Laminas are often represented by a two-dimensional region in a plane. The geometric center of such a region is called itscentroid. Since we have assumed the density of the lamina is constant, the center of mass of the lamina depends only onthe shape of the corresponding region in the plane; it does not depend on the density. In this case, the center of mass of thelamina corresponds to the centroid of the delineated region in the plane. As with systems of point masses, we need to findthe total mass of the lamina, as well as the moments of the lamina with respect to the x- and y-axes.
We first consider a lamina in the shape of a rectangle. Recall that the center of mass of a lamina is the point where the laminabalances. For a rectangle, that point is both the horizontal and vertical center of the rectangle. Based on this understanding,it is clear that the center of mass of a rectangular lamina is the point where the diagonals intersect, which is a result of thesymmetry principle, and it is stated here without proof.
Theorem 2.11: The Symmetry Principle
If a region R is symmetric about a line l, then the centroid of R lies on l.


Let’s turn to more general laminas. Suppose we have a lamina bounded above by the graph of a continuous function f (x),
below by the x-axis, and on the left and right by the lines x = a and x = b, respectively, as shown in the following figure.


Figure 2.65 A region in the plane representing a lamina.


As with systems of point masses, to find the center of mass of the lamina, we need to find the total mass of the lamina, aswell as the moments of the lamina with respect to the x- and y-axes. As we have done many times before, we approximatethese quantities by partitioning the interval ⎡⎣a, b⎤⎦ and constructing rectangles.
For i = 0, 1, 2,…, n, let P = {xi} be a regular partition of ⎡⎣a, b⎤⎦. Recall that we can choose any point within the
interval [xi − 1, xi] as our xi* . In this case, we want xi* to be the x-coordinate of the centroid of our rectangles. Thus, for
i = 1, 2,…, n, we select xi* ∈ [xi − 1, xi] such that xi* is the midpoint of the interval. That is, xi* = (xi − 1 + xi)/2.
Now, for i = 1, 2,…, n, construct a rectangle of height f ⎛⎝xi* ⎞⎠ on [xi − 1, xi]. The center of mass of this rectangle is


Chapter 2 | Applications of Integration 205





⎝xi* ,



⎝ f (xi* )



⎠/2

⎠, as shown in the following figure.


Figure 2.66 A representative rectangle of the lamina.


Next, we need to find the total mass of the rectangle. Let ρ represent the density of the lamina (note that ρ is a constant).
In this case, ρ is expressed in terms of mass per unit area. Thus, to find the total mass of the rectangle, we multiply the area
of the rectangle by ρ. Then, the mass of the rectangle is given by ρ f (xi* )Δx.
To get the approximate mass of the lamina, we add the masses of all the rectangles to get


m ≈ ∑
i = 1


n


ρ f (xi* )Δx.


This is a Riemann sum. Taking the limit as n → ∞ gives the exact mass of the lamina:
m = lim


n → ∞

i = 1


n


ρ f (xi* )Δx = ρ∫ a
b
f (x)dx.


Next, we calculate the moment of the lamina with respect to the x-axis. Returning to the representative rectangle, recall its
center of mass is ⎛⎝xi* , ⎛⎝ f (xi* )⎞⎠/2⎞⎠. Recall also that treating the rectangle as if it is a point mass located at the center of
mass does not change the moment. Thus, the moment of the rectangle with respect to the x-axis is given by the mass of
the rectangle, ρ f (xi* )Δx, multiplied by the distance from the center of mass to the x-axis: ⎛⎝ f (xi* )⎞⎠/2. Therefore, the
moment with respect to the x-axis of the rectangle is ρ⎛⎝⎡⎣ f (xi* )⎤⎦2/2⎞⎠Δx. Adding the moments of the rectangles and taking
the limit of the resulting Riemann sum, we see that the moment of the lamina with respect to the x-axis is


Mx = limn → ∞∑
i = 1


n


ρ

⎣ f (xi* )




2


2
Δx = ρ∫


a


b ⎡
⎣ f (x)⎤⎦2


2
dx.


We derive the moment with respect to the y-axis similarly, noting that the distance from the center of mass of the rectangleto the y-axis is xi* . Then the moment of the lamina with respect to the y-axis is given by


My = limn → ∞∑
i = 1


n


ρxi* f (xi* )Δx = ρ∫a
b
x f (x)dx.


We find the coordinates of the center of mass by dividing the moments by the total mass to give
x– = My/m and y


– = Mx/m. If we look closely at the expressions for Mx, My, andm, we notice that the constant ρ
cancels out when x– and y– are calculated.
We summarize these findings in the following theorem.
Theorem 2.12: Center of Mass of a Thin Plate in the xy-Plane
Let R denote a region bounded above by the graph of a continuous function f (x), below by the x-axis, and on the left


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and right by the lines x = a and x = b, respectively. Let ρ denote the density of the associated lamina. Then we
can make the following statements:


i. The mass of the lamina is
(2.18)


m = ρ∫
a


b
f (x)dx.


ii. The moments Mx and My of the lamina with respect to the x- and y-axes, respectively, are
(2.19)


Mx = ρ∫
a


b ⎡
⎣ f (x)⎤⎦2


2
dx andMy = ρ∫


a


b
x f (x)dx.


iii. The coordinates of the center of mass ⎛⎝ x– , y– ⎞⎠ are
(2.20)


x– =
My
m and y


– = Mxm .


In the next example, we use this theorem to find the center of mass of a lamina.
Example 2.31
Finding the Center of Mass of a Lamina
Let R be the region bounded above by the graph of the function f (x) = x and below by the x-axis over the
interval [0, 4]. Find the centroid of the region.
Solution
The region is depicted in the following figure.


Figure 2.67 Finding the center of mass of a lamina.


Since we are only asked for the centroid of the region, rather than the mass or moments of the associatedlamina, we know the density constant ρ cancels out of the calculations eventually. Therefore, for the sake of
convenience, let’s assume ρ = 1.
First, we need to calculate the total mass:


m = ρ∫
a


b
f (x)dx = ∫


0


4
x dx


= 2
3
x3/2|0


4
= 2


3
[8 − 0] = 16


3
.


Chapter 2 | Applications of Integration 207




2.31


Next, we compute the moments:
Mx = ρ∫


a


b ⎡
⎣ f (x)⎤⎦2


2
dx


= ∫
0


4
x
2
dx = 1


4
x2|0


4
= 4


and
My = ρ∫


a


b
x f (x)dx


= ∫
0


4
x xdx = ∫


0


4
x3/2dx


= 2
5
x5/2|0


4
= 2


5
[32 − 0] = 64


5
.


Thus, we have
x– =


My
m =


64/5
16/3


= 64
5


· 3
16


= 12
5


and y– = Mxy =
4


16/3
= 4 · 3


16
= 3


4
.


The centroid of the region is (12/5, 3/4).


Let R be the region bounded above by the graph of the function f (x) = x2 and below by the x-axis over
the interval [0, 2]. Find the centroid of the region.


We can adapt this approach to find centroids of more complex regions as well. Suppose our region is bounded above by thegraph of a continuous function f (x), as before, but now, instead of having the lower bound for the region be the x-axis,
suppose the region is bounded below by the graph of a second continuous function, g(x), as shown in the following figure.


Figure 2.68 A region between two functions.


Again, we partition the interval ⎡⎣a, b⎤⎦ and construct rectangles. A representative rectangle is shown in the following figure.


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Figure 2.69 A representative rectangle of the region betweentwo functions.


Note that the centroid of this rectangle is ⎛⎝xi* , ⎛⎝ f (xi* ) + g(xi* )⎞⎠/2⎞⎠. We won’t go through all the details of the Riemann
sum development, but let’s look at some of the key steps. In the development of the formulas for the mass of the laminaand the moment with respect to the y-axis, the height of each rectangle is given by f (xi* ) − g(xi* ), which leads to the
expression f (x) − g(x) in the integrands.
In the development of the formula for the moment with respect to the x-axis, the moment of each rectangle is found
by multiplying the area of the rectangle, ρ⎡⎣ f (xi* ) − g(xi* )⎤⎦Δx, by the distance of the centroid from the x-axis,

⎝ f (xi* ) + g(xi* )



⎠/2, which gives ρ(1/2)⎧⎩⎨⎡⎣ f (xi* )⎤⎦2 − ⎡⎣g(xi* )⎤⎦2⎫⎭⎬Δx. Summarizing these findings, we arrive at the


following theorem.
Theorem 2.13: Center of Mass of a Lamina Bounded by Two Functions
Let R denote a region bounded above by the graph of a continuous function f (x), below by the graph of the
continuous function g(x), and on the left and right by the lines x = a and x = b, respectively. Let ρ denote the
density of the associated lamina. Then we can make the following statements:


i. The mass of the lamina is
(2.21)


m = ρ∫
a


b

⎣ f (x) − g(x)⎤⎦dx.


ii. The moments Mx and My of the lamina with respect to the x- and y-axes, respectively, are
(2.22)


Mx = ρ∫
a


b
1
2



⎣ f (x)⎤⎦2 − ⎡⎣g(x)⎤⎦2⎞⎠dx andMy = ρ∫a


b
x⎡⎣ f (x) − g(x)⎤⎦dx.


iii. The coordinates of the center of mass ⎛⎝ x– , y– ⎞⎠ are
(2.23)


x– =
My
m and y


– = Mxm .


We illustrate this theorem in the following example.
Example 2.32
Finding the Centroid of a Region Bounded by Two Functions
Let R be the region bounded above by the graph of the function f (x) = 1 − x2 and below by the graph of the
function g(x) = x − 1. Find the centroid of the region.


Chapter 2 | Applications of Integration 209




Solution
The region is depicted in the following figure.


Figure 2.70 Finding the centroid of a region between twocurves.


The graphs of the functions intersect at (−2, −3) and (1, 0), so we integrate from −2 to 1. Once again, for the
sake of convenience, assume ρ = 1.
First, we need to calculate the total mass:


m = ρ∫
a


b

⎣ f (x) − g(x)⎤⎦dx


= ∫
−2


1

⎣1 − x


2 − (x − 1)⎤⎦dx = ∫
−2


1
(2 − x2 − x)dx


= ⎡⎣2x −
1
3
x3 − 1


2
x2⎤⎦ |−2


1
= ⎡⎣2 −


1
3
− 1


2

⎦−

⎣−4 +


8
3
− 2⎤⎦ =


9
2
.


Next, we compute the moments:
Mx = ρ∫


a


b
1
2



⎣ f (x)⎤⎦2 − ⎡⎣g(x)⎤⎦2⎞⎠dx


= 1
2


−2


1 ⎛


⎝1 − x


2⎞


2
− (x − 1)2



⎠dx =


1
2


−2


1

⎝x


4 − 3x2 + 2x⎞⎠dx


= 1
2


x5


5
− x3 + x2



⎦ |−2


1


= − 27
10


and
My = ρ∫


a


b
x⎡⎣ f (x) − g(x)⎤⎦dx


= ∫
−2


1
x⎡⎣(1 − x


2) − (x − 1)⎤⎦dx = ∫
−2


1
x⎡⎣2 − x


2 − x⎤⎦dx = ∫
−2


1

⎝2x − x


4 − x2⎞⎠dx


=

⎣x


2 − x
5


5
− x


3


3

⎦ |−2


1


= − 9
4
.


Therefore, we have
x– =


My
m = −


9
4
· 2
9
= − 1


2
and y– = Mxy = −


27
10


· 2
9
= − 3


5
.


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2.32


The centroid of the region is ⎛⎝−(1/2), −(3/5)⎞⎠.


Let R be the region bounded above by the graph of the function f (x) = 6 − x2 and below by the graph
of the function g(x) = 3 − 2x. Find the centroid of the region.


The Symmetry Principle
We stated the symmetry principle earlier, when we were looking at the centroid of a rectangle. The symmetry principle canbe a great help when finding centroids of regions that are symmetric. Consider the following example.
Example 2.33
Finding the Centroid of a Symmetric Region
Let R be the region bounded above by the graph of the function f (x) = 4 − x2 and below by the x-axis. Find the
centroid of the region.
Solution
The region is depicted in the following figure.


Figure 2.71 We can use the symmetry principle to help findthe centroid of a symmetric region.


The region is symmetric with respect to the y-axis. Therefore, the x-coordinate of the centroid is zero. We needonly calculate y– . Once again, for the sake of convenience, assume ρ = 1.
First, we calculate the total mass:


m = ρ∫
a


b
f (x)dx


= ∫
−2


2

⎝4 − x


2⎞
⎠dx


=

⎣4x −


x3


3

⎦ |−2


2


= 32
3
.


Chapter 2 | Applications of Integration 211




2.33


Next, we calculate the moments. We only need Mx :
Mx = ρ∫


a


b ⎡
⎣ f (x)⎤⎦2


2
dx


= 1
2


−2


2

⎣4 − x


2⎤


2
dx = 1


2


−2


2

⎝16 − 8x


2 + x4⎞⎠dx


= 1
2


x5


5
− 8x


3


3
+ 16x

⎦ |−2


2


= 256
15


.


Then we have
y– = Mxy =


256
15


· 3
32


= 8
5
.


The centroid of the region is (0, 8/5).


Let R be the region bounded above by the graph of the function f (x) = 1 − x2 and below by x-axis.
Find the centroid of the region.


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The Grand Canyon Skywalk
The Grand Canyon Skywalk opened to the public on March 28, 2007. This engineering marvel is a horseshoe-shapedobservation platform suspended 4000 ft above the Colorado River on the West Rim of the Grand Canyon. Its crystal-clear glass floor allows stunning views of the canyon below (see the following figure).


Figure 2.72 The Grand Canyon Skywalk offers magnificent views of the canyon. (credit: 10da_ralta, WikimediaCommons)


The Skywalk is a cantilever design, meaning that the observation platform extends over the rim of the canyon, with novisible means of support below it. Despite the lack of visible support posts or struts, cantilever structures are engineeredto be very stable and the Skywalk is no exception. The observation platform is attached firmly to support posts thatextend 46 ft down into bedrock. The structure was built to withstand 100-mph winds and an 8.0-magnitude earthquakewithin 50 mi, and is capable of supporting more than 70,000,000 lb.
One factor affecting the stability of the Skywalk is the center of gravity of the structure. We are going to calculatethe center of gravity of the Skywalk, and examine how the center of gravity changes when tourists walk out onto theobservation platform.
The observation platform is U-shaped. The legs of the U are 10 ft wide and begin on land, under the visitors’ center,48 ft from the edge of the canyon. The platform extends 70 ft over the edge of the canyon.
To calculate the center of mass of the structure, we treat it as a lamina and use a two-dimensional region in the xy-planeto represent the platform. We begin by dividing the region into three subregions so we can consider each subregion


Chapter 2 | Applications of Integration 213




separately. The first region, denoted R1, consists of the curved part of the U. We model R1 as a semicircular annulus,
with inner radius 25 ft and outer radius 35 ft, centered at the origin (see the following figure).


Figure 2.73 We model the Skywalk with three sub-regions.


The legs of the platform, extending 35 ft between R1 and the canyon wall, comprise the second sub-region, R2. Last,
the ends of the legs, which extend 48 ft under the visitor center, comprise the third sub-region, R3. Assume the density
of the lamina is constant and assume the total weight of the platform is 1,200,000 lb (not including the weight of the
visitor center; we will consider that later). Use g = 32 ft/sec2.


1. Compute the area of each of the three sub-regions. Note that the areas of regions R2 and R3 should include
the areas of the legs only, not the open space between them. Round answers to the nearest square foot.


2. Determine the mass associated with each of the three sub-regions.
3. Calculate the center of mass of each of the three sub-regions.
4. Now, treat each of the three sub-regions as a point mass located at the center of mass of the correspondingsub-region. Using this representation, calculate the center of mass of the entire platform.
5. Assume the visitor center weighs 2,200,000 lb, with a center of mass corresponding to the center of mass of


R3. Treating the visitor center as a point mass, recalculate the center of mass of the system. How does the
center of mass change?


6. Although the Skywalk was built to limit the number of people on the observation platform to 120, the platformis capable of supporting up to 800 people weighing 200 lb each. If all 800 people were allowed on the platform,and all of them went to the farthest end of the platform, how would the center of gravity of the system beaffected? (Include the visitor center in the calculations and represent the people by a point mass located at thefarthest edge of the platform, 70 ft from the canyon wall.)


Theorem of Pappus
This section ends with a discussion of the theorem of Pappus for volume, which allows us to find the volume of particular


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kinds of solids by using the centroid. (There is also a theorem of Pappus for surface area, but it is much less useful than thetheorem for volume.)
Theorem 2.14: Theorem of Pappus for Volume
Let R be a region in the plane and let l be a line in the plane that does not intersect R. Then the volume of the solid ofrevolution formed by revolving R around l is equal to the area of Rmultiplied by the distance d traveled by the centroidof R.


Proof
We can prove the case when the region is bounded above by the graph of a function f (x) and below by the graph of a
function g(x) over an interval ⎡⎣a, b⎤⎦, and for which the axis of revolution is the y-axis. In this case, the area of the region is
A = ∫


a


b

⎣ f (x) − g(x)⎤⎦dx. Since the axis of rotation is the y-axis, the distance traveled by the centroid of the region depends


only on the x-coordinate of the centroid, x– , which is
x– =


My
m ,


where
m = ρ∫


a


b

⎣ f (x) − g(x)⎤⎦dx andMy = ρ∫


a


b
x⎡⎣ f (x) − g(x)⎤⎦dx.


Then,


d = 2π
ρ∫


a


b
x⎡⎣ f (x) − g(x)⎤⎦dx


ρ∫
a


b

⎣ f (x) − g(x)⎤⎦dx


and thus
d · A = 2π∫


a


b
x⎡⎣ f (x) − g(x)⎤⎦dx.


However, using the method of cylindrical shells, we have
V = 2π∫


a


b
x⎡⎣ f (x) − g(x)⎤⎦dx.


So,
V = d · A


and the proof is complete.

Example 2.34
Using the Theorem of Pappus for Volume
Let R be a circle of radius 2 centered at (4, 0). Use the theorem of Pappus for volume to find the volume of the
torus generated by revolving R around the y-axis.


Chapter 2 | Applications of Integration 215




2.34


Solution
The region and torus are depicted in the following figure.


Figure 2.74 Determining the volume of a torus by using the theorem of Pappus. (a) Acircular region R in the plane; (b) the torus generated by revolving R about the y-axis.


The region R is a circle of radius 2, so the area of R is A = 4π units2. By the symmetry principle, the centroid of
R is the center of the circle. The centroid travels around the y-axis in a circular path of radius 4, so the centroid
travels d = 8π units. Then, the volume of the torus is A · d = 32π2 units3.


Let R be a circle of radius 1 centered at (3, 0). Use the theorem of Pappus for volume to find the
volume of the torus generated by revolving R around the y-axis.


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2.6 EXERCISES
For the following exercises, calculate the center of mass forthe collection of masses given.
254. m1 = 2 at x1 = 1 and m2 = 4 at x2 = 2
255. m1 = 1 at x1 = −1 and m2 = 3 at x2 = 2
256. m = 3 at x = 0, 1, 2, 6
257. Unit masses at (x, y) = (1, 0), (0, 1), (1, 1)
258. m1 = 1 at (1, 0) and m2 = 4 at (0, 1)
259. m1 = 1 at (1, 0) and m2 = 3 at (2, 2)
For the following exercises, compute the center of mass
x– .


260. ρ = 1 for x ∈ (−1, 3)
261. ρ = x2 for x ∈ (0, L)
262. ρ = 1 for x ∈ (0, 1) and ρ = 2 for x ∈ (1, 2)
263. ρ = sin x for x ∈ (0, π)
264. ρ = cos x for x ∈ ⎛⎝0, π2⎞⎠
265. ρ = ex for x ∈ (0, 2)
266. ρ = x3 + xe−x for x ∈ (0, 1)
267. ρ = x sin x for x ∈ (0, π)
268. ρ = x for x ∈ (1, 4)
269. ρ = ln x for x ∈ (1, e)
For the following exercises, compute the center of mass

⎝ x– , y– ⎞⎠. Use symmetry to help locate the center of mass
whenever possible.
270. ρ = 7 in the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1
271. ρ = 3 in the triangle with vertices (0, 0), (a, 0),
and (0, b)
272. ρ = 2 for the region bounded by y = cos(x),
y = −cos(x), x = − π


2
, and x = π


2


For the following exercises, use a calculator to draw theregion, then compute the center of mass ⎛⎝ x– , y– ⎞⎠. Use
symmetry to help locate the center of mass wheneverpossible.
273. [T] The region bounded by y = cos(2x),
x = − π


4
, and x = π


4


274. [T] The region between y = 2x2, y = 0, x = 0,
and x = 1
275. [T] The region between y = 5


4
x2 and y = 5


276. [T] Region between y = x, y = ln(x), x = 1,
and x = 4
277. [T] The region bounded by y = 0, x2


4
+
y2


9
= 1


278. [T] The region bounded by y = 0, x = 0, and
x2
4


+
y2


9
= 1


279. [T] The region bounded by y = x2 and y = x4 in
the first quadrant
For the following exercises, use the theorem of Pappus todetermine the volume of the shape.
280. Rotating y = mx around the x -axis between x = 0
and x = 1
281. Rotating y = mx around the y -axis between x = 0
and x = 1
282. A general cone created by rotating a triangle withvertices (0, 0), (a, 0), and (0, b) around the y -axis.
Does your answer agree with the volume of a cone?
283. A general cylinder created by rotating a rectanglewith vertices (0, 0), (a, 0), (0, b), and (a, b) around
the y -axis. Does your answer agree with the volume of a
cylinder?
284. A sphere created by rotating a semicircle with radius
a around the y -axis. Does your answer agree with the
volume of a sphere?
For the following exercises, use a calculator to draw theregion enclosed by the curve. Find the area M and the


Chapter 2 | Applications of Integration 217




centroid ⎛⎝ x– , y– ⎞⎠ for the given shapes. Use symmetry to
help locate the center of mass whenever possible.
285. [T] Quarter-circle: y = 1 − x2, y = 0, and
x = 0


286. [T] Triangle: y = x, y = 2 − x, and y = 0
287. [T] Lens: y = x2 and y = x
288. [T] Ring: y2 + x2 = 1 and y2 + x2 = 4
289. [T] Half-ring: y2 + x2 = 1, y2 + x2 = 4, and
y = 0


290. Find the generalized center of mass in the sliver
between y = xa and y = xb with a > b. Then, use the
Pappus theorem to find the volume of the solid generatedwhen revolving around the y-axis.
291. Find the generalized center of mass between
y = a2 − x2, x = 0, and y = 0. Then, use the Pappus
theorem to find the volume of the solid generated whenrevolving around the y-axis.
292. Find the generalized center of mass between
y = b sin(ax), x = 0, and x = πa . Then, use the
Pappus theorem to find the volume of the solid generatedwhen revolving around the y-axis.
293. Use the theorem of Pappus to find the volume ofa torus (pictured here). Assume that a disk of radius a
is positioned with the left end of the circle at x = b,
b > 0, and is rotated around the y-axis.


294. Find the center of mass ⎛⎝ x– , y– ⎞⎠ for a thin wire along
the semicircle y = 1 − x2 with unit mass. (Hint: Use the
theorem of Pappus.)


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2.7 | Integrals, Exponential Functions, and Logarithms
Learning Objectives


2.7.1 Write the definition of the natural logarithm as an integral.
2.7.2 Recognize the derivative of the natural logarithm.
2.7.3 Integrate functions involving the natural logarithmic function.
2.7.4 Define the number e through an integral.
2.7.5 Recognize the derivative and integral of the exponential function.
2.7.6 Prove properties of logarithms and exponential functions using integrals.
2.7.7 Express general logarithmic and exponential functions in terms of natural logarithms andexponentials.


We already examined exponential functions and logarithms in earlier chapters. However, we glossed over some key detailsin the previous discussions. For example, we did not study how to treat exponential functions with exponents that areirrational. The definition of the number e is another area where the previous development was somewhat incomplete. Wenow have the tools to deal with these concepts in a more mathematically rigorous way, and we do so in this section.
For purposes of this section, assume we have not yet defined the natural logarithm, the number e, or any of the integrationand differentiation formulas associated with these functions. By the end of the section, we will have studied these conceptsin a mathematically rigorous way (and we will see they are consistent with the concepts we learned earlier).
We begin the section by defining the natural logarithm in terms of an integral. This definition forms the foundation forthe section. From this definition, we derive differentiation formulas, define the number e, and expand these concepts to
logarithms and exponential functions of any base.
The Natural Logarithm as an Integral
Recall the power rule for integrals:


∫ xndx = x
n + 1


n + 1
+ C, n ≠ −1.


Clearly, this does not work when n = −1, as it would force us to divide by zero. So, what do we do with ∫ 1xdx? Recall
from the Fundamental Theorem of Calculus that ∫


1


x
1
t dt is an antiderivative of 1/x. Therefore, we can make the following


definition.
Definition
For x > 0, define the natural logarithm function by


(2.24)
ln x = ∫


1


x
1
t dt.


For x > 1, this is just the area under the curve y = 1/t from 1 to x. For x < 1, we have ∫
1


x
1
t dt = −∫x


1
1
t dt, so in


this case it is the negative of the area under the curve from x to 1 (see the following figure).


Chapter 2 | Applications of Integration 219




Figure 2.75 (a) When x > 1, the natural logarithm is the area under the
curve y = 1/t from 1 to x. (b) When x < 1, the natural logarithm is the
negative of the area under the curve from x to 1.


Notice that ln 1 = 0. Furthermore, the function y = 1/t > 0 for x > 0. Therefore, by the properties of integrals, it is clear
that ln x is increasing for x > 0.
Properties of the Natural Logarithm
Because of the way we defined the natural logarithm, the following differentiation formula falls out immediately as a resultof to the Fundamental Theorem of Calculus.
Theorem 2.15: Derivative of the Natural Logarithm
For x > 0, the derivative of the natural logarithm is given by


d
dx


ln x = 1x .


Theorem 2.16: Corollary to the Derivative of the Natural Logarithm
The function ln x is differentiable; therefore, it is continuous.


A graph of ln x is shown in Figure 2.76. Notice that it is continuous throughout its domain of (0, ∞).


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2.35


Figure 2.76 The graph of f (x) = ln x shows that it is a
continuous function.


Example 2.35
Calculating Derivatives of Natural Logarithms
Calculate the following derivatives:


a. d
dx


ln⎛⎝5x
3 − 2⎞⎠


b. d
dx

⎝ln(3x)⎞⎠2


Solution
We need to apply the chain rule in both cases.


a. d
dx


ln⎛⎝5x
3 − 2⎞⎠ =


15x2


5x3 − 2


b. d
dx

⎝ln(3x)⎞⎠2 =


2⎛⎝ln(3x)⎞⎠ · 3
3x


= 2

⎝ln(3x)⎞⎠
x


Calculate the following derivatives:
a. d


dx
ln⎛⎝2x


2 + x⎞⎠


b. d
dx

⎝ln

⎝x


3⎞




2


Note that if we use the absolute value function and create a new function ln |x|, we can extend the domain of the natural
logarithm to include x < 0. Then ⎛⎝d/(dx)⎞⎠ln |x| = 1/x. This gives rise to the familiar integration formula.


Theorem 2.17: Integral of (1/u) du
The natural logarithm is the antiderivative of the function f (u) = 1/u:


Chapter 2 | Applications of Integration 221




2.36


∫ 1udu = ln |u| + C.


Example 2.36
Calculating Integrals Involving Natural Logarithms
Calculate the integral ∫ x


x2 + 4
dx.


Solution
Using u -substitution, let u = x2 + 4. Then du = 2x dx and we have


∫ x
x2 + 4


dx = 1
2
∫ 1udu12ln |u| + C =


1
2
ln |x2 + 4| + C = 12ln⎛⎝x2 + 4⎞⎠+ C.


Calculate the integral ∫ x2
x3 + 6


dx.


Although we have called our function a “logarithm,” we have not actually proved that any of the properties of logarithmshold for this function. We do so here.
Theorem 2.18: Properties of the Natural Logarithm
If a, b > 0 and r is a rational number, then


i. ln 1 = 0
ii. ln(ab) = ln a + ln b
iii. ln⎛⎝ab⎞⎠ = ln a − ln b
iv. ln(ar) = r ln a


Proof
i. By definition, ln 1 = ∫


1


1
1
t dt = 0.


ii. We have
ln(ab) = ∫


1


ab
1
t dt = ∫1


a
1
t dt + ∫a


ab
1
t dt.


Use u-substitution on the last integral in this expression. Let u = t/a. Then du = (1/a)dt. Furthermore, when
t = a, u = 1, and when t = ab, u = b. So we get


ln(ab) = ∫
1


a
1
t dt + ∫a


ab
1
t dt = ∫1


a
1
t dt + ∫1


ab
a
t ·


1
adt = ∫1


a
1
t dt + ∫1


b
1
udu = ln a + ln b.


iii. Note that


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2.37


d
dx


ln(xr) = rx
r − 1


xr
= rx.


Furthermore,
d
dx


(r ln x) = rx.


Since the derivatives of these two functions are the same, by the Fundamental Theorem of Calculus, they must differby a constant. So we have
ln(xr) = r ln x + C


for some constant C. Taking x = 1, we get
ln(1r) = r ln(1) + C


0 = r(0) + C
C = 0.


Thus ln(xr) = r ln x and the proof is complete. Note that we can extend this property to irrational values of r later
in this section.Part iii. follows from parts ii. and iv. and the proof is left to you.



Example 2.37
Using Properties of Logarithms
Use properties of logarithms to simplify the following expression into a single logarithm:


ln 9 − 2 ln 3 + ln⎛⎝
1
3

⎠.


Solution
We have


ln 9 − 2 ln 3 + ln⎛⎝
1
3

⎠ = ln



⎝3


2⎞
⎠− 2 ln 3 + ln



⎝3


−1⎞
⎠ = 2 ln 3 − 2 ln 3 − ln 3 = −ln 3.


Use properties of logarithms to simplify the following expression into a single logarithm:
ln 8 − ln 2 − ln⎛⎝


1
4

⎠.


Defining the Number e
Now that we have the natural logarithm defined, we can use that function to define the number e.


Definition
The number e is defined to be the real number such that


ln e = 1.


To put it another way, the area under the curve y = 1/t between t = 1 and t = e is 1 (Figure 2.77). The proof that such
a number exists and is unique is left to you. (Hint: Use the Intermediate Value Theorem to prove existence and the fact that


Chapter 2 | Applications of Integration 223




ln x is increasing to prove uniqueness.)


Figure 2.77 The area under the curve from 1 to e is equal
to one.


The number e can be shown to be irrational, although we won’t do so here (see the Student Project in Taylor and
Maclaurin Series). Its approximate value is given by


e ≈ 2.71828182846.The Exponential Function
We now turn our attention to the function ex. Note that the natural logarithm is one-to-one and therefore has an inverse
function. For now, we denote this inverse function by exp x. Then,


exp(ln x) = x for x > 0 and ln(exp x) = x for all x.


The following figure shows the graphs of exp x and ln x.


Figure 2.78 The graphs of ln x and exp x.


We hypothesize that exp x = ex. For rational values of x, this is easy to show. If x is rational, then we have
ln(ex) = x ln e = x. Thus, when x is rational, ex = exp x. For irrational values of x, we simply define ex as the
inverse function of ln x.


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Definition
For any real number x, define y = ex to be the number for which


(2.25)ln y = ln(ex) = x.
Then we have ex = exp(x) for all x, and thus


(2.26)eln x = x for x > 0 and ln(ex) = x
for all x.
Properties of the Exponential Function
Since the exponential function was defined in terms of an inverse function, and not in terms of a power of e, we must
verify that the usual laws of exponents hold for the function ex.


Theorem 2.19: Properties of the Exponential Function
If p and q are any real numbers and r is a rational number, then


i. e p eq = e p + q
ii. e p


eq
= e


p − q


iii. (e p)r = e pr


Proof
Note that if p and q are rational, the properties hold. However, if p or q are irrational, we must apply the inverse
function definition of ex and verify the properties. Only the first property is verified here; the other two are left to you. We
have


ln(e p eq) = ln(e p) + ln(eq) = p + q = ln⎛⎝e
p + q⎞
⎠.


Since ln x is one-to-one, then
e p eq = e


p + q
.



As with part iv. of the logarithm properties, we can extend property iii. to irrational values of r, and we do so by the end
of the section.
We also want to verify the differentiation formula for the function y = ex. To do this, we need to use implicit
differentiation. Let y = ex. Then


ln y = x
d
dx


ln y = d
dx


x


1
y
dy
dx


= 1


dy
dx


= y.


Thus, we see


Chapter 2 | Applications of Integration 225




2.38


d
dx


ex = ex


as desired, which leads immediately to the integration formula
∫ ex dx = ex + C.


We apply these formulas in the following examples.
Example 2.38
Using Properties of Exponential Functions
Evaluate the following derivatives:


a. d
dt
e3t et


2


b. d
dx


e3x
2


Solution
We apply the chain rule as necessary.


a. d
dt
e3t et


2
= d


dt
e3t + t


2
= e3t + t


2
(3 + 2t)


b. d
dx


e3x
2
= e3x


2
6x


Evaluate the following derivatives:
a. d


dx





⎜e


x2


e5x







b. d
dt

⎝e


2t⎞


3


Example 2.39
Using Properties of Exponential Functions
Evaluate the following integral: ∫ 2xe−x2dx.
Solution
Using u -substitution, let u = −x2. Then du = −2x dx, and we have


∫ 2xe−x2dx = −∫ eudu = −eu + C = −e−x2 + C.


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2.39 Evaluate the following integral: ∫ 4
e3x


dx.


General Logarithmic and Exponential Functions
We close this section by looking at exponential functions and logarithms with bases other than e. Exponential functions
are functions of the form f (x) = ax. Note that unless a = e, we still do not have a mathematically rigorous definition
of these functions for irrational exponents. Let’s rectify that here by defining the function f (x) = ax in terms of the
exponential function ex. We then examine logarithms with bases other than e as inverse functions of exponential
functions.
Definition
For any a > 0, and for any real number x, define y = ax as follows:


y = ax = ex ln a.


Now ax is defined rigorously for all values of x. This definition also allows us to generalize property iv. of logarithms and
property iii. of exponential functions to apply to both rational and irrational values of r. It is straightforward to show that
properties of exponents hold for general exponential functions defined in this way.
Let’s now apply this definition to calculate a differentiation formula for ax. We have


d
dx


ax = d
dx


ex ln a = ex ln a ln a = ax ln a.


The corresponding integration formula follows immediately.
Theorem 2.20: Derivatives and Integrals Involving General Exponential Functions
Let a > 0. Then,


d
dx


ax = ax ln a


and
∫ ax dx = 1


ln a
ax + C.


If a ≠ 1, then the function ax is one-to-one and has a well-defined inverse. Its inverse is denoted by loga x. Then,
y = loga x if and only if x = a


y.


Note that general logarithm functions can be written in terms of the natural logarithm. Let y = loga x. Then, x = ay.
Taking the natural logarithm of both sides of this second equation, we get


ln x = ln(ay)
ln x = y ln a


y = ln x
ln a


log x = ln x
ln a


.


Thus, we see that all logarithmic functions are constant multiples of one another. Next, we use this formula to find adifferentiation formula for a logarithm with base a. Again, let y = loga x. Then,


Chapter 2 | Applications of Integration 227




2.40


dy
dx


= d
dx

⎝loga x





= d
dx


ln x
ln a



= ⎛⎝
1


ln a


d
dx


(ln x)


= 1
ln a


· 1x


= 1
x ln a


.


Theorem 2.21: Derivatives of General Logarithm Functions
Let a > 0. Then,


d
dx


loga x = 1x ln a
.


Example 2.40
Calculating Derivatives of General Exponential and Logarithm Functions
Evaluate the following derivatives:


a. d
dt

⎝4


t · 2t
2⎞


b. d
dx


log8

⎝7x


2 + 4⎞⎠


Solution
We need to apply the chain rule as necessary.


a. d
dt

⎝4


t · 2t
2⎞
⎠ =


d
dt

⎝2


2t · 2t
2⎞
⎠ =


d
dt

⎝2


2t + t2⎞
⎠ = 2


2t + t2 ln(2)(2 + 2t)


b. d
dx


log8

⎝7x


2 + 4⎞⎠ =
1



⎝7x


2 + 4⎞⎠(ln 8)
(14x)


Evaluate the following derivatives:
a. d


dt
4t


4


b. d
dx


log3

⎝ x


2 + 1⎞⎠


Example 2.41
Integrating General Exponential Functions


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2.41


Evaluate the following integral: ∫ 3
23x


dx.


Solution
Use u-substitution and let u = −3x. Then du = −3dx and we have


∫ 3
23x


dx = ∫ 3 · 2−3xdx = −∫ 2udu = − 1
ln 2


2u + C = − 1
ln 2


2−3x + C.


Evaluate the following integral: ∫ x22x3dx.


Chapter 2 | Applications of Integration 229




2.7 EXERCISES
For the following exercises, find the derivative dy


dx
.


295. y = ln(2x)
296. y = ln(2x + 1)
297. y = 1


ln x


For the following exercises, find the indefinite integral.
298. ∫ dt


3t


299. ∫ dx
1 + x


For the following exercises, find the derivative dy/dx.
(You can use a calculator to plot the function and thederivative to confirm that it is correct.)
300. [T] y = ln(x)x
301. [T] y = x ln(x)
302. [T] y = log10 x
303. [T] y = ln(sin x)
304. [T] y = ln(ln x)
305. [T] y = 7 ln(4x)
306. [T] y = ln⎛⎝(4x)7⎞⎠
307. [T] y = ln(tan x)
308. [T] y = ln(tan(3x))
309. [T] y = ln⎛⎝cos2 x⎞⎠
For the following exercises, find the definite or indefiniteintegral.
310. ∫


0


1
dx


3 + x


311. ∫
0


1
dt


3 + 2t


312. ∫
0


2
x dx
x2 + 1


313. ∫
0


2
x3dx
x2 + 1


314. ∫
2


e
dx


x ln x


315. ∫
2


e
dx



⎝x ln(x)⎞⎠2


316. ∫ cos x dx
sin x


317. ∫
0


π/4
tan x dx


318. ∫ cot(3x)dx


319. ∫ (ln x)2dxx
For the following exercises, compute dy/dx by
differentiating ln y.
320. y = x2 + 1
321. y = x2 + 1 x2 − 1
322. y = esin x
323. y = x−1/x


324. y = e(ex)
325. y = xe


326. y = x(ex)


327. y = x x3 x6


328. y = x−1/ln x


329. y = e−ln x
For the following exercises, evaluate by any method.


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330. ∫
5


10
dt
t − ∫5x


10x
dt
t


331. ∫
1



dx
x + ∫−2


−1
dx
x


332. d
dx

x


1
dt
t


333. d
dx

x


x2
dt
t


334. d
dx


ln(sec x + tan x)


For the following exercises, use the function ln x. If you
are unable to find intersection points analytically, use acalculator.
335. Find the area of the region enclosed by x = 1 and
y = 5 above y = ln x.
336. [T] Find the arc length of ln x from x = 1 to
x = 2.


337. Find the area between ln x and the x-axis from
x = 1 to x = 2.


338. Find the volume of the shape created when rotatingthis curve from x = 1 to x = 2 around the x-axis, as
pictured here.


339. [T] Find the surface area of the shape created whenrotating the curve in the previous exercise from x = 1 to
x = 2 around the x-axis.
If you are unable to find intersection points analytically inthe following exercises, use a calculator.
340. Find the area of the hyperbolic quarter-circleenclosed by x = 2 and y = 2 above y = 1/x.
341. [T] Find the arc length of y = 1/x from
x = 1 to x = 4.


342. Find the area under y = 1/x and above the x-axis
from x = 1 to x = 4.
For the following exercises, verify the derivatives andantiderivatives.
343. d


dx
ln⎛⎝x + x


2 + 1⎞⎠ =
1


1 + x2


344. d
dx


ln⎛⎝
x − a
x + a



⎠ =


2a

⎝x


2 − a2⎞⎠


345. d
dx


ln



⎜1 + 1 − x


2


x



⎟ = − 1


x 1 − x2


346. d
dx


ln⎛⎝x + x
2 − a2⎞⎠ =


1
x2 − a2


347. ∫ dx
x ln(x)ln(ln x)


= ln⎛⎝ln(ln x)⎞⎠+ C


Chapter 2 | Applications of Integration 231




2.8 | Exponential Growth and Decay
Learning Objectives


2.8.1 Use the exponential growth model in applications, including population growth andcompound interest.
2.8.2 Explain the concept of doubling time.
2.8.3 Use the exponential decay model in applications, including radioactive decay and Newton’slaw of cooling.
2.8.4 Explain the concept of half-life.


One of the most prevalent applications of exponential functions involves growth and decay models. Exponential growthand decay show up in a host of natural applications. From population growth and continuously compounded interest toradioactive decay and Newton’s law of cooling, exponential functions are ubiquitous in nature. In this section, we examineexponential growth and decay in the context of some of these applications.
Exponential Growth Model
Many systems exhibit exponential growth. These systems follow a model of the form y = y0 ekt, where y0 represents
the initial state of the system and k is a positive constant, called the growth constant. Notice that in an exponential growth
model, we have


(2.27)y′ = ky0 ekt = ky.
That is, the rate of growth is proportional to the current function value. This is a key feature of exponential growth.Equation 2.27 involves derivatives and is called a differential equation. We learn more about differential equations inIntroduction to Differential Equations.
Rule: Exponential Growth Model
Systems that exhibit exponential growth increase according to the mathematical model


y = y0 e
kt,


where y0 represents the initial state of the system and k > 0 is a constant, called the growth constant.


Population growth is a common example of exponential growth. Consider a population of bacteria, for instance. It seemsplausible that the rate of population growth would be proportional to the size of the population. After all, the more bacteriathere are to reproduce, the faster the population grows. Figure 2.79 and Table 2.1 represent the growth of a populationof bacteria with an initial population of 200 bacteria and a growth constant of 0.02. Notice that after only 2 hours (120
minutes), the population is 10 times its original size!


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Figure 2.79 An example of exponential growth for bacteria.
Time (min) Population Size (no. of bacteria)
10 244


20 298


30 364


40 445


50 544


60 664


70 811


80 991


90 1210


100 1478


110 1805


120 2205


Table 2.1 Exponential Growth of a Bacterial Population
Note that we are using a continuous function to model what is inherently discrete behavior. At any given time, the real-worldpopulation contains a whole number of bacteria, although the model takes on noninteger values. When using exponential


Chapter 2 | Applications of Integration 233




2.42


growth models, we must always be careful to interpret the function values in the context of the phenomenon we aremodeling.
Example 2.42
Population Growth
Consider the population of bacteria described earlier. This population grows according to the function
f (t) = 200e0.02t, where t is measured in minutes. How many bacteria are present in the population after 5
hours (300 minutes)? When does the population reach 100,000 bacteria?
Solution
We have f (t) = 200e0.02t. Then


f (300) = 200e0.02(300) ≈ 80,686.


There are 80,686 bacteria in the population after 5 hours.
To find when the population reaches 100,000 bacteria, we solve the equation


100,000 = 200e0.02t


500 = e0.02t


ln 500 = 0.02t


t = ln 500
0.02


≈ 310.73.


The population reaches 100,000 bacteria after 310.73 minutes.


Consider a population of bacteria that grows according to the function f (t) = 500e0.05t, where t is
measured in minutes. How many bacteria are present in the population after 4 hours? When does the populationreach 100 million bacteria?


Let’s now turn our attention to a financial application: compound interest. Interest that is not compounded is called simpleinterest. Simple interest is paid once, at the end of the specified time period (usually 1 year). So, if we put $1000 in a
savings account earning 2% simple interest per year, then at the end of the year we have


1000(1 + 0.02) = $1020.


Compound interest is paid multiple times per year, depending on the compounding period. Therefore, if the bankcompounds the interest every 6 months, it credits half of the year’s interest to the account after 6 months. During the
second half of the year, the account earns interest not only on the initial $1000, but also on the interest earned during the
first half of the year. Mathematically speaking, at the end of the year, we have


1000⎛⎝1 +
0.02
2



2
= $1020.10.


Similarly, if the interest is compounded every 4 months, we have
1000⎛⎝1 +


0.02
3



3
= $1020.13,


and if the interest is compounded daily (365 times per year), we have $1020.20. If we extend this concept, so that the
interest is compounded continuously, after t years we have


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2.43


1000 lim
n → ∞



⎝1 +


0.02
n



nt
.


Now let’s manipulate this expression so that we have an exponential growth function. Recall that the number e can be
expressed as a limit:


e = lim
m → ∞



⎝1 +


1
m



m
.


Based on this, we want the expression inside the parentheses to have the form (1 + 1/m). Let n = 0.02m. Note that as
n → ∞, m → ∞ as well. Then we get


1000 lim
n → ∞



⎝1 +


0.02
n



nt
= 1000 lim


m → ∞

⎝1 +


0.02
0.02m





0.02mt
= 1000



⎣ limm → ∞



⎝1 +


1
m



m⎤


0.02t


.


We recognize the limit inside the brackets as the number e. So, the balance in our bank account after t years is given by
1000e0.02t. Generalizing this concept, we see that if a bank account with an initial balance of $P earns interest at a rate
of r%, compounded continuously, then the balance of the account after t years is


Balance = Pert.


Example 2.43
Compound Interest
A 25-year-old student is offered an opportunity to invest some money in a retirement account that pays 5%
annual interest compounded continuously. How much does the student need to invest today to have $1 million
when she retires at age 65? What if she could earn 6% annual interest compounded continuously instead?
Solution
We have


1,000,000 = Pe0.05(40)


P = 135,335.28.


She must invest $135,335.28 at 5% interest.
If, instead, she is able to earn 6%, then the equation becomes


1,000,000 = Pe0.06(40)


P = 90,717.95.


In this case, she needs to invest only $90,717.95. This is roughly two-thirds the amount she needs to invest at
5%. The fact that the interest is compounded continuously greatly magnifies the effect of the 1% increase in
interest rate.


Suppose instead of investing at age 25 b2 − 4ac, the student waits until age 35. How much would
she have to invest at 5%? At 6%?


If a quantity grows exponentially, the time it takes for the quantity to double remains constant. In other words, it takes thesame amount of time for a population of bacteria to grow from 100 to 200 bacteria as it does to grow from 10,000 to
20,000 bacteria. This time is called the doubling time. To calculate the doubling time, we want to know when the quantity


Chapter 2 | Applications of Integration 235




2.44


reaches twice its original size. So we have
2y0 = y0 e


kt


2 = ekt


ln 2 = kt


t = ln 2
k


.


Definition
If a quantity grows exponentially, the doubling time is the amount of time it takes the quantity to double. It is givenby


Doubling time = ln 2
k


.


Example 2.44
Using the Doubling Time
Assume a population of fish grows exponentially. A pond is stocked initially with 500 fish. After 6 months,
there are 1000 fish in the pond. The owner will allow his friends and neighbors to fish on his pond after the fish
population reaches 10,000. When will the owner’s friends be allowed to fish?
Solution
We know it takes the population of fish 6 months to double in size. So, if t represents time in months,
by the doubling-time formula, we have 6 = (ln 2)/k. Then, k = (ln 2)/6. Thus, the population is given by
y = 500e



⎝(ln 2)/6⎞⎠t. To figure out when the population reaches 10,000 fish, we must solve the following


equation:
10,000 = 500e(ln 2/6)t


20 = e(ln 2/6)t


ln 20 = ⎛⎝
ln 2
6

⎠t


t = 6(ln 20)
ln 2


≈ 25.93.


The owner’s friends have to wait 25.93 months (a little more than 2 years) to fish in the pond.


Suppose it takes 9 months for the fish population in Example 2.44 to reach 1000 fish. Under these
circumstances, how long do the owner’s friends have to wait?


Exponential Decay Model
Exponential functions can also be used to model populations that shrink (from disease, for example), or chemicalcompounds that break down over time. We say that such systems exhibit exponential decay, rather than exponential growth.The model is nearly the same, except there is a negative sign in the exponent. Thus, for some positive constant k, we have
y = y0 e


−kt.


As with exponential growth, there is a differential equation associated with exponential decay. We have


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y′ = −ky0 e
−kt = −ky.


Rule: Exponential Decay Model
Systems that exhibit exponential decay behave according to the model


y = y0 e
−kt,


where y0 represents the initial state of the system and k > 0 is a constant, called the decay constant.


The following figure shows a graph of a representative exponential decay function.


Figure 2.80 An example of exponential decay.


Let’s look at a physical application of exponential decay. Newton’s law of cooling says that an object cools at a rateproportional to the difference between the temperature of the object and the temperature of the surroundings. In other words,if T represents the temperature of the object and Ta represents the ambient temperature in a room, then
T′ = −k(T − Ta).


Note that this is not quite the right model for exponential decay. We want the derivative to be proportional to the function,and this expression has the additional Ta term. Fortunately, we can make a change of variables that resolves this issue. Let
y(t) = T(t) − Ta. Then y′(t) = T′(t) − 0 = T′(t), and our equation becomes


y′ = −ky.


From our previous work, we know this relationship between y and its derivative leads to exponential decay. Thus,
y = y0 e


−kt,


and we see that
T − Ta =



⎝T0 − Ta



⎠e−kt


T = ⎛⎝T0 − Ta

⎠e−kt + Ta


where T0 represents the initial temperature. Let’s apply this formula in the following example.
Example 2.45
Newton’s Law of Cooling


Chapter 2 | Applications of Integration 237




2.45


According to experienced baristas, the optimal temperature to serve coffee is between 155°F and 175°F.
Suppose coffee is poured at a temperature of 200°F, and after 2 minutes in a 70°F room it has cooled to
180°F. When is the coffee first cool enough to serve? When is the coffee too cold to serve? Round answers to
the nearest half minute.
Solution
We have


T = ⎛⎝T0 − Ta

⎠e−kt + Ta


180 = (200 − 70)e−k(2) + 70


110 = 130e−2k


11
13


= e−2k


ln 11
13


= −2k


ln 11 − ln 13 = −2k


k = ln 13 − ln 11
2


.


Then, the model is
T = 130e(ln 11 − ln 13/2)t + 70.


The coffee reaches 175°F when
175 = 130e(ln 11 − ln 13/2)t + 70


105 = 130e(ln 11 − ln 13/2)t


21
26


= e(ln 11 − ln 13/2)t


ln 21
26


= ln 11 − ln 13
2


t


ln 21 − ln 26 = ln 11 − ln 13
2


t


t = 2(ln 21 − ln 26)
ln 11 − ln 13


≈ 2.56.


The coffee can be served about 2.5 minutes after it is poured. The coffee reaches 155°F at
155 = 130e(ln 11 − ln 13/2)t + 70


85 = 130e(ln 11 − ln 13)t


17
26


= e(ln 11 − ln 13)t


ln 17 − ln 26 = ⎛⎝
ln 11 − ln 13


2

⎠t


t = 2(ln 17 − ln 26)
ln 11 − ln 13


≈ 5.09.


The coffee is too cold to be served about 5 minutes after it is poured.


Suppose the room is warmer (75°F) and, after 2 minutes, the coffee has cooled only to 185°F. When
is the coffee first cool enough to serve? When is the coffee be too cold to serve? Round answers to the nearesthalf minute.


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Just as systems exhibiting exponential growth have a constant doubling time, systems exhibiting exponential decay have aconstant half-life. To calculate the half-life, we want to know when the quantity reaches half its original size. Therefore, wehave
y0
2


= y0 e
−kt


1
2


= e−kt


−ln 2 = −kt


t = ln 2
k


.


Note: This is the same expression we came up with for doubling time.
Definition
If a quantity decays exponentially, the half-life is the amount of time it takes the quantity to be reduced by half. It isgiven by


Half-life = ln 2
k


.


Example 2.46
Radiocarbon Dating
One of the most common applications of an exponential decay model is carbon dating. Carbon-14 decays (emits
a radioactive particle) at a regular and consistent exponential rate. Therefore, if we know how much carbon wasoriginally present in an object and how much carbon remains, we can determine the age of the object. The half-life of carbon-14 is approximately 5730 years—meaning, after that many years, half the material has converted
from the original carbon-14 to the new nonradioactive nitrogen-14. If we have 100 g carbon-14 today, how
much is left in 50 years? If an artifact that originally contained 100 g of carbon now contains 10 g of carbon,
how old is it? Round the answer to the nearest hundred years.
Solution
We have


5730 = ln 2
k


k = ln 2
5730


.


So, the model says
y = 100e−(ln 2/5730)t.


In 50 years, we have
y = 100e−(ln 2/5730)(50)


≈ 99.40.


Therefore, in 50 years, 99.40 g of carbon-14 remains.
To determine the age of the artifact, we must solve


Chapter 2 | Applications of Integration 239




2.46


10 = 100e−(ln 2/5730)t


1
10


= e−(ln 2/5730)t


t ≈ 19035.


The artifact is about 19,000 years old.


If we have 100 g of carbon-14, how much is left after. years? If an artifact that originally contained
100 g of carbon now contains 20g of carbon, how old is it? Round the answer to the nearest hundred years.


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2.8 EXERCISES
True or False? If true, prove it. If false, find the true answer.
348. The doubling time for y = ect is ⎛⎝ln (2)⎞⎠/⎛⎝ln (c)⎞⎠.
349. If you invest $500, an annual rate of interest of
3% yields more money in the first year than a 2.5%
continuous rate of interest.
350. If you leave a 100°C pot of tea at room temperature
(25°C) and an identical pot in the refrigerator (5°C),
with k = 0.02, the tea in the refrigerator reaches a
drinkable temperature (70°C) more than 5 minutes
before the tea at room temperature.
351. If given a half-life of t years, the constant k for
y = ekt is calculated by k = ln (1/2)/t.
For the following exercises, use y = y0 ekt.
352. If a culture of bacteria doubles in 3 hours, how many
hours does it take to multiply by 10?
353. If bacteria increase by a factor of 10 in 10 hours,
how many hours does it take to increase by 100?
354. How old is a skull that contains one-fifth as muchradiocarbon as a modern skull? Note that the half-life ofradiocarbon is 5730 years.
355. If a relic contains 90% as much radiocarbon as
new material, can it have come from the time of Christ(approximately 2000 years ago)? Note that the half-life of
radiocarbon is 5730 years.
356. The population of Cairo grew from 5 million to
10 million in 20 years. Use an exponential model to find
when the population was 8 million.
357. The populations of New York and Los Angeles aregrowing at 1% and 1.4% a year, respectively. Starting
from 8 million (New York) and 6 million (Los Angeles),
when are the populations equal?
358. Suppose the value of $1 in Japanese yen decreases
at 2% per year. Starting from $1 = ¥250, when will
$1 = ¥1?


359. The effect of advertising decays exponentially. If
40% of the population remembers a new product after 3
days, how long will 20% remember it?


360. If y = 1000 at t = 3 and y = 3000 at t = 4,
what was y0 at t = 0?
361. If y = 100 at t = 4 and y = 10 at t = 8, when
does y = 1?
362. If a bank offers annual interest of 7.5% or
continuous interest of 7.25%, which has a better annual
yield?
363. What continuous interest rate has the same yield asan annual rate of 9%?
364. If you deposit $5000 at 8% annual interest, how
many years can you withdraw $500 (starting after the first
year) without running out of money?
365. You are trying to save $50,000 in 20 years for
college tuition for your child. If interest is a continuous
10%, how much do you need to invest initially?
366. You are cooling a turkey that was taken out of theoven with an internal temperature of 165°F. After 10
minutes of resting the turkey in a 70°F apartment, the
temperature has reached 155°F. What is the temperature
of the turkey 20 minutes after taking it out of the oven?
367. You are trying to thaw some vegetables that areat a temperature of 1°F. To thaw vegetables safely, you
must put them in the refrigerator, which has an ambienttemperature of 44°F. You check on your vegetables 2
hours after putting them in the refrigerator to find that theyare now 12°F. Plot the resulting temperature curve and use
it to determine when the vegetables reach 33°F.
368. You are an archaeologist and are given a bone that isclaimed to be from a Tyrannosaurus Rex. You know thesedinosaurs lived during the Cretaceous Era (146 million
years to 65 million years ago), and you find by
radiocarbon dating that there is 0.000001% the amount of
radiocarbon. Is this bone from the Cretaceous?
369. The spent fuel of a nuclear reactor containsplutonium-239, which has a half-life of 24,000 years. If 1
barrel containing 10 kg of plutonium-239 is sealed, how
many years must pass until only 10g of plutonium-239 is
left?
For the next set of exercises, use the following table, whichfeatures the world population by decade.


Chapter 2 | Applications of Integration 241




Years since 1950 Population (millions)
0 2,556


10 3,039


20 3,706


30 4,453


40 5,279


50 6,083


60 6,849


Source: http://www.factmonster.com/ipka/A0762181.html.
370. [T] The best-fit exponential curve to the data of the
form P(t) = aebt is given by P(t) = 2686e0.01604t. Use
a graphing calculator to graph the data and the exponentialcurve together.
371. [T] Find and graph the derivative y′ of your
equation. Where is it increasing and what is the meaning ofthis increase?
372. [T] Find and graph the second derivative of yourequation. Where is it increasing and what is the meaning ofthis increase?
373. [T] Find the predicted date when the populationreaches 10 billion. Using your previous answers about
the first and second derivatives, explain why exponentialgrowth is unsuccessful in predicting the future.
For the next set of exercises, use the following table, whichshows the population of San Francisco during the 19thcentury.


Years since1850 Population(thousands)
0 21.00


10 56.80


20 149.5


30 234.0


Source: http://www.sfgenealogy.com/sf/history/hgpop.htm.
374. [T] The best-fit exponential curve to the data of the
form P(t) = aebt is given by P(t) = 35.26e0.06407t. Use
a graphing calculator to graph the data and the exponentialcurve together.
375. [T] Find and graph the derivative y′ of your
equation. Where is it increasing? What is the meaning ofthis increase? Is there a value where the increase ismaximal?
376. [T] Find and graph the second derivative of yourequation. Where is it increasing? What is the meaning ofthis increase?


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2.9 | Calculus of the Hyperbolic Functions
Learning Objectives


2.9.1 Apply the formulas for derivatives and integrals of the hyperbolic functions.
2.9.2 Apply the formulas for the derivatives of the inverse hyperbolic functions and theirassociated integrals.
2.9.3 Describe the common applied conditions of a catenary curve.


We were introduced to hyperbolic functions in Introduction to Functions and Graphs (http://cnx.org/content/m53472/latest/) , along with some of their basic properties. In this section, we look at differentiation and integrationformulas for the hyperbolic functions and their inverses.
Derivatives and Integrals of the Hyperbolic Functions
Recall that the hyperbolic sine and hyperbolic cosine are defined as


sinh x = e
x − e−x


2
and cosh x = e


x + e−x


2
.


The other hyperbolic functions are then defined in terms of sinh x and cosh x. The graphs of the hyperbolic functions are
shown in the following figure.


Figure 2.81 Graphs of the hyperbolic functions.


It is easy to develop differentiation formulas for the hyperbolic functions. For example, looking at sinh x we have


Chapter 2 | Applications of Integration 243




d
dx


(sinh x) = d
dx


ex − e−x


2



= 1
2


d
dx


(ex) − d
dx


(e−x)⎤⎦


= 1
2
[ex + e−x] = cosh x.


Similarly, (d/dx)cosh x = sinh x. We summarize the differentiation formulas for the hyperbolic functions in the following
table.


f(x)
d
dx


f(x)


sinh x cosh x


cosh x sinh x


tanh x sech2 x


coth x −csch2 x


sech x −sech x tanh x


csch x −csch x coth x


Table 2.2 Derivatives of theHyperbolic Functions
Let’s take a moment to compare the derivatives of the hyperbolic functions with the derivatives of the standardtrigonometric functions. There are a lot of similarities, but differences as well. For example, the derivatives of the sinefunctions match: (d/dx)sin x = cos x and (d/dx)sinh x = cosh x. The derivatives of the cosine functions, however, differ
in sign: (d/dx)cos x = −sin x, but (d/dx)cosh x = sinh x. As we continue our examination of the hyperbolic functions,
we must be mindful of their similarities and differences to the standard trigonometric functions.
These differentiation formulas for the hyperbolic functions lead directly to the following integral formulas.


∫ sinh u du = cosh u + C ∫ csch2 u du = −coth u + C


∫ cosh u du = sinh u + C ∫ sech u tanh u du = −sech u + C


∫ sech2u du = tanh u + C ∫ csch u coth u du = −csch u + C


Example 2.47
Differentiating Hyperbolic Functions
Evaluate the following derivatives:


a. d
dx

⎝sinh



⎝x


2⎞




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2.47


b. d
dx


(cosh x)2


Solution
Using the formulas in Table 2.2 and the chain rule, we get


a. d
dx

⎝sinh



⎝x


2⎞


⎠ = cosh



⎝x


2⎞
⎠ · 2x


b. d
dx


(cosh x)2 = 2 cosh x sinh x


Evaluate the following derivatives:
a. d


dx

⎝tanh



⎝x


2 + 3x⎞⎠



b. d
dx





⎜ 1
(sinh x)2







Example 2.48
Integrals Involving Hyperbolic Functions
Evaluate the following integrals:


a. ∫ x cosh⎛⎝x2⎞⎠dx
b. ∫ tanh x dx


Solution
We can use u-substitution in both cases.


a. Let u = x2. Then, du = 2x dx and
∫ x cosh⎛⎝x2⎞⎠dx = ∫ 12cosh u du =


1
2
sinh u + C = 1


2
sinh⎛⎝x


2⎞
⎠+ C.


b. Let u = cosh x. Then, du = sinh x dx and
∫ tanh x dx = ∫ sinh x


cosh x
dx = ∫ 1udu = ln|u| + C = ln|cosh x| + C.


Note that cosh x > 0 for all x, so we can eliminate the absolute value signs and obtain
∫ tanh x dx = ln(cosh x) + C.


Chapter 2 | Applications of Integration 245




2.48 Evaluate the following integrals:
a. ∫ sinh3 x cosh x dx
b. ∫ sech2 (3x)dx


Calculus of Inverse Hyperbolic Functions
Looking at the graphs of the hyperbolic functions, we see that with appropriate range restrictions, they all have inverses.Most of the necessary range restrictions can be discerned by close examination of the graphs. The domains and ranges ofthe inverse hyperbolic functions are summarized in the following table.


Function Domain Range
sinh−1 x (−∞, ∞) (−∞, ∞)


cosh−1 x (1, ∞)

⎣0, ∞)


tanh−1 x (−1, 1) (−∞, ∞)


coth−1 x (−∞, −1) ∪ (1, ∞) (−∞, 0) ∪ (0, ∞)


sech−1 x (0, 1)

⎣0, ∞)


csch−1 x (−∞, 0) ∪ (0, ∞) (−∞, 0) ∪ (0, ∞)


Table 2.3 Domains and Ranges of the Inverse HyperbolicFunctions
The graphs of the inverse hyperbolic functions are shown in the following figure.


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Figure 2.82 Graphs of the inverse hyperbolic functions.


To find the derivatives of the inverse functions, we use implicit differentiation. We have
y = sinh−1 x


sinh y = x
d
dx


sinh y = d
dx


x


cosh y
dy
dx


= 1.


Recall that cosh2 y − sinh2 y = 1, so cosh y = 1 + sinh2 y. Then,
dy
dx


= 1
cosh y


= 1
1 + sinh2 y


= 1
1 + x2


.


We can derive differentiation formulas for the other inverse hyperbolic functions in a similar fashion. These differentiationformulas are summarized in the following table.


Chapter 2 | Applications of Integration 247




f(x)
d
dx


f(x)


sinh−1 x


1
1 + x2


cosh−1 x
1


x2 − 1


tanh−1 x
1


1 − x2


coth−1 x
1


1 − x2


sech−1 x
−1


x 1 − x2


csch−1 x


−1


|x| 1 + x2


Table 2.4 Derivatives of theInverse Hyperbolic Functions
Note that the derivatives of tanh−1 x and coth−1 x are the same. Thus, when we integrate 1/⎛⎝1 − x2⎞⎠, we need to select
the proper antiderivative based on the domain of the functions and the values of x. Integration formulas involving the
inverse hyperbolic functions are summarized as follows.


∫ 1
1 + u2


du = sinh−1u + C ∫ 1
u 1 − u2


du = −sech−1 |u| + C


∫ 1
u2 − 1


du = cosh−1u + C ∫ 1
u 1 + u2


du = −csch−1 |u| + C


∫ 1
1 − u2


du =



⎨tanh


−1u + C if |u| < 1


coth−1u + C if |u| > 1


Example 2.49
Differentiating Inverse Hyperbolic Functions
Evaluate the following derivatives:


a. d
dx

⎝sinh


−1 ⎛

x
3





b. d
dx

⎝tanh


−1 x⎞⎠
2


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2.49


2.50


Solution
Using the formulas in Table 2.4 and the chain rule, we obtain the following results:


a. d
dx

⎝sinh


−1 ⎛

x
3



⎠ =


1


3 1 + x
2


9


= 1
9 + x2


b. d
dx

⎝tanh


−1 x⎞⎠
2
=


2⎛⎝tanh
−1 x⎞⎠


1 − x2


Evaluate the following derivatives:
a. d


dx

⎝cosh


−1 (3x)⎞⎠


b. d
dx

⎝coth


−1 x⎞⎠
3


Example 2.50
Integrals Involving Inverse Hyperbolic Functions
Evaluate the following integrals:


a. ∫ 1
4x2 − 1


dx


b. ∫ 1
2x 1 − 9x2


dx


Solution
We can use u-substitution in both cases.


a. Let u = 2x. Then, du = 2dx and we have
∫ 1


4x2 − 1
dx = ∫ 1


2 u2 − 1
du = 1


2
cosh−1u + C = 1


2
cosh−1 (2x) + C.


b. Let u = 3x. Then, du = 3dx and we obtain
∫ 1


2x 1 − 9x2
dx = 1


2
∫ 1
u 1 − u2


du = − 1
2
sech−1 |u| + C = − 12


sech−1 |3x| + C.


Evaluate the following integrals:
a. ∫ 1


x2 − 4
dx, x > 2


b. ∫ 1
1 − e2x


dx


Chapter 2 | Applications of Integration 249




Applications
One physical application of hyperbolic functions involves hanging cables. If a cable of uniform density is suspendedbetween two supports without any load other than its own weight, the cable forms a curve called a catenary. High-voltagepower lines, chains hanging between two posts, and strands of a spider’s web all form catenaries. The following figureshows chains hanging from a row of posts.


Figure 2.83 Chains between these posts take the shape of a catenary. (credit: modification of work by OKFoundryCompany,Flickr)


Hyperbolic functions can be used to model catenaries. Specifically, functions of the form y = a cosh(x/a) are catenaries.
Figure 2.84 shows the graph of y = 2 cosh(x/2).


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2.51


Figure 2.84 A hyperbolic cosine function forms the shape ofa catenary.


Example 2.51
Using a Catenary to Find the Length of a Cable
Assume a hanging cable has the shape 10 cosh(x/10) for −15 ≤ x ≤ 15, where x is measured in feet.
Determine the length of the cable (in feet).
Solution
Recall from Section 6.4 that the formula for arc length is


Arc Length = ∫
a


b
1 + ⎡⎣ f ′(x)⎤⎦2 dx.


We have f (x) = 10 cosh(x/10), so f ′(x) = sinh(x/10). Then
Arc Length = ∫


a


b
1 + ⎡⎣ f ′(x)⎤⎦2 dx


= ∫
−15


15
1 + sinh2 ⎛⎝


x
10

⎠dx.


Now recall that 1 + sinh2 x = cosh2 x, so we have
Arc Length = ∫


−15


15
1 + sinh2 ⎛⎝


x
10

⎠dx


= ∫
−15


15
cosh⎛⎝


x
10

⎠dx


= 10 sinh⎛⎝
x
10

⎠|−15
15


= 10⎡⎣sinh


3
2

⎠− sinh



⎝−


3
2



⎦ = 20 sinh




3
2



≈ 42.586 ft.


Assume a hanging cable has the shape 15 cosh(x/15) for −20 ≤ x ≤ 20. Determine the length of the
cable (in feet).


Chapter 2 | Applications of Integration 251




2.9 EXERCISES
377. [T] Find expressions for cosh x + sinh x and
cosh x − sinh x. Use a calculator to graph these functions
and ensure your expression is correct.
378. From the definitions of cosh(x) and sinh(x), find
their antiderivatives.
379. Show that cosh(x) and sinh(x) satisfy y″ = y.
380. Use the quotient rule to verify that
tanh(x)′ = sech2 (x).


381. Derive cosh2 (x) + sinh2 (x) = cosh(2x) from the
definition.
382. Take the derivative of the previous expression to findan expression for sinh(2x).
383. Prove
sinh(x + y) = sinh(x)cosh(y) + cosh(x)sinh(y) by
changing the expression to exponentials.
384. Take the derivative of the previous expression to findan expression for cosh(x + y).
For the following exercises, find the derivatives of thegiven functions and graph along with the function to ensureyour answer is correct.
385. [T] cosh(3x + 1)
386. [T] sinh⎛⎝x2⎞⎠
387. [T] 1


cosh(x)


388. [T] sinh⎛⎝ln(x)⎞⎠
389. [T] cosh2 (x) + sinh2 (x)
390. [T] cosh2 (x) − sinh2 (x)
391. [T] tanh⎛⎝ x2 + 1⎞⎠
392. [T] 1 + tanh(x)


1 − tanh(x)


393. [T] sinh6 (x)
394. [T] ln⎛⎝sech(x) + tanh(x)⎞⎠


For the following exercises, find the antiderivatives for thegiven functions.
395. cosh(2x + 1)
396. tanh(3x + 2)
397. x cosh⎛⎝x2⎞⎠
398. 3x3 tanh⎛⎝x4⎞⎠
399. cosh2 (x)sinh(x)
400. tanh2 (x)sech2 (x)
401. sinh(x)


1 + cosh(x)


402. coth(x)
403. cosh(x) + sinh(x)
404. ⎛⎝cosh(x) + sinh(x)⎞⎠n
For the following exercises, find the derivatives for thefunctions.
405. tanh−1 (4x)
406. sinh−1 ⎛⎝x2⎞⎠
407. sinh−1 ⎛⎝cosh(x)⎞⎠
408. cosh−1 ⎛⎝x3⎞⎠
409. tanh−1 (cos(x))


410. esinh−1 (x)
411. ln⎛⎝tanh−1 (x)⎞⎠
For the following exercises, find the antiderivatives for thefunctions.
412. ∫ dx


4 − x2


413. ∫ dx
a2 − x2


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414. ∫ dx
x2 + 1


415. ∫ x dx
x2 + 1


416. ∫ − dx
x 1 − x2


417. ∫ ex
e2x − 1


418. ∫ − 2x
x4 − 1


For the following exercises, use the fact that a falling bodywith friction equal to velocity squared obeys the equation
dv/dt = g − v2.


419. Show that v(t) = g tanh⎛⎝ gt⎞⎠ satisfies this
equation.
420. Derive the previous expression for v(t) by
integrating dv


g − v2
= dt.


421. [T] Estimate how far a body has fallen in 12 seconds
by finding the area underneath the curve of v(t).
For the following exercises, use this scenario: A cablehanging under its own weight has a slope S = dy/dx that
satisfies dS/dx = c 1 + S2. The constant c is the ratio of
cable density to tension.
422. Show that S = sinh(cx) satisfies this equation.
423. Integrate dy/dx = sinh(cx) to find the cable height
y(x) if y(0) = 1/c.
424. Sketch the cable and determine how far down it sagsat x = 0.
For the following exercises, solve each problem.
425. [T] A chain hangs from two posts 2 m apart to form
a catenary described by the equation y = 2 cosh(x/2) − 1.
Find the slope of the catenary at the left fence post.
426. [T] A chain hangs from two posts four meters apartto form a catenary described by the equation
y = 4 cosh(x/4) − 3. Find the total length of the catenary
(arc length).


427. [T] A high-voltage power line is a catenary describedby y = 10 cosh(x/10). Find the ratio of the area under the
catenary to its arc length. What do you notice?
428. A telephone line is a catenary described by
y = a cosh(x/a). Find the ratio of the area under the
catenary to its arc length. Does this confirm your answerfor the previous question?
429. Prove the formula for the derivative of
y = sinh−1(x) by differentiating x = sinh(y). (Hint: Use
hyperbolic trigonometric identities.)
430. Prove the formula for the derivative of
y = cosh−1(x) by differentiating x = cosh(y). (Hint:
Use hyperbolic trigonometric identities.)
431. Prove the formula for the derivative of
y = sech−1(x) by differentiating x = sech(y). (Hint: Use
hyperbolic trigonometric identities.)
432. Prove that

⎝cosh(x) + sinh(x)⎞⎠n = cosh(nx) + sinh(nx).


433. Prove the expression for sinh−1 (x). Multiply
x = sinh(y) = (1/2)⎛⎝e


y − e
−y⎞
⎠ by 2ey and solve for y.


Does your expression match the textbook?
434. Prove the expression for cosh−1 (x). Multiply
x = cosh(y) = (1/2)⎛⎝e


y − e
−y⎞
⎠ by 2ey and solve for y.


Does your expression match the textbook?


Chapter 2 | Applications of Integration 253




arc length
catenary
center of mass
centroid


cross-section
density function


disk method
doubling time
exponential decay
exponential growth
frustum
half-life
Hooke’s law
hydrostatic pressure
lamina
method of cylindrical shells


moment


slicing method


solid of revolution


CHAPTER 2 REVIEW
KEY TERMS


the arc length of a curve can be thought of as the distance a person would travel along the path of the curve
a curve in the shape of the function y = a cosh(x/a) is a catenary; a cable of uniform density suspended


between two supports assumes the shape of a catenary
the point at which the total mass of the system could be concentrated without changing the moment


the centroid of a region is the geometric center of the region; laminas are often represented by regions in theplane; if the lamina has a constant density, the center of mass of the lamina depends only on the shape of thecorresponding planar region; in this case, the center of mass of the lamina corresponds to the centroid of therepresentative region
the intersection of a plane and a solid object
a density function describes how mass is distributed throughout an object; it can be a linear density,expressed in terms of mass per unit length; an area density, expressed in terms of mass per unit area; or a volumedensity, expressed in terms of mass per unit volume; weight-density is also used to describe weight (rather than mass)per unit volume


a special case of the slicing method used with solids of revolution when the slices are disks
if a quantity grows exponentially, the doubling time is the amount of time it takes the quantity to double,and is given by (ln 2)/k


systems that exhibit exponential decay follow a model of the form y = y0 e−kt
systems that exhibit exponential growth follow a model of the form y = y0 ekt


a portion of a cone; a frustum is constructed by cutting the cone with a plane parallel to the base
if a quantity decays exponentially, the half-life is the amount of time it takes the quantity to be reduced by half. Itis given by (ln 2)/k


this law states that the force required to compress (or elongate) a spring is proportional to the distance thespring has been compressed (or stretched) from equilibrium; in other words, F = kx, where k is a constant
the pressure exerted by water on a submerged object


a thin sheet of material; laminas are thin enough that, for mathematical purposes, they can be treated as if they aretwo-dimensional
a method of calculating the volume of a solid of revolution by dividing the solid intonested cylindrical shells; this method is different from the methods of disks or washers in that we integrate withrespect to the opposite variable


if n masses are arranged on a number line, the moment of the system with respect to the origin is given by
M = ∑


i = 1


n


mi xi; if, instead, we consider a region in the plane, bounded above by a function f (x) over an interval

⎣a, b⎤⎦, then the moments of the region with respect to the x- and y-axes are given by Mx = ρ∫


a


b ⎡
⎣ f (x)⎤⎦2


2
dx and


My = ρ∫
a


b
x f (x)dx, respectively
a method of calculating the volume of a solid that involves cutting the solid into pieces, estimating thevolume of each piece, then adding these estimates to arrive at an estimate of the total volume; as the number of slicesgoes to infinity, this estimate becomes an integral that gives the exact value of the volume


a solid generated by revolving a region in a plane around a line in that plane


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surface area
symmetry principle
theorem of Pappus for volume


washer method
work


the surface area of a solid is the total area of the outer layer of the object; for objects such as cubes orbricks, the surface area of the object is the sum of the areas of all of its faces
the symmetry principle states that if a region R is symmetric about a line l, then the centroid of Rlies on l


this theorem states that the volume of a solid of revolution formed by revolving aregion around an external axis is equal to the area of the region multiplied by the distance traveled by the centroid ofthe region
a special case of the slicing method used with solids of revolution when the slices are washers


the amount of energy it takes to move an object; in physics, when a force is constant, work is expressed as theproduct of force and distance
KEY EQUATIONS


• Area between two curves, integrating on the x-axis
A = ∫


a


b

⎣ f (x) − g(x)⎤⎦dx


• Area between two curves, integrating on the y-axis
A = ∫


c


d

⎣u(y) − v(y)⎤⎦dy


• Disk Method along the x-axis
V = ∫


a


b
π⎡⎣ f (x)⎤⎦2dx


• Disk Method along the y-axis
V = ∫


c


d
π⎡⎣g(y)⎤⎦2dy


• Washer Method
V = ∫


a


b
π⎡⎣

⎝ f (x)⎞⎠2 − ⎛⎝g(x)⎞⎠2⎤⎦dx


• Method of Cylindrical Shells
V = ∫


a


b

⎝2πx f (x)⎞⎠dx


• Arc Length of a Function of x
Arc Length = ∫


a


b
1 + ⎡⎣ f ′(x)⎤⎦2 dx


• Arc Length of a Function of y
Arc Length = ∫


c


d
1 + ⎡⎣g′(y)⎤⎦2 dy


• Surface Area of a Function of x
Surface Area = ∫


a


b⎛
⎝2π f (x) 1 +



⎝ f ′(x)⎞⎠2



⎠dx


• Mass of a one-dimensional object
m = ∫


a


b
ρ(x)dx


• Mass of a circular object
m = ∫


0


r
2πxρ(x)dx


Chapter 2 | Applications of Integration 255




• Work done on an object
W = ∫


a


b
F(x)dx


• Hydrostatic force on a plate
F = ∫


a


b
ρw(x)s(x)dx


• Mass of a lamina
m = ρ∫


a


b
f (x)dx


• Moments of a lamina
Mx = ρ∫


a


b ⎡
⎣ f (x)⎤⎦2


2
dx andMy = ρ∫


a


b
x f (x)dx


• Center of mass of a lamina
x– =


My
m and y


– = Mxm


• Natural logarithm function
• ln x = ∫


1


x
1
t dt Z


• Exponential function y = ex
• ln y = ln(ex) = x Z


KEY CONCEPTS
2.1 Areas between Curves


• Just as definite integrals can be used to find the area under a curve, they can also be used to find the area betweentwo curves.
• To find the area between two curves defined by functions, integrate the difference of the functions.
• If the graphs of the functions cross, or if the region is complex, use the absolute value of the difference of thefunctions. In this case, it may be necessary to evaluate two or more integrals and add the results to find the area ofthe region.
• Sometimes it can be easier to integrate with respect to y to find the area. The principles are the same regardless ofwhich variable is used as the variable of integration.


2.2 Determining Volumes by Slicing
• Definite integrals can be used to find the volumes of solids. Using the slicing method, we can find a volume byintegrating the cross-sectional area.
• For solids of revolution, the volume slices are often disks and the cross-sections are circles. The method of disksinvolves applying the method of slicing in the particular case in which the cross-sections are circles, and using theformula for the area of a circle.
• If a solid of revolution has a cavity in the center, the volume slices are washers. With the method of washers, thearea of the inner circle is subtracted from the area of the outer circle before integrating.


2.3 Volumes of Revolution: Cylindrical Shells
• The method of cylindrical shells is another method for using a definite integral to calculate the volume of a solid ofrevolution. This method is sometimes preferable to either the method of disks or the method of washers because weintegrate with respect to the other variable. In some cases, one integral is substantially more complicated than the


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other.
• The geometry of the functions and the difficulty of the integration are the main factors in deciding which integrationmethod to use.


2.4 Arc Length of a Curve and Surface Area
• The arc length of a curve can be calculated using a definite integral.
• The arc length is first approximated using line segments, which generates a Riemann sum. Taking a limit then givesus the definite integral formula. The same process can be applied to functions of y.
• The concepts used to calculate the arc length can be generalized to find the surface area of a surface of revolution.
• The integrals generated by both the arc length and surface area formulas are often difficult to evaluate. It may benecessary to use a computer or calculator to approximate the values of the integrals.


2.5 Physical Applications
• Several physical applications of the definite integral are common in engineering and physics.
• Definite integrals can be used to determine the mass of an object if its density function is known.
• Work can also be calculated from integrating a force function, or when counteracting the force of gravity, as in apumping problem.
• Definite integrals can also be used to calculate the force exerted on an object submerged in a liquid.


2.6 Moments and Centers of Mass
• Mathematically, the center of mass of a system is the point at which the total mass of the system could beconcentrated without changing the moment. Loosely speaking, the center of mass can be thought of as the balancingpoint of the system.
• For point masses distributed along a number line, the moment of the system with respect to the origin is


M = ∑
i = 1


n


mi xi. For point masses distributed in a plane, the moments of the system with respect to the x- and


y-axes, respectively, are Mx = ∑
i = 1


n


mi yi and My = ∑
i = 1


n


mi xi, respectively.
• For a lamina bounded above by a function f (x), the moments of the system with respect to the x- and y-axes,
respectively, are Mx = ρ∫


a


b ⎡
⎣ f (x)⎤⎦2


2
dx and My = ρ∫


a


b
x f (x)dx.


• The x- and y-coordinates of the center of mass can be found by dividing the moments around the y-axis and aroundthe x-axis, respectively, by the total mass. The symmetry principle says that if a region is symmetric with respect toa line, then the centroid of the region lies on the line.
• The theorem of Pappus for volume says that if a region is revolved around an external axis, the volume of theresulting solid is equal to the area of the region multiplied by the distance traveled by the centroid of the region.


2.7 Integrals, Exponential Functions, and Logarithms
• The earlier treatment of logarithms and exponential functions did not define the functions precisely and formally.This section develops the concepts in a mathematically rigorous way.
• The cornerstone of the development is the definition of the natural logarithm in terms of an integral.
• The function ex is then defined as the inverse of the natural logarithm.
• General exponential functions are defined in terms of ex, and the corresponding inverse functions are general
logarithms.


Chapter 2 | Applications of Integration 257




• Familiar properties of logarithms and exponents still hold in this more rigorous context.
2.8 Exponential Growth and Decay


• Exponential growth and exponential decay are two of the most common applications of exponential functions.
• Systems that exhibit exponential growth follow a model of the form y = y0 ekt.
• In exponential growth, the rate of growth is proportional to the quantity present. In other words, y′ = ky.
• Systems that exhibit exponential growth have a constant doubling time, which is given by (ln 2)/k.
• Systems that exhibit exponential decay follow a model of the form y = y0 e−kt.
• Systems that exhibit exponential decay have a constant half-life, which is given by (ln 2)/k.


2.9 Calculus of the Hyperbolic Functions
• Hyperbolic functions are defined in terms of exponential functions.
• Term-by-term differentiation yields differentiation formulas for the hyperbolic functions. These differentiationformulas give rise, in turn, to integration formulas.
• With appropriate range restrictions, the hyperbolic functions all have inverses.
• Implicit differentiation yields differentiation formulas for the inverse hyperbolic functions, which in turn give riseto integration formulas.
• The most common physical applications of hyperbolic functions are calculations involving catenaries.


CHAPTER 2 REVIEW EXERCISES
True or False? Justify your answer with a proof or acounterexample.
435. The amount of work to pump the water out of a half-full cylinder is half the amount of work to pump the waterout of the full cylinder.
436. If the force is constant, the amount of work to movean object from x = a to x = b is F(b − a).
437. The disk method can be used in any situation inwhich the washer method is successful at finding thevolume of a solid of revolution.
438. If the half-life of seaborgium-266 is 360 ms, then
k = ⎛⎝ln(2)⎞⎠/360.


For the following exercises, use the requested method todetermine the volume of the solid.
439. The volume that has a base of the ellipse
x2/4 + y2/9 = 1 and cross-sections of an equilateral
triangle perpendicular to the y-axis. Use the method of
slicing.
440. y = x2 − x, from x = 1 to x = 4, rotated around
they-axis using the washer method


441. x = y2 and x = 3y rotated around the y-axis using
the washer method
442. x = 2y2 − y3, x = 0, and y = 0 rotated around the
x-axis using cylindrical shells
For the following exercises, find


a. the area of the region,
b. the volume of the solid when rotated around thex-axis, and
c. the volume of the solid when rotated around they-axis. Use whichever method seems mostappropriate to you.


443. y = x3, x = 0, y = 0, and x = 2


444. y = x2 − x and x = 0
445. [T] y = ln(x) + 2 and y = x


446. y = x2 and y = x


447. y = 5 + x, y = x2, x = 0, and x = 1


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448. Below x2 + y2 = 1 and above y = 1 − x
449. Find the mass of ρ = e−x on a disk centered at the
origin with radius 4.
450. Find the center of mass for ρ = tan2 x on
x ∈ ⎛⎝−


π
4
, π
4

⎠.


451. Find the mass and the center of mass of ρ = 1 on
the region bounded by y = x5 and y = x.
For the following exercises, find the requested arc lengths.
452. The length of x for y = cosh(x) from
x = 0 to x = 2.


453. The length of y for x = 3 − y from y = 0 to
y = 4


For the following exercises, find the surface area andvolume when the given curves are revolved around thespecified axis.
454. The shape created by revolving the region between
y = 4 + x, y = 3 − x, x = 0, and x = 2 rotated
around the y-axis.
455. The loudspeaker created by revolving y = 1/x from
x = 1 to x = 4 around the x-axis.
For the following exercises, consider the Karun-3 dam inIran. Its shape can be approximated as an isosceles trianglewith height 205 m and width 388 m. Assume the current
depth of the water is 180 m. The density of water is 1000
kg/m 3.
456. Find the total force on the wall of the dam.
457. You are a crime scene investigator attempting todetermine the time of death of a victim. It is noon and
45°F outside and the temperature of the body is 78°F.
You know the cooling constant is k = 0.00824°F/min.
When did the victim die, assuming that a human’stemperature is 98°F ?
For the following exercise, consider the stock market crashin 1929 in the United States. The table lists the Dow Jones
industrial average per year leading up to the crash.


Years after 1920 Value ($)
1 63.90


3 100


5 110


7 160


9 381.17


Source: http://stockcharts.com/freecharts/historical/djia19201940.html
458. [T] The best-fit exponential curve to these data isgiven by y = 40.71 + 1.224x. Why do you think the gains
of the market were unsustainable? Use first and secondderivatives to help justify your answer. What would thismodel predict the Dow Jones industrial average to be in
2014 ?
For the following exercises, consider the catenoid, the onlysolid of revolution that has a minimal surface, or zeromean curvature. A catenoid in nature can be found whenstretching soap between two rings.
459. Find the volume of the catenoid y = cosh(x) from
x = −1 to x = 1 that is created by rotating this curve
around the x-axis, as shown here.


460. Find surface area of the catenoid y = cosh(x) from
x = −1 to x = 1 that is created by rotating this curve
around the x -axis.


Chapter 2 | Applications of Integration 259




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3 | TECHNIQUES OFINTEGRATION


Figure 3.1 Careful planning of traffic signals can prevent or reduce the number of accidents at busy intersections. (credit:modification of work by David McKelvey, Flickr)


Chapter Outline
3.1 Integration by Parts
3.2 Trigonometric Integrals
3.3 Trigonometric Substitution
3.4 Partial Fractions
3.5 Other Strategies for Integration
3.6 Numerical Integration
3.7 Improper Integrals


Introduction
In a large city, accidents occurred at an average rate of one every three months at a particularly busy intersection. Afterresidents complained, changes were made to the traffic lights at the intersection. It has now been eight months since thechanges were made and there have been no accidents. Were the changes effective or is the eight-month interval withoutan accident a result of chance? We explore this question later in this chapter and see that integration is an essential part ofdetermining the answer (see Example 3.49).
We saw in the previous chapter how important integration can be for all kinds of different topics—from calculations ofvolumes to flow rates, and from using a velocity function to determine a position to locating centers of mass. It is nosurprise, then, that techniques for finding antiderivatives (or indefinite integrals) are important to know for everyone who


Chapter 3 | Techniques of Integration 261




uses them. We have already discussed some basic integration formulas and the method of integration by substitution. Inthis chapter, we study some additional techniques, including some ways of approximating definite integrals when normaltechniques do not work.
3.1 | Integration by Parts


Learning Objectives
3.1.1 Recognize when to use integration by parts.
3.1.2 Use the integration-by-parts formula to solve integration problems.
3.1.3 Use the integration-by-parts formula for definite integrals.


By now we have a fairly thorough procedure for how to evaluate many basic integrals. However, although we can integrate
∫ xsin(x2)dx by using the substitution, u = x2, something as simple looking as ∫ xsinx dx defies us. Many students
want to know whether there is a product rule for integration. There isn’t, but there is a technique based on the product rulefor differentiation that allows us to exchange one integral for another. We call this technique integration by parts.
The Integration-by-Parts Formula
If, h(x) = f (x)g(x), then by using the product rule, we obtain h′(x) = f ′(x)g(x) + g′(x) f (x). Although at first it may
seem counterproductive, let’s now integrate both sides of this equation: ∫ h′(x)dx = ∫ ⎛⎝g(x) f ′(x) + f (x)g′(x)⎞⎠dx.
This gives us


h(x) = f (x)g(x) = ∫ g(x) f ′(x)dx + ∫ f (x)g′(x)dx.


Now we solve for ∫ f (x)g′(x)dx :
∫ f (x)g′(x)dx = f (x)g(x) − ∫ g(x) f ′(x)dx.


By making the substitutions u = f (x) and v = g(x), which in turn make du = f ′(x)dx and dv = g′(x)dx, we have the
more compact form


∫ u dv = uv − ∫ v du.


Theorem 3.1: Integration by Parts
Let u = f (x) and v = g(x) be functions with continuous derivatives. Then, the integration-by-parts formula for the
integral involving these two functions is:


(3.1)∫ u dv = uv − ∫ v du.


The advantage of using the integration-by-parts formula is that we can use it to exchange one integral for another, possiblyeasier, integral. The following example illustrates its use.
Example 3.1
Using Integration by Parts
Use integration by parts with u = x and dv = sinx dx to evaluate ∫ xsinx dx.


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3.1


Solution
By choosing u = x, we have du = 1dx. Since dv = sinx dx, we get v = ∫ sinx dx = −cosx. It is handy to
keep track of these values as follows:


u = x dv = sinx dx


du = 1dx v = ∫ sinx dx = −cosx.


Applying the integration-by-parts formula results in
∫ xsinx dx = (x)(−cosx) − ∫ (−cosx)(1dx) Substitute.


= −xcosx + ∫ cosx dx Simplify.


= −xcosx + sinx + C. Use ∫ cosx dx = sinx + C.


Analysis
At this point, there are probably a few items that need clarification. First of all, you may be curious aboutwhat would have happened if we had chosen u = sinx and dv = x. If we had done so, then we would
have du = cosx and v = 1


2
x2. Thus, after applying integration by parts, we have


∫ xsinx dx = 1
2
x2 sinx − ∫ 1


2
x2 cosx dx. Unfortunately, with the new integral, we are in no better position


than before. It is important to keep in mind that when we apply integration by parts, we may need to try severalchoices for u and dv before finding a choice that works.
Second, you may wonder why, when we find v = ∫ sinx dx = −cosx, we do not use v = −cosx + K. To see
that it makes no difference, we can rework the problem using v = −cosx + K:


∫ xsinx dx = (x)(−cosx + K) − ∫ (−cosx + K)(1dx)


= −xcosx + Kx + ∫ cosx dx − ∫ Kdx
= −xcosx + Kx + sinx − Kx + C
= −xcosx + sinx + C.


As you can see, it makes no difference in the final solution.
Last, we can check to make sure that our antiderivative is correct by differentiating −xcosx + sinx + C:


d
dx


(−xcosx + sinx + C) = (−1)cosx + (−x)(−sinx) + cosx


= xsinx.


Therefore, the antiderivative checks out.


Watch this video (http://www.openstaxcollege.org/l/20_intbyparts1) and visit this website(http://www.openstaxcollege.org/l/20_intbyparts2) for examples of integration by parts.


Evaluate ∫ xe2x dx using the integration-by-parts formula with u = x and dv = e2x dx.


The natural question to ask at this point is: How do we know how to choose u and dv? Sometimes it is a matter of trial
and error; however, the acronym LIATE can often help to take some of the guesswork out of our choices. This acronym


Chapter 3 | Techniques of Integration 263




stands for Logarithmic Functions, Inverse Trigonometric Functions, Algebraic Functions, Trigonometric Functions, andExponential Functions. This mnemonic serves as an aid in determining an appropriate choice for u.
The type of function in the integral that appears first in the list should be our first choice of u. For example, if an integral
contains a logarithmic function and an algebraic function, we should choose u to be the logarithmic function, because L
comes before A in LIATE. The integral in Example 3.1 has a trigonometric function (sinx) and an algebraic function
(x). Because A comes before T in LIATE, we chose u to be the algebraic function. When we have chosen u, dv is
selected to be the remaining part of the function to be integrated, together with dx.
Why does this mnemonic work? Remember that whatever we pick to be dv must be something we can integrate. Since we
do not have integration formulas that allow us to integrate simple logarithmic functions and inverse trigonometric functions,it makes sense that they should not be chosen as values for dv. Consequently, they should be at the head of the list as
choices for u. Thus, we put LI at the beginning of the mnemonic. (We could just as easily have started with IL, since
these two types of functions won’t appear together in an integration-by-parts problem.) The exponential and trigonometricfunctions are at the end of our list because they are fairly easy to integrate and make good choices for dv. Thus, we have
TE at the end of our mnemonic. (We could just as easily have used ET at the end, since when these types of functions appeartogether it usually doesn’t really matter which one is u and which one is dv.) Algebraic functions are generally easy both
to integrate and to differentiate, and they come in the middle of the mnemonic.
Example 3.2
Using Integration by Parts
Evaluate ∫ lnx


x3
dx.


Solution
Begin by rewriting the integral:


∫ lnx
x3


dx = ∫ x−3 lnx dx.


Since this integral contains the algebraic function x−3 and the logarithmic function lnx, choose u = lnx,
since L comes before A in LIATE. After we have chosen u = lnx, we must choose dv = x−3dx.
Next, since u = lnx, we have du = 1xdx. Also, v = ∫ x−3dx = − 12x−2. Summarizing,


u = lnx dv = x−3dx


du = 1xdx v = ∫ x−3dx = − 12x
−2.


Substituting into the integration-by-parts formula (Equation 3.1) gives
∫ lnx


x3
dx = ∫ x−3 lnx dx = ⎛⎝lnx)(− 12x


−2⎞
⎠− ∫



⎝−


1
2
x−2⎞⎠(


1
xdx)


= − 1
2
x−2 lnx + ∫ 1


2
x−3dx Simplify.


= − 1
2
x−2 lnx − 1


4
x−2 + C Integrate.


= − 1
2x2


lnx − 1
4x2


+ C. Rewrite with positive integers.


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3.2 Evaluate ∫ x lnx dx.


In some cases, as in the next two examples, it may be necessary to apply integration by parts more than once.
Example 3.3
Applying Integration by Parts More Than Once
Evaluate ∫ x2 e3x dx.
Solution
Using LIATE, choose u = x2 and dv = e3x dx. Thus, du = 2x dx and v = ∫ e3x dx = ⎛⎝13⎞⎠e3x. Therefore,


u = x2 dv = e3x dx


du = 2x dx v = ∫ e3x dx = 1
3
e3x.


Substituting into Equation 3.1 produces
∫ x2 e3x dx = 1


3
x2 e3x − ∫ 2


3
xe3x dx.


We still cannot integrate ∫ 2
3
xe3x dx directly, but the integral now has a lower power on x. We can evaluate this


new integral by using integration by parts again. To do this, choose u = x and dv = 2
3
e3x dx. Thus, du = dx


and v = ∫ ⎛⎝23⎞⎠e3x dx = ⎛⎝29⎞⎠e3x. Now we have
u = x dv = 2


3
e3x dx


du = dx v = ∫ 2
3
e3x dx = 2


9
e3x.


Substituting back into the previous equation yields
∫ x2 e3x dx = 1


3
x2 e3x − ⎛⎝


2
9
xe3x − ∫ 2


9
e3x dx⎞⎠.


After evaluating the last integral and simplifying, we obtain
∫ x2 e3x dx = 1


3
x2 e3x − 2


9
xe3x + 2


27
e3x + C.


Example 3.4
Applying Integration by Parts When LIATE Doesn’t Quite Work
Evaluate ∫ t3 et2dt.


Chapter 3 | Techniques of Integration 265




Solution
If we use a strict interpretation of the mnemonic LIATE to make our choice of u, we end up with u = t3 and
dv = et


2
dt. Unfortunately, this choice won’t work because we are unable to evaluate ∫ et2dt. However, since


we can evaluate ∫ tet2dx, we can try choosing u = t2 and dv = tet2dt. With these choices we have
u = t2 dv = tet


2
dt


du = 2t dt v = ∫ tet2dt = 1
2
et


2
.


Thus, we obtain
∫ t3 et2dt = 1


2
t2 et


2
− ∫ 1


2
et


2
2tdt


= 1
2
t2 et


2
− 1


2
et


2
+ C.


Example 3.5
Applying Integration by Parts More Than Once
Evaluate ∫ sin(lnx)dx.
Solution
This integral appears to have only one function—namely, sin(lnx) —however, we can always use the constant
function 1 as the other function. In this example, let’s choose u = sin(lnx) and dv = 1dx. (The decision to
use u = sin(lnx) is easy. We can’t choose dv = sin(lnx)dx because if we could integrate it, we wouldn’t be
using integration by parts in the first place!) Consequently, du = (1/x)cos(lnx)dx and v = ∫ 1dx = x. After
applying integration by parts to the integral and simplifying, we have


∫ sin(lnx)dx = xsin(lnx) − ∫ cos(lnx)dx.


Unfortunately, this process leaves us with a new integral that is very similar to the original. However, let’s seewhat happens when we apply integration by parts again. This time let’s choose u = cos(lnx) and dv = 1dx,
making du = −(1/x)sin(lnx)dx and v = ∫ 1dx = x. Substituting, we have


∫ sin(lnx)dx = xsin(lnx) − ⎛⎝xcos(lnx) — ∫ − sin(lnx)dx

⎠.


After simplifying, we obtain
∫ sin(lnx)dx = xsin(lnx) − xcos(lnx) − ∫ sin(lnx)dx.


The last integral is now the same as the original. It may seem that we have simply gone in a circle, but now we
can actually evaluate the integral. To see how to do this more clearly, substitute I = ∫ sin(lnx)dx. Thus, the


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3.3


equation becomes
I = xsin(lnx) − xcos(lnx) − I.


First, add I to both sides of the equation to obtain
2I = xsin(lnx) − xcos(lnx).


Next, divide by 2:
I = 1


2
xsin(lnx) − 1


2
xcos(lnx).


Substituting I = ∫ sin(lnx)dx again, we have
∫ sin(lnx)dx = 1


2
xsin(lnx) − 1


2
xcos(lnx).


From this we see that (1/2)xsin(lnx) − (1/2)xcos(lnx) is an antiderivative of sin(lnx)dx. For the most general
antiderivative, add +C:


∫ sin(lnx)dx = 1
2
xsin(lnx) − 1


2
xcos(lnx) + C.


Analysis
If this method feels a little strange at first, we can check the answer by differentiation:


d
dx


1
2
xsin(lnx) − 1


2
xcos(lnx)⎞⎠


= 1
2
(sin(lnx)) + cos(lnx) · 1x ·


1
2
x − ⎛⎝


1
2
cos(lnx) − sin(lnx) · 1x ·


1
2
x⎞⎠


= sin(lnx).


Evaluate ∫ x2 sinx dx.


Integration by Parts for Definite Integrals
Now that we have used integration by parts successfully to evaluate indefinite integrals, we turn our attention to definiteintegrals. The integration technique is really the same, only we add a step to evaluate the integral at the upper and lowerlimits of integration.
Theorem 3.2: Integration by Parts for Definite Integrals
Let u = f (x) and v = g(x) be functions with continuous derivatives on [a, b]. Then


(3.2)

a


b
u dv = uv|a


b − ∫
a


b
v du.


Example 3.6
Finding the Area of a Region


Chapter 3 | Techniques of Integration 267




Find the area of the region bounded above by the graph of y = tan−1 x and below by the x -axis over the interval
[0, 1].


Solution
This region is shown in Figure 3.2. To find the area, we must evaluate ∫


0


1


tan−1 x dx.


Figure 3.2 To find the area of the shaded region, we have touse integration by parts.


For this integral, let’s choose u = tan−1 x and dv = dx, thereby making du = 1
x2 + 1


dx and v = x. After
applying the integration-by-parts formula (Equation 3.2) we obtain


Area = x tan−1 x|0
1 − ∫


0


1
x


x2 + 1
dx.


Use u-substitution to obtain

0


1
x


x2 + 1
dx = 1


2
ln|x2 + 1|0


1
.


Thus,
Area = x tan−1 x|0


1
− 1


2
ln|x2 + 1||0


1
= π


4
− 1


2
ln 2.


At this point it might not be a bad idea to do a “reality check” on the reasonableness of our solution. Since
π
4
− 1


2
ln2 ≈ 0.4388, and from Figure 3.2 we expect our area to be slightly less than 0.5, this solution appears


to be reasonable.


Example 3.7


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3.4


Finding a Volume of Revolution
Find the volume of the solid obtained by revolving the region bounded by the graph of f (x) = e−x, the x-axis,
the y-axis, and the line x = 1 about the y-axis.
Solution
The best option to solving this problem is to use the shell method. Begin by sketching the region to be revolved,along with a typical rectangle (see the following graph).


Figure 3.3 We can use the shell method to find a volume of revolution.


To find the volume using shells, we must evaluate 2π∫
0


1
xe−x dx. To do this, let u = x and dv = e−x. These


choices lead to du = dx and v = ∫ e−x = −e−x. Substituting into Equation 3.2, we obtain


Volume = 2π∫
0


1


xe−x dx = 2π(−xe−x |0
1


+ ∫
0


1


e−x dx) Use integration by parts.


= −2πxe−x |0
1 − 2πe−x |0


1
Evaluate ∫


0


1


e−x dx = −e−x |0
1


.


= 2π − 4πe . Evaluate and simplify.


Analysis
Again, it is a good idea to check the reasonableness of our solution. We observe that the solid has a volumeslightly less than that of a cylinder of radius 1 and height of 1/e added to the volume of a cone of base radius
1 and height of 1 − 1


3
. Consequently, the solid should have a volume a bit less than


π(1)2 1e +


π
3

⎠(1)


2 ⎛
⎝1 −


1
e

⎠ =



3e


− π
3
≈ 1.8177.


Since 2π − 4πe ≈ 1.6603, we see that our calculated volume is reasonable.


Evaluate ∫
0


π/2
xcosx dx.


Chapter 3 | Techniques of Integration 269




3.1 EXERCISES
In using the technique of integration by parts, you mustcarefully choose which expression is u. For each of thefollowing problems, use the guidelines in this section tochoose u. Do not evaluate the integrals.
1. ∫ x3 e2x dx


2. ∫ x3 ln(x)dx


3. ∫ y3 cosydx


4. ∫ x2 arctanx dx


5. ∫ e3x sin(2x)dx
Find the integral by using the simplest method. Not allproblems require integration by parts.
6. ∫ vsinvdv


7. ∫ lnx dx (Hint: ∫ lnx dx is equivalent to
∫ 1 · ln(x)dx.)


8. ∫ xcosx dx


9. ∫ tan−1 x dx


10. ∫ x2ex dx


11. ∫ xsin(2x)dx


12. ∫ xe4x dx


13. ∫ xe−x dx


14. ∫ xcos3x dx


15. ∫ x2cosx dx


16. ∫ x lnx dx


17. ∫ ln(2x + 1)dx


18. ∫ x2 e4xdx


19. ∫ ex sinx dx


20. ∫ ex cosx dx


21. ∫ xe−x2dx


22. ∫ x2 e−x dx


23. ∫ sin(ln(2x))dx


24. ∫ cos(lnx)dx


25. ∫ (lnx)2dx


26. ∫ ln(x2)dx


27. ∫ x2 lnx dx


28. ∫ sin−1 x dx


29. ∫ cos−1(2x)dx


30. ∫ xarctanx dx


31. ∫ x2 sinx dx


32. ∫ x3 cosx dx


33. ∫ x3 sinx dx


34. ∫ x3 ex dx


35. ∫ xsec−1 x dx


36. ∫ xsec2 x dx


37. ∫ xcoshx dx


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Compute the definite integrals. Use a graphing utility toconfirm your answers.
38. ∫


1/e


1
lnx dx


39. ∫
0


1
xe−2x dx (Express the answer in exact form.)


40. ∫
0


1
e x dx(let u = x)


41. ∫
1


e
ln(x2)dx


42. ∫
0


π
xcosx dx


43. ∫
−π


π
xsinx dx (Express the answer in exact form.)


44. ∫
0


3
ln(x2 + 1)dx (Express the answer in exact form.)


45. ∫
0


π/2
x2 sinx dx (Express the answer in exact form.)


46. ∫
0


1
x5x dx (Express the answer using five significant


digits.)
47. Evaluate ∫ cosx ln(sinx)dx
Derive the following formulas using the technique ofintegration by parts. Assume that n is a positive integer.These formulas are called reduction formulas because theexponent in the x term has been reduced by one in eachcase. The second integral is simpler than the originalintegral.
48. ∫ xn ex dx = xn ex − n∫ xn − 1 ex dx


49. ∫ xn cosx dx = xn sinx − n∫ xn − 1 sinx dx


50. ∫ xn sinx dx = ______


51. Integrate ∫ 2x 2x − 3dx using two methods:
a. Using parts, letting dv = 2x − 3dx
b. Substitution, letting u = 2x − 3


State whether you would use integration by parts to


evaluate the integral. If so, identify u and dv. If not,describe the technique used to perform the integrationwithout actually doing the problem.
52. ∫ x lnx dx


53. ∫ ln2 xx dx


54. ∫ xex dx


55. ∫ xex2 − 3dx


56. ∫ x2 sinx dx


57. ∫ x2 sin(3x3 + 2)dx
Sketch the region bounded above by the curve, the x-axis,and x = 1, and find the area of the region. Provide the
exact form or round answers to the number of placesindicated.
58. y = 2xe−x (Approximate answer to four decimal
places.)
59. y = e−x sin(πx) (Approximate answer to five
decimal places.)
Find the volume generated by rotating the region boundedby the given curves about the specified line. Express theanswers in exact form or approximate to the number ofdecimal places indicated.
60. y = sinx, y = 0, x = 2π, x = 3π about the y-axis
(Express the answer in exact form.)
61. y = e−x y = 0, x = −1x = 0; about x = 1
(Express the answer in exact form.)
62. A particle moving along a straight line has a velocity
of v(t) = t2 e−t after t sec. How far does it travel in the
first 2 sec? (Assume the units are in feet and express theanswer in exact form.)
63. Find the area under the graph of y = sec3 x from
x = 0to x = 1. (Round the answer to two significant
digits.)
64. Find the area between y = (x − 2)ex and the x-axis
from x = 2 to x = 5. (Express the answer in exact form.)


Chapter 3 | Techniques of Integration 271




65. Find the area of the region enclosed by the curve
y = xcosx and the x-axis for 11π


2
≤ x ≤ 13π


2
. (Express


the answer in exact form.)
66. Find the volume of the solid generated by revolvingthe region bounded by the curve y = lnx, the x-axis,
and the vertical line x = e2 about the x-axis. (Express the
answer in exact form.)
67. Find the volume of the solid generated by revolvingthe region bounded by the curve y = 4cosx and the
x-axis, π


2
≤ x ≤ 3π


2
, about the x-axis. (Express the


answer in exact form.)
68. Find the volume of the solid generated by revolvingthe region in the first quadrant bounded by y = ex and
the x-axis, from x = 0 to x = ln(7), about the y-axis.
(Express the answer in exact form.)


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3.5


3.2 | Trigonometric Integrals
Learning Objectives


3.2.1 Solve integration problems involving products and powers of sinx and cosx.
3.2.2 Solve integration problems involving products and powers of tanx and secx.
3.2.3 Use reduction formulas to solve trigonometric integrals.


In this section we look at how to integrate a variety of products of trigonometric functions. These integrals are calledtrigonometric integrals. They are an important part of the integration technique called trigonometric substitution, which isfeatured in Trigonometric Substitution. This technique allows us to convert algebraic expressions that we may not beable to integrate into expressions involving trigonometric functions, which we may be able to integrate using the techniquesdescribed in this section. In addition, these types of integrals appear frequently when we study polar, cylindrical, andspherical coordinate systems later. Let’s begin our study with products of sinx and cosx.
Integrating Products and Powers of sinx and cosx
A key idea behind the strategy used to integrate combinations of products and powers of sinx and cosx involves rewriting
these expressions as sums and differences of integrals of the form ∫ sin j xcosx dx or ∫ cos j xsinx dx. After rewriting
these integrals, we evaluate them using u-substitution. Before describing the general process in detail, let’s take a look atthe following examples.
Example 3.8
Integrating ∫cos j xsinxdx
Evaluate ∫ cos3 xsinx dx.
Solution
Use u -substitution and let u = cosx. In this case, du = −sinx dx. Thus,


∫ cos3 xsinx dx = −∫ u3 du


= − 1
4
u4 + C


= − 1
4
cos4 x + C.


Evaluate ∫ sin4 xcosx dx.


Example 3.9
A Preliminary Example: Integrating ∫cos j xsinkxdx Where k is Odd


Chapter 3 | Techniques of Integration 273




3.6


Evaluate ∫ cos2 xsin3 x dx.
Solution
To convert this integral to integrals of the form ∫ cos j xsinx dx, rewrite sin3 x = sin2 xsinx and make the
substitution sin2 x = 1 − cos2 x. Thus,


∫ cos2 xsin3 x dx = ∫ cos2 x(1 − cos2 x)sinx dx Let u = cosx; then du = −sinx dx.


= −∫ u2 ⎛⎝1 − u2⎞⎠du


= ∫ ⎛⎝u4 − u2⎞⎠du


= 1
5
u5 − 1


3
u3 + C


= 1
5
cos5 x − 1


3
cos3 x + C.


Evaluate ∫ cos3 xsin2 x dx.


In the next example, we see the strategy that must be applied when there are only even powers of sinx and cosx. For
integrals of this type, the identities


sin2 x = 1
2
− 1


2
cos(2x) = 1 − cos(2x)


2


and
cos2 x = 1


2
+ 1


2
cos(2x) = 1 + cos(2x)


2


are invaluable. These identities are sometimes known as power-reducing identities and they may be derived from the
double-angle identity cos(2x) = cos2 x − sin2 x and the Pythagorean identity cos2 x + sin2 x = 1.
Example 3.10
Integrating an Even Power of sinx
Evaluate ∫ sin2 x dx.
Solution
To evaluate this integral, let’s use the trigonometric identity sin2 x = 1


2
− 1


2
cos(2x). Thus,


∫ sin2 x dx = ∫ ⎛⎝12 −
1
2
cos(2x)⎞⎠dx


= 1
2
x − 1


4
sin(2x) + C.


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3.7 Evaluate ∫ cos2 x dx.


The general process for integrating products of powers of sinx and cosx is summarized in the following set of guidelines.


Problem-Solving Strategy: Integrating Products and Powers of sin x and cos x
To integrate ∫ cos j xsink x dx use the following strategies:


1. If k is odd, rewrite sink x = sink − 1 xsinx and use the identity sin2 x = 1 − cos2 x to rewrite sink − 1 x in
terms of cosx. Integrate using the substitution u = cosx. This substitution makes du = −sinx dx.


2. If j is odd, rewrite cos j x = cos j − 1 xcosx and use the identity cos2 x = 1 − sin2 x to rewrite cos j − 1 x
in terms of sinx. Integrate using the substitution u = sinx. This substitution makes du = cosx dx. (Note: If
both j and k are odd, either strategy 1 or strategy 2 may be used.)


3. If both j and k are even, use sin2 x = (1/2) − (1/2)cos(2x) and cos2 x = (1/2) + (1/2)cos(2x). After
applying these formulas, simplify and reapply strategies 1 through 3 as appropriate.


Example 3.11
Integrating ∫cos j xsinkxdx where k is Odd
Evaluate ∫ cos8 xsin5 x dx.
Solution
Since the power on sinx is odd, use strategy 1. Thus,


∫ cos8 xsin5 x dx = ∫ cos8 xsin4 xsinx dx Break off sinx.


= ∫ cos8 x(sin2 x)2 sinx dx Rewrite sin4 x = (sin2 x)2.


= ∫ cos8 x(1 − cos2 x)2 sinx dx Substitute sin2 x = 1 − cos2 x.


= ∫ u8 (1 − u2)2(−du) Let u = cosx and du = −sinx dx.


= ∫ ⎛⎝−u8 + 2u10 − u12⎞⎠du Expand.


= − 1
9
u9 + 2


11
u11 − 1


13
u13 + C Evaluate the integral.


= − 1
9
cos9 x + 2


11
cos11 x − 1


13
cos13 x + C. Substitute u = cosx.


Example 3.12


Chapter 3 | Techniques of Integration 275




3.8


3.9


Integrating ∫cos j xsinkxdx where k and j are Even
Evaluate ∫ sin4 x dx.
Solution
Since the power on sinx is even (k = 4) and the power on cosx is even ⎛⎝ j = 0⎞⎠, we must use strategy 3.
Thus,


∫ sin4 x dx = ∫ ⎛⎝sin2 x⎞⎠
2
dx Rewrite sin4 x = ⎛⎝sin


2 x⎞⎠
2
.


= ∫ ⎛⎝12 −
1
2
cos(2x)⎞⎠


2
dx Substitute sin2 x = 1


2
− 1


2
cos(2x).


= ∫ ⎛⎝14 −
1
2
cos(2x) + 1


4
cos2(2x)⎞⎠dx Expand




1
2
− 1


2
cos(2x)⎞⎠


2
.


= ∫ ⎛⎝14 −
1
2
cos(2x) + 1


4
(1
2
+ 1


2
cos(4x)⎞⎠dx.


Since cos2(2x) has an even power, substitute cos2(2x) = 1
2
+ 1


2
cos(4x):


= ∫ ⎛⎝38 −
1
2
cos(2x) + 1


8
cos(4x)⎞⎠dx Simplify.


= 3
8
x − 1


4
sin(2x) + 1


32
sin(4x) + C Evaluate the integral.


Evaluate ∫ cos3 x dx.


Evaluate ∫ cos2(3x)dx.


In some areas of physics, such as quantum mechanics, signal processing, and the computation of Fourier series, it is oftennecessary to integrate products that include sin(ax), sin(bx), cos(ax), and cos(bx). These integrals are evaluated by
applying trigonometric identities, as outlined in the following rule.
Rule: Integrating Products of Sines and Cosines of Different Angles
To integrate products involving sin(ax), sin(bx), cos(ax), and cos(bx), use the substitutions


(3.3)sin(ax)sin(bx) = 1
2
cos((a − b)x) − 1


2
cos((a + b)x)


(3.4)sin(ax)cos(bx) = 1
2
sin⎛⎝(a − b)x⎞⎠+ 12


sin((a + b)x)


(3.5)cos(ax)cos(bx) = 1
2
cos((a − b)x) + 1


2
cos((a + b)x)


These formulas may be derived from the sum-of-angle formulas for sine and cosine.


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3.10


Example 3.13
Evaluating ∫ sin(ax)cos(bx)dx
Evaluate ∫ sin(5x)cos(3x)dx.
Solution
Apply the identity sin(5x)cos(3x) = 1


2
sin(2x) − 1


2
cos(8x). Thus,


∫ sin(5x)cos(3x)dx = ∫ 1
2
sin(2x) − 1


2
cos(8x)dx


= − 1
4
cos(2x) − 1


16
sin(8x) + C.


Evaluate ∫ cos(6x)cos(5x)dx.


Integrating Products and Powers of tanx and secx
Before discussing the integration of products and powers of tanx and secx, it is useful to recall the integrals involving
tanx and secx we have already learned:
1. ∫ sec2 x dx = tanx + C
2. ∫ secx tanx dx = secx + C
3. ∫ tanx dx = ln|secx| + C
4. ∫ secx dx = ln|secx + tanx| + C.


For most integrals of products and powers of tanx and secx, we rewrite the expression we wish to integrate as the sum
or difference of integrals of the form ∫ tan j xsec2 x dx or ∫ sec j x tanx dx. As we see in the following example, we can
evaluate these new integrals by using u-substitution.
Example 3.14
Evaluating ∫ sec j xtanxdx
Evaluate ∫ sec5 x tanx dx.
Solution
Start by rewriting sec5 x tanx as sec4 xsecx tanx.


Chapter 3 | Techniques of Integration 277




3.11


∫ sec5 x tanx dx = ∫ sec4 xsecx tanx dx Let u = secx; then, du = secx tanx dx.


= ∫ u4du Evaluate the integral.


= 1
5
u5 + C Substitute secx = u.


= 1
5
sec5 x + C


You can read some interesting information at this website (http://www.openstaxcollege.org/l/20_intseccube) to learn about a common integral involving the secant.


Evaluate ∫ tan5 xsec2 x dx.


We now take a look at the various strategies for integrating products and powers of secx and tanx.


Problem-Solving Strategy: Integrating ∫ tankxsec j xdx
To integrate ∫ tank xsec j x dx, use the following strategies:


1. If j is even and j ≥ 2, rewrite sec j x = sec j − 2 xsec2 x and use sec2 x = tan2 x + 1 to rewrite sec j − 2 x
in terms of tanx. Let u = tanx and du = sec2 x.


2. If k is odd and j ≥ 1, rewrite tank xsec j x = tank − 1 xsec j − 1 xsecx tanx and use tan2 x = sec2 x − 1 to
rewrite tank − 1 x in terms of secx. Let u = secx and du = secx tanx dx. (Note: If j is even and k is odd,
then either strategy 1 or strategy 2 may be used.)


3. If k is odd where k ≥ 3 and j = 0, rewrite
tank x = tank − 2 x tan2 x = tank − 2 x(sec2 x − 1) = tank − 2 xsec2 x − tank − 2 x. It may be necessary to
repeat this process on the tank − 2 x term.


4. If k is even and j is odd, then use tan2 x = sec2 x − 1 to express tank x in terms of secx. Use integration
by parts to integrate odd powers of secx.


Example 3.15
Integrating ∫ tankxsec j xdx when j is Even
Evaluate ∫ tan6 xsec4 x dx.


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Solution
Since the power on secx is even, rewrite sec4 x = sec2 xsec2 x and use sec2 x = tan2 x + 1 to rewrite the first
sec2 x in terms of tanx. Thus,


∫ tan6 xsec4 x dx = ∫ tan6 x⎛⎝tan2 x + 1⎞⎠sec2 x dx Let u = tanx and du = sec2 x.


= ∫ u6 ⎛⎝u2 + 1⎞⎠du Expand.


= ∫ (u8 + u6)du Evaluate the integral.


= 1
9
u9 + 1


7
u7 + C Substitute tanx = u.


= 1
9
tan9 x + 1


7
tan7 x + C.


Example 3.16
Integrating ∫ tankxsec j xdx when k is Odd
Evaluate ∫ tan5 xsec3 x dx.
Solution
Since the power on tanx is odd, begin by rewriting tan5 xsec3 x = tan4 xsec2 xsecx tanx. Thus,


tan5 xsec3 x = tan4 xsec2 xsecx tanx. Write tan4 x = (tan2 x)2.


∫ tan5 xsec3 x dx = ∫ (tan2 x)2 sec2 xsecx tanx dx Use tan2 x = sec2 x − 1.


= ∫ (sec2 x − 1)2 sec2 xsecx tanx dx Let u = secx and du = secx tanx dx.


= ∫ (u2 − 1)2u2du Expand.


= ∫ ⎛⎝u6 − 2u4 + u2⎞⎠du Integrate.


= 1
7
u7 − 2


5
u5 + 1


3
u3 + C Substitute secx = u.


= 1
7
sec7 x − 2


5
sec5 x + 1


3
sec3 x + C.


Example 3.17
Integrating ∫ tankxdx where k is Odd and k ≥ 3
Evaluate ∫ tan3 x dx.


Chapter 3 | Techniques of Integration 279




3.12


Solution
Begin by rewriting tan3 x = tanx tan2 x = tanx⎛⎝sec2 x − 1⎞⎠ = tanxsec2 x − tanx. Thus,


∫ tan3 x dx = ∫ ⎛⎝tanxsec2 x − tanx⎞⎠dx


= ∫ tanxsec2 x dx − ∫ tanx dx


= 1
2
tan2 x − ln|secx| + C.


For the first integral, use the substitution u = tanx. For the second integral, use the formula.


Example 3.18
Integrating ∫ sec3 xdx
Integrate ∫ sec3 x dx.
Solution
This integral requires integration by parts. To begin, let u = secx and dv = sec2 x. These choices make
du = secx tanx and v = tanx. Thus,


∫ sec3 x dx = secx tanx − ∫ tanxsecx tanx dx


= secx tanx − ∫ tan2 xsecx dx Simplify.


= secx tanx − ∫ ⎛⎝sec2 x − 1⎞⎠secx dx Substitute tan2 x = sec2 x − 1.


= secx tanx + ∫ secx dx − ∫ sec3 x dx Rewrite.


= secx tanx + ln|secx + tanx| − ∫ sec3 x dx. Evaluate∫ secx dx.


We now have
∫ sec3 x dx = secx tanx + ln|secx + tanx| − ∫ sec3 x dx.


Since the integral ∫ sec3 x dx has reappeared on the right-hand side, we can solve for ∫ sec3 x dx by adding it
to both sides. In doing so, we obtain


2∫ sec3 x dx = secx tanx + ln|secx + tanx|.


Dividing by 2, we arrive at
∫ sec3 x dx = 1


2
secx tanx + 1


2
ln|secx + tanx| + C.


Evaluate ∫ tan3 xsec7 x dx.


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Reduction Formulas
Evaluating ∫ secn x dx for values of n where n is odd requires integration by parts. In addition, we must also know
the value of ∫ secn − 2 x dx to evaluate ∫ secn x dx. The evaluation of ∫ tann x dx also requires being able to integrate
∫ tann − 2 x dx. To make the process easier, we can derive and apply the following power reduction formulas. These
rules allow us to replace the integral of a power of secx or tanx with the integral of a lower power of secx or tanx.


Rule: Reduction Formulas for ∫ secnxdx and ∫ tannxdx
(3.6)∫ secn x dx = 1


n − 1
secn − 2 x tanx + n − 2


n − 1
∫ secn − 2 x dx


(3.7)∫ tann x dx = 1
n − 1


tann − 1 x − ∫ tann − 2 x dx


The first power reduction rule may be verified by applying integration by parts. The second may be verified byfollowing the strategy outlined for integrating odd powers of tanx.


Example 3.19
Revisiting ∫ sec3 xdx
Apply a reduction formula to evaluate ∫ sec3 x dx.
Solution
By applying the first reduction formula, we obtain


∫ sec3 x dx = 1
2
secx tanx + 1


2
∫ secx dx


= 1
2
secx tanx + 1


2
ln|secx + tanx| + C.


Example 3.20
Using a Reduction Formula
Evaluate ∫ tan4 x dx.
Solution
Applying the reduction formula for ∫ tan4 x dx we have


Chapter 3 | Techniques of Integration 281




3.13


∫ tan4 x dx = 1
3
tan3 x − ∫ tan2 x dx


= 1
3
tan3 x − (tanx − ∫ tan0 x dx) Apply the reduction formula to∫ tan2 x dx.


= 1
3
tan3 x − tanx + ∫ 1 dx Simplify.


= 1
3
tan3 x − tanx + x + C. Evaluate∫ 1dx.


Apply the reduction formula to ∫ sec5 x dx.


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3.2 EXERCISES
Fill in the blank to make a true statement.
69. sin2 x + _______ = 1
70. sec2 x − 1 = _______
Use an identity to reduce the power of the trigonometricfunction to a trigonometric function raised to the firstpower.
71. sin2 x = _______
72. cos2 x = _______
Evaluate each of the following integrals by u-substitution.
73. ∫ sin3 xcosx dx


74. ∫ cosxsinx dx


75. ∫ tan5(2x)sec2(2x)dx


76. ∫ sin7(2x)cos(2x)dx


77. ∫ tan⎛⎝x2⎞⎠sec2 ⎛⎝x2⎞⎠dx


78. ∫ tan2 xsec2 x dx
Compute the following integrals using the guidelines forintegrating powers of trigonometric functions. Use a CASto check the solutions. (Note: Some of the problems may bedone using techniques of integration learned previously.)
79. ∫ sin3 x dx


80. ∫ cos3 x dx


81. ∫ sinxcosx dx


82. ∫ cos5 x dx


83. ∫ sin5 xcos2 x dx


84. ∫ sin3 xcos3 x dx


85. ∫ sinxcosx dx


86. ∫ sinxcos3 x dx


87. ∫ secx tanx dx


88. ∫ tan(5x)dx


89. ∫ tan2 xsecx dx


90. ∫ tanxsec3 x dx


91. ∫ sec4 x dx


92. ∫ cotx dx


93. ∫ cscx dx


94. ∫ tan3 xsecxdx
For the following exercises, find a general formula for theintegrals.
95. ∫ sin2axcosax dx


96. ∫ sinaxcosax dx.
Use the double-angle formulas to evaluate the followingintegrals.
97. ∫


0


π
sin2 x dx


98. ∫
0


π
sin4 x dx


99. ∫ cos23x dx


100. ∫ sin2 xcos2 x dx


101. ∫ sin2 x dx + ∫ cos2 x dx


102. ∫ sin2 xcos2(2x)dx


Chapter 3 | Techniques of Integration 283




For the following exercises, evaluate the definite integrals.Express answers in exact form whenever possible.
103. ∫


0



cosxsin2x dx


104. ∫
0


π
sin3xsin5x dx


105. ∫
0


π
cos(99x)sin(101x)dx


106. ∫
−π


π
cos2(3x)dx


107. ∫
0



sinxsin(2x)sin(3x)dx


108. ∫
0



cos(x/2)sin(x/2)dx


109. ∫
π/6


π/3
cos3 x
sinx


dx (Round this answer to three decimal
places.)
110. ∫


−π/3


π/3
sec2 x − 1dx


111. ∫
0


π/2
1 − cos(2x)dx


112. Find the area of the region bounded by the graphs of
the equations y = sinx, y = sin3 x, x = 0, and x = π


2
.


113. Find the area of the region bounded by the graphsof the equations
y = cos2 x, y = sin2 x, x = − π


4
, and x = π


4
.


114. A particle moves in a straight line with the velocity
function v(t) = sin(ωt)cos2 (ωt). Find its position
function x = f (t) if f (0) = 0.
115. Find the average value of the function
f (x) = sin2 xcos3 x over the interval [−π, π].
For the following exercises, solve the differentialequations.
116. dy


dx
= sin2 x. The curve passes through point


(0, 0).


117. dy


= sin4 (πθ)


118. Find the length of the curve
y = ln(cscx), π


4
≤ x ≤ π


2
.


119. Find the length of the curve
y = ln(sinx), π


3
≤ x ≤ π


2
.


120. Find the volume generated by revolving the curve
y = cos(3x) about the x-axis, 0 ≤ x ≤ π


36
.


For the following exercises, use this information: The innerproduct of two functions f and g over [a, b] is defined
by f (x) · g(x) = 〈 f , g 〉 = ∫


a


b
f · gdx. Two distinct


functions f and g are said to be orthogonal if
〈 f , g 〉 = 0.


121. Show that {sin(2x), cos(3x)} are orthogonal over
the interval [−π, π].
122. Evaluate ∫


−π


π
sin(mx)cos(nx)dx.


123. Integrate y′ = tanxsec4 x.
For each pair of integrals, determine which one is moredifficult to evaluate. Explain your reasoning.
124. ∫ sin456 xcosx dx or ∫ sin2 xcos2 x dx


125. ∫ tan350 xsec2 x dx or ∫ tan350 xsecx dx


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3.3 | Trigonometric Substitution
Learning Objectives


3.3.1 Solve integration problems involving the square root of a sum or difference of two squares.
In this section, we explore integrals containing expressions of the form a2 − x2, a2 + x2, and x2 − a2, where the
values of a are positive. We have already encountered and evaluated integrals containing some expressions of this type, but
many still remain inaccessible. The technique of trigonometric substitution comes in very handy when evaluating theseintegrals. This technique uses substitution to rewrite these integrals as trigonometric integrals.
Integrals Involving a2− x2
Before developing a general strategy for integrals containing a2 − x2, consider the integral ∫ 9 − x2dx. This integral
cannot be evaluated using any of the techniques we have discussed so far. However, if we make the substitution
x = 3sinθ, we have dx = 3cosθdθ. After substituting into the integral, we have


∫ 9 − x2dx = ∫ 9 − (3sinθ)23cosθdθ.


After simplifying, we have
∫ 9 − x2dx = ∫ 9 1 − sin2 θcosθdθ.


Letting 1 − sin2 θ = cos2 θ, we now have
∫ 9 − x2dx = ∫ 9 cos2 θcosθdθ.


Assuming that cosθ ≥ 0, we have
∫ 9 − x2dx = ∫ 9cos2 θdθ.


At this point, we can evaluate the integral using the techniques developed for integrating powers and products oftrigonometric functions. Before completing this example, let’s take a look at the general theory behind this idea.
To evaluate integrals involving a2 − x2, we make the substitution x = asinθ and dx = acosθ. To see that this
actually makes sense, consider the following argument: The domain of a2 − x2 is [−a, a]. Thus, −a ≤ x ≤ a.
Consequently, −1 ≤ xa ≤ 1. Since the range of sinx over ⎡⎣−(π/2), π/2⎤⎦ is [−1, 1], there is a unique angle θ satisfying
−(π/2) ≤ θ ≤ π/2 so that sinθ = x/a, or equivalently, so that x = asinθ. If we substitute x = asinθ into a2 − x2,
we get


a2 − x2 = a2 − (asinθ)2 Let x = asinθwhere − π
2
≤ θ ≤ π


2
. Simplify.


= a2 − a2 sin2 θ Factor out a2.


= a2(1 − sin2 θ) Substitute 1 − sin2 x = cos2 x.


= a2 cos2 θ Take the square root.


= |acosθ|
= acosθ.


Since cosx ≥ 0 on −π
2
≤ θ ≤ π


2
and a > 0, |acosθ| = acosθ. We can see, from this discussion, that by making the


substitution x = asinθ, we are able to convert an integral involving a radical into an integral involving trigonometric
functions. After we evaluate the integral, we can convert the solution back to an expression involving x. To see how to


Chapter 3 | Techniques of Integration 285




do this, let’s begin by assuming that 0 < x < a. In this case, 0 < θ < π
2
. Since sinθ = xa, we can draw the reference


triangle in Figure 3.4 to assist in expressing the values of cosθ, tanθ, and the remaining trigonometric functions in
terms of x. It can be shown that this triangle actually produces the correct values of the trigonometric functions evaluated
at θ for all θ satisfying −π


2
≤ θ ≤ π


2
. It is useful to observe that the expression a2 − x2 actually appears as the length


of one side of the triangle. Last, should θ appear by itself, we use θ = sin−1 ⎛⎝xa⎞⎠.


Figure 3.4 A reference triangle can help express thetrigonometric functions evaluated at θ in terms of x.


The essential part of this discussion is summarized in the following problem-solving strategy.


Problem-Solving Strategy: Integrating Expressions Involving a2− x2
1. It is a good idea to make sure the integral cannot be evaluated easily in another way. For example, although


this method can be applied to integrals of the form ∫ 1
a2 − x2


dx, ∫ x
a2 − x2


dx, and ∫ x a2 − x2dx,
they can each be integrated directly either by formula or by a simple u-substitution.


2. Make the substitution x = asinθ and dx = acosθdθ. Note: This substitution yields a2 − x2 = acosθ.
3. Simplify the expression.
4. Evaluate the integral using techniques from the section on trigonometric integrals.
5. Use the reference triangle from Figure 3.4 to rewrite the result in terms of x. You may also need to use some


trigonometric identities and the relationship θ = sin−1 ⎛⎝xa⎞⎠.


The following example demonstrates the application of this problem-solving strategy.
Example 3.21
Integrating an Expression Involving a2− x2
Evaluate ∫ 9 − x2dx.
Solution
Begin by making the substitutions x = 3sinθ and dx = 3cosθdθ. Since sinθ = x


3
, we can construct the


reference triangle shown in the following figure.


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Figure 3.5 A reference triangle can be constructed forExample 3.21.


Thus,
∫ 9 − x2dx = ∫ 9 − (3sinθ)23cosθdθ Substitute x = 3sinθ and dx = 3cosθdθ.


= ∫ 9(1 − sin2 θ)3cosθdθ Simplify.


= ∫ 9cos2 θ3cosθdθ Substitute cos2 θ = 1 − sin2 θ.


= ∫ 3|cosθ|3cosθdθ Take the square root.


= ∫ 9cos2 θdθ Simplify. Since −
π
2
≤ θ ≤ π


2
, cosθ ≥ 0 and


|cosθ| = cosθ.


= ∫ 9⎛⎝12 +
1
2
cos(2θ)⎞⎠dθ


Use the strategy for integrating an even power


of cosθ.


= 9
2
θ + 9


4
sin(2θ) + C Evaluate the integral.


= 9
2
θ + 9


4
(2sinθcosθ) + C Substitute sin(2θ) = 2sinθcosθ.


= 9
2
sin−1 ⎛⎝


x
3

⎠+


9
2
· x
3
· 9 − x


2


3
+ C


Substitute sin−1 ⎛⎝
x
3

⎠ = θ and sinθ =


x
3
. Use


the reference triangle to see that


cosθ = 9 − x
2


3
and make this substitution.


= 9
2
sin−1 ⎛⎝


x
3

⎠+


x 9 − x2
2


+ C. Simplify.


Example 3.22
Integrating an Expression Involving a2− x2


Evaluate ∫ 4 − x2x dx.
Solution
First make the substitutions x = 2sinθ and dx = 2cosθdθ. Since sinθ = x


2
, we can construct the reference


triangle shown in the following figure.


Chapter 3 | Techniques of Integration 287




Figure 3.6 A reference triangle can be constructed forExample 3.22.


Thus,
∫ 4 − x2x dx = ∫


4 − (2sinθ)2


2sinθ
2cosθdθ Substitute x = 2sinθ and = 2cosθdθ.


= ∫ 2cos2 θ
sinθ


dθ Substitute cos2 θ = 1 − sin2 θ and simplify.


= ∫ 2(1 − sin
2 θ)


sinθ
dθ Substitute sin2 θ = 1 − cos2 θ.


= ∫ (2cscθ − 2sinθ)dθ
Separate the numerator, simplify, and use


cscθ = 1
sinθ


.


= 2ln|cscθ − cotθ| + 2cosθ + C Evaluate the integral.


= 2ln|2x − 4 − x2x | + 4 − x2 + C. Use the reference triangle to rewrite theexpression in terms of x and simplify.


In the next example, we see that we sometimes have a choice of methods.
Example 3.23
Integrating an Expression Involving a2− x2 Two Ways
Evaluate ∫ x3 1 − x2dx two ways: first by using the substitution u = 1 − x2 and then by using a
trigonometric substitution.
Solution
Method 1
Let u = 1 − x2 and hence x2 = 1 − u. Thus, du = −2xdx. In this case, the integral becomes


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3.14


∫ x3 1 − x2dx = − 1
2
∫ x2 1 − x2(−2xdx) Make the substitution.


= − 1
2
∫ (1 − u) udu Expand the expression.


= − 1
2
∫ ⎛⎝u1/2 − u3/2⎞⎠du Evaluate the integral.


= − 1
2


2
3
u3/2 − 2


5
u5/2⎞⎠+ C Rewrite in terms of x.


= − 1
3

⎝1 − x


2⎞


3/2
+ 1


5

⎝1 − x


2⎞


5/2
+ C.


Method 2
Let x = sinθ. In this case, dx = cosθdθ. Using this substitution, we have


∫ x3 1 − x2dx = ∫ sin3 θcos2 θdθ


= ∫ ⎛⎝1 − cos2 θ⎞⎠cos2 θsinθdθ Let u = cosθ. Thus, du = −sinθdθ.


= ∫ ⎛⎝u4 − u2⎞⎠du


= 1
5
u5 − 1


3
u3 + C Substitute cosθ = u.


= 1
5
cos5 θ − 1


3
cos3 θ + C


Use a reference triangle to see that


cosθ = 1 − x2.


= 1
5

⎝1 − x


2⎞


5/2
− 1


3

⎝1 − x


2⎞


3/2
+ C.


Rewrite the integral ∫ x3
25 − x2


dx using the appropriate trigonometric substitution (do not evaluate
the integral).


Integrating Expressions Involving a2+ x2
For integrals containing a2 + x2, let’s first consider the domain of this expression. Since a2 + x2 is defined for all
real values of x, we restrict our choice to those trigonometric functions that have a range of all real numbers. Thus, our
choice is restricted to selecting either x = a tanθ or x = acotθ. Either of these substitutions would actually work, but
the standard substitution is x = a tanθ or, equivalently, tanθ = x/a. With this substitution, we make the assumption that
−(π/2) < θ < π/2, so that we also have θ = tan−1 (x/a). The procedure for using this substitution is outlined in the
following problem-solving strategy.


Problem-Solving Strategy: Integrating Expressions Involving a2+ x2
1. Check to see whether the integral can be evaluated easily by using another method. In some cases, it is moreconvenient to use an alternative method.
2. Substitute x = a tanθ and dx = asec2 θdθ. This substitution yields


a2 + x2 = a2 + (a tanθ)2 = a2(1 + tan2 θ) = a2 sec2 θ = |asecθ| = asecθ. (Since −π2 < θ < π2 and
secθ > 0 over this interval, |asecθ| = asecθ.)


Chapter 3 | Techniques of Integration 289




3. Simplify the expression.
4. Evaluate the integral using techniques from the section on trigonometric integrals.
5. Use the reference triangle from Figure 3.7 to rewrite the result in terms of x. You may also need to use


some trigonometric identities and the relationship θ = tan−1 ⎛⎝xa⎞⎠. (Note: The reference triangle is based on the
assumption that x > 0; however, the trigonometric ratios produced from the reference triangle are the same as
the ratios for which x ≤ 0.)


Figure 3.7 A reference triangle can be constructed to expressthe trigonometric functions evaluated at θ in terms of x.


Example 3.24
Integrating an Expression Involving a2+ x2
Evaluate ∫ dx


1 + x2
and check the solution by differentiating.


Solution
Begin with the substitution x = tanθ and dx = sec2 θdθ. Since tanθ = x, draw the reference triangle in the
following figure.


Figure 3.8 The reference triangle for Example 3.24.


Thus,
∫ dx


1 + x2
= ∫ sec2 θ


secθ


Substitute x = tanθ and dx = sec2 θdθ. This


substitution makes 1 + x2 = secθ. Simplify.


= ∫ secθdθ Evaluate the integral.


= ln|secθ + tanθ| + C
Use the reference triangle to express the result


in terms of x.


= ln| 1 + x2 + x| + C.


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To check the solution, differentiate:
d
dx

⎝ln| 1 + x2 + x|⎞⎠ = 11 + x2 + x ·





⎜ x


1 + x2
+ 1





= 1
1 + x2 + x


· x + 1 + x
2


1 + x2


= 1
1 + x2


.


Since 1 + x2 + x > 0 for all values of x, we could rewrite ln| 1 + x2 + x| + C = ln⎛⎝ 1 + x2 + x⎞⎠+ C, if
desired.


Example 3.25
Evaluating ∫ dx


1 + x2
Using a Different Substitution


Use the substitution x = sinhθ to evaluate ∫ dx
1 + x2


.


Solution
Because sinhθ has a range of all real numbers, and 1 + sinh2 θ = cosh2 θ, we may also use the substitution
x = sinhθ to evaluate this integral. In this case, dx = coshθdθ. Consequently,


∫ dx
1 + x2


= ∫ coshθ
1 + sinh2 θ



Substitute x = sinhθ and dx = coshθdθ.


Substitute 1 + sinh2 θ = cosh2 θ.


= ∫ coshθ
cosh2 θ


dθ cosh2 θ = |coshθ|


= ∫ coshθ
|coshθ|


dθ |coshθ| = coshθ since coshθ > 0 for all θ.


= ∫ coshθ
coshθ


dθ Simplify.


= ∫ 1dθ Evaluate the integral.
= θ + C Since x = sinhθ, we know θ = sinh−1 x.


= sinh−1 x + C.


Analysis
This answer looks quite different from the answer obtained using the substitution x = tanθ. To see that the
solutions are the same, set y = sinh−1 x. Thus, sinhy = x. From this equation we obtain:


ey − e
−y


2
= x.


After multiplying both sides by 2ey and rewriting, this equation becomes:


Chapter 3 | Techniques of Integration 291




e
2y


− 2xey − 1 = 0.


Use the quadratic equation to solve for ey :
ey = 2x ± 4x


2 + 4
2


.


Simplifying, we have:
ey = x ± x2 + 1.


Since x − x2 + 1 < 0, it must be the case that ey = x + x2 + 1. Thus,
y = ln⎛⎝x + x


2 + 1⎞⎠.


Last, we obtain
sinh−1 x = ln⎛⎝x + x


2 + 1⎞⎠.


After we make the final observation that, since x + x2 + 1 > 0,
ln⎛⎝x + x


2 + 1⎞⎠ = ln| 1 + x2 + x|,
we see that the two different methods produced equivalent solutions.


Example 3.26
Finding an Arc Length
Find the length of the curve y = x2 over the interval [0, 1


2
].


Solution
Because dy


dx
= 2x, the arc length is given by



0


1/2
1 + (2x)2dx = ∫


0


1/2
1 + 4x2dx.


To evaluate this integral, use the substitution x = 1
2
tanθ and dx = 1


2
sec2 θdθ. We also need to change the limits


of integration. If x = 0, then θ = 0 and if x = 1
2
, then θ = π


4
. Thus,


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3.15



0


1/2
1 + 4x2dx = ∫


0


π/4
1 + tan2 θ1


2
sec2 θdθ


After substitution,


1 + 4x2 = tanθ. Substitute


1 + tan2 θ = sec2 θ and simplify.


= 1
2


0


π/4
sec3 θdθ


We derived this integral in the


previous section.


= 1
2


1
2
secθ tanθ + ln|secθ + tanθ|⎞⎠|0π/4 Evaluate and simplify.


= 1
4
( 2 + ln( 2 + 1)).


Rewrite ∫ x3 x2 + 4dx by using a substitution involving tanθ.


Integrating Expressions Involving x2− a2
The domain of the expression x2 − a2 is (−∞, −a] ∪ [a, +∞). Thus, either x < −a or x > a. Hence, xa ≤ − 1
or xa ≥ 1. Since these intervals correspond to the range of secθ on the set ⎡⎣0, π2⎞⎠ ∪ ⎛⎝π2, π⎤⎦, it makes sense to use the
substitution secθ = xa or, equivalently, x = asecθ, where 0 ≤ θ < π2 or π2 < θ ≤ π. The corresponding substitution
for dx is dx = asecθ tanθdθ. The procedure for using this substitution is outlined in the following problem-solving
strategy.


Problem-Solving Strategy: Integrals Involving x2− a2
1. Check to see whether the integral cannot be evaluated using another method. If so, we may wish to considerapplying an alternative technique.
2. Substitute x = asecθ and dx = asecθ tanθdθ. This substitution yields


x2 − a2 = (asecθ)2 − a2 = a2(sec2 θ + 1) = a2 tan2 θ = |a tanθ|.


For x ≥ a, |a tanθ| = a tanθ and for x ≤ − a, |a tanθ| = −a tanθ.
3. Simplify the expression.
4. Evaluate the integral using techniques from the section on trigonometric integrals.
5. Use the reference triangles from Figure 3.9 to rewrite the result in terms of x. You may also need to use some


trigonometric identities and the relationship θ = sec−1 ⎛⎝xa⎞⎠. (Note: We need both reference triangles, since the
values of some of the trigonometric ratios are different depending on whether x > a or x < −a.)


Chapter 3 | Techniques of Integration 293




Figure 3.9 Use the appropriate reference triangle to express the trigonometric functions evaluated at θ in terms of x.


Example 3.27
Finding the Area of a Region
Find the area of the region between the graph of f (x) = x2 − 9 and the x-axis over the interval [3, 5].
Solution
First, sketch a rough graph of the region described in the problem, as shown in the following figure.


Figure 3.10 Calculating the area of the shaded region requiresevaluating an integral with a trigonometric substitution.


We can see that the area is A = ∫
3


5
x2 − 9dx. To evaluate this definite integral, substitute x = 3secθ and


dx = 3secθ tanθdθ. We must also change the limits of integration. If x = 3, then 3 = 3secθ and hence
θ = 0. If x = 5, then θ = sec−1 ⎛⎝53⎞⎠. After making these substitutions and simplifying, we have


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3.16


Area = ∫
3


5
x2 − 9dx


= ∫
0


sec−1 (5/3)


9tan2 θsecθdθ Use tan2 θ = 1 − sec2 θ.


= ∫
0


sec−1 (5/3)


9(sec2 θ − 1)secθdθ Expand.


= ∫
0


sec−1 (5/3)


9(sec3 θ − secθ)dθ Evaluate the integral.


= ⎛⎝
9
2
ln|secθ + tanθ| + 92secθ tanθ



⎠− 9ln|secθ + tanθ||0sec


−1 (5/3)
Simplify.


= 9
2
secθ tanθ − 9


2
ln|secθ + tanθ||0sec


−1 (5/3) Evaluate. Use sec

⎝sec


−1 5
3

⎠ =


5
3


and tan⎛⎝sec
−1 5


3

⎠ =


4
3
.


= 9
2
· 5
3
· 4
3
− 9


2
ln|53 + 43| − ⎛⎝92 · 1 · 0 − 92ln|1 + 0|⎞⎠


= 10 − 9
2
ln3.


Evaluate ∫ dx
x2 − 4


. Assume that x > 2.


Chapter 3 | Techniques of Integration 295




3.3 EXERCISES
Simplify the following expressions by writing each oneusing a single trigonometric function.
126. 4 − 4sin2 θ
127. 9sec2 θ − 9
128. a2 + a2 tan2 θ
129. a2 + a2 sinh2 θ
130. 16cosh2 θ − 16
Use the technique of completing the square to express eachtrinomial as the square of a binomial.
131. 4x2 − 4x + 1
132. 2x2 − 8x + 3
133. −x2 − 2x + 4
Integrate using the method of trigonometric substitution.Express the final answer in terms of the variable.
134. ∫ dx


4 − x2


135. ∫ dx
x2 − a2


136. ∫ 4 − x2dx


137. ∫ dx
1 + 9x2


138. ∫ x2dx
1 − x2


139. ∫ dx
x2 1 − x2


140. ∫ dx
(1 + x2)2


141. ∫ x2 + 9dx


142. ∫ x2 − 25x dx


143. ∫ θ3dθ
9 − θ2




144. ∫ dx
x6 − x2


145. ∫ x6 − x8dx


146. ∫ dx

⎝1 + x


2⎞


3/2


147. ∫ dx

⎝x


2 − 9⎞⎠
3/2


148. ∫ 1 + x2dxx
149. ∫ x2dx


x2 − 1


150. ∫ x2dx
x2 + 4


151. ∫ dx
x2 x2 + 1


152. ∫ x2dx
1 + x2


153. ∫
−1


1
(1 − x2)3/2dx


In the following exercises, use the substitutions
x = sinhθ, coshθ, or tanhθ. Express the final answers
in terms of the variable x.
154. ∫ dx


x2 − 1


155. ∫ dx
x 1 − x2


156. ∫ x2 − 1dx


157. ∫ x2 − 1
x2


dx


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158. ∫ dx
1 − x2


159. ∫ 1 + x2
x2


dx


Use the technique of completing the square to evaluate thefollowing integrals.
160. ∫ 1


x2 − 6x
dx


161. ∫ 1
x2 + 2x + 1


dx


162. ∫ 1
−x2 + 2x + 8


dx


163. ∫ 1
−x2 + 10x


dx


164. ∫ 1
x2 + 4x − 12


dx


165. Evaluate the integral without using calculus:


−3


3
9 − x2dx.


166. Find the area enclosed by the ellipse x2
4


+
y2


9
= 1.


167. Evaluate the integral ∫ dx
1 − x2


using two different
substitutions. First, let x = cosθ and evaluate using
trigonometric substitution. Second, let x = sinθ and use
trigonometric substitution. Are the answers the same?
168. Evaluate the integral ∫ dx


x x2 − 1
using the


substitution x = secθ. Next, evaluate the same integral
using the substitution x = cscθ. Show that the results are
equivalent.
169. Evaluate the integral ∫ x


x2 + 1
dx using the form


∫ 1udu. Next, evaluate the same integral using x = tanθ.
Are the results the same?
170. State the method of integration you would use to
evaluate the integral ∫ x x2 + 1dx. Why did you choose
this method?


171. State the method of integration you would use to
evaluate the integral ∫ x2 x2 − 1dx. Why did you
choose this method?
172. Evaluate ∫


−1


1
xdx
x2 + 1


173. Find the length of the arc of the curve over thespecified interval: y = lnx, [1, 5]. Round the answer to
three decimal places.
174. Find the surface area of the solid generated byrevolving the region bounded by the graphs of
y = x2, y = 0, x = 0, and x = 2 about the x-axis.
(Round the answer to three decimal places).
175. The region bounded by the graph of f (x) = 1


1 + x2


and the x-axis between x = 0 and x = 1 is revolved about
the x-axis. Find the volume of the solid that is generated.
Solve the initial-value problem for y as a function of x.
176. ⎛⎝x2 + 36⎞⎠dydx = 1, y(6) = 0


177. ⎛⎝64 − x2⎞⎠dydx = 1, y(0) = 3
178. Find the area bounded by
y = 2


64 − 4x2
, x = 0, y = 0, and x = 2.


179. An oil storage tank can be described as the volumegenerated by revolving the area bounded by
y = 16


64 + x2
, x = 0, y = 0, x = 2 about the x-axis. Find


the volume of the tank (in cubic meters).
180. During each cycle, the velocity v (in feet per second)
of a robotic welding device is given by v = 2t − 14


4 + t2
,


where t is time in seconds. Find the expression for thedisplacement s (in feet) as a function of t if s = 0 when
t = 0.


181. Find the length of the curve y = 16 − x2 between
x = 0 and x = 2.


Chapter 3 | Techniques of Integration 297




3.4 | Partial Fractions
Learning Objectives


3.4.1 Integrate a rational function using the method of partial fractions.
3.4.2 Recognize simple linear factors in a rational function.
3.4.3 Recognize repeated linear factors in a rational function.
3.4.4 Recognize quadratic factors in a rational function.


We have seen some techniques that allow us to integrate specific rational functions. For example, we know that
∫ duu = ln|u| + C and ∫ duu2 + a2


= 1atan
−1 ⎛

u
a

⎠+ C.


However, we do not yet have a technique that allows us to tackle arbitrary quotients of this type. Thus, it is not immediately
obvious how to go about evaluating ∫ 3x


x2 − x − 2
dx. However, we know from material previously developed that


∫ ⎛⎝ 1x + 1 +
2


x − 2

⎠dx = ln|x + 1| + 2ln|x − 2| + C.


In fact, by getting a common denominator, we see that
1


x + 1
+ 2
x − 2


= 3x
x2 − x − 2


.


Consequently,
∫ 3x
x2 − x − 2


dx = ∫ ⎛⎝ 1x + 1 +
2


x − 2

⎠dx.


In this section, we examine the method of partial fraction decomposition, which allows us to decompose rational functionsinto sums of simpler, more easily integrated rational functions. Using this method, we can rewrite an expression such as:
3x


x2 − x − 2
as an expression such as 1


x + 1
+ 2
x − 2


.


The key to the method of partial fraction decomposition is being able to anticipate the form that the decomposition of arational function will take. As we shall see, this form is both predictable and highly dependent on the factorization of thedenominator of the rational function. It is also extremely important to keep in mind that partial fraction decomposition
can be applied to a rational function P(x)


Q(x)
only if deg(P(x)) < deg⎛⎝Q(x)⎞⎠. In the case when deg(P(x)) ≥ deg⎛⎝Q(x)⎞⎠, we


must first perform long division to rewrite the quotient P(x)
Q(x)


in the form A(x) + R(x)
Q(x)


, where deg(R(x)) < deg⎛⎝Q(x)⎞⎠.
We then do a partial fraction decomposition on R(x)


Q(x)
. The following example, although not requiring partial fraction


decomposition, illustrates our approach to integrals of rational functions of the form ∫ P(x)
Q(x)


dx, where
deg(P(x)) ≥ deg⎛⎝Q(x)⎞⎠.


Example 3.28
Integrating ∫ P(x)


Q(x)
dx, where deg(P(x)) ≥ deg⎛⎝Q(x)⎞⎠


Evaluate ∫ x2 + 3x + 5
x + 1


dx.


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3.17


Solution
Since deg⎛⎝x2 + 3x + 5⎞⎠ ≥ deg(x + 1), we perform long division to obtain


x2 + 3x + 5
x + 1


= x + 2 + 3
x + 1


.


Thus,
∫ x


2 + 3x + 5
x + 1


dx = ∫ ⎛⎝x + 2 + 3x + 1

⎠dx


= 1
2
x2 + 2x + 3ln|x + 1| + C.


Visit this website (http://www.openstaxcollege.org/l/20_polylongdiv) for a review of long division ofpolynomials.


Evaluate ∫ x − 3
x + 2


dx.


To integrate ∫ P(x)
Q(x)


dx, where deg(P(x)) < deg⎛⎝Q(x)⎞⎠, we must begin by factoring Q(x).
Nonrepeated Linear Factors
If Q(x) can be factored as ⎛⎝a1 x + b1⎞⎠⎛⎝a2 x + b2⎞⎠…⎛⎝an x + bn⎞⎠, where each linear factor is distinct, then it is possible to
find constants A1, A2 ,… An satisfying


P(x)
Q(x)


=
A1


a1 x + b1
+


A2
a2 x + b2


+ ⋯ + An
an x + bn


.


The proof that such constants exist is beyond the scope of this course.
In this next example, we see how to use partial fractions to integrate a rational function of this type.
Example 3.29
Partial Fractions with Nonrepeated Linear Factors
Evaluate ∫ 3x + 2


x3 − x2 − 2x
dx.


Solution
Since deg(3x + 2) < deg⎛⎝x3 − x2 − 2x⎞⎠, we begin by factoring the denominator of 3x + 2


x3 − x2 − 2x
. We can see


that x3 − x2 − 2x = x(x − 2)(x + 1). Thus, there are constants A, B, and C satisfying
3x + 2


x(x − 2)(x + 1)
= Ax +


B
x − 2


+ C
x + 1


.


Chapter 3 | Techniques of Integration 299




(3.8)


We must now find these constants. To do so, we begin by getting a common denominator on the right. Thus,
3x + 2


x(x − 2)(x + 1)
= A(x − 2)(x + 1) + Bx(x + 1) + Cx(x − 2)


x(x − 2)(x + 1)
.


Now, we set the numerators equal to each other, obtaining
3x + 2 = A(x − 2)(x + 1) + Bx(x + 1) + Cx(x − 2).


There are two different strategies for finding the coefficients A, B, and C. We refer to these as the method of
equating coefficients and the method of strategic substitution.
Rule: Method of Equating Coefficients
Rewrite Equation 3.8 in the form


3x + 2 = (A + B + C)x2 + (−A + B − 2C)x + (−2A).


Equating coefficients produces the system of equations
A + B + C = 0


−A + B − 2C = 3


−2A = 2.


To solve this system, we first observe that −2A = 2 ⇒ A = −1. Substituting this value into the first two
equations gives us the system


B + C = 1
B − 2C = 2.


Multiplying the second equation by −1 and adding the resulting equation to the first produces
−3C = 1,


which in turn implies that C = − 1
3
. Substituting this value into the equation B + C = 1 yields B = 4


3
.


Thus, solving these equations yields A = −1, B = 4
3
, and C = − 1


3
.


It is important to note that the system produced by this method is consistent if and only if we have set up thedecomposition correctly. If the system is inconsistent, there is an error in our decomposition.


Rule: Method of Strategic Substitution
The method of strategic substitution is based on the assumption that we have set up the decompositioncorrectly. If the decomposition is set up correctly, then there must be values of A, B, and C that satisfy
Equation 3.8 for all values of x. That is, this equation must be true for any value of x we care to substitute
into it. Therefore, by choosing values of x carefully and substituting them into the equation, we may find
A, B, and C easily. For example, if we substitute x = 0, the equation reduces to 2 = A(−2)(1).
Solving for A yields A = −1. Next, by substituting x = 2, the equation reduces to 8 = B(2)(3),
or equivalently B = 4/3. Last, we substitute x = −1 into the equation and obtain −1 = C(−1)(−3).
Solving, we have C = − 1


3
.


It is important to keep in mind that if we attempt to use this method with a decomposition that has not been


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set up correctly, we are still able to find values for the constants, but these constants are meaningless. If wedo opt to use the method of strategic substitution, then it is a good idea to check the result by recombiningthe terms algebraically.


Now that we have the values of A, B, and C, we rewrite the original integral:
∫ 3x + 2
x3 − x2 − 2x


dx = ∫ ⎛⎝−
1
x +


4
3
· 1
(x − 2)


− 1
3
· 1
(x + 1)



⎠dx.


Evaluating the integral gives us
∫ 3x + 2
x3 − x2 − 2x


dx = −ln|x| + 43
ln|x − 2| − 13


ln|x + 1| + C.


In the next example, we integrate a rational function in which the degree of the numerator is not less than the degree of thedenominator.
Example 3.30
Dividing before Applying Partial Fractions
Evaluate ∫ x2 + 3x + 1


x2 − 4
dx.


Solution
Since degree(x2 + 3x + 1) ≥ degree(x2 − 4), we must perform long division of polynomials. This results in


x2 + 3x + 1
x2 − 4


= 1 + 3x + 5
x2 − 4


.


Next, we perform partial fraction decomposition on 3x + 5
x2 − 4


= 3x + 5
(x + 2)(x − 2)


. We have
3x + 5


(x − 2)(x + 2)
= A


x − 2
+ B
x + 2


.


Thus,
3x + 5 = A(x + 2) + B(x − 2).


Solving for A and B using either method, we obtain A = 11/4 and B = 1/4.
Rewriting the original integral, we have


∫ x
2 + 3x + 1
x2 − 4


dx = ∫ ⎛⎝1 + 114 ·
1


x − 2
+ 1


4
· 1
x + 2

⎠dx.


Evaluating the integral produces
∫ x


2 + 3x + 1
x2 − 4


dx = x + 11
4
ln|x − 2| + 14


ln|x + 2| + C.


Chapter 3 | Techniques of Integration 301




3.18


As we see in the next example, it may be possible to apply the technique of partial fraction decomposition to a nonrationalfunction. The trick is to convert the nonrational function to a rational function through a substitution.
Example 3.31
Applying Partial Fractions after a Substitution
Evaluate ∫ cosx


sin2 x − sinx
dx.


Solution
Let’s begin by letting u = sinx. Consequently, du = cosxdx. After making these substitutions, we have


∫ cosx
sin2 x − sinx


dx = ∫ du
u2 − u


= ∫ du
u(u − 1)


.


Applying partial fraction decomposition to 1/u(u − 1) gives 1
u(u − 1)


= − 1u +
1


u − 1
.


Thus,
∫ cosx


sin2 x − sinx
dx = −ln|u| + ln|u − 1| + C


= −ln|sinx| + ln|sinx − 1| + C.


Evaluate ∫ x + 1
(x + 3)(x − 2)


dx.


Repeated Linear Factors
For some applications, we need to integrate rational expressions that have denominators with repeated linear factors—thatis, rational functions with at least one factor of the form (ax + b)n, where n is a positive integer greater than or equal to
2. If the denominator contains the repeated linear factor (ax + b)n, then the decomposition must contain


A1
ax + b


+
A2


(ax + b)2
+ ⋯ + An


(ax + b)n
.


As we see in our next example, the basic technique used for solving for the coefficients is the same, but it requires morealgebra to determine the numerators of the partial fractions.
Example 3.32
Partial Fractions with Repeated Linear Factors
Evaluate ∫ x − 2


(2x − 1)2(x − 1)
dx.


Solution
We have degree(x − 2) < degree⎛⎝(2x − 1)2 (x − 1)⎞⎠, so we can proceed with the decomposition. Since


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(3.9)


3.19


(2x − 1)2 is a repeated linear factor, include A
2x − 1


+ B
(2x − 1)2


in the decomposition. Thus,
x − 2


(2x − 1)2(x − 1)
= A


2x − 1
+ B


(2x − 1)2
+ C
x − 1


.


After getting a common denominator and equating the numerators, we have
x − 2 = A(2x − 1)(x − 1) + B(x − 1) + C(2x − 1)2.


We then use the method of equating coefficients to find the values of A, B, and C.
x − 2 = (2A + 4C)x2 + (−3A + B − 4C)x + (A − B + C).


Equating coefficients yields 2A + 4C = 0, −3A + B − 4C = 1, and A − B + C = −2. Solving this system
yields A = 2, B = 3, and C = −1.
Alternatively, we can use the method of strategic substitution. In this case, substituting x = 1 and x = 1/2 into
Equation 3.9 easily produces the values B = 3 and C = −1. At this point, it may seem that we have run out
of good choices for x, however, since we already have values for B and C, we can substitute in these values
and choose any value for x not previously used. The value x = 0 is a good option. In this case, we obtain the
equation −2 = A(−1)(−1) + 3(−1) + (−1)(−1)2 or, equivalently, A = 2.
Now that we have the values for A, B, and C, we rewrite the original integral and evaluate it:


∫ x − 2
(2x − 1)2(x − 1)


dx = ∫



⎜ 2
2x − 1


+ 3
(2x − 1)2


− 1
x − 1





⎟dx


= ln|2x − 1| − 32(2x − 1)
− ln|x − 1| + C.


Set up the partial fraction decomposition for ∫ x + 2
(x + 3)3 (x − 4)2


dx. (Do not solve for the coefficients
or complete the integration.)


The General Method
Now that we are beginning to get the idea of how the technique of partial fraction decomposition works, let’s outline thebasic method in the following problem-solving strategy.
Problem-Solving Strategy: Partial Fraction Decomposition
To decompose the rational function P(x)/Q(x), use the following steps:


1. Make sure that degree(P(x)) < degree(Q(x)). If not, perform long division of polynomials.
2. Factor Q(x) into the product of linear and irreducible quadratic factors. An irreducible quadratic is a quadratic


that has no real zeros.
3. Assuming that deg(P(x)) < deg(Q(x)), the factors of Q(x) determine the form of the decomposition of


P(x)/Q(x).


a. If Q(x) can be factored as ⎛⎝a1 x + b1⎞⎠⎛⎝a2 x + b2⎞⎠…⎛⎝an x + bn⎞⎠, where each linear factor is distinct,


Chapter 3 | Techniques of Integration 303




then it is possible to find constants A1, A2, ...An satisfying
P(x)
Q(x)


=
A1


a1 x + b1
+


A2
a2 x + b2


+ ⋯ + An
an x + bn


.


b. If Q(x) contains the repeated linear factor (ax + b)n, then the decomposition must contain
A1


ax + b
+


A2
(ax + b)2


+ ⋯ + An
(ax + b)n


.


c. For each irreducible quadratic factor ax2 + bx + c that Q(x) contains, the decomposition must
include


Ax + B
ax2 + bx + c


.


d. For each repeated irreducible quadratic factor ⎛⎝ax2 + bx + c⎞⎠n, the decomposition must include
A1 x + B1


ax2 + bx + c
+


A2 x + B2
(ax2 + bx + c)2


+ ⋯ + An x + Bn
(ax2 + bx + c)n


.


e. After the appropriate decomposition is determined, solve for the constants.
f. Last, rewrite the integral in its decomposed form and evaluate it using previously developed techniquesor integration formulas.


Simple Quadratic Factors
Now let’s look at integrating a rational expression in which the denominator contains an irreducible quadratic factor. Recall
that the quadratic ax2 + bx + c is irreducible if ax2 + bx + c = 0 has no real zeros—that is, if b2 − 4ac < 0.
Example 3.33
Rational Expressions with an Irreducible Quadratic Factor
Evaluate ∫ 2x − 3


x3 + x
dx.


Solution
Since deg(2x − 3) < deg(x3 + x), factor the denominator and proceed with partial fraction decomposition.
Since x3 + x = x(x2 + 1) contains the irreducible quadratic factor x2 + 1, include Ax + B


x2 + 1
as part of the


decomposition, along with Cx for the linear term x. Thus, the decomposition has the form
2x − 3


x(x2 + 1)
= Ax + B


x2 + 1
+ Cx .


After getting a common denominator and equating the numerators, we obtain the equation
2x − 3 = (Ax + B)x + C⎛⎝x


2 + 1⎞⎠.


Solving for A, B, and C, we get A = 3, B = 2, and C = −3.
Thus,


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2x − 3
x3 + x


= 3x + 2
x2 + 1


− 3x .


Substituting back into the integral, we obtain
∫ 2x − 3


x3 + x
dx = ∫ ⎛⎝


3x + 2
x2 + 1


− 3x

⎠dx


= 3∫ x
x2 + 1


dx + 2∫ 1
x2 + 1


dx − 3∫ 1xdx Split up the integral.


= 3
2
ln|x2 + 1| + 2tan−1 x − 3ln|x| + C. Evaluate each integral.


Note: We may rewrite ln|x2 + 1| = ln(x2 + 1), if we wish to do so, since x2 + 1 > 0.


Example 3.34
Partial Fractions with an Irreducible Quadratic Factor
Evaluate ∫ dx


x3 − 8
.


Solution
We can start by factoring x3 − 8 = (x − 2)(x2 + 2x + 4). We see that the quadratic factor x2 + 2x + 4 is
irreducible since 22 − 4(1)(4) = −12 < 0. Using the decomposition described in the problem-solving strategy,
we get


1
(x − 2)(x2 + 2x + 4)


= A
x − 2


+ Bx + C
x2 + 2x + 4


.


After obtaining a common denominator and equating the numerators, this becomes
1 = A⎛⎝x


2 + 2x + 4⎞⎠+ (Bx + C)(x − 2).


Applying either method, we get A = 1
12


, B = − 1
12


, andC = − 1
3
.


Rewriting ∫ dx
x3 − 8


, we have
∫ dx
x3 − 8


= 1
12
∫ 1
x − 2


dx − 1
12
∫ x + 4
x2 + 2x + 4


dx.


We can see that
∫ 1
x − 2


dx = ln|x − 2| + C, but ∫ x + 4
x2 + 2x + 4


dx requires a bit more effort. Let’s begin by completing the
square on x2 + 2x + 4 to obtain


x2 + 2x + 4 = (x + 1)2 + 3.


Chapter 3 | Techniques of Integration 305




By letting u = x + 1 and consequently du = dx, we see that
∫ x + 4
x2 + 2x + 4


dx = ∫ x + 4
(x + 1)2 + 3


dx
Complete the square on the


denominator.


= ∫ u + 3
u2 + 3


du
Substitute u = x + 1, x = u − 1,
and du = dx.


= ∫ u
u2 + 3


du + ∫ 3
u2 + 3


du Split the numerator apart.


= 1
2
ln|u2 + 3| + 33tan


−1 u
3
+ C Evaluate each integral.


= 1
2
ln|x2 + 2x + 4| + 3tan−1 ⎛⎝x + 13



⎠+ C.


Rewrite in terms of x and
simplify.


Substituting back into the original integral and simplifying gives
∫ dx


x3 − 8
= 1


12
ln|x − 2| − 124


ln|x2 + 2x + 4| − 312tan−1


x + 1


3

⎠+ C.


Here again, we can drop the absolute value if we wish to do so, since x2 + 2x + 4 > 0 for all x.


Example 3.35
Finding a Volume
Find the volume of the solid of revolution obtained by revolving the region enclosed by the graph of
f (x) = x


2



⎝x


2 + 1⎞⎠
2
and the x-axis over the interval [0, 1] about the y-axis.


Solution
Let’s begin by sketching the region to be revolved (see Figure 3.11). From the sketch, we see that the shellmethod is a good choice for solving this problem.


Figure 3.11 We can use the shell method to find the volumeof revolution obtained by revolving the region shown about they-axis.


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3.20


The volume is given by
V = 2π∫


0


1
x · x


2



⎝x


2 + 1⎞⎠
2
dx = 2π∫


0


1
x3


(x2 + 1)2
dx.


Since deg⎛⎝⎛⎝x2 + 1⎞⎠
2⎞
⎠ = 4 > 3 = deg(x


3), we can proceed with partial fraction decomposition. Note that
(x2 + 1)2 is a repeated irreducible quadratic. Using the decomposition described in the problem-solving strategy,
we get


x3


(x2 + 1)2
= Ax + B


x2 + 1
+ Cx + D


(x2 + 1)2
.


Finding a common denominator and equating the numerators gives
x3 = (Ax + B)⎛⎝x


2 + 1⎞⎠+ Cx + D.


Solving, we obtain A = 1, B = 0, C = −1, and D = 0. Substituting back into the integral, we have
V = 2π∫


0


1
x3


(x2 + 1)2
dx


= 2π∫
0


1 ⎛



⎜ x
x2 + 1


− x
(x2 + 1)2





⎟dx


= 2π


1
2
ln(x2 + 1) + 1


2
· 1
x2 + 1



⎠|01


= π⎛⎝ln2 −
1
2

⎠.


Set up the partial fraction decomposition for ∫ x2 + 3x + 1
(x + 2)(x − 3)2 (x2 + 4)2


dx.


Chapter 3 | Techniques of Integration 307




3.4 EXERCISES
Express the rational function as a sum or difference of twosimpler rational expressions.
182. 1


(x − 3)(x − 2)


183. x2 + 1
x(x + 1)(x + 2)


184. 1
x3 − x


185. 3x + 1
x2


186. 3x2
x2 + 1


(Hint: Use long division first.)


187. 2x4
x2 − 2x


188. 1
(x − 1)(x2 + 1)


189. 1
x2(x − 1)


190. x
x2 − 4


191. 1
x(x − 1)(x − 2)(x − 3)


192. 1
x4 − 1


= 1
(x + 1)(x − 1)⎛⎝x


2 + 1⎞⎠


193. 3x2
x3 − 1


= 3x
2


(x − 1)(x2 + x + 1)


194. 2x
(x + 2)2


195. 3x4 + x3 + 20x2 + 3x + 31
(x + 1)⎛⎝x


2 + 4⎞⎠
2


Use the method of partial fractions to evaluate each of thefollowing integrals.
196. ∫ dx


(x − 3)(x − 2)


197. ∫ 3x
x2 + 2x − 8


dx


198. ∫ dx
x3 − x


199. ∫ x
x2 − 4


dx


200. ∫ dx
x(x − 1)(x − 2)(x − 3)


201. ∫ 2x2 + 4x + 22
x2 + 2x + 10


dx


202. ∫ dx
x2 − 5x + 6


203. ∫ 2 − x
x2 + x


dx


204. ∫ 2
x2 − x − 6


dx


205. ∫ dx
x3 − 2x2 − 4x + 8


206. ∫ dx
x4 − 10x2 + 9


Evaluate the following integrals, which have irreduciblequadratic factors.
207. ∫ 2


(x − 4)⎛⎝x
2 + 2x + 6⎞⎠


dx


208. ∫ x2
x3 − x2 + 4x − 4


dx


209. ∫ x3 + 6x2 + 3x + 6
x3 + 2x2


dx


210. ∫ x
(x − 1)⎛⎝x


2 + 2x + 2⎞⎠
2
dx


Use the method of partial fractions to evaluate thefollowing integrals.
211. ∫ 3x + 4⎛


⎝x
2 + 4⎞⎠(3 − x)


dx


212. ∫ 2
(x + 2)2(2 − x)


dx


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213. ∫ 3x + 4
x3 − 2x − 4


dx (Hint: Use the rational root
theorem.)
Use substitution to convert the integrals to integrals ofrational functions. Then use partial fractions to evaluate theintegrals.
214. ∫


0


1
ex


36 − e2x
dx (Give the exact answer and the


decimal equivalent. Round to five decimal places.)
215. ∫ ex dx


e2x − ex
dx


216. ∫ sinxdx
1 − cos2 x


217. ∫ sinx
cos2 x + cosx − 6


dx


218. ∫ 1 − x
1 + x


dx


219. ∫ dt

⎝e
t − e−t⎞⎠


2


220. ∫ 1 + ex
1 − ex


dx


221. ∫ dx
1 + x + 1


222. ∫ dx
x + x4


223. ∫ cosx
sinx(1 − sinx)


dx


224. ∫ ex

⎝e


2x − 4⎞⎠
2
dx


225. ∫
1


2
1


x2 4 − x2
dx


226. ∫ 1
2 + e−x


dx


227. ∫ 1
1 + ex


dx


Use the given substitution to convert the integral to anintegral of a rational function, then evaluate.


228. ∫ 1
t − t3


dt t = x3


229. ∫ 1
x + x3


dx; x = u6


230. Graph the curve y = x
1 + x


over the interval ⎡⎣0, 5⎤⎦.
Then, find the area of the region bounded by the curve, thex-axis, and the line x = 4.


231. Find the volume of the solid generated when the
region bounded by y = 1/ x(3 − x), y = 0, x = 1,
and x = 2 is revolved about the x-axis.
232. The velocity of a particle moving along a line is a
function of time given by v(t) = 88t2


t2 + 1
. Find the distance


that the particle has traveled after t = 5 sec.
Solve the initial-value problem for x as a function of t.
233. ⎛⎝t2 − 7t + 12⎞⎠dxdt = 1, ⎛⎝t > 4, x(5) = 0⎞⎠


234. (t + 5)dx
dt


= x2 + 1, t > −5, x(1) = tan1


235. ⎛⎝2t3 − 2t2 + t − 1⎞⎠dxdt = 3, x(2) = 0
236. Find the x-coordinate of the centroid of the area
bounded by y⎛⎝x2 − 9⎞⎠ = 1, y = 0, x = 4, and x = 5.
(Round the answer to two decimal places.)
237. Find the volume generated by revolving the area
bounded by y = 1


x3 + 7x2 + 6x
x = 1, x = 7, and y = 0


about the y-axis.
238. Find the area bounded by y = x − 12


x2 − 8x − 20
,


y = 0, x = 2, and x = 4. (Round the answer to the
nearest hundredth.)


Chapter 3 | Techniques of Integration 309




239. Evaluate the integral ∫ dx
x3 + 1


.


For the following problems, use the substitutions
tan⎛⎝


x
2

⎠ = t, dx =


2
1 + t2


dt, sinx = 2t
1 + t2


, and
cosx = 1 − t


2


1 + t2
.


240. ∫ dx
3 − 5sinx


241. Find the area under the curve y = 1
1 + sinx


between
x = 0 and x = π. (Assume the dimensions are in inches.)
242. Given tan⎛⎝x2⎞⎠ = t, derive the formulas
dx = 2


1 + t2
dt, sinx = 2t


1 + t2
, and cosx = 1 − t2


1 + t2
.


243. Evaluate ∫ x − 83 x dx.


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3.5 | Other Strategies for Integration
Learning Objectives


3.5.1 Use a table of integrals to solve integration problems.
3.5.2 Use a computer algebra system (CAS) to solve integration problems.


In addition to the techniques of integration we have already seen, several other tools are widely available to assist with theprocess of integration. Among these tools are integration tables, which are readily available in many books, including theappendices to this one. Also widely available are computer algebra systems (CAS), which are found on calculators and inmany campus computer labs, and are free online.
Tables of Integrals
Integration tables, if used in the right manner, can be a handy way either to evaluate or check an integral quickly. Keep inmind that when using a table to check an answer, it is possible for two completely correct solutions to look very different.For example, in Trigonometric Substitution, we found that, by using the substitution x = tanθ, we can arrive at


∫ dx
1 + x2


= ln⎛⎝x + x
2 + 1⎞⎠+ C.


However, using x = sinhθ, we obtained a different solution—namely,
∫ dx


1 + x2
= sinh−1 x + C.


We later showed algebraically that the two solutions are equivalent. That is, we showed that sinh−1 x = ln⎛⎝x + x2 + 1⎞⎠.
In this case, the two antiderivatives that we found were actually equal. This need not be the case. However, as long as thedifference in the two antiderivatives is a constant, they are equivalent.
Example 3.36
Using a Formula from a Table to Evaluate an Integral
Use the table formula


∫ a2 − u2
u2


du = − a
2 − u2
u − sin


−1 u
a + C


to evaluate ∫ 16 − e2x
ex


dx.


Solution
If we look at integration tables, we see that several formulas contain expressions of the form a2 − u2. This
expression is actually similar to 16 − e2x, where a = 4 and u = ex. Keep in mind that we must also have
du = ex. Multiplying the numerator and the denominator of the given integral by ex should help to put this
integral in a useful form. Thus, we now have


∫ 16 − e
2x


ex
dx = ∫ 16 − e


2x


e2x
ex dx.


Chapter 3 | Techniques of Integration 311




Substituting u = ex and du = ex produces ∫ a2 − u2
u2


du. From the integration table (#88 in Appendix A),


∫ a2 − u2
u2


du = − a
2 − u2
u − sin


−1 u
a + C.


Thus,
∫ 16 − e


2x


ex
dx = ∫ 16 − e


2x


e2x
ex dx Substitute u = ex and du = ex dx.


= ∫ 42 − u2
u2


du Apply the formula using a = 4.


= − 4
2 − u2
u − sin


−1 u
4
+ C Substitute u = ex.


= − 16 − e
2x


u − sin
−1 ⎛

ex
4

⎠+ C.


Computer Algebra Systems
If available, a CAS is a faster alternative to a table for solving an integration problem. Many such systems are widelyavailable and are, in general, quite easy to use.
Example 3.37
Using a Computer Algebra System to Evaluate an Integral
Use a computer algebra system to evaluate ∫ dx


x2 − 4
. Compare this result with ln| x2 − 42 + x2| + C, a result


we might have obtained if we had used trigonometric substitution.
Solution
Using Wolfram Alpha, we obtain


∫ dx
x2 − 4


= ln| x2 − 4 + x| + C.
Notice that


ln| x2 − 42 + x2| + C = ln| x2 − 4 + x2 | + C = ln| x2 − 4 + x| − ln2 + C.
Since these two antiderivatives differ by only a constant, the solutions are equivalent. We could have alsodemonstrated that each of these antiderivatives is correct by differentiating them.


You can access an integral calculator (http://www.openstaxcollege.org/l/20_intcalc) for more examples.


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3.21


Example 3.38
Using a CAS to Evaluate an Integral
Evaluate ∫ sin3 xdx using a CAS. Compare the result to 1


3
cos3 x − cosx + C, the result we might have


obtained using the technique for integrating odd powers of sinx discussed earlier in this chapter.
Solution
Using Wolfram Alpha, we obtain


∫ sin3 xdx = 1
12


(cos(3x) − 9cosx) + C.


This looks quite different from 1
3
cos3 x − cosx + C. To see that these antiderivatives are equivalent, we can


make use of a few trigonometric identities:
1
12


(cos(3x) − 9cosx) = 1
12


(cos(x + 2x) − 9cosx)


= 1
12


(cos(x)cos(2x) − sin(x)sin(2x) − 9cosx)


= 1
12


(cosx⎛⎝2cos
2 x − 1⎞⎠− sinx(2sinxcosx) − 9cosx)


= 1
12


(2cos x − cosx − 2cosx⎛⎝1 − cos
2 x⎞⎠− 9cosx)


= 1
12


(4cos x − 12cosx)


= 1
3
cos x − cosx.


Thus, the two antiderivatives are identical.
We may also use a CAS to compare the graphs of the two functions, as shown in the following figure.


Figure 3.12 The graphs of y = 1
3
cos3 x − cosx and


y = 1
12


(cos(3x) − 9cosx) are identical.


Use a CAS to evaluate ∫ dx
x2 + 4


.


Chapter 3 | Techniques of Integration 313




3.5 EXERCISES
Use a table of integrals to evaluate the following integrals.


244. ∫
0


4
x


1 + 2x
dx


245. ∫ x + 3
x2 + 2x + 2


dx


246. ∫ x3 1 + 2x2dx


247. ∫ 1
x2 + 6x


dx


248. ∫ x
x + 1


dx


249. ∫ x · 2x2dx


250. ∫ 1
4x2 + 25


dx


251. ∫ dy
4 − y2


252. ∫ sin3(2x)cos(2x)dx


253. ∫ csc(2w)cot(2w)dw


254. ∫ 2ydy


255. ∫
0


1
3xdx


x2 + 8


256. ∫
−1/4


1/4
sec2(πx)tan(πx)dx


257. ∫
0


π/2
tan2 ⎛⎝


x
2

⎠dx


258. ∫ cos3 xdx


259. ∫ tan5 (3x)dx


260. ∫ sin2 ycos3 ydy
Use a CAS to evaluate the following integrals. Tables can


also be used to verify the answers.
261. [T] ∫ dw


1 + sec⎛⎝
w
2



262. [T] ∫ dw
1 − cos(7w)


263. [T] ∫
0


t
dt


4cos t + 3sin t


264. [T] ∫ x2 − 9
3x


dx


265. [T] ∫ dx
x1/2 + x1/3


266. [T] ∫ dx
x x − 1


267. [T] ∫ x3 sinxdx


268. [T] ∫ x x4 − 9dx


269. [T] ∫ x
1 + e−x


2
dx


270. [T] ∫ 3 − 5x
2x


dx


271. [T] ∫ dx
x x − 1


272. [T] ∫ ex cos−1(ex)dx
Use a calculator or CAS to evaluate the following integrals.
273. [T] ∫


0


π/4
cos(2x)dx


274. [T] ∫
0


1
x · e−x


2
dx


275. [T] ∫
0


8
2x


x2 + 36
dx


276. [T] ∫
0


2/ 3
1


4 + 9x2
dx


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277. [T] ∫ dx
x2 + 4x + 13


278. [T] ∫ dx
1 + sinx


Use tables to evaluate the integrals. You may need tocomplete the square or change variables to put the integralinto a form given in the table.
279. ∫ dx


x2 + 2x + 10


280. ∫ dx
x2 − 6x


281. ∫ ex
e2x − 4


dx


282. ∫ cosx
sin2 x + 2sinx


dx


283. ∫ arctan

⎝x


3⎞


x4
dx


284. ∫ ln|x|arcsin(ln|x|)x dx
Use tables to perform the integration.
285. ∫ dx


x2 + 16


286. ∫ 3x
2x + 7


dx


287. ∫ dx
1 − cos(4x)


288. ∫ dx
4x + 1


289. Find the area bounded by
y⎛⎝4 + 25x


2⎞
⎠ = 5, x = 0, y = 0, and x = 4. Use a table of


integrals or a CAS.
290. The region bounded between the curve
y = 1


1 + cosx
, 0.3 ≤ x ≤ 1.1, and the x-axis is


revolved about the x-axis to generate a solid. Use a table ofintegrals to find the volume of the solid generated. (Roundthe answer to two decimal places.)


291. Use substitution and a table of integrals to find thearea of the surface generated by revolving the curve
y = ex, 0 ≤ x ≤ 3, about the x-axis. (Round the answer
to two decimal places.)
292. [T] Use an integral table and a calculator to findthe area of the surface generated by revolving the curve
y = x


2


2
, 0 ≤ x ≤ 1, about the x-axis. (Round the answer


to two decimal places.)
293. [T]Use a CAS or tables to find the area of the surfacegenerated by revolving the curve y = cosx, 0 ≤ x ≤ π


2
,


about the x-axis. (Round the answer to two decimalplaces.)
294. Find the length of the curve y = x2


4
over [0, 8].


295. Find the length of the curve y = ex over ⎡⎣0, ln(2)⎤⎦.
296. Find the area of the surface formed by revolvingthe graph of y = 2 x over the interval [0, 9] about the
x-axis.
297. Find the average value of the function
f (x) = 1


x2 + 1
over the interval [−3, 3].


298. Approximate the arc length of the curve y = tan(πx)
over the interval ⎡⎣0, 14⎤⎦. (Round the answer to three
decimal places.)


Chapter 3 | Techniques of Integration 315




3.6 | Numerical Integration
Learning Objectives


3.6.1 Approximate the value of a definite integral by using the midpoint and trapezoidal rules.
3.6.2 Determine the absolute and relative error in using a numerical integration technique.
3.6.3 Estimate the absolute and relative error using an error-bound formula.
3.6.4 Recognize when the midpoint and trapezoidal rules over- or underestimate the true valueof an integral.
3.6.5 Use Simpson’s rule to approximate the value of a definite integral to a given accuracy.


The antiderivatives of many functions either cannot be expressed or cannot be expressed easily in closed form (that is,in terms of known functions). Consequently, rather than evaluate definite integrals of these functions directly, we resortto various techniques of numerical integration to approximate their values. In this section we explore several of thesetechniques. In addition, we examine the process of estimating the error in using these techniques.
The Midpoint Rule
Earlier in this text we defined the definite integral of a function over an interval as the limit of Riemann sums. In general,
any Riemann sum of a function f (x) over an interval [a, b] may be viewed as an estimate of ∫


a


b
f (x)dx. Recall that a


Riemann sum of a function f (x) over an interval [a, b] is obtained by selecting a partition
P = {x0, x1, x2 ,…, xn}, where a = x0 < x1 < x2 < ⋯ < xn = b


and a set
S =





⎨x1* , x2* ,…, xn*





⎬, where xi − 1 ≤ xi* ≤ xi for all i.


The Riemann sum corresponding to the partition P and the set S is given by ∑
i = 1


n


f (xi* )Δxi, where Δxi = xi − xi − 1,
the length of the ith subinterval.
Themidpoint rule for estimating a definite integral uses a Riemann sum with subintervals of equal width and the midpoints,
mi, of each subinterval in place of xi* . Formally, we state a theorem regarding the convergence of the midpoint rule as
follows.
Theorem 3.3: The Midpoint Rule
Assume that f (x) is continuous on ⎡⎣a, b⎤⎦. Let n be a positive integer and Δx = b − an . If ⎡⎣a, b⎤⎦ is divided into n
subintervals, each of length Δx, and mi is the midpoint of the ith subinterval, set


(3.10)
Mn = ∑


i = 1


n


f (mi)Δx.


Then lim
n → ∞


Mn = ∫
a


b
f (x)dx.


As we can see in Figure 3.13, if f (x) ≥ 0 over [a, b], then ∑
i = 1


n


f (mi)Δx corresponds to the sum of the areas of
rectangles approximating the area between the graph of f (x) and the x-axis over ⎡⎣a, b⎤⎦. The graph shows the rectangles
corresponding to M4 for a nonnegative function over a closed interval [a, b].


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Figure 3.13 The midpoint rule approximates the area betweenthe graph of f (x) and the x-axis by summing the areas of
rectangles with midpoints that are points on f (x).


Example 3.39
Using the Midpoint Rule with M4
Use the midpoint rule to estimate ∫


0


1
x2dx using four subintervals. Compare the result with the actual value of


this integral.
Solution
Each subinterval has length Δx = 1 − 0


4
= 1


4
. Therefore, the subintervals consist of



⎣0,


1
4

⎦,


1
4
, 1
2

⎦,


1
2
, 3
4

⎦, and




3
4
, 1⎤⎦.


The midpoints of these subintervals are ⎧

⎨1
8
, 3
8
, 5
8
, 7
8



⎬. Thus,


M4 =
1
4
f ⎛⎝
1
8

⎠+


1
4
f ⎛⎝
3
8

⎠+


1
4
f ⎛⎝
5
8

⎠+


1
4
f ⎛⎝
7
8

⎠ =


1
4
· 1
64


+ 1
4
· 9
64


+ 1
4
· 25
64


+ 1
4
· 21
64


= 21
64


.


Since


0


1
x2dx = 1


3
and |13 − 2164| = 1192 ≈ 0.0052,


we see that the midpoint rule produces an estimate that is somewhat close to the actual value of the definiteintegral.


Example 3.40
Using the Midpoint Rule with M6
Use M6 to estimate the length of the curve y = 12x2 on [1, 4].


Chapter 3 | Techniques of Integration 317




3.22


Solution
The length of y = 1


2
x2 on [1, 4] is



1


4
1 +


dy
dx



2


dx.


Since dy
dx


= x, this integral becomes ∫
1


4
1 + x2dx.


If [1, 4] is divided into six subintervals, then each subinterval has length Δx = 4 − 1
6


= 1
2
and the midpoints


of the subintervals are ⎧

⎨5
4
, 7
4
, 9
4
, 11


4
, 13


4
, 15


4



⎬. If we set f (x) = 1 + x2,


M6 =
1
2
f ⎛⎝
5
4

⎠+


1
2
f ⎛⎝
7
4

⎠+


1
2
f ⎛⎝
9
4

⎠+


1
2
f ⎛⎝
11
4

⎠+


1
2
f ⎛⎝
13
4

⎠+


1
2
f ⎛⎝
15
4



≈ 1
2
(1.6008 + 2.0156 + 2.4622 + 2.9262 + 3.4004 + 3.8810) = 8.1431.


Use the midpoint rule with n = 2 to estimate ∫
1


2
1
xdx.


The Trapezoidal Rule
We can also approximate the value of a definite integral by using trapezoids rather than rectangles. In Figure 3.14, the areabeneath the curve is approximated by trapezoids rather than by rectangles.


Figure 3.14 Trapezoids may be used to approximate the areaunder a curve, hence approximating the definite integral.


The trapezoidal rule for estimating definite integrals uses trapezoids rather than rectangles to approximate the area undera curve. To gain insight into the final form of the rule, consider the trapezoids shown in Figure 3.14. We assume that thelength of each subinterval is given by Δx. First, recall that the area of a trapezoid with a height of h and bases of length
b1 and b2 is given by Area = 12h(b1 + b2). We see that the first trapezoid has a height Δx and parallel bases of length
f (x0) and f (x1). Thus, the area of the first trapezoid in Figure 3.14 is


1
2
Δx( f (x0) + f (x1)).


The areas of the remaining three trapezoids are


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1
2
Δx( f (x1) + f (x2)),


1
2
Δx( f (x2) + f (x3)), and


1
2
Δx( f (x3) + f (x4)).


Consequently,

a


b
f (x)dx ≈ 1


2
Δx( f (x0) + f (x1)) +


1
2
Δx( f (x1) + f (x2)) +


1
2
Δx( f (x2) + f (x3)) +


1
2
Δx( f (x3) + f (x4)).


After taking out a common factor of 1
2
Δx and combining like terms, we have



a


b
f (x)dx ≈ 1


2
Δx⎛⎝ f (x0) + 2 f (x1) + 2 f (x2) + 2 f (x3) + f (x4)



⎠.


Generalizing, we formally state the following rule.
Theorem 3.4: The Trapezoidal Rule
Assume that f (x) is continuous over ⎡⎣a, b⎤⎦. Let n be a positive integer and Δx = b − an . Let ⎡⎣a, b⎤⎦ be divided into
n subintervals, each of length Δx, with endpoints at P = ⎧⎩⎨x0, x1, x2…, xn⎫⎭⎬. Set


(3.11)Tn = 12Δx⎛⎝ f (x0) + 2 f (x1) + 2 f (x2) + ⋯ + 2 f (xn − 1) + f (xn)⎞⎠.
Then, lim


n → +∞
Tn = ∫


a


b
f (x)dx.


Before continuing, let’s make a few observations about the trapezoidal rule. First of all, it is useful to note that
Tn = 12


(Ln + Rn) where Ln = ∑
i = 1


n


f (xi − 1)Δx and Rn = ∑
i = 1


n


f (xi)Δx.


That is, Ln and Rn approximate the integral using the left-hand and right-hand endpoints of each subinterval, respectively.
In addition, a careful examination of Figure 3.15 leads us to make the following observations about using the trapezoidalrules and midpoint rules to estimate the definite integral of a nonnegative function. The trapezoidal rule tends tooverestimate the value of a definite integral systematically over intervals where the function is concave up and tounderestimate the value of a definite integral systematically over intervals where the function is concave down. On the otherhand, the midpoint rule tends to average out these errors somewhat by partially overestimating and partially underestimatingthe value of the definite integral over these same types of intervals. This leads us to hypothesize that, in general, themidpoint rule tends to be more accurate than the trapezoidal rule.


Figure 3.15 The trapezoidal rule tends to be less accurate than the midpoint rule.


Chapter 3 | Techniques of Integration 319




3.23


Example 3.41
Using the Trapezoidal Rule
Use the trapezoidal rule to estimate ∫


0


1
x2dx using four subintervals.


Solution
The endpoints of the subintervals consist of elements of the set P = ⎧



⎨0, 1


4
, 1
2
, 3
4
, 1



⎬ and Δx = 1 − 0


4
= 1


4
.


Thus,


0


1
x2dx ≈ 1


2
· 1
4

⎝ f (0) + 2 f




1
4

⎠+ 2 f




1
2

⎠+ 2 f




3
4

⎠+ f (1)





= 1
8

⎝0 + 2 ·


1
16


+ 2 · 1
4
+ 2 · 9


16
+ 1⎞⎠


= 11
32


.


Use the trapezoidal rule with n = 2 to estimate ∫
1


2
1
xdx.


Absolute and Relative Error
An important aspect of using these numerical approximation rules consists of calculating the error in using them forestimating the value of a definite integral. We first need to define absolute error and relative error.
Definition
If B is our estimate of some quantity having an actual value of A, then the absolute error is given by |A − B|. The
relative error is the error as a percentage of the absolute value and is given by |A − BA | = |A − BA | · 100%.


Example 3.42
Calculating Error in the Midpoint Rule
Calculate the absolute and relative error in the estimate of ∫


0


1
x2dx using the midpoint rule, found in Example


3.39.
Solution
The calculated value is ∫


0


1
x2dx = 1


3
and our estimate from the example is M4 = 2164. Thus, the absolute error


is given by |⎛⎝13⎞⎠− ⎛⎝2164⎞⎠| = 1192 ≈ 0.0052. The relative error is


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3.24


1/192
1/3


= 1
64


≈ 0.015625 ≈ 1.6%.


Example 3.43
Calculating Error in the Trapezoidal Rule
Calculate the absolute and relative error in the estimate of ∫


0


1
x2dx using the trapezoidal rule, found in


Example 3.41.
Solution
The calculated value is ∫


0


1
x2dx = 1


3
and our estimate from the example is T4 = 1132. Thus, the absolute error


is given by |13 − 1132| = 196 ≈ 0.0104. The relative error is given by
1/96
1/3


= 0.03125 ≈ 3.1%.


In an earlier checkpoint, we estimated ∫
1


2
1
xdx to be 2435 using T2. The actual value of this integral is


ln2. Using 24
35


≈ 0.6857 and ln2 ≈ 0.6931, calculate the absolute error and the relative error.


In the two previous examples, we were able to compare our estimate of an integral with the actual value of the integral;however, we do not typically have this luxury. In general, if we are approximating an integral, we are doing so because wecannot compute the exact value of the integral itself easily. Therefore, it is often helpful to be able to determine an upperbound for the error in an approximation of an integral. The following theorem provides error bounds for the midpoint andtrapezoidal rules. The theorem is stated without proof.
Theorem 3.5: Error Bounds for the Midpoint and Trapezoidal Rules
Let f (x) be a continuous function over ⎡⎣a, b⎤⎦, having a second derivative f ″(x) over this interval. If M is the
maximum value of | f ″(x)| over [a, b], then the upper bounds for the error in using Mn and Tn to estimate

a


b
f (x)dx are


(3.12)
Error inMn ≤


M(b − a)3


24n2


and
(3.13)


Error in Tn ≤
M(b − a)3


12n2
.


Chapter 3 | Techniques of Integration 321




3.25


We can use these bounds to determine the value of n necessary to guarantee that the error in an estimate is less than a
specified value.
Example 3.44
Determining the Number of Intervals to Use
What value of n should be used to guarantee that an estimate of ∫


0


1
ex


2
dx is accurate to within 0.01 if we use


the midpoint rule?
Solution
We begin by determining the value of M, the maximum value of | f ″(x)| over [0, 1] for f (x) = ex2. Since
f ′ (x) = 2xex


2
, we have


f ″ (x) = 2ex
2
+ 4x2 ex


2
.


Thus,
| f ″(x)| = 2ex


2 ⎛
⎝1 + 2x


2⎞
⎠ ≤ 2 · e · 3 = 6e.


From the error-bound Equation 3.12, we have
Error inMn ≤


M(b − a)3


24n2
≤ 6e(1 − 0)


3


24n2
= 6e


24n2
.


Now we solve the following inequality for n:
6e


24n2
≤ 0.01.


Thus, n ≥ 600e
24


≈ 8.24. Since n must be an integer satisfying this inequality, a choice of n = 9 would
guarantee that |∫01ex2dx − Mn| < 0.01.
Analysis
We might have been tempted to round 8.24 down and choose n = 8, but this would be incorrect because we
must have an integer greater than or equal to 8.24. We need to keep in mind that the error estimates provide an
upper bound only for the error. The actual estimate may, in fact, be a much better approximation than is indicatedby the error bound.


Use Equation 3.13 to find an upper bound for the error in using M4 to estimate ∫
0


1
x2dx.


Simpson’s Rule
With the midpoint rule, we estimated areas of regions under curves by using rectangles. In a sense, we approximated thecurve with piecewise constant functions. With the trapezoidal rule, we approximated the curve by using piecewise linearfunctions. What if we were, instead, to approximate a curve using piecewise quadratic functions? With Simpson’s rule,we do just this. We partition the interval into an even number of subintervals, each of equal width. Over the first pair


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of subintervals we approximate ∫
x0


x2
f (x)dx with ∫


x0


x2
p(x)dx, where p(x) = Ax2 + Bx + C is the quadratic function


passing through (x0, f (x0)), (x1, f (x1)), and (x2, f (x2)) (Figure 3.16). Over the next pair of subintervals we
approximate ∫


x2


x4
f (x)dx with the integral of another quadratic function passing through (x2, f (x2)), (x3, f (x3)), and


(x4, f (x4)). This process is continued with each successive pair of subintervals.


Figure 3.16 With Simpson’s rule, we approximate a definite integral by integrating a piecewise quadratic function.


To understand the formula that we obtain for Simpson’s rule, we begin by deriving a formula for this approximation overthe first two subintervals. As we go through the derivation, we need to keep in mind the following relationships:
f (x0) = p(x0) = Ax0


2 + Bx0 + C


f (x1) = p(x1) = Ax1
2 + Bx1 + C


f (x2) = p(x2) = Ax2
2 + Bx2 + C


x2 − x0 = 2Δx, where Δx is the length of a subinterval.
x2 + x0 = 2x1, since x1 =


(x2 + x0)
2


.


Thus,


Chapter 3 | Techniques of Integration 323





x0


x2
f (x)dx ≈ ∫


x0


x2
p(x)dx


= ∫
x0


x2
(Ax2 + Bx + C)dx


= A
3
x3 + B


2
x2 + Cx|


x2
x0


Find the antiderivative.


= A
3

⎝x2


3 − x0
3⎞
⎠+


B
2

⎝x2


2 − x0
2⎞
⎠+ C(x2 − x0) Evaluate the antiderivative.


= A
3
(x2 − x0)



⎝x2


2 + x2 x0 + x0
2⎞


+B
2
(x2 − x0)(x2 + x0) + C(x2 − x0)


=
x2 − x0


6

⎝2A

⎝x2


2 + x2 x0 + x0
2⎞
⎠+ 3B(x2 + x0) + 6C



⎠ Factor out


x2 − x0
6


.


= Δx
3



⎝Ax2


2 + Bx2 + C

⎠+ (Ax0


2 + Bx0 + C



+A⎛⎝x2
2 + 2x2 x0 + x0


2⎞
⎠+ 2B(x2 + x0) + 4C)


= Δx
3

⎝ f (x2) + f (x0) + A(x2 + x0)


2 + 2B(x2 + x0) + 4C

⎠ Rearrange the terms.


Factor and substitute.


f (x2) = Ax0
2 + Bx0 + C and


f (x0) = Ax0
2 + Bx0 + C.


= Δx
3

⎝ f (x2) + f (x0) + A



⎝2x1




2 + 2B⎛⎝2x1



⎠+ 4C



⎠ Substitute x2 + x0 = 2x1.


= Δx
3

⎝ f (x2) + 4 f (x1) + f (x0)



⎠.


Expand and substitute


f (x1) = Ax1
2 + Bx1+.


If we approximate ∫
x2


x4
f (x)dx using the same method, we see that we have



x0


x4
f (x)dx ≈ Δx


3

⎝ f (x4) + 4 f (x3) + f (x2)



⎠.


Combining these two approximations, we get

x0


x4
f (x)dx = Δx


3

⎝ f (x0) + 4 f (x1) + 2 f (x2) + 4 f (x3) + f (x4)



⎠.


The pattern continues as we add pairs of subintervals to our approximation. The general rule may be stated as follows.
Theorem 3.6: Simpson’s Rule
Assume that f (x) is continuous over ⎡⎣a, b⎤⎦. Let n be a positive even integer and Δx = b − an . Let ⎡⎣a, b⎤⎦ be divided
into n subintervals, each of length Δx, with endpoints at P = ⎧⎩⎨x0, x1, x2 ,…, xn⎫⎭⎬. Set


(3.14)Sn = Δx3 ⎛⎝ f (x0) + 4 f (x1) + 2 f (x2) + 4 f (x3) + 2 f (x4) + ⋯ + 2 f (xn − 2) + 4 f (xn − 1) + f (xn)⎞⎠.
Then,


lim
n → +∞


Sn = ∫
a


b
f (x)dx.


Just as the trapezoidal rule is the average of the left-hand and right-hand rules for estimating definite integrals, Simpson’s


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rule may be obtained from the midpoint and trapezoidal rules by using a weighted average. It can be shown that
S2n =




2
3

⎠Mn +




1
3

⎠Tn.


It is also possible to put a bound on the error when using Simpson’s rule to approximate a definite integral. The bound inthe error is given by the following rule:
Rule: Error Bound for Simpson’s Rule
Let f (x) be a continuous function over [a, b] having a fourth derivative, f (4)(x), over this interval. If M is the
maximum value of | f (4)(x)| over [a, b], then the upper bound for the error in using Sn to estimate ∫a


b
f (x)dx is


given by
(3.15)


Error in Sn ≤
M(b − a)5


180n4
.


Example 3.45
Applying Simpson’s Rule 1
Use S2 to approximate ∫


0


1
x3dx. Estimate a bound for the error in S2.


Solution
Since [0, 1] is divided into two intervals, each subinterval has length Δx = 1 − 0


2
= 1


2
. The endpoints of these


subintervals are ⎧

⎨0, 1


2
, 1



⎬. If we set f (x) = x3, then


S4 =
1
3
· 1
2

⎝ f (0) + 4 f




1
2

⎠+ f (1)



⎠ =


1
6

⎝0 + 4 ·


1
8
+ 1⎞⎠ =


1
4
. Since f (4) (x) = 0 and consequently M = 0, we


see that
Error in S2 ≤


0(1)5


180 ⋅ 24
= 0.


This bound indicates that the value obtained through Simpson’s rule is exact. A quick check will verify that, in
fact, ∫


0


1
x3dx = 1


4
.


Example 3.46
Applying Simpson’s Rule 2
Use S6 to estimate the length of the curve y = 12x2 over [1, 4].


Solution


Chapter 3 | Techniques of Integration 325




3.26


The length of y = 1
2
x2 over [1, 4] is ∫


1


4
1 + x2dx. If we divide [1, 4] into six subintervals, then each


subinterval has length Δx = 4 − 1
6


= 1
2
, and the endpoints of the subintervals are ⎧



⎨1, 3


2
, 2, 5


2
, 3, 7


2
, 4



⎬.


Setting f (x) = 1 + x2,
S6 =


1
3
· 1
2

⎝ f (1) + 4 f




3
2

⎠+ 2 f (2) + 4 f




5
2

⎠+ 2 f (3) + 4 f




7
2

⎠+ f (4)



⎠.


After substituting, we have
S6 =


1
6
(1.4142 + 4 · 1.80278 + 2 · 2.23607 + 4 · 2.69258 + 2 · 3.16228 + 4 · 3.64005 + 4.12311)


≈ 8.14594.


Use S2 to estimate ∫
1


2
1
xdx.


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3.6 EXERCISES
Approximate the following integrals using either themidpoint rule, trapezoidal rule, or Simpson’s rule asindicated. (Round answers to three decimal places.)
299. ∫


1


2
dx
x ; trapezoidal rule; n = 5


300. ∫
0


3
4 + x3dx; trapezoidal rule; n = 6


301. ∫
0


3
4 + x3dx; Simpson’s rule; n = 3


302. ∫
0


12
x2dx; midpoint rule; n = 6


303. ∫
0


1
sin2 (πx)dx; midpoint rule; n = 3


304. Use the midpoint rule with eight subdivisions to
estimate ∫


2


4
x2dx.


305. Use the trapezoidal rule with four subdivisions to
estimate ∫


2


4
x2dx.


306. Find the exact value of ∫
2


4
x2dx. Find the error


of approximation between the exact value and the valuecalculated using the trapezoidal rule with four subdivisions.Draw a graph to illustrate.
Approximate the integral to three decimal places using theindicated rule.
307. ∫


0


1
sin2 (πx)dx; trapezoidal rule; n = 6


308. ∫
0


3
1


1 + x3
dx; trapezoidal rule; n = 6


309. ∫
0


3
1


1 + x3
dx; Simpson’s rule; n = 3


310. ∫
0


0.8
e−x


2
dx; trapezoidal rule; n = 4


311. ∫
0


0.8
e−x


2
dx; Simpson’s rule; n = 4


312. ∫
0


0.4
sin(x2)dx; trapezoidal rule; n = 4


313. ∫
0


0.4
sin(x2)dx; Simpson’s rule; n = 4


314. ∫
0.1


0.5
cosx
x dx; trapezoidal rule; n = 4


315. ∫
0.1


0.5
cosx
x dx; Simpson’s rule; n = 4


316. Evaluate ∫
0


1
dx


1 + x2
exactly and show that the result


is π/4. Then, find the approximate value of the integral
using the trapezoidal rule with n = 4 subdivisions. Use the
result to approximate the value of π.


317. Approximate ∫
2


4
1
lnx


dx using the midpoint rule
with four subdivisions to four decimal places.
318. Approximate ∫


2


4
1
lnx


dx using the trapezoidal rule
with eight subdivisions to four decimal places.
319. Use the trapezoidal rule with four subdivisions to
estimate ∫


0


0.8
x3dx to four decimal places.


320. Use the trapezoidal rule with four subdivisions to
estimate ∫


0


0.8
x3dx. Compare this value with the exact


value and find the error estimate.
321. Using Simpson’s rule with four subdivisions, find


0


π/2
cos(x)dx.


322. Show that the exact value of ∫
0


1
xe−x dx = 1 − 2e .


Find the absolute error if you approximate the integralusing the midpoint rule with 16 subdivisions.
323. Given ∫


0


1
xe−x dx = 1 − 2e , use the trapezoidal


rule with 16 subdivisions to approximate the integral andfind the absolute error.


Chapter 3 | Techniques of Integration 327




324. Find an upper bound for the error in estimating


0


3
(5x + 4)dx using the trapezoidal rule with six steps.


325. Find an upper bound for the error in estimating


4


5
1


(x − 1)2
dx using the trapezoidal rule with seven


subdivisions.
326. Find an upper bound for the error in estimating


0


3
(6x2 − 1)dx using Simpson’s rule with n = 10 steps.


327. Find an upper bound for the error in estimating


2


5
1


x − 1
dx using Simpson’s rule with n = 10 steps.


328. Find an upper bound for the error in estimating


0


π
2xcos(x)dx using Simpson’s rule with four steps.


329. Estimate the minimum number of subintervals
needed to approximate the integral ∫


1


4

⎝5x


2 + 8⎞⎠dx with
an error magnitude of less than 0.0001 using the trapezoidalrule.
330. Determine a value of n such that the trapezoidal rule
will approximate ∫


0


1
1 + x2dx with an error of no more


than 0.01.
331. Estimate the minimum number of subintervals
needed to approximate the integral ∫


2


3

⎝2x


3 + 4x⎞⎠dx with
an error of magnitude less than 0.0001 using the trapezoidalrule.
332. Estimate the minimum number of subintervals
needed to approximate the integral ∫


3


4
1


(x − 1)2
dx with an


error magnitude of less than 0.0001 using the trapezoidalrule.
333. Use Simpson’s rule with four subdivisions toapproximate the area under the probability density function
y = 1



e−x


2/2 from x = 0 to x = 0.4.
334. Use Simpson’s rule with n = 14 to approximate (to
three decimal places) the area of the region bounded by thegraphs of y = 0, x = 0, and x = π/2.


335. The length of one arch of the curve y = 3sin(2x) is
given by L = ∫


0


π/2
1 + 36cos2(2x)dx. Estimate L using


the trapezoidal rule with n = 6.
336. The length of the ellipse
x = acos(t), y = bsin(t), 0 ≤ t ≤ 2π is given by
L = 4a∫


0


π/2
1 − e2 cos2(t)dt, where e is the


eccentricity of the ellipse. Use Simpson’s rule with n = 6
subdivisions to estimate the length of the ellipse when
a = 2 and e = 1/3.
337. Estimate the area of the surface generated byrevolving the curve y = cos(2x), 0 ≤ x ≤ π


4
about the


x-axis. Use the trapezoidal rule with six subdivisions.
338. Estimate the area of the surface generated by
revolving the curve y = 2x2, 0 ≤ x ≤ 3 about the
x-axis. Use Simpson’s rule with n = 6.
339. The growth rate of a certain tree (in feet) is given by
y = 2


t + 1
+ e−t


2 /2, where t is time in years. Estimate the
growth of the tree through the end of the second year byusing Simpson’s rule, using two subintervals. (Round theanswer to the nearest hundredth.)
340. [T] Use a calculator to approximate ∫


0


1
sin(πx)dx


using the midpoint rule with 25 subdivisions. Compute therelative error of approximation.
341. [T] Given ∫


1


5

⎝3x


2 − 2x⎞⎠dx = 100, approximate
the value of this integral using the midpoint rule with 16subdivisions and determine the absolute error.
342. Given that we know the Fundamental Theorem ofCalculus, why would we want to develop numericalmethods for definite integrals?


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343. The table represents the coordinates (x, y) that give
the boundary of a lot. The units of measurement are meters.Use the trapezoidal rule to estimate the number of squaremeters of land that is in this lot.


x y x y
0 125 600 95
100 125 700 88
200 120 800 75
300 112 900 35
400 90 1000 0
500 90


344. Choose the correct answer. When Simpson’s rule isused to approximate the definite integral, it is necessary thatthe number of partitions be____a. an even numberb. odd numberc. either an even or an odd numberd. a multiple of 4
345. The “Simpson” sum is based on the area under a____.
346. The error formula for Simpson’s rule dependson___.a. f (x)


b. f ′(x)
c. f (4)(x)
d. the number of steps


Chapter 3 | Techniques of Integration 329




3.7 | Improper Integrals
Learning Objectives


3.7.1 Evaluate an integral over an infinite interval.
3.7.2 Evaluate an integral over a closed interval with an infinite discontinuity within the interval.
3.7.3 Use the comparison theorem to determine whether a definite integral is convergent.


Is the area between the graph of f (x) = 1x and the x-axis over the interval [1, +∞) finite or infinite? If this same region
is revolved about the x-axis, is the volume finite or infinite? Surprisingly, the area of the region described is infinite, but thevolume of the solid obtained by revolving this region about the x-axis is finite.
In this section, we define integrals over an infinite interval as well as integrals of functions containing a discontinuity onthe interval. Integrals of these types are called improper integrals. We examine several techniques for evaluating improperintegrals, all of which involve taking limits.
Integrating over an Infinite Interval
How should we go about defining an integral of the type ∫


a


+∞
f (x)dx? We can integrate ∫


a


t
f (x)dx for any value of


t, so it is reasonable to look at the behavior of this integral as we substitute larger values of t. Figure 3.17 shows that

a


t
f (x)dx may be interpreted as area for various values of t. In other words, we may define an improper integral as a


limit, taken as one of the limits of integration increases or decreases without bound.


Figure 3.17 To integrate a function over an infinite interval, we consider the limit of the integral as the upper limit increaseswithout bound.


Definition
1. Let f (x) be continuous over an interval of the form [a, +∞). Then


(3.16)

a


+∞
f (x)dx = lim


t → +∞

a


t
f (x)dx,


provided this limit exists.
2. Let f (x) be continuous over an interval of the form (−∞, b]. Then


(3.17)


−∞


b
f (x)dx = lim


t → −∞

t


b
f (x)dx,


provided this limit exists.In each case, if the limit exists, then the improper integral is said to converge. If the limit does not exist, thenthe improper integral is said to diverge.
3. Let f (x) be continuous over (−∞, +∞). Then


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(3.18)


−∞


+∞
f (x)dx = ∫


−∞


0
f (x)dx + ∫


0


+∞
f (x)dx,


provided that ∫
−∞


0
f (x)dx and ∫


0


+∞
f (x)dx both converge. If either of these two integrals diverge, then



−∞


+∞
f (x)dx diverges. (It can be shown that, in fact, ∫


−∞


+∞
f (x)dx = ∫


−∞


a
f (x)dx + ∫


a


+∞
f (x)dx for any


value of a.)


In our first example, we return to the question we posed at the start of this section: Is the area between the graph of
f (x) = 1x and the x -axis over the interval [1, +∞) finite or infinite?
Example 3.47
Finding an Area
Determine whether the area between the graph of f (x) = 1x and the x-axis over the interval [1, +∞) is finite or
infinite.
Solution
We first do a quick sketch of the region in question, as shown in the following graph.


Figure 3.18 We can find the area between the curve
f (x) = 1/x and the x-axis on an infinite interval.


We can see that the area of this region is given by A = ∫
1



1
xdx. Then we have


Chapter 3 | Techniques of Integration 331




A = ∫
1



1
xdx


= lim
t → +∞



1


t
1
xdx Rewrite the improper integral as a limit.


= lim
t → +∞


ln|x||1t Find the antiderivative.
= lim


t → +∞
(ln|t| − ln1) Evaluate the antiderivative.


= +∞. Evaluate the limit.


Since the improper integral diverges to +∞, the area of the region is infinite.


Example 3.48
Finding a Volume
Find the volume of the solid obtained by revolving the region bounded by the graph of f (x) = 1x and the x-axis
over the interval [1, +∞) about the x -axis.
Solution
The solid is shown in Figure 3.19. Using the disk method, we see that the volume V is


V = π∫
1


+∞
1
x2
dx.


Figure 3.19 The solid of revolution can be generated by rotating an infinite area about thex-axis.


Then we have


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V = π∫
1


+∞
1
x2
dx


= π lim
t → +∞



1


t
1
x2
dx Rewrite as a limit.


= π lim
t → +∞


− 1x |1t Find the antiderivative.
= π lim


t → +∞

⎝−


1
t + 1



⎠ Evaluate the antiderivative.


= π.


The improper integral converges to π. Therefore, the volume of the solid of revolution is π.


In conclusion, although the area of the region between the x-axis and the graph of f (x) = 1/x over the interval [1, +∞)
is infinite, the volume of the solid generated by revolving this region about the x-axis is finite. The solid generated is knownas Gabriel’s Horn.


Visit this website (http://www.openstaxcollege.org/l/20_GabrielsHorn) to read more about Gabriel’sHorn.


Example 3.49
Chapter Opener: Traffic Accidents in a City


Figure 3.20 (credit: modification of work by DavidMcKelvey, Flickr)


In the chapter opener, we stated the following problem: Suppose that at a busy intersection, traffic accidents occurat an average rate of one every three months. After residents complained, changes were made to the traffic lightsat the intersection. It has now been eight months since the changes were made and there have been no accidents.Were the changes effective or is the 8-month interval without an accident a result of chance?
Probability theory tells us that if the average time between events is k, the probability that X, the time between
events, is between a and b is given by


P(a ≤ x ≤ b) = ∫
a


b
f (x)dxwhere f (x) =







0 if x < 0


ke−kx if x ≥ 0
.


Thus, if accidents are occurring at a rate of one every 3 months, then the probability that X, the time between
accidents, is between a and b is given by


Chapter 3 | Techniques of Integration 333




P(a ≤ x ≤ b) = ∫
a


b
f (x)dxwhere f (x) =







0 if x < 0


3e−3x if x ≥ 0
.


To answer the question, we must compute P(X ≥ 8) = ∫
8


+∞
3e−3x dx and decide whether it is likely that 8


months could have passed without an accident if there had been no improvement in the traffic situation.
Solution
We need to calculate the probability as an improper integral:


P(X ≥ 8) = ∫
8


+∞
3e−3x dx


= lim
t → +∞



8


t
3e−3x dx


= lim
t → +∞


−e−3x|8
t


= lim
t → +∞


(−e−3t + e−24)


≈ 3.8 × 10−11.


The value 3.8 × 10−11 represents the probability of no accidents in 8 months under the initial conditions. Since
this value is very, very small, it is reasonable to conclude the changes were effective.


Example 3.50
Evaluating an Improper Integral over an Infinite Interval
Evaluate ∫


−∞


0
1


x2 + 4
dx. State whether the improper integral converges or diverges.


Solution
Begin by rewriting ∫


−∞


0
1


x2 + 4
dx as a limit using Equation 3.17 from the definition. Thus,



−∞


0
1


x2 + 4
dx = lim


t → −∞

t


0
1


x2 + 4
dx Rewrite as a limit.


= lim
t → −∞


tan−1 x
2|t0 Find the antiderivative.


= lim
t → −∞


(tan−10 − tan−1 t
2
) Evaluate the antiderivative.


= π
2
. Evaluate the limit and simplify.


The improper integral converges to π
2
.


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3.27


Example 3.51
Evaluating an Improper Integral on (−∞, +∞)


Evaluate ∫
−∞


+∞
xex dx. State whether the improper integral converges or diverges.


Solution
Start by splitting up the integral:



−∞


+∞
xex dx = ∫


−∞


0
xex dx + ∫


0


+∞
xex dx.


If either ∫
−∞


0
xex dx or ∫


0


+∞
xex dx diverges, then ∫


−∞


+∞
xex dx diverges. Compute each integral separately.


For the first integral,


−∞


0
xex dx = lim


t → −∞

t


0
xex dx Rewrite as a limit.


= lim
t → −∞


(xex − ex)|t0 Use integration by parts to find heantiderivative. (Here u = x and dv = ex.)
= lim


t → −∞

⎝−1 − te


t + et⎞⎠ Evaluate the antiderivative.


= −1.


Evaluate the limit. Note: lim
t → −∞


tet is


indeterminate of the form 0 ·∞. Thus,


lim
t → −∞


tet = lim
t → −∞


t
e−t


= lim
t → −∞


−1
e−t


= lim
t → −∞


− et = 0 by


L’Hôpital’s Rule.


The first improper integral converges. For the second integral,


0


+∞
xex dx = lim


t → +∞


0


t
xex dx Rewrite as a limit.


= lim
t → +∞


(xex − ex)|0t Find the antiderivative.
= lim


t → +∞

⎝te


t − et + 1⎞⎠ Evaluate the antiderivative.


= lim
t → +∞



⎝(t − 1)e


t + 1⎞⎠ Rewrite. (te
t − et is indeterminate.)


= +∞. Evaluate the limit.


Thus, ∫
0


+∞
xex dx diverges. Since this integral diverges, ∫


−∞


+∞
xex dx diverges as well.


Evaluate ∫
−3


+∞
e−x dx. State whether the improper integral converges or diverges.


Integrating a Discontinuous Integrand
Now let’s examine integrals of functions containing an infinite discontinuity in the interval over which the integration


Chapter 3 | Techniques of Integration 335




occurs. Consider an integral of the form ∫
a


b
f (x)dx, where f (x) is continuous over [a, b) and discontinuous at b. Since


the function f (x) is continuous over [a, t] for all values of t satisfying a < t < b, the integral ∫
a


t
f (x)dx is defined


for all such values of t. Thus, it makes sense to consider the values of ∫
a


t
f (x)dx as t approaches b for a < t < b. That


is, we define ∫
a


b
f (x)dx = lim


t → b−

a


t
f (x)dx, provided this limit exists. Figure 3.21 illustrates ∫


a


t
f (x)dx as areas of


regions for values of t approaching b.


Figure 3.21 As t approaches b from the left, the value of the area from a to t approaches the area from a to b.


We use a similar approach to define ∫
a


b
f (x)dx, where f (x) is continuous over (a, b] and discontinuous at a. We now


proceed with a formal definition.
Definition


1. Let f (x) be continuous over [a, b). Then,
(3.19)



a


b
f (x)dx = lim


t → b−

a


t
f (x)dx.


2. Let f (x) be continuous over (a, b]. Then,
(3.20)



a


b
f (x)dx = lim


t → a+

t


b
f (x)dx.


In each case, if the limit exists, then the improper integral is said to converge. If the limit does not exist, thenthe improper integral is said to diverge.
3. If f (x) is continuous over [a, b] except at a point c in (a, b), then


(3.21)

a


b
f (x)dx = ∫


a


c
f (x)dx + ∫


c


b
f (x)dx,


provided both ∫
a


c
f (x)dx and ∫


c


b
f (x)dx converge. If either of these integrals diverges, then ∫


a


b
f (x)dx


diverges.


The following examples demonstrate the application of this definition.


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Example 3.52
Integrating a Discontinuous Integrand
Evaluate ∫


0


4
1


4 − x
dx, if possible. State whether the integral converges or diverges.


Solution
The function f (x) = 1


4 − x
is continuous over [0, 4) and discontinuous at 4. Using Equation 3.19 from the


definition, rewrite ∫
0


4
1


4 − x
dx as a limit:



0


4
1


4 − x
dx = lim


t → 4−


0


t
1


4 − x
dx Rewrite as a limit.


= lim
t → 4−



⎝−2 4 − x



⎠|0t Find the antiderivative.


= lim
t → 4−



⎝−2 4 − t + 4



⎠ Evaluate the antiderivative.


= 4. Evaluate the limit.


The improper integral converges.


Example 3.53
Integrating a Discontinuous Integrand
Evaluate ∫


0


2
x lnxdx. State whether the integral converges or diverges.


Solution
Since f (x) = x lnx is continuous over (0, 2] and is discontinuous at zero, we can rewrite the integral in limit
form using Equation 3.20:



0


2
x lnxdx = lim


t → 0+

t


2
x lnxdx Rewrite as a limit.


= lim
t → 0+




1
2
x2 lnx − 1


4
x2⎞⎠|t2 Evaluate ∫ x lnxdx using integration by partswith u = lnx and dv = x.


= lim
t → 0+



⎝2ln2 − 1 −


1
2
t2 ln t + 1


4
t2⎞⎠. Evaluate the antiderivative.


= 2ln2 − 1.


Evaluate the limit. lim
t → 0+


t2 ln t is indeterminate.


To evaluate it, rewrite as a quotient and apply


L’Hôpital’s rule.


The improper integral converges.


Chapter 3 | Techniques of Integration 337




3.28


Example 3.54
Integrating a Discontinuous Integrand
Evaluate ∫


−1


1
1
x3
dx. State whether the improper integral converges or diverges.


Solution
Since f (x) = 1/x3 is discontinuous at zero, using Equation 3.21, we can write



−1


1
1
x3
dx = ∫


−1


0
1
x3
dx + ∫


0


1
1
x3
dx.


If either of the two integrals diverges, then the original integral diverges. Begin with ∫
−1


0
1
x3
dx :



−1


0
1
x3
dx = lim


t → 0−


−1


t
1
x3
dx Rewrite as a limit.


= lim
t → 0−

⎝−


1
2x2

⎠|−1t Find the antiderivative.


= lim
t → 0−

⎝−


1
2t2


+ 1
2

⎠ Evaluate the antiderivative.


= +∞. Evaluate the limit.


Therefore, ∫
−1


0
1
x3
dx diverges. Since ∫


−1


0
1
x3
dx diverges, ∫


−1


1
1
x3
dx diverges.


Evaluate ∫
0


2
1
xdx. State whether the integral converges or diverges.


A Comparison Theorem
It is not always easy or even possible to evaluate an improper integral directly; however, by comparing it with anothercarefully chosen integral, it may be possible to determine its convergence or divergence. To see this, consider twocontinuous functions f (x) and g(x) satisfying 0 ≤ f (x) ≤ g(x) for x ≥ a (Figure 3.22). In this case, we may view
integrals of these functions over intervals of the form [a, t] as areas, so we have the relationship


0 ≤ ∫
a


t
f (x)dx ≤ ∫


a


t
g(x)dx for t ≥ a.


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Figure 3.22 If 0 ≤ f (x) ≤ g(x) for x ≥ a, then for
t ≥ a, ∫


a


t
f (x)dx ≤ ∫


a


t
g(x)dx.


Thus, if

a


+∞
f (x)dx = lim


t → +∞

a


t
f (x)dx = +∞,


then

a


+∞
g(x)dx = lim


t → +∞

a


t
g(x)dx = +∞ as well. That is, if the area of the region between the graph of f (x) and the x-axis


over [a, +∞) is infinite, then the area of the region between the graph of g(x) and the x-axis over [a, +∞) is infinite
too.
On the other hand, if

a


+∞
g(x)dx = lim


t → +∞

a


t
g(x)dx = L for some real number L, then



a


+∞
f (x)dx = lim


t → +∞

a


t
f (x)dx must converge to some value less than or equal to L, since ∫


a


t
f (x)dx increases as t


increases and ∫
a


t
f (x)dx ≤ L for all t ≥ a.


If the area of the region between the graph of g(x) and the x-axis over [a, +∞) is finite, then the area of the region
between the graph of f (x) and the x-axis over [a, +∞) is also finite.
These conclusions are summarized in the following theorem.
Theorem 3.7: A Comparison Theorem
Let f (x) and g(x) be continuous over [a, +∞). Assume that 0 ≤ f (x) ≤ g(x) for x ≥ a.


i. If ∫
a


+∞
f (x)dx = lim


t → +∞

a


t
f (x)dx = +∞, then ∫


a


+∞
g(x)dx = lim


t → +∞

a


t
g(x)dx = +∞.


ii. If ∫
a


+∞
g(x)dx = lim


t → +∞

a


t
g(x)dx = L, where L is a real number, then



a


+∞
f (x)dx = lim


t → +∞

a


t
f (x)dx = M for some real number M ≤ L.


Chapter 3 | Techniques of Integration 339




3.29


Example 3.55
Applying the Comparison Theorem
Use a comparison to show that ∫


1


+∞
1
xex


dx converges.


Solution
We can see that


0 ≤ 1
xex


≤ 1
ex


= e−x,


so if ∫
1


+∞
e−x dx converges, then so does ∫


1


+∞
1
xex


dx. To evaluate ∫
1


+∞
e−x dx, first rewrite it as a limit:



1


+∞
e−xdx = lim


t → +∞


1


t
e−x dx


= lim
t → +∞


(−e−x)| t1
= lim


t → +∞

⎝−e


−t + e1⎞⎠


= e1.


Since ∫
1


+∞
e−x dx converges, so does ∫


1


+∞
1
xex


dx.


Example 3.56
Applying the Comparison Theorem
Use the comparison theorem to show that ∫


1


+∞
1
x p


dx diverges for all p < 1.


Solution
For p < 1, 1/x ≤ 1/(x p) over [1, +∞). In Example 3.47, we showed that ∫


1


+∞
1
xdx = +∞. Therefore,



1


+∞
1
x p


dx diverges for all p < 1.


Use a comparison to show that ∫
e


+∞
lnx
x dx diverges.


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Laplace Transforms
In the last few chapters, we have looked at several ways to use integration for solving real-world problems. For thisnext project, we are going to explore a more advanced application of integration: integral transforms. Specifically, wedescribe the Laplace transform and some of its properties. The Laplace transform is used in engineering and physics tosimplify the computations needed to solve some problems. It takes functions expressed in terms of time and transformsthem to functions expressed in terms of frequency. It turns out that, in many cases, the computations needed to solveproblems in the frequency domain are much simpler than those required in the time domain.
The Laplace transform is defined in terms of an integral as


L⎧⎩⎨ f (t)



⎬ = F(s) = ∫


0



e−st f (t)dt.


Note that the input to a Laplace transform is a function of time, f (t), and the output is a function of frequency, F(s).
Although many real-world examples require the use of complex numbers (involving the imaginary number i = −1),
in this project we limit ourselves to functions of real numbers.
Let’s start with a simple example. Here we calculate the Laplace transform of f (t) = t . We have


L{t} = ∫
0



te−st dt.


This is an improper integral, so we express it in terms of a limit, which gives
L{t} = ∫


0



te−st dt = lim


z → ∞


0


z
te−st dt.


Now we use integration by parts to evaluate the integral. Note that we are integrating with respect to t, so we treat thevariable s as a constant. We have
u = t dv = e−st dt


du = dt v = −1se
−st.


Then we obtain
lim


z → ∞


0


z
te−st dt = lim


z → ∞



⎣− tse


−st⎤
⎦|0
z
+ 1s∫0


z
e−st dt



= lim
z → ∞



⎣−


z
se


−sz + 0se
−0s⎤
⎦+


1
s∫0


z
e−st dt



= lim
z → ∞



⎣−


z
se


−sz + 0⎤⎦− 1s


e−st
s

⎦|0
z⎤


= lim
z → ∞



⎣−


z
se


−sz⎤
⎦− 1


s2

⎣e


−sz − 1⎤⎦



= lim
z → ∞

⎣−


z
sesz

⎦− limz → ∞





1
s2 esz

⎦+ limz → ∞


1
s2


= 0 − 0 + 1
s2


= 1
s2


.


1. Calculate the Laplace transform of f (t) = 1.


Chapter 3 | Techniques of Integration 341




2. Calculate the Laplace transform of f (t) = e−3t.
3. Calculate the Laplace transform of f (t) = t2. (Note, you will have to integrate by parts twice.)


Laplace transforms are often used to solve differential equations. Differential equations are not covered indetail until later in this book; but, for now, let’s look at the relationship between the Laplace transform of afunction and the Laplace transform of its derivative.Let’s start with the definition of the Laplace transform. We have
L⎧⎩⎨ f (t)





⎬ = ∫


0



e−st f (t)dt = lim


z → ∞


0


z
e−st f (t)dt.


4. Use integration by parts to evaluate lim
z → ∞



0


z
e−st f (t)dt. (Let u = f (t) and dv = e−st dt.)


After integrating by parts and evaluating the limit, you should see that
L⎧⎩⎨ f (t)





⎬ =


f (0)
s +


1
s

⎣L



⎨ f ′(t)⎫⎭⎬⎤⎦.


Then,
L⎧⎩⎨ f ′(t)





⎬ = sL⎧⎩⎨ f (t)





⎬ − f (0).


Thus, differentiation in the time domain simplifies to multiplication by s in the frequency domain.The final thing we look at in this project is how the Laplace transforms of f (t) and its antiderivative are
related. Let g(t) = ∫


0


t
f (u)du. Then,


L⎧⎩⎨g(t)



⎬ = ∫


0



e−st g(t)dt = lim


z → ∞


0


z
e−st g(t)dt.


5. Use integration by parts to evaluate lim
z → ∞



0


z
e−st g(t)dt. (Let u = g(t) and dv = e−st dt. Note, by the way,


that we have defined g(t), du = f (t)dt.)
As you might expect, you should see that


L⎧⎩⎨g(t)



⎬ = 1s · L





⎨ f (t)⎫⎭⎬.


Integration in the time domain simplifies to division by s in the frequency domain.


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3.7 EXERCISES
Evaluate the following integrals. If the integral is notconvergent, answer “divergent.”
347. ∫


2


4
dx


(x − 3)2


348. ∫
0



1


4 + x2
dx


349. ∫
0


2
1


4 − x2
dx


350. ∫
1



1


x lnx
dx


351. ∫
1



xe−x dx


352. ∫
−∞



x


x2 + 1
dx


353. Without integrating, determine whether the integral


1



1


x3 + 1
dx converges or diverges by comparing the


function f (x) = 1
x3 + 1


with g(x) = 1
x3


.


354. Without integrating, determine whether the integral


1



1


x + 1
dx converges or diverges.


Determine whether the improper integrals converge ordiverge. If possible, determine the value of the integrals thatconverge.
355. ∫


0



e−x cosxdx


356. ∫
1



lnx
x dx


357. ∫
0


1
lnx
x dx


358. ∫
0


1
lnxdx


359. ∫
−∞



1


x2 + 1
dx


360. ∫
1


5
dx
x − 1


361. ∫
−2


2
dx


(1 + x)2


362. ∫
0



e−x dx


363. ∫
0



sinxdx


364. ∫
−∞



ex


1 + e2x
dx


365. ∫
0


1
dx
x3


366. ∫
0


2
dx
x3


367. ∫
−1


2
dx
x3


368. ∫
0


1
dx


1 − x2


369. ∫
0


3
1


x − 1
dx


370. ∫
1



5
x3
dx


371. ∫
3


5
5


(x − 4)2
dx


Determine the convergence of each of the followingintegrals by comparison with the given integral. If theintegral converges, find the number to which it converges.
372. ∫


1



dx


x2 + 4x
; compare with ∫


1



dx
x2


.


373. ∫
1



dx
x + 1


; compare with ∫
1



dx
2 x


.


Evaluate the integrals. If the integral diverges, answer“diverges.”


Chapter 3 | Techniques of Integration 343




374. ∫
1



dx
xe


375. ∫
0


1
dx


376. ∫
0


1
dx
1 − x


377. ∫
0


1
dx


1 − x


378. ∫
−∞


0
dx


x2 + 1


379. ∫
−1


1
dx


1 − x2


380. ∫
0


1
lnx
x dx


381. ∫
0


e
ln(x)dx


382. ∫
0



xe−x dx


383. ∫
−∞



x



⎝x


2 + 1⎞⎠
2
dx


384. ∫
0



e−x dx


Evaluate the improper integrals. Each of these integralshas an infinite discontinuity either at an endpoint or at aninterior point of the interval.
385. ∫


0


9
dx
9 − x


386. ∫
−27


1
dx
x2/3


387. ∫
0


3
dx


9 − x2


388. ∫
6


24
dt


t t2 − 36


389. ∫
0


4
x ln(4x)dx


390. ∫
0


3
x


9 − x2
dx


391. Evaluate ∫
.5


t
dx


1 − x2
. (Be careful!) (Express your


answer using three decimal places.)
392. Evaluate ∫


1


4
dx


x2 − 1
. (Express the answer in exact


form.)
393. Evaluate ∫


2



dx


(x2 − 1)3/2
.


394. Find the area of the region in the first quadrant
between the curve y = e−6x and the x-axis.
395. Find the area of the region bounded by the curve
y = 7


x2
, the x-axis, and on the left by x = 1.


396. Find the area under the curve y = 1
(x + 1)3/2


,


bounded on the left by x = 3.
397. Find the area under y = 5


1 + x2
in the first


quadrant.
398. Find the volume of the solid generated by revolving
about the x-axis the region under the curve y = 3x from
x = 1 to x = ∞.
399. Find the volume of the solid generated by revolving
about the y-axis the region under the curve y = 6e−2x in
the first quadrant.
400. Find the volume of the solid generated by revolvingabout the x-axis the area under the curve y = 3e−x in the
first quadrant.
The Laplace transform of a continuous function over the
interval [0, ∞) is defined by F(s) = ∫


0



e−sx f (x)dx


(see the Student Project). This definition is used to solvesome important initial-value problems in differentialequations, as discussed later. The domain of F is the setof all real numbers s such that the improper integralconverges. Find the Laplace transform F of each of thefollowing functions and give the domain of F.


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401. f (x) = 1
402. f (x) = x
403. f (x) = cos(2x)
404. f (x) = eax
405. Use the formula for arc length to show that the
circumference of the circle x2 + y2 = 1 is 2π.
A function is a probability density function if it satisfies
the following definition: ∫


−∞



f (t)dt = 1. The probability


that a random variable x lies between a and b is given by
P(a ≤ x ≤ b) = ∫


a


b
f (t)dt.


406. Show that f (x) = ⎧



0if x < 0


7e−7x if x ≥ 0
is a probability


density function.
407. Find the probability that x is between 0 and 0.3. (Usethe function defined in the preceding problem.) Use four-place decimal accuracy.


Chapter 3 | Techniques of Integration 345




absolute error


computer algebra system (CAS)
improper integral


integration by parts


integration table
midpoint rule


numerical integration
partial fraction decomposition
power reduction formula
relative error
Simpson’s rule


trigonometric integral
trigonometric substitution


CHAPTER 3 REVIEW
KEY TERMS


if B is an estimate of some quantity having an actual value of A, then the absolute error is given by
|A − B|


technology used to perform many mathematical tasks, including integration
an integral over an infinite interval or an integral of a function containing an infinite discontinuity onthe interval; an improper integral is defined in terms of a limit. The improper integral converges if this limit is a finitereal number; otherwise, the improper integral diverges
a technique of integration that allows the exchange of one integral for another using the formula


∫ u dv = uv − ∫ v du


a table that lists integration formulas
a rule that uses a Riemann sum of the form Mn = ∑


i = 1


n


f (mi)Δx, where mi is the midpoint of the ith


subinterval to approximate ∫
a


b
f (x)dx


the variety of numerical methods used to estimate the value of a definite integral, including themidpoint rule, trapezoidal rule, and Simpson’s rule
a technique used to break down a rational function into the sum of simple rationalfunctions


a rule that allows an integral of a power of a trigonometric function to be exchanged for anintegral involving a lower power
error as a percentage of the absolute value, given by |A − BA | = |A − BA | · 100%


a rule that approximates ∫
a


b
f (x)dx using the integrals of a piecewise quadratic function. The


approximation Sn to ∫
a


b
f (x)dx is given by Sn = Δx3 ⎛⎝


f (x0) + 4 f (x1) + 2 f (x2) + 4 f (x3) + 2 f (x4) + 4 f (x5)


+ ⋯ + 2 f (xn − 2) + 4 f (xn − 1) + f (xn)



trapezoidal rule a rule that approximates ∫
a


b
f (x)dx using trapezoids


an integral involving powers and products of trigonometric functions
an integration technique that converts an algebraic integral containing expressions of the


form a2 − x2, a2 + x2, or x2 − a2 into a trigonometric integral
KEY EQUATIONS


• Integration by parts formula
∫ u dv = uv − ∫ v du


• Integration by parts for definite integrals

a


b
u dv = uv|a


b − ∫
a


b
v du


To integrate products involving sin(ax), sin(bx), cos(ax), and cos(bx), use the substitutions.


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• Sine Products
sin(ax)sin(bx) = 1


2
cos((a − b)x) − 1


2
cos((a + b)x)


• Sine and Cosine Products
sin(ax)cos(bx) = 1


2
sin⎛⎝(a − b)x⎞⎠+ 12


sin((a + b)x)


• Cosine Products
cos(ax)cos(bx) = 1


2
cos((a − b)x) + 1


2
cos((a + b)x)


• Power Reduction Formula
∫ secn x dx = 1


n − 1
secn − 1 x + n − 2


n − 1
∫ secn − 2 x dx


• Power Reduction Formula
∫ tann x dx = 1


n − 1
tann − 1 x − ∫ tann − 2 x dx


• Midpoint rule
Mn = ∑


i = 1


n


f (mi)Δx


• Trapezoidal rule
Tn = 12


Δx⎛⎝ f (x0) + 2 f (x1) + 2 f (x2) + ⋯ + 2 f (xn − 1) + f (xn)



• Simpson’s rule
Sn = Δx3



⎝ f (x0) + 4 f (x1) + 2 f (x2) + 4 f (x3) + 2 f (x4) + 4 f (x5) + ⋯ + 2 f (xn − 2) + 4 f (xn − 1) + f (xn)





• Error bound for midpoint rule
Error inMn ≤


M(b − a)3


24n2


• Error bound for trapezoidal rule
Error in Tn ≤


M(b − a)3


12n2


• Error bound for Simpson’s rule
Error in Sn ≤


M(b − a)5


180n4


• Improper integrals

a


+∞
f (x)dx = lim


t → +∞

a


t
f (x)dx



−∞


b
f (x)dx = lim


t → −∞

t


b
f (x)dx



−∞


+∞
f (x)dx = ∫


−∞


0
f (x)dx + ∫


0


+∞
f (x)dx


KEY CONCEPTS
3.1 Integration by Parts


• The integration-by-parts formula allows the exchange of one integral for another, possibly easier, integral.
• Integration by parts applies to both definite and indefinite integrals.


Chapter 3 | Techniques of Integration 347




3.2 Trigonometric Integrals
• Integrals of trigonometric functions can be evaluated by the use of various strategies. These strategies include


1. Applying trigonometric identities to rewrite the integral so that it may be evaluated by u-substitution
2. Using integration by parts
3. Applying trigonometric identities to rewrite products of sines and cosines with different arguments as thesum of individual sine and cosine functions
4. Applying reduction formulas


3.3 Trigonometric Substitution
• For integrals involving a2 − x2, use the substitution x = asinθ and dx = acosθdθ.
• For integrals involving a2 + x2, use the substitution x = a tanθ and dx = asec2 θdθ.
• For integrals involving x2 − a2, substitute x = asecθ and dx = asecθ tanθdθ.


3.4 Partial Fractions
• Partial fraction decomposition is a technique used to break down a rational function into a sum of simple rationalfunctions that can be integrated using previously learned techniques.
• When applying partial fraction decomposition, we must make sure that the degree of the numerator is lessthan the degree of the denominator. If not, we need to perform long division before attempting partial fractiondecomposition.
• The form the decomposition takes depends on the type of factors in the denominator. The types of factorsinclude nonrepeated linear factors, repeated linear factors, nonrepeated irreducible quadratic factors, and repeatedirreducible quadratic factors.


3.5 Other Strategies for Integration
• An integration table may be used to evaluate indefinite integrals.
• A CAS (or computer algebra system) may be used to evaluate indefinite integrals.
• It may require some effort to reconcile equivalent solutions obtained using different methods.


3.6 Numerical Integration
• We can use numerical integration to estimate the values of definite integrals when a closed form of the integral isdifficult to find or when an approximate value only of the definite integral is needed.
• The most commonly used techniques for numerical integration are the midpoint rule, trapezoidal rule, andSimpson’s rule.
• The midpoint rule approximates the definite integral using rectangular regions whereas the trapezoidal ruleapproximates the definite integral using trapezoidal approximations.
• Simpson’s rule approximates the definite integral by first approximating the original function using piecewisequadratic functions.


3.7 Improper Integrals
• Integrals of functions over infinite intervals are defined in terms of limits.
• Integrals of functions over an interval for which the function has a discontinuity at an endpoint may be defined interms of limits.


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• The convergence or divergence of an improper integral may be determined by comparing it with the value of animproper integral for which the convergence or divergence is known.
CHAPTER 3 REVIEW EXERCISES
For the following exercises, determine whether thestatement is true or false. Justify your answer with a proofor a counterexample.
408. ∫ ex sin(x)dx cannot be integrated by parts.


409. ∫ 1
x4 + 1


dx cannot be integrated using partial
fractions.
410. In numerical integration, increasing the number ofpoints decreases the error.
411. Integration by parts can always yield the integral.
For the following exercises, evaluate the integral using thespecified method.
412. ∫ x2 sin(4x)dx using integration by parts


413. ∫ 1
x2 x2 + 16


dx using trigonometric substitution


414. ∫ x ln(x)dx using integration by parts


415. ∫ 3x
x3 + 2x2 − 5x − 6


dx using partial fractions


416. ∫ x5

⎝4x


2 + 4⎞⎠
5/2


dx using trigonometric substitution


417. ∫ 4 − sin2(x)
sin2(x)


cos(x)dx using a table of integrals or
a CAS
For the following exercises, integrate using whatevermethod you choose.
418. ∫ sin2(x)cos2(x)dx


419. ∫ x3 x2 + 2dx


420. ∫ 3x2 + 1
x4 − 2x3 − x2 + 2x


dx


421. ∫ 1
x4 + 4


dx


422. ∫ 3 + 16x4
x4


dx


For the following exercises, approximate the integralsusing the midpoint rule, trapezoidal rule, and Simpson’srule using four subintervals, rounding to three decimals.
423. [T] ∫


1


2
x5 + 2dx


424. [T] ∫
0


π
e
−sin(x2)


dx


425. [T] ∫
1


4ln(1/x)
x dx


For the following exercises, evaluate the integrals, ifpossible.
426. ∫


1



1
xn
dx, for what values of n does this integral


converge or diverge?


427. ∫
1



e−x
x dx


For the following exercises, consider the gamma function
given by Γ(a) = ∫


0



e
−y


ya − 1dy.


428. Show that Γ(a) = (a − 1)Γ(a − 1).


Chapter 3 | Techniques of Integration 349




429. Extend to show that Γ(a) = (a − 1)!, assuming a
is a positive integer.
The fastest car in the world, the Bugati Veyron, can reach atop speed of 408 km/h. The graph represents its velocity.


430. [T] Use the graph to estimate the velocity every20 sec and fit to a graph of the form
v(t) = aexpbx sin(cx) + d. (Hint: Consider the time
units.)
431. [T] Using your function from the previous problem,find exactly how far the Bugati Veyron traveled in the 1 min40 sec included in the graph.


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4 | INTRODUCTION TODIFFERENTIAL EQUATIONS


Figure 4.1 The white-tailed deer (Odocoileus virginianus) of the eastern United States. Differential equations can be used tostudy animal populations. (credit: modification of work by Rachel Kramer, Flickr)


Chapter Outline
4.1 Basics of Differential Equations
4.2 Direction Fields and Numerical Methods
4.3 Separable Equations
4.4 The Logistic Equation
4.5 First-order Linear Equations


Introduction
Many real-world phenomena can be modeled mathematically by using differential equations. Population growth,radioactive decay, predator-prey models, and spring-mass systems are four examples of such phenomena. In this chapter westudy some of these applications.
Suppose we wish to study a population of deer over time and determine the total number of animals in a given area. Wecan first observe the population over a period of time, estimate the total number of deer, and then use various assumptionsto derive a mathematical model for different scenarios. Some factors that are often considered are environmental impact,threshold population values, and predators. In this chapter we see how differential equations can be used to predictpopulations over time (see Example 4.14).
Another goal of this chapter is to develop solution techniques for different types of differential equations. As the equationsbecome more complicated, the solution techniques also become more complicated, and in fact an entire course couldbe dedicated to the study of these equations. In this chapter we study several types of differential equations and theircorresponding methods of solution.


Chapter 4 | Introduction to Differential Equations 351




4.1 | Basics of Differential Equations
Learning Objectives


4.1.1 Identify the order of a differential equation.
4.1.2 Explain what is meant by a solution to a differential equation.
4.1.3 Distinguish between the general solution and a particular solution of a differential equation.
4.1.4 Identify an initial-value problem.
4.1.5 Identify whether a given function is a solution to a differential equation or an initial-valueproblem.


Calculus is the mathematics of change, and rates of change are expressed by derivatives. Thus, one of the most commonways to use calculus is to set up an equation containing an unknown function y = f (x) and its derivative, known as
a differential equation. Solving such equations often provides information about how quantities change and frequentlyprovides insight into how and why the changes occur.
Techniques for solving differential equations can take many different forms, including direct solution, use of graphs, orcomputer calculations. We introduce the main ideas in this chapter and describe them in a little more detail later in thecourse. In this section we study what differential equations are, how to verify their solutions, some methods that are usedfor solving them, and some examples of common and useful equations.
General Differential Equations
Consider the equation y′ = 3x2, which is an example of a differential equation because it includes a derivative. There is a
relationship between the variables x and y: y is an unknown function of x. Furthermore, the left-hand side of the equation
is the derivative of y. Therefore we can interpret this equation as follows: Start with some function y = f (x) and take its
derivative. The answer must be equal to 3x2. What function has a derivative that is equal to 3x2? One such function is
y = x3, so this function is considered a solution to a differential equation.


Definition
A differential equation is an equation involving an unknown function y = f (x) and one or more of its derivatives.
A solution to a differential equation is a function y = f (x) that satisfies the differential equation when f and its
derivatives are substituted into the equation.


Go to this website (http://www.openstaxcollege.org/l/20_Differential) to explore more on this topic.


Some examples of differential equations and their solutions appear in Table 4.1.


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4.1


Equation Solution
y′ = 2x y = x2


y′ + 3y = 6x + 11 y = e−3x + 2x + 3


y′′ − 3y′ + 2y = 24e−2x y = 3ex − 4e2x + 2e−2x


Table 4.1 Examples of Differential Equations and TheirSolutions
Note that a solution to a differential equation is not necessarily unique, primarily because the derivative of a constant is
zero. For example, y = x2 + 4 is also a solution to the first differential equation in Table 4.1. We will return to this idea a
little bit later in this section. For now, let’s focus on what it means for a function to be a solution to a differential equation.
Example 4.1
Verifying Solutions of Differential Equations
Verify that the function y = e−3x + 2x + 3 is a solution to the differential equation y′ + 3y = 6x + 11.
Solution
To verify the solution, we first calculate y′ using the chain rule for derivatives. This gives y′ = −3e−3x + 2.
Next we substitute y and y′ into the left-hand side of the differential equation:


(−3e−2x + 2) + 3(e−2x + 2x + 3).


The resulting expression can be simplified by first distributing to eliminate the parentheses, giving
−3e−2x + 2 + 3e−2x + 6x + 9.


Combining like terms leads to the expression 6x + 11, which is equal to the right-hand side of the differential
equation. This result verifies that y = e−3x + 2x + 3 is a solution of the differential equation.


Verify that y = 2e3x − 2x − 2 is a solution to the differential equation y′ − 3y = 6x + 4.


It is convenient to define characteristics of differential equations that make it easier to talk about them and categorize them.The most basic characteristic of a differential equation is its order.
Definition
The order of a differential equation is the highest order of any derivative of the unknown function that appears in theequation.


Chapter 4 | Introduction to Differential Equations 353




4.2


Example 4.2
Identifying the Order of a Differential Equation
What is the order of each of the following differential equations?


a. y′ − 4y = x2 − 3x + 4
b. x2 y‴ − 3xy″ + xy′ − 3y = sinx
c. 4xy(4) − 6x2y″ + 12x4y = x


3 − 3x2 + 4x − 12


Solution
a. The highest derivative in the equation is y′, so the order is 1.
b. The highest derivative in the equation is y‴, so the order is 3.
c. The highest derivative in the equation is y(4), so the order is 4.


What is the order of the following differential equation?

⎝x


4 − 3x⎞⎠y
(5) − ⎛⎝3x


2 + 1⎞⎠y′ + 3y = sinxcosx


General and Particular Solutions
We already noted that the differential equation y′ = 2x has at least two solutions: y = x2 and y = x2 + 4. The only
difference between these two solutions is the last term, which is a constant. What if the last term is a different constant?
Will this expression still be a solution to the differential equation? In fact, any function of the form y = x2 + C, where C
represents any constant, is a solution as well. The reason is that the derivative of x2 + C is 2x, regardless of the value of
C. It can be shown that any solution of this differential equation must be of the form y = x2 + C. This is an example of a
general solution to a differential equation. A graph of some of these solutions is given in Figure 4.2. (Note: in this graphwe used even integer values for C ranging between −4 and 4. In fact, there is no restriction on the value of C; it can be
an integer or not.)


Figure 4.2 Family of solutions to the differential equation
y′ = 2x.


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4.3


In this example, we are free to choose any solution we wish; for example, y = x2 − 3 is a member of the family of solutions
to this differential equation. This is called a particular solution to the differential equation. A particular solution can oftenbe uniquely identified if we are given additional information about the problem.
Example 4.3
Finding a Particular Solution
Find the particular solution to the differential equation y′ = 2x passing through the point (2, 7).
Solution
Any function of the form y = x2 + C is a solution to this differential equation. To determine the value of C,
we substitute the values x = 2 and y = 7 into this equation and solve for C:


y = x2 + C


7 = 22 + C = 4 + C


C = 3.


Therefore the particular solution passing through the point (2, 7) is y = x2 + 3.


Find the particular solution to the differential equation
y′ = 4x + 3


passing through the point (1, 7), given that y = 2x2 + 3x + C is a general solution to the differential
equation.


Initial-Value Problems
Usually a given differential equation has an infinite number of solutions, so it is natural to ask which one we want to use.To choose one solution, more information is needed. Some specific information that can be useful is an initial value, whichis an ordered pair that is used to find a particular solution.
A differential equation together with one or more initial values is called an initial-value problem. The general rule isthat the number of initial values needed for an initial-value problem is equal to the order of the differential equation. Forexample, if we have the differential equation y′ = 2x, then y(3) = 7 is an initial value, and when taken together, these
equations form an initial-value problem. The differential equation y″ − 3y′ + 2y = 4ex is second order, so we need two
initial values. With initial-value problems of order greater than one, the same value should be used for the independentvariable. An example of initial values for this second-order equation would be y(0) = 2 and y′(0) = −1. These two initial
values together with the differential equation form an initial-value problem. These problems are so named because often theindependent variable in the unknown function is t, which represents time. Thus, a value of t = 0 represents the beginning
of the problem.
Example 4.4
Verifying a Solution to an Initial-Value Problem


Chapter 4 | Introduction to Differential Equations 355




4.4


Verify that the function y = 2e−2t + et is a solution to the initial-value problem
y′ + 2y = 3et, y(0) = 3.


Solution
For a function to satisfy an initial-value problem, it must satisfy both the differential equation and the initialcondition. To show that y satisfies the differential equation, we start by calculating y′. This gives
y′ = −4e−2t + et. Next we substitute both y and y′ into the left-hand side of the differential equation and
simplify:


y′ + 2y = ⎛⎝−4e
−2t + et⎞⎠+ 2



⎝2e


−2t + et⎞⎠


= −4e−2t + et + 4e−2t + 2et


= 3et.


This is equal to the right-hand side of the differential equation, so y = 2e−2t + et solves the differential equation.
Next we calculate y(0):


y(0) = 2e−2(0) + e0


= 2 + 1
= 3.


This result verifies the initial value. Therefore the given function satisfies the initial-value problem.


Verify that y = 3e2t + 4sin t is a solution to the initial-value problem
y′ − 2y = 4cos t − 8sin t, y(0) = 3.


In Example 4.4, the initial-value problem consisted of two parts. The first part was the differential equation
y′ + 2y = 3ex, and the second part was the initial value y(0) = 3. These two equations together formed the initial-value
problem.
The same is true in general. An initial-value problem will consists of two parts: the differential equation and the initialcondition. The differential equation has a family of solutions, and the initial condition determines the value of C. The
family of solutions to the differential equation in Example 4.4 is given by y = 2e−2t + Cet. This family of solutions is
shown in Figure 4.3, with the particular solution y = 2e−2t + et labeled.


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(4.1)


Figure 4.3 A family of solutions to the differential equation
y′ + 2y = 3et. The particular solution y = 2e−2t + et is
labeled.


Example 4.5
Solving an Initial-value Problem
Solve the following initial-value problem:


y′ = 3ex + x2 − 4, y(0) = 5.


Solution
The first step in solving this initial-value problem is to find a general family of solutions. To do this, we find anantiderivative of both sides of the differential equation


∫ y′dx = ∫ ⎛⎝3ex + x2 − 4⎞⎠dx,


namely,
y + C1 = 3e


x + 1
3
x3 − 4x + C2.


We are able to integrate both sides because the y term appears by itself. Notice that there are two integrationconstants: C1 and C2. Solving Equation 4.1 for y gives
y = 3ex + 1


3
x3 − 4x + C2 − C1.


Because C1 and C2 are both constants, C2 − C1 is also a constant. We can therefore define C = C2 − C1,
which leads to the equation


y = 3ex + 1
3
x3 − 4x + C.


Next we determine the value of C. To do this, we substitute x = 0 and y = 5 into Equation 4.1 and solve for
C:


5 = 3e0 + 1
3
03 − 4(0) + C


5 = 3 + C
C = 2.


Chapter 4 | Introduction to Differential Equations 357




4.5


Now we substitute the value C = 2 into Equation 4.1. The solution to the initial-value problem is
y = 3ex + 1


3
x3 − 4x + 2.


Analysis
The difference between a general solution and a particular solution is that a general solution involves a family offunctions, either explicitly or implicitly defined, of the independent variable. The initial value or values determinewhich particular solution in the family of solutions satisfies the desired conditions.


Solve the initial-value problem
y′ = x2 − 4x + 3 − 6ex, y(0) = 8.


In physics and engineering applications, we often consider the forces acting upon an object, and use this information tounderstand the resulting motion that may occur. For example, if we start with an object at Earth’s surface, the primary forceacting upon that object is gravity. Physicists and engineers can use this information, along with Newton’s second law ofmotion (in equation form F = ma, where F represents force, m represents mass, and a represents acceleration), to
derive an equation that can be solved.


Figure 4.4 For a baseball falling in air, the only force actingon it is gravity (neglecting air resistance).


In Figure 4.4 we assume that the only force acting on a baseball is the force of gravity. This assumption ignores airresistance. (The force due to air resistance is considered in a later discussion.) The acceleration due to gravity at Earth’s
surface, g, is approximately 9.8 m/s2. We introduce a frame of reference, where Earth’s surface is at a height of 0 meters.
Let v(t) represent the velocity of the object in meters per second. If v(t) > 0, the ball is rising, and if v(t) < 0, the ball
is falling (Figure 4.5).


Figure 4.5 Possible velocities for the rising/falling baseball.


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Our goal is to solve for the velocity v(t) at any time t. To do this, we set up an initial-value problem. Suppose the mass
of the ball is m, where m is measured in kilograms. We use Newton’s second law, which states that the force acting on
an object is equal to its mass times its acceleration (F = ma). Acceleration is the derivative of velocity, so a(t) = v′(t).
Therefore the force acting on the baseball is given by F = m v′(t). However, this force must be equal to the force of gravity
acting on the object, which (again using Newton’s second law) is given by Fg = −mg, since this force acts in a downward
direction. Therefore we obtain the equation F = Fg, which becomes m v′(t) = −mg. Dividing both sides of the equation
by m gives the equation


v′(t) = −g.


Notice that this differential equation remains the same regardless of the mass of the object.
We now need an initial value. Because we are solving for velocity, it makes sense in the context of the problem to assumethat we know the initial velocity, or the velocity at time t = 0. This is denoted by v(0) = v0.
Example 4.6
Velocity of a Moving Baseball
A baseball is thrown upward from a height of 3 meters above Earth’s surface with an initial velocity of 10 m/s,
and the only force acting on it is gravity. The ball has a mass of 0.15 kg at Earth’s surface.


a. Find the velocity v(t) of the baseball at time t.
b. What is its velocity after 2 seconds?


Solution
a. From the preceding discussion, the differential equation that applies in this situation is


v′(t) = −g,


where g = 9.8 m/s2. The initial condition is v(0) = v0, where v0 = 10 m/s. Therefore the initial-
value problem is v′(t) = −9.8 m/s2, v(0) = 10 m/s.
The first step in solving this initial-value problem is to take the antiderivative of both sides of thedifferential equation. This gives


∫ v′ (t)dt = ∫ −9.8dt
v(t) = −9.8t + C.


The next step is to solve for C. To do this, substitute t = 0 and v(0) = 10:
v(t) = −9.8t + C


v(0) = −9.8(0) + C


10 = C.


Therefore C = 10 and the velocity function is given by v(t) = −9.8t + 10.
b. To find the velocity after 2 seconds, substitute t = 2 into v(t).


Chapter 4 | Introduction to Differential Equations 359




4.6


v(t) = −9.8t + 10


v(2) = −9.8(2) + 10


v(2) = −9.6.


The units of velocity are meters per second. Since the answer is negative, the object is falling at a speedof 9.6 m/s.


Suppose a rock falls from rest from a height of 100 meters and the only force acting on it is gravity. Find
an equation for the velocity v(t) as a function of time, measured in meters per second.


A natural question to ask after solving this type of problem is how high the object will be above Earth’s surface at a givenpoint in time. Let s(t) denote the height above Earth’s surface of the object, measured in meters. Because velocity is the
derivative of position (in this case height), this assumption gives the equation s′ (t) = v(t). An initial value is necessary;
in this case the initial height of the object works well. Let the initial height be given by the equation s(0) = s0. Together
these assumptions give the initial-value problem


s′ (t) = v(t), s(0) = s0.


If the velocity function is known, then it is possible to solve for the position function as well.
Example 4.7
Height of a Moving Baseball
A baseball is thrown upward from a height of 3 meters above Earth’s surface with an initial velocity of 10 m/s,
and the only force acting on it is gravity. The ball has a mass of 0.15 kilogram at Earth’s surface.


a. Find the position s(t) of the baseball at time t.
b. What is its height after 2 seconds?


Solution
a. We already know the velocity function for this problem is v(t) = −9.8t + 10. The initial height of the


baseball is 3 meters, so s0 = 3. Therefore the initial-value problem for this example is
To solve the initial-value problem, we first find the antiderivatives:


∫ s′ (t)dt = ∫ −9.8t + 10dt


s(t) = −4.9t2 + 10t + C.


Next we substitute t = 0 and solve for C:
s(t) = −4.9t2 + 10t + C


s(0) = −4.9(0)2 + 10(0) + C


3 = C.


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Therefore the position function is s(t) = −4.9t2 + 10t + 3.
b. The height of the baseball after 2 s is given by s(2):


s(2) = −4.9(2)2 + 10(2) + 3


= −4.9(4) + 23
= 3.4.


Therefore the baseball is 3.4 meters above Earth’s surface after 2 seconds. It is worth noting that the
mass of the ball cancelled out completely in the process of solving the problem.


Chapter 4 | Introduction to Differential Equations 361




4.1 EXERCISES
Determine the order of the following differential equations.
1. y′ + y = 3y2
2. (y′)2 = y′ + 2y
3. y‴ + y″y′ = 3x2
4. y′ = y″ + 3t2


5. dy
dt


= t


6. dy
dx


+
d2 y


dx2
= 3x4


7. ⎛⎝dydt ⎞⎠
2


+ 8
dy
dt


+ 3y = 4t


Verify that the following functions are solutions to thegiven differential equation.
8. y = x3


3
solves y′ = x2


9. y = 2e−x + x − 1 solves y′ = x − y
10. y = e3x − ex


2
solves y′ = 3y + ex


11. y = 1
1 − x


solves y′ = y2


12. y = ex2/2 solves y′ = xy
13. y = 4 + lnx solves xy′ = 1
14. y = 3 − x + x lnx solves y′ = lnx
15. y = 2ex − x − 1 solves y′ = y + x
16. y = ex + sinx


2
− cosx


2
solves y′ = cosx + y


17. y = πe−cosx solves y′ = ysinx
Verify the following general solutions and find theparticular solution.


18. Find the particular solution to the differential equation
y′ = 4x2 that passes through (−3, −30), given that
y = C + 4x


3


3
is a general solution.


19. Find the particular solution to the differential equation
y′ = 3x3 that passes through (1, 4.75), given that
y = C + 3x


4


4
is a general solution.


20. Find the particular solution to the differential equation
y′ = 3x2 y that passes through (0, 12), given that
y = Cex


3 is a general solution.
21. Find the particular solution to the differential equation
y′ = 2xy that passes through ⎛⎝0, 12⎞⎠, given that
y = Cex


2 is a general solution.
22. Find the particular solution to the differential equation
y′ = ⎛⎝2xy⎞⎠2 that passes through ⎛⎝1, − 12⎞⎠, given that
y = − 3


C + 4x3
is a general solution.


23. Find the particular solution to the differential equation
y′ x2 = y that passes through ⎛⎝1, 2e⎞⎠, given that
y = Ce−1/x is a general solution.
24. Find the particular solution to the differential equation
8dx
dt


= −2cos(2t) − cos(4t) that passes through (π, π),
given that x = C − 1


8
sin(2t) − 1


32
sin(4t) is a general


solution.
25. Find the particular solution to the differential equation
du
dt


= tanu that passes through ⎛⎝1, π2⎞⎠, given that
u = sin−1 ⎛⎝e


C + t⎞
⎠ is a general solution.


26. Find the particular solution to the differential equation
dy
dt


= e
(t + y) that passes through (1, 0), given that


y = −ln(C − et) is a general solution.
27. Find the particular solution to the differential equation
y′(1 − x2) = 1 + y that passes through (0, −2), given
that y = C x + 1


1 − x
− 1 is a general solution.


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For the following problems, find the general solution to thedifferential equation.
28. y′ = 3x + ex
29. y′ = lnx + tanx
30. y′ = sinxecosx
31. y′ = 4x
32. y′ = sin−1 (2x)
33. y′ = 2t t2 + 16
34. x′ = coth t + ln t + 3t2
35. x′ = t 4 + t
36. y′ = y
37. y′ = yx
Solve the following initial-value problems starting from
y(t = 0) = 1 and y(t = 0) = −1. Draw both solutions on
the same graph.
38. dy


dt
= 2t


39. dy
dt


= −t


40. dy
dt


= 2y


41. dy
dt


= −y


42. dy
dt


= 2


Solve the following initial-value problems starting from
y0 = 10. At what time does y increase to 100 or drop to
1?


43. dy
dt


= 4t


44. dy
dt


= 4y


45. dy
dt


= −2y


46. dy
dt


= e4t


47. dy
dt


= e−4t


Recall that a family of solutions includes solutions to adifferential equation that differ by a constant. For thefollowing problems, use your calculator to graph a familyof solutions to the given differential equation. Use initialconditions from y(t = 0) = −10 to y(t = 0) = 10
increasing by 2. Is there some critical point where the
behavior of the solution begins to change?
48. [T] y′ = y(x)
49. [T] xy′ = y
50. [T] y′ = t3
51. [T] y′ = x + y (Hint: y = Cex − x − 1 is the
general solution)
52. [T] y′ = x lnx + sinx
53. Find the general solution to describe the velocity of aball of mass 1 lb that is thrown upward at a rate a ft/sec.
54. In the preceding problem, if the initial velocity of theball thrown into the air is a = 25 ft/s, write the particular
solution to the velocity of the ball. Solve to find the timewhen the ball hits the ground.
55. You throw two objects with differing masses m1 and
m2 upward into the air with the same initial velocity a ft/
s. What is the difference in their velocity after 1 second?
56. [T] You throw a ball of mass 1 kilogram upward
with a velocity of a = 25 m/s on Mars, where the force
of gravity is g = −3.711 m/s2. Use your calculator to
approximate how much longer the ball is in the air onMars.
57. [T] For the previous problem, use your calculator toapproximate how much higher the ball went on Mars.
58. [T] A car on the freeway accelerates according to
a = 15cos(πt), where t is measured in hours. Set up
and solve the differential equation to determine the velocityof the car if it has an initial speed of 51 mph. After 40
minutes of driving, what is the driver’s velocity?


Chapter 4 | Introduction to Differential Equations 363




59. [T] For the car in the preceding problem, find theexpression for the distance the car has traveled in time t,
assuming an initial distance of 0. How long does it take
the car to travel 100 miles? Round your answer to hours
and minutes.
60. [T] For the previous problem, find the total distancetraveled in the first hour.
61. Substitute y = Be3t into y′ − y = 8e3t to find a
particular solution.
62. Substitute y = acos(2t) + bsin(2t) into
y′ + y = 4sin(2t) to find a particular solution.
63. Substitute y = a + bt + ct2 into y′ + y = 1 + t2 to
find a particular solution.
64. Substitute y = aet cos t + bet sin t into
y′ = 2et cos t to find a particular solution.
65. Solve y′ = ekt with the initial condition y(0) = 0
and solve y′ = 1 with the same initial condition. As k
approaches 0, what do you notice?


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4.2 | Direction Fields and Numerical Methods
Learning Objectives


4.2.1 Draw the direction field for a given first-order differential equation.
4.2.2 Use a direction field to draw a solution curve of a first-order differential equation.
4.2.3 Use Euler’s Method to approximate the solution to a first-order differential equation.


For the rest of this chapter we will focus on various methods for solving differential equations and analyzing the behaviorof the solutions. In some cases it is possible to predict properties of a solution to a differential equation without knowingthe actual solution. We will also study numerical methods for solving differential equations, which can be programmed byusing various computer languages or even by using a spreadsheet program, such as Microsoft Excel.
Creating Direction Fields
Direction fields (also called slope fields) are useful for investigating first-order differential equations. In particular, weconsider a first-order differential equation of the form


y′ = f (x, y).


An applied example of this type of differential equation appears in Newton’s law of cooling, which we will solve explicitlylater in this chapter. First, though, let us create a direction field for the differential equation
T′ (t) = −0.4(T − 72).


Here T(t) represents the temperature (in degrees Fahrenheit) of an object at time t, and the ambient temperature is 72°F.
Figure 4.6 shows the direction field for this equation.


Figure 4.6 Direction field for the differential equation
T′ (t) = −0.4(T − 72). Two solutions are plotted: one with
initial temperature less than 72°F and the other with initial
temperature greater than 72°F.


The idea behind a direction field is the fact that the derivative of a function evaluated at a given point is the slope of thetangent line to the graph of that function at the same point. Other examples of differential equations for which we can createa direction field include


Chapter 4 | Introduction to Differential Equations 365




y′ = 3x + 2y − 4
y′ = x2 − y2


y′ = 2x + 4
y − 2


.


To create a direction field, we start with the first equation: y′ = 3x + 2y − 4. We let (x0, y0) be any ordered pair, and we
substitute these numbers into the right-hand side of the differential equation. For example, if we choose x = 1 and y = 2,
substituting into the right-hand side of the differential equation yields


y′ = 3x + 2y − 4
= 3(1) + 2(2) − 4 = 3.


This tells us that if a solution to the differential equation y′ = 3x + 2y − 4 passes through the point (1, 2), then the
slope of the solution at that point must equal 3. To start creating the direction field, we put a short line segment at the
point (1, 2) having slope 3. We can do this for any point in the domain of the function f (x, y) = 3x + 2y − 4, which
consists of all ordered pairs (x, y) in ℝ2. Therefore any point in the Cartesian plane has a slope associated with it,
assuming that a solution to the differential equation passes through that point. The direction field for the differential equation
y′ = 3x + 2y − 4 is shown in Figure 4.7.


Figure 4.7 Direction field for the differential equation
y′ = 3x + 2y − 4.


We can generate a direction field of this type for any differential equation of the form y′ = f (x, y).


Definition
A direction field (slope field) is a mathematical object used to graphically represent solutions to a first-orderdifferential equation. At each point in a direction field, a line segment appears whose slope is equal to the slope of asolution to the differential equation passing through that point.


Using Direction Fields
We can use a direction field to predict the behavior of solutions to a differential equation without knowing the actualsolution. For example, the direction field in Figure 4.7 serves as a guide to the behavior of solutions to the differentialequation y′ = 3x + 2y − 4.


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4.7


To use a direction field, we start by choosing any point in the field. The line segment at that point serves as a signposttelling us what direction to go from there. For example, if a solution to the differential equation passes through the point
(0, 1), then the slope of the solution passing through that point is given by y′ = 3(0) + 2(1) − 4 = −2. Now let x
increase slightly, say to x = 0.1. Using the method of linear approximations gives a formula for the approximate value of
y for x = 0.1. In particular,


L(x) = y0 + f ′ (x0)(x − x0)


= 1 − 2(x0 − 0)


= 1 − 2x0.


Substituting x0 = 0.1 into L(x) gives an approximate y value of 0.8.
At this point the slope of the solution changes (again according to the differential equation). We can keep progressing,recalculating the slope of the solution as we take small steps to the right, and watching the behavior of the solution. Figure4.8 shows a graph of the solution passing through the point (0, 1).


Figure 4.8 Direction field for the differential equation
y′ = 3x + 2y − 4 with the solution passing through the point
(0, 1).


The curve is the graph of the solution to the initial-value problem
y′ = 3x + 2y − 4, y(0) = 1.


This curve is called a solution curve passing through the point (0, 1). The exact solution to this initial-value problem is
y = − 3


2
x + 5


4
− 1


4
e2x,


and the graph of this solution is identical to the curve in Figure 4.8.
Create a direction field for the differential equation y′ = x2 − y2 and sketch a solution curve passing


through the point (−1, 2).


Go to this Java applet (http://www.openstaxcollege.org/l/20_DifferEq) and this website(http://www.openstaxcollege.org/l/20_SlopeFields) to see more about slope fields.


Chapter 4 | Introduction to Differential Equations 367




Now consider the direction field for the differential equation y′ = (x − 3)(y2 − 4), shown in Figure 4.9. This direction
field has several interesting properties. First of all, at y = −2 and y = 2, horizontal dashes appear all the way across the
graph. This means that if y = −2, then y′ = 0. Substituting this expression into the right-hand side of the differential
equation gives


(x − 3)(y2 − 4) = (x − 3)((−2 − 4)


= (x − 3)(0)
= 0
= y′.


Therefore y = −2 is a solution to the differential equation. Similarly, y = 2 is a solution to the differential equation. These
are the only constant-valued solutions to the differential equation, as we can see from the following argument. Suppose
y = k is a constant solution to the differential equation. Then y′ = 0. Substituting this expression into the differential
equation yields 0 = (x − 3)⎛⎝k2 − 4⎞⎠. This equation must be true for all values of x, so the second factor must equal zero.
This result yields the equation k2 − 4 = 0. The solutions to this equation are k = −2 and k = 2, which are the constant
solutions already mentioned. These are called the equilibrium solutions to the differential equation.


Figure 4.9 Direction field for the differential equation
y′ = (x − 3)(y2 − 4) showing two solutions. These solutions
are very close together, but one is barely above the equilibriumsolution x = −2 and the other is barely below the same
equilibrium solution.


Definition
Consider the differential equation y′ = f (x, y). An equilibrium solution is any solution to the differential equation
of the form y = c, where c is a constant.


To determine the equilibrium solutions to the differential equation y′ = f (x, y), set the right-hand side equal to zero. An
equilibrium solution of the differential equation is any function of the form y = k such that f (x, k) = 0 for all values of
x in the domain of f .
An important characteristic of equilibrium solutions concerns whether or not they approach the line y = k as an asymptote


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for large values of x.


Definition
Consider the differential equation y′ = f (x, y), and assume that all solutions to this differential equation are defined
for x ≥ x0. Let y = k be an equilibrium solution to the differential equation.


1. y = k is an asymptotically stable solution to the differential equation if there exists ε > 0 such that for any
value c ∈ (k − ε, k + ε) the solution to the initial-value problem


y′ = f (x, y), y(x0) = c


approaches k as x approaches infinity.
2. y = k is an asymptotically unstable solution to the differential equation if there exists ε > 0 such that for


any value c ∈ (k − ε, k + ε) the solution to the initial-value problem
y′ = f (x, y), y(x0) = c


never approaches k as x approaches infinity.
3. y = k is an asymptotically semi-stable solution to the differential equation if it is neither asymptotically


stable nor asymptotically unstable.


Now we return to the differential equation y′ = (x − 3)(y2 − 4), with the initial condition y(0) = 0.5. The direction field
for this initial-value problem, along with the corresponding solution, is shown in Figure 4.10.


Figure 4.10 Direction field for the initial-value problem
y′ = (x − 3)(y2 − 4), y(0) = 0.5.


The values of the solution to this initial-value problem stay between y = −2 and y = 2, which are the equilibrium
solutions to the differential equation. Furthermore, as x approaches infinity, y approaches 2. The behavior of solutions
is similar if the initial value is higher than 2, for example, y(0) = 2.3. In this case, the solutions decrease and approach
y = 2 as x approaches infinity. Therefore y = 2 is an asymptotically stable solution to the differential equation.


Chapter 4 | Introduction to Differential Equations 369




What happens when the initial value is below y = −2? This scenario is illustrated in Figure 4.11, with the initial value
y(0) = −3.


Figure 4.11 Direction field for the initial-value problem
y′ = (x − 3)(y2 − 4), y(0) = −3.


The solution decreases rapidly toward negative infinity as x approaches infinity. Furthermore, if the initial value is slightly
higher than −2, then the solution approaches 2, which is the other equilibrium solution. Therefore in neither case does
the solution approach y = −2, so y = −2 is called an asymptotically unstable, or unstable, equilibrium solution.
Example 4.8
Stability of an Equilibrium Solution
Create a direction field for the differential equation y′ = (y − 3)2(y2 + y − 2) and identify any equilibrium
solutions. Classify each of the equilibrium solutions as stable, unstable, or semi-stable.
Solution
The direction field is shown in Figure 4.12.


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4.8


Figure 4.12 Direction field for the differential equation
y′ = (y − 3)2(y2 + y − 2).


The equilibrium solutions are y = −2, y = 1, and y = 3. To classify each of the solutions, look at an arrow
directly above or below each of these values. For example, at y = −2 the arrows directly below this solution
point up, and the arrows directly above the solution point down. Therefore all initial conditions close to y = −2
approach y = −2, and the solution is stable. For the solution y = 1, all initial conditions above and below
y = 1 are repelled (pushed away) from y = 1, so this solution is unstable. The solution y = 3 is semi-stable,
because for initial conditions slightly greater than 3, the solution approaches infinity, and for initial conditions
slightly less than 3, the solution approaches y = 1.
Analysis
It is possible to find the equilibrium solutions to the differential equation by setting the right-hand side equal tozero and solving for y. This approach gives the same equilibrium solutions as those we saw in the direction field.


Create a direction field for the differential equation y′ = (x + 5)(y + 2)(y2 − 4y + 4) and identify any
equilibrium solutions. Classify each of the equilibrium solutions as stable, unstable, or semi-stable.


Euler’s Method
Consider the initial-value problem


y′ = 2x − 3, y(0) = 3.


Integrating both sides of the differential equation gives y = x2 − 3x + C, and solving for C yields the particular solution
y = x2 − 3x + 3. The solution for this initial-value problem appears as the parabola in Figure 4.13.


Chapter 4 | Introduction to Differential Equations 371




Figure 4.13 Euler’s Method for the initial-value problem
y′ = 2x − 3, y(0) = 3.


The red graph consists of line segments that approximate the solution to the initial-value problem. The graph starts atthe same initial value of (0, 3). Then the slope of the solution at any point is determined by the right-hand side of the
differential equation, and the length of the line segment is determined by increasing the x value by 0.5 each time (the step
size). This approach is the basis of Euler’s Method.
Before we state Euler’s Method as a theorem, let’s consider another initial-value problem:


y′ = x2 − y2, y(−1) = 2.


The idea behind direction fields can also be applied to this problem to study the behavior of its solution. For example, at
the point (−1, 2), the slope of the solution is given by y′ = (−1)2 − 22 = −3, so the slope of the tangent line to the
solution at that point is also equal to −3. Now we define x0 = −1 and y0 = 2. Since the slope of the solution at this
point is equal to −3, we can use the method of linear approximation to approximate y near (−1, 2).


L(x) = y0 + f ′ (x0)(x − x0).


Here x0 = −1, y0 = 2, and f ′ (x0) = −3, so the linear approximation becomes
L(x) = 2 − 3⎛⎝x − (−1)⎞⎠


= 2 − 3x − 3
= −3x − 1.


Now we choose a step size. The step size is a small value, typically 0.1 or less, that serves as an increment for x; it is
represented by the variable h. In our example, let h = 0.1. Incrementing x0 by h gives our next x value:


x1 = x0 + h = −1 + 0.1 = −0.9.


We can substitute x1 = −0.9 into the linear approximation to calculate y1.
y1 = L(x1)


= −3(−0.9) − 1
= 1.7.


Therefore the approximate y value for the solution when x = −0.9 is y = 1.7. We can then repeat the process, using
x1 = −0.9 and y1 = 1.7 to calculate x2 and y2. The new slope is given by y′ = (−0.9)2 − (1.7)2 = −2.08. First,
x2 = x1 + h = −0.9 + 0.1 = −0.8. Using linear approximation gives


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L(x) = y1 + f ′ (x1)(x − x1)


= 1.7 − 2.08(x − (−0.9))
= 1.7 − 2.08x − 1.872
= −2.08x − 0.172.


Finally, we substitute x2 = −0.8 into the linear approximation to calculate y2.
y2 = L(x2)


= −2.08x2 − 0.172


= −2.08(−0.8) − 0.172
= 1.492.


Therefore the approximate value of the solution to the differential equation is y = 1.492 when x = −0.8.
What we have just shown is the idea behind Euler’s Method. Repeating these steps gives a list of values for the solution.These values are shown in Table 4.2, rounded off to four decimal places.


n 0 1 2 3 4 5


xn −1 −0.9 −0.8 −0.7 −0.6 −0.5


yn 2 1.7 1.492 1.3334 1.2046 1.0955


n 6 7 8 9 10


xn −0.4 −0.3 −0.2 −0.1 0


yn 1.0004 1.9164 1.8414 1.7746 1.7156


Table 4.2 Using Euler’s Method to Approximate Solutions to a DifferentialEquation


Theorem 4.1: Euler’s Method
Consider the initial-value problem
y′ = f (x, y), y(x0) = y0.


To approximate a solution to this problem using Euler’s method, define
(4.2)xn = x0 + nh


yn = yn − 1 + h f (xn − 1, yn − 1).


Here h > 0 represents the step size and n is an integer, starting with 1. The number of steps taken is counted by the
variable n.


Typically h is a small value, say 0.1 or 0.05. The smaller the value of h, the more calculations are needed. The higher
the value of h, the fewer calculations are needed. However, the tradeoff results in a lower degree of accuracy for larger
step size, as illustrated in Figure 4.14.


Chapter 4 | Introduction to Differential Equations 373




Figure 4.14 Euler’s method for the initial-value problem y′ = 2x − 3, y(0) = 3 with (a) a step size of
h = 0.5; and (b) a step size of h = 0.25.


Example 4.9
Using Euler’s Method
Consider the initial-value problem


y′ = 3x2 − y2 + 1, y(0) = 2.


Use Euler’s method with a step size of 0.1 to generate a table of values for the solution for values of x between
0 and 1.
Solution
We are given h = 0.1 and f (x, y) = 3x2 − y2 + 1. Furthermore, the initial condition y(0) = 2 gives x0 = 0
and y0 = 2. Using Equation 4.2 with n = 0, we can generate Table 4.3.


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n xn yn = yn − 1 + h f (xn − 1, yn − 1)


0 0 2


1 0.1 y1 = y0 + h f (x0, y0) = 1.7


2 0.2 y2 = y1 + h f (x1, y1) = 1.514


3 0.3 y3 = y2 + h f (x2, y2) = 1.3968


4 0.4 y4 = y3 + h f (x3, y3) = 1.3287


5 0.5 y5 = y4 + h f (x4, y4) = 1.3001


6 0.6 y6 = y5 + h f (x5, y5) = 1.3061


7 0.7 y7 = y6 + h f (x6, y6) = 1.3435


8 0.8 y8 = y7 + h f (x7, y7) = 1.4100


9 0.9 y9 = y8 + h f (x8, y8) = 1.5032


10 1.0 y10 = y9 + h f (x9, y9) = 1.6202


Table 4.3Using Euler’s Method to Approximate Solutions to aDifferential Equation
With ten calculations, we are able to approximate the values of the solution to the initial-value problem for valuesof x between 0 and 1.


Go to thiswebsite (http://www.openstaxcollege.org/l/20_EulersMethod) for more information on Euler’smethod.


Chapter 4 | Introduction to Differential Equations 375




4.9 Consider the initial-value problem
y′ = x3 + y2, y(1) = −2.


Using a step size of 0.1, generate a table with approximate values for the solution to the initial-value problem
for values of x between 1 and 2.


Visit this website (http://www.openstaxcollege.org/l/20_EulerMethod2) for a practical application of thematerial in this section.


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4.2 EXERCISES
For the following problems, use the direction field belowfrom the differential equation y′ = −2y. Sketch the graph
of the solution for the given initial conditions.


66. y(0) = 1
67. y(0) = 0
68. y(0) = −1
69. Are there any equilibria? What are their stabilities?
For the following problems, use the direction field below
from the differential equation y′ = y2 − 2y. Sketch the
graph of the solution for the given initial conditions.


70. y(0) = 3
71. y(0) = 1
72. y(0) = −1
73. Are there any equilibria? What are their stabilities?


Draw the direction field for the following differentialequations, then solve the differential equation. Draw yoursolution on top of the direction field. Does your solutionfollow along the arrows on your direction field?
74. y′ = t3
75. y′ = et


76. dy
dx


= x2 cosx


77. dy
dt


= tet


78. dx
dt


= cosh(t)


Draw the directional field for the following differentialequations. What can you say about the behavior of thesolution? Are there equilibria? What stability do theseequilibria have?
79. y′ = y2 − 1
80. y′ = y − x
81. y′ = 1 − y2 − x2
82. y′ = t2 siny
83. y′ = 3y + xy
Match the direction field with the given differentialequations. Explain your selections.


Chapter 4 | Introduction to Differential Equations 377




84. y′ = −3y
85. y′ = −3t
86. y′ = et


87. y′ = 1
2
y + t


88. y′ = −ty
Match the direction field with the given differentialequations. Explain your selections.


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89. y′ = t siny
90. y′ = −tcosy
91. y′ = t tany
92. y′ = sin2 y
93. y′ = y2 t3
Estimate the following solutions using Euler’s method with
n = 5 steps over the interval t = [0, 1]. If you are able
to solve the initial-value problem exactly, compare yoursolution with the exact solution. If you are unable to solvethe initial-value problem, the exact solution will beprovided for you to compare with Euler’s method. Howaccurate is Euler’s method?
94. y′ = −3y, y(0) = 1
95. y′ = t2
96. y′ = 3t − y, y(0) = 1. Exact solution is
y = 3t + 4e−t − 3


97. y′ = y + t2, y(0) = 3. Exact solution is
y = 5et − 2 − t2 − 2t


98. y′ = 2t, y(0) = 0
99. [T] y′ = e(x + y), y(0) = −1. Exact solution is
y = −ln(e + 1 − ex)


Chapter 4 | Introduction to Differential Equations 379




100. y′ = y2 ln(x + 1), y(0) = 1. Exact solution is
y = − 1


(x + 1)(ln(x + 1) − 1)


101. y′ = 2x, y(0) = 0, Exact solution is y = 2x − 1
ln(2)


102. y′ = y, y(0) = −1. Exact solution is y = −ex.
103. y′ = −5t, y(0) = −2. Exact solution is
y = − 5


2
t2 − 2


Differential equations can be used to model diseaseepidemics. In the next set of problems, we examine thechange of size of two sub-populations of people living ina city: individuals who are infected and individuals whoare susceptible to infection. S represents the size of the
susceptible population, and I represents the size of the
infected population. We assume that if a susceptible personinteracts with an infected person, there is a probability
c that the susceptible person will become infected. Each
infected person recovers from the infection at a rate r
and becomes susceptible again. We consider the case ofinfluenza, where we assume that no one dies from thedisease, so we assume that the total population size ofthe two sub-populations is a constant number, N. The
differential equations that model these population sizes are


S′ = rI − cSI and
I′ = cSI − rI.


Here c represents the contact rate and r is the recovery
rate.
104. Show that, by our assumption that the totalpopulation size is constant (S + I = N), you can reduce
the system to a single differential equation in
I: I′ = c(N − I)I − rI.


105. Assuming the parameters are c = 0.5, N = 5, and
r = 0.5, draw the resulting directional field.
106. [T] Use computational software or a calculator tocompute the solution to the initial-value problem
y′ = ty, y(0) = 2 using Euler’s Method with the given
step size h. Find the solution at t = 1. For a hint, here
is “pseudo-code” for how to write a computer programto perform Euler’s Method for y′ = f (t, y), y(0) = 2:
Create function f (t, y) Define parameters
y(1) = y0, t(0) = 0, step size h, and total number
of steps, N Write a for loop: for k = 1 to N
fn = f⎛⎝t(k), y(k)⎞⎠ y(k+1) = y(k) + h*fn


t(k+1) = t(k) + h


107. Solve the initial-value problem for the exactsolution.
108. Draw the directional field
109. h = 1
110. [T] h = 10
111. [T] h = 100
112. [T] h = 1000
113. [T] Evaluate the exact solution at t = 1. Make a
table of errors for the relative error between the Euler’smethod solution and the exact solution. How much does theerror change? Can you explain?
Consider the initial-value problem y′ = −2y, y(0) = 2.
114. Show that y = 2e−2x solves this initial-value
problem.
115. Draw the directional field of this differentialequation.
116. [T] By hand or by calculator or computer,approximate the solution using Euler’s Method at t = 10
using h = 5.
117. [T] By calculator or computer, approximate thesolution using Euler’s Method at t = 10 using h = 100.
118. [T] Plot exact answer and each Euler approximation(for h = 5 and h = 100) at each h on the directional
field. What do you notice?


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4.3 | Separable Equations
Learning Objectives


4.3.1 Use separation of variables to solve a differential equation.
4.3.2 Solve applications using separation of variables.


We now examine a solution technique for finding exact solutions to a class of differential equations known as separabledifferential equations. These equations are common in a wide variety of disciplines, including physics, chemistry, andengineering. We illustrate a few applications at the end of the section.
Separation of Variables
We start with a definition and some examples.
Definition
A separable differential equation is any equation that can be written in the form


(4.3)y′ = f (x)g(y).
The term ‘separable’ refers to the fact that the right-hand side of the equation can be separated into a function of x times a
function of y. Examples of separable differential equations include


y′ = ⎛⎝x
2 − 4⎞⎠



⎝3y + 2⎞⎠


y′ = 6x2 + 4x


y′ = secy + tany
y′ = xy + 3x − 2y − 6.


The second equation is separable with f (x) = 6x2 + 4x and g(y) = 1, the third equation is separable with f (x) = 1 and
g(y) = secy + tany, and the right-hand side of the fourth equation can be factored as (x + 3)⎛⎝y − 2⎞⎠, so it is separable
as well. The third equation is also called an autonomous differential equation because the right-hand side of the equationis a function of y alone. If a differential equation is separable, then it is possible to solve the equation using the method of
separation of variables.
Problem-Solving Strategy: Separation of Variables


1. Check for any values of y that make g(y) = 0. These correspond to constant solutions.
2. Rewrite the differential equation in the form dy


g(y)
= f (x)dx.


3. Integrate both sides of the equation.
4. Solve the resulting equation for y if possible.
5. If an initial condition exists, substitute the appropriate values for x and y into the equation and solve for the


constant.
Note that Step 4. states “Solve the resulting equation for y if possible.” It is not always possible to obtain y as an
explicit function of x. Quite often we have to be satisfied with finding y as an implicit function of x.


Chapter 4 | Introduction to Differential Equations 381




Example 4.10
Using Separation of Variables
Find a general solution to the differential equation y′ = ⎛⎝x2 − 4⎞⎠⎛⎝3y + 2⎞⎠ using the method of separation of
variables.
Solution
Follow the five-step method of separation of variables.


1. In this example, f (x) = x2 − 4 and g(y) = 3y + 2. Setting g(y) = 0 gives y = − 2
3
as a constant


solution.
2. Rewrite the differential equation in the form


dy
3y + 2


= (x2 − 4)dx.


3. Integrate both sides of the equation:



dy
3y + 2


= ∫ ⎛⎝x2 − 4⎞⎠dx.


Let u = 3y + 2. Then du = 3dy
dx


dx, so the equation becomes
1
3∫


1
udu =


1
3
x3 − 4x + C


1
3
ln|u| = 13


x3 − 4x + C


1
3
ln|3y + 2| = 13x


3 − 4x + C.


4. To solve this equation for y, first multiply both sides of the equation by 3.
ln|3y + 2| = x3 − 12x + 3C


Now we use some logic in dealing with the constant C. Since C represents an arbitrary constant, 3C
also represents an arbitrary constant. If we call the second arbitrary constant C1, the equation becomes


ln|3y + 2| = x3 − 12x + C1.


Now exponentiate both sides of the equation (i.e., make each side of the equation the exponent for thebase e).
e
ln|3y + 2| = e


x3 − 12x + C1


|3y + 2| = e
C1 ex


3 − 12x


Again define a new constant C2 = ec1 (note that C2 > 0):
|3y + 2| = C2 ex


3 − 12x.


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4.10


(4.4)


This corresponds to two separate equations: 3y + 2 = C2 ex3 − 12x and 3y + 2 = −C2 ex3 − 12x.
The solution to either equation can be written in the form y = −2 ± C2 ex


3 − 12x


3
.


Since C2 > 0, it does not matter whether we use plus or minus, so the constant can actually have either
sign. Furthermore, the subscript on the constant C is entirely arbitrary, and can be dropped. Therefore
the solution can be written as


y = −2 + Ce
x3 − 12x


3
.


5. No initial condition is imposed, so we are finished.


Use the method of separation of variables to find a general solution to the differential equation
y′ = 2xy + 3y − 4x − 6.


Example 4.11
Solving an Initial-Value Problem
Using the method of separation of variables, solve the initial-value problem


y′ = (2x + 3)(y2 − 4), y(0) = −3.


Solution
Follow the five-step method of separation of variables.


1. In this example, f (x) = 2x + 3 and g(y) = y2 − 4. Setting g(y) = 0 gives y = ± 2 as constant
solutions.


2. Divide both sides of the equation by y2 − 4 and multiply by dx. This gives the equation
dy


y2 − 4
= (2x + 3)dx.


3. Next integrate both sides:



1
y2 − 4


dy = ∫ (2x + 3)dx.


To evaluate the left-hand side, use the method of partial fraction decomposition. This leads to the identity
1


y2 − 4
= 1


4



1
y − 2


− 1
y + 2

⎠.


Then Equation 4.4 becomes


Chapter 4 | Introduction to Differential Equations 383




1
4





1
y − 2


− 1
y + 2

⎠dy = ∫ (2x + 3)dx


1
4

⎝ln|y − 2| − ln|y + 2|⎞⎠ = x2 + 3x + C.


Multiplying both sides of this equation by 4 and replacing 4C with C1 gives
ln|y − 2| − ln|y + 2| = 4x2 + 12x + C1


ln|y − 2y + 2| = 4x2 + 12x + C1.
4. It is possible to solve this equation for y. First exponentiate both sides of the equation and define


C2 = e
C1 :


|y − 2y + 2| = C2 e4x2 + 12x.
Next we can remove the absolute value and let C2 be either positive or negative. Then multiply both
sides by y + 2.


y − 2 = C2

⎝y + 2⎞⎠e4x


2 + 12x


y − 2 = C2 ye
4x2 + 12x + 2C2 e


4x2 + 12x.


Now collect all terms involving y on one side of the equation, and solve for y:
y − C2 ye


4x2 + 12x = 2 + 2C2 e
4x2 + 12x


y(1 − C2 e
4x2 + 12x) = 2 + 2C2 e


4x2 + 12x


y =
2 + 2C2 e


4x2 + 12x


1 − C2 e
4x2 + 12x


.


5. To determine the value of C2, substitute x = 0 and y = −1 into the general solution. Alternatively,
we can put the same values into an earlier equation, namely the equation y − 2


y + 2
= C2 e


4x2 + 12. This is
much easier to solve for C2 :


y − 2
y + 2


= C2 e
4x2 + 12x


−1 − 2
−1 + 2


= C2 e
4(0)2 + 12(0)


C2 = −3.


Therefore the solution to the initial-value problem is
y = 2 − 6e


4x2 + 12x


1 + 3e4x
2 + 12x


.


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4.11


A graph of this solution appears in Figure 4.15.


Figure 4.15 Graph of the solution to the initial-value problem
y′ = (2x + 3)⎛⎝y


2 − 4⎞⎠, y(0) = −3.


Find the solution to the initial-value problem
6y′ = (2x + 1)⎛⎝y


2 − 2y − 8⎞⎠, y(0) = −3


using the method of separation of variables.


Applications of Separation of Variables
Many interesting problems can be described by separable equations. We illustrate two types of problems: solutionconcentrations and Newton’s law of cooling.
Solution concentrations
Consider a tank being filled with a salt solution. We would like to determine the amount of salt present in the tank as afunction of time. We can apply the process of separation of variables to solve this problem and similar problems involvingsolution concentrations.
Example 4.12
Determining Salt Concentration over Time
A tank containing 100 L of a brine solution initially has 4 kg of salt dissolved in the solution. At time t = 0,
another brine solution flows into the tank at a rate of 2 L/min. This brine solution contains a concentration of
0.5 kg/L of salt. At the same time, a stopcock is opened at the bottom of the tank, allowing the combined solution
to flow out at a rate of 2 L/min, so that the level of liquid in the tank remains constant (Figure 4.16). Find the
amount of salt in the tank as a function of time (measured in minutes), and find the limiting amount of salt in thetank, assuming that the solution in the tank is well mixed at all times.


Chapter 4 | Introduction to Differential Equations 385




(4.5)


Figure 4.16 A brine tank with an initial amount of saltsolution accepts an input flow and delivers an output flow. Howdoes the amount of salt change with time?
Solution
First we define a function u(t) that represents the amount of salt in kilograms in the tank as a function of time.
Then du


dt
represents the rate at which the amount of salt in the tank changes as a function of time. Also, u(0)


represents the amount of salt in the tank at time t = 0, which is 4 kilograms.
The general setup for the differential equation we will solve is of the form


du
dt


= INFLOW RATE − OUTFLOW RATE.


INFLOW RATE represents the rate at which salt enters the tank, and OUTFLOW RATE represents the rate atwhich salt leaves the tank. Because solution enters the tank at a rate of 2 L/min, and each liter of solution
contains 0.5 kilogram of salt, every minute 2(0.5) = 1 kilogram of salt enters the tank. Therefore INFLOW
RATE = 1.
To calculate the rate at which salt leaves the tank, we need the concentration of salt in the tank at any point intime. Since the actual amount of salt varies over time, so does the concentration of salt. However, the volume ofthe solution remains fixed at 100 liters. The number of kilograms of salt in the tank at time t is equal to u(t).
Thus, the concentration of salt is u(t)


100
kg/L, and the solution leaves the tank at a rate of 2 L/min. Therefore


salt leaves the tank at a rate of u(t)
100


· 2 = u(t)
50


kg/min, and OUTFLOW RATE is equal to u(t)
50


. Therefore the
differential equation becomes du


dt
= 1 − u


50
, and the initial condition is u(0) = 4. The initial-value problem to


be solved is
du
dt


= 1 − u
50


, u(0) = 4.


The differential equation is a separable equation, so we can apply the five-step strategy for solution.
Step 1. Setting 1 − u


50
= 0 gives u = 50 as a constant solution. Since the initial amount of salt in the tank is 4


kilograms, this solution does not apply.
Step 2. Rewrite the equation as


du
dt


= 50 − u
50


.


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4.12


Then multiply both sides by dt and divide both sides by 50 − u:
du


50 − u
= dt


50
.


Step 3. Integrate both sides:



du
50 − u


= ⌠

dt
50


−ln|50 − u| = t50 + C.


Step 4. Solve for u(t):
ln|50 − u| = − t50 − C


e
ln|50 − u| = e−(t/50) − C


|50 − u| = C1 e−t/50.


Eliminate the absolute value by allowing the constant to be either positive or negative:
50 − u = C1 e


−t/50.


Finally, solve for u(t):
u(t) = 50 − C1 e


−t/50.


Step 5. Solve for C1 :
u(0) = 50 − C1 e


−0/50


4 = 50 − C1
C1 = 46.


The solution to the initial value problem is u(t) = 50 − 46e−t/50. To find the limiting amount of salt in the tank,
take the limit as t approaches infinity:


lim
t → ∞


u(t) = 50 − 46e−t/50


= 50 − 46(0)
= 50.


Note that this was the constant solution to the differential equation. If the initial amount of salt in the tank is 50
kilograms, then it remains constant. If it starts at less than 50 kilograms, then it approaches 50 kilograms overtime.


A tank contains 3 kilograms of salt dissolved in 75 liters of water. A salt solution of 0.4 kg salt/L is
pumped into the tank at a rate of 6 L/min and is drained at the same rate. Solve for the salt concentration at
time t. Assume the tank is well mixed at all times.


Newton’s law of cooling
Newton’s law of cooling states that the rate of change of an object’s temperature is proportional to the difference betweenits own temperature and the ambient temperature (i.e., the temperature of its surroundings). If we let T(t) represent
the temperature of an object as a function of time, then dT


dt
represents the rate at which that temperature changes. The


Chapter 4 | Introduction to Differential Equations 387




temperature of the object’s surroundings can be represented by Ts. Then Newton’s law of cooling can be written in the
form


dT
dt


= k⎛⎝T(t) − Ts



or simply
(4.6)dT


dt
= k(T − Ts).


The temperature of the object at the beginning of any experiment is the initial value for the initial-value problem. We callthis temperature T0. Therefore the initial-value problem that needs to be solved takes the form
(4.7)dT


dt
= k(T − Ts), T(0) = T0,


where k is a constant that needs to be either given or determined in the context of the problem. We use these equations in
Example 4.13.
Example 4.13
Waiting for a Pizza to Cool
A pizza is removed from the oven after baking thoroughly, and the temperature of the oven is 350°F. The
temperature of the kitchen is 75°F, and after 5 minutes the temperature of the pizza is 340°F. We would like
to wait until the temperature of the pizza reaches 300°F before cutting and serving it (Figure 4.17). How much
longer will we have to wait?


Figure 4.17 From Newton’s law of cooling, if the pizza cools
10°F in 5 minutes, how long before it cools to 300°F?


Solution
The ambient temperature (surrounding temperature) is 75°F, so Ts = 75. The temperature of the pizza when
it comes out of the oven is 350°F, which is the initial temperature (i.e., initial value), so T0 = 350. Therefore
Equation 4.4 becomes


dT
dt


= k(T − 75), T(0) = 350.


To solve the differential equation, we use the five-step technique for solving separable equations.
1. Setting the right-hand side equal to zero gives T = 75 as a constant solution. Since the pizza starts at


350°F, this is not the solution we are seeking.


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2. Rewrite the differential equation by multiplying both sides by dt and dividing both sides by T − 75:
dT


T − 75
= kdt.


3. Integrate both sides:



dT
T − 75


= ∫ kdt
ln|T − 75| = kt + C.


4. Solve for T by first exponentiating both sides:
e
ln|T − 75| = ekt + C


|T − 75| = C1 ekt


T − 75 = C1 e
kt


T(t) = 75 + C1 e
kt.


5. Solve for C1 by using the initial condition T(0) = 350:
T(t) = 75 + C1 e


kt


T(0) = 75 + C1 e
k(0)


350 = 75 + C1
C1 = 275.


Therefore the solution to the initial-value problem is
T(t) = 75 + 275ekt.


To determine the value of k, we need to use the fact that after 5 minutes the temperature of the pizza
is 340°F. Therefore T(5) = 340. Substituting this information into the solution to the initial-value
problem, we have


T(t) = 75 + 275ekt


T(5) = 340 = 75 + 275e5k


265 = 275e5k


e5k = 53
55


lne5k = ln⎛⎝
53
55



5k = ln⎛⎝
53
55



k = 1
5
ln⎛⎝


53
55

⎠ ≈ − 0.007408.


So now we have T(t) = 75 + 275e−0.007048t. When is the temperature 300°F? Solving for t, we find


Chapter 4 | Introduction to Differential Equations 389




4.13


T(t) = 75 + 275e−0.007048t


300 = 75 + 275e−0.007048t


225 = 275e−0.007048t


e−0.007048t = 9
11


lne−0.007048t = ln 9
11


−0.007048t = ln 9
11


t = − 1
0.007048


ln 9
11


≈ 28.5.


Therefore we need to wait an additional 23.5 minutes (after the temperature of the pizza reached
340°F). That should be just enough time to finish this calculation.


A cake is removed from the oven after baking thoroughly, and the temperature of the oven is 450°F.
The temperature of the kitchen is 70°F, and after 10 minutes the temperature of the cake is 430°F.


a. Write the appropriate initial-value problem to describe this situation.
b. Solve the initial-value problem for T(t).
c. How long will it take until the temperature of the cake is within 5°F of room temperature?


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4.3 EXERCISES
Solve the following initial-value problems with the initialcondition y0 = 0 and graph the solution.
119. dy


dt
= y + 1


120. dy
dt


= y − 1


121. dy
dt


= y + 1


122. dy
dt


= −y − 1


Find the general solution to the differential equation.
123. x2 y′ = (x + 1)y
124. y′ = tan(y)x
125. y′ = 2xy2


126. dy
dt


= ycos(3t + 2)


127. 2xdy
dx


= y2


128. y′ = ey x2
129. (1 + x)y′ = (x + 2)⎛⎝y − 1⎞⎠
130. dx


dt
= 3t2 ⎛⎝x


2 + 4⎞⎠


131. tdy
dt


= 1 − y2


132. y′ = ex ey
Find the solution to the initial-value problem.
133. y′ = ey − x, y(0) = 0
134. y′ = y2(x + 1), y(0) = 2
135. dy


dx
= y3 xex


2
, y(0) = 1


136. dy
dt


= y2 ex sin(3x), y(0) = 1


137. y′ = x
sech2 y


, y(0) = 0


138. y′ = 2xy(1 + 2y), y(0) = −1
139. dx


dt
= ln(t) 1 − x2, x(0) = 0


140. y′ = 3x2(y2 + 4), y(0) = 0
141. y′ = ey5x, y(0) = ln(ln(5))
142. y′ = −2x tan(y), y(0) = π


2


For the following problems, use a software program or yourcalculator to generate the directional fields. Solve explicitlyand draw solution curves for several initial conditions. Arethere some critical initial conditions that change thebehavior of the solution?
143. [T] y′ = 1 − 2y
144. [T] y′ = y2 x3


145. [T] y′ = y3 ex
146. [T] y′ = ey
147. [T] y′ = y ln(x)
148. Most drugs in the bloodstream decay according tothe equation y′ = cy, where y is the concentration of
the drug in the bloodstream. If the half-life of a drug is
2 hours, what fraction of the initial dose remains after 6
hours?
149. A drug is administered intravenously to a patientat a rate r mg/h and is cleared from the body at a rate
proportional to the amount of drug still present in the body,
d Set up and solve the differential equation, assuming
there is no drug initially present in the body.
150. [T] How often should a drug be taken if its dose is
3 mg, it is cleared at a rate c = 0.1 mg/h, and 1 mg is
required to be in the bloodstream at all times?
151. A tank contains 1 kilogram of salt dissolved in 100
liters of water. A salt solution of 0.1 kg salt/L is pumped
into the tank at a rate of 2 L/min and is drained at the same
rate. Solve for the salt concentration at time t. Assume the
tank is well mixed.


Chapter 4 | Introduction to Differential Equations 391




152. A tank containing 10 kilograms of salt dissolved
in 1000 liters of water has two salt solutions pumped in.
The first solution of 0.2 kg salt/L is pumped in at a rate
of 20 L/min and the second solution of 0.05 kg salt/L is
pumped in at a rate of 5 L/min. The tank drains at 25
L/min. Assume the tank is well mixed. Solve for the saltconcentration at time t.
153. [T] For the preceding problem, find how much salt isin the tank 1 hour after the process begins.
154. Torricelli’s law states that for a water tank with ahole in the bottom that has a cross-section of A and with
a height of water h above the bottom of the tank, the
rate of change of volume of water flowing from the tankis proportional to the square root of the height of water,
according to dV


dt
= −A 2gh, where g is the acceleration


due to gravity. Note that dV
dt


= Adh
dt


. Solve the resulting
initial-value problem for the height of the water, assuminga tank with a hole of radius 2 ft. The initial height of water
is 100 ft.
155. For the preceding problem, determine how long ittakes the tank to drain.
For the following problems, use Newton’s law of cooling.
156. The liquid base of an ice cream has an initialtemperature of 200°F before it is placed in a freezer with
a constant temperature of 0°F. After 1 hour, the
temperature of the ice-cream base has decreased to 140°F.
Formulate and solve the initial-value problem to determinethe temperature of the ice cream.
157. [T] The liquid base of an ice cream has an initialtemperature of 210°F before it is placed in a freezer with
a constant temperature of 20°F. After 2 hours, the
temperature of the ice-cream base has decreased to 170°F.
At what time will the ice cream be ready to eat? (Assume
30°F is the optimal eating temperature.)
158. [T] You are organizing an ice cream social. Theoutside temperature is 80°F and the ice cream is at 10°F.
After 10 minutes, the ice cream temperature has risen by
10°F. How much longer can you wait before the ice cream
melts at 40°F?
159. You have a cup of coffee at temperature 70°C and
the ambient temperature in the room is 20°C. Assuming
a cooling rate k of 0.125, write and solve the differential
equation to describe the temperature of the coffee withrespect to time.


160. [T] You have a cup of coffee at temperature 70°C
that you put outside, where the ambient temperature is
0°C. After 5 minutes, how much colder is the coffee?
161. You have a cup of coffee at temperature 70°C and
you immediately pour in 1 part milk to 5 parts coffee.
The milk is initially at temperature 1°C. Write and solve
the differential equation that governs the temperature ofthis coffee.
162. You have a cup of coffee at temperature 70°C,
which you let cool 10 minutes before you pour in the same
amount of milk at 1°C as in the preceding problem. How
does the temperature compare to the previous cup after 10
minutes?
163. Solve the generic problem y′ = ay + b with initial
condition y(0) = c.
164. Prove the basic continual compounded interestequation. Assuming an initial deposit of P0 and an interest
rate of r, set up and solve an equation for continually
compounded interest.
165. Assume an initial nutrient amount of I kilograms
in a tank with L liters. Assume a concentration of c kg/
L being pumped in at a rate of r L/min. The tank is well
mixed and is drained at a rate of r L/min. Find the equation
describing the amount of nutrient in the tank.
166. Leaves accumulate on the forest floor at a rate of
2 g/cm2/yr and also decompose at a rate of 90% per
year. Write a differential equation governing the number ofgrams of leaf litter per square centimeter of forest floor,assuming at time 0 there is no leaf litter on the ground.
Does this amount approach a steady value? What is thatvalue?
167. Leaves accumulate on the forest floor at a rate of 4
g/cm2/yr. These leaves decompose at a rate of 10% per
year. Write a differential equation governing the number ofgrams of leaf litter per square centimeter of forest floor.Does this amount approach a steady value? What is thatvalue?


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4.4 | The Logistic Equation
Learning Objectives


4.4.1 Describe the concept of environmental carrying capacity in the logistic model of populationgrowth.
4.4.2 Draw a direction field for a logistic equation and interpret the solution curves.
4.4.3 Solve a logistic equation and interpret the results.


Differential equations can be used to represent the size of a population as it varies over time. We saw this in an earlierchapter in the section on exponential growth and decay, which is the simplest model. A more realistic model includes otherfactors that affect the growth of the population. In this section, we study the logistic differential equation and see how itapplies to the study of population dynamics in the context of biology.
Population Growth and Carrying Capacity
To model population growth using a differential equation, we first need to introduce some variables and relevant terms. Thevariable t. will represent time. The units of time can be hours, days, weeks, months, or even years. Any given problem
must specify the units used in that particular problem. The variable P will represent population. Since the population varies
over time, it is understood to be a function of time. Therefore we use the notation P(t) for the population as a function
of time. If P(t) is a differentiable function, then the first derivative dP


dt
represents the instantaneous rate of change of the


population as a function of time.
In Exponential Growth and Decay, we studied the exponential growth and decay of populations and radioactive
substances. An example of an exponential growth function is P(t) = P0 ert. In this function, P(t) represents the
population at time t, P0 represents the initial population (population at time t = 0), and the constant r > 0 is called
the growth rate. Figure 4.18 shows a graph of P(t) = 100e0.03t. Here P0 = 100 and r = 0.03.


Figure 4.18 An exponential growth model of population.


We can verify that the function P(t) = P0 ert satisfies the initial-value problem
dP
dt


= rP, P(0) = P0.


This differential equation has an interesting interpretation. The left-hand side represents the rate at which the populationincreases (or decreases). The right-hand side is equal to a positive constant multiplied by the current population. Thereforethe differential equation states that the rate at which the population increases is proportional to the population at that pointin time. Furthermore, it states that the constant of proportionality never changes.
One problem with this function is its prediction that as time goes on, the population grows without bound. This is unrealisticin a real-world setting. Various factors limit the rate of growth of a particular population, including birth rate, death rate,food supply, predators, and so on. The growth constant r usually takes into consideration the birth and death rates but
none of the other factors, and it can be interpreted as a net (birth minus death) percent growth rate per unit time. A naturalquestion to ask is whether the population growth rate stays constant, or whether it changes over time. Biologists have foundthat in many biological systems, the population grows until a certain steady-state population is reached. This possibility is


Chapter 4 | Introduction to Differential Equations 393




not taken into account with exponential growth. However, the concept of carrying capacity allows for the possibility that ina given area, only a certain number of a given organism or animal can thrive without running into resource issues.
Definition
The carrying capacity of an organism in a given environment is defined to be the maximum population of thatorganism that the environment can sustain indefinitely.


We use the variable K to denote the carrying capacity. The growth rate is represented by the variable r. Using these
variables, we can define the logistic differential equation.
Definition
Let K represent the carrying capacity for a particular organism in a given environment, and let r be a real number
that represents the growth rate. The function P(t) represents the population of this organism as a function of time t,
and the constant P0 represents the initial population (population of the organism at time t = 0). Then the logistic
differential equation is


(4.8)dP
dt


= rP⎛⎝1 −
P
K

⎠− = rP.


See this website (http://www.openstaxcollege.org/l/20_logisticEq) for more information on the logisticequation.


The logistic equation was first published by Pierre Verhulst in 1845. This differential equation can be coupled with the
initial condition P(0) = P0 to form an initial-value problem for P(t).
Suppose that the initial population is small relative to the carrying capacity. Then P


K
is small, possibly close to zero. Thus,


the quantity in parentheses on the right-hand side of Equation 4.8 is close to 1, and the right-hand side of this equation
is close to rP. If r > 0, then the population grows rapidly, resembling exponential growth.
However, as the population grows, the ratio P


K
also grows, because K is constant. If the population remains below the


carrying capacity, then P
K
is less than 1, so 1 − P


K
> 0. Therefore the right-hand side of Equation 4.8 is still positive,


but the quantity in parentheses gets smaller, and the growth rate decreases as a result. If P = K then the right-hand side is
equal to zero, and the population does not change.
Now suppose that the population starts at a value higher than the carrying capacity. Then P


K
> 1, and 1 − P


K
< 0.


Then the right-hand side of Equation 4.8 is negative, and the population decreases. As long as P > K, the population
decreases. It never actually reaches K because dP


dt
will get smaller and smaller, but the population approaches the carrying


capacity as t approaches infinity. This analysis can be represented visually by way of a phase line. A phase line describes
the general behavior of a solution to an autonomous differential equation, depending on the initial condition. For the caseof a carrying capacity in the logistic equation, the phase line is as shown in Figure 4.19.


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Figure 4.19 A phase line for the differential equation
dP
dt


= rP⎛⎝1 −
P
K

⎠.


This phase line shows that when P is less than zero or greater than K, the population decreases over time. When P is
between 0 and K, the population increases over time.
Example 4.14
Chapter Opener: Examining the Carrying Capacity of a Deer Population


Figure 4.20 (credit: modification of work by Rachel Kramer,Flickr)


Let’s consider the population of white-tailed deer (Odocoileus virginianus) in the state of Kentucky. TheKentucky Department of Fish and Wildlife Resources (KDFWR) sets guidelines for hunting and fishing inthe state. Before the hunting season of 2004, it estimated a population of 900,000 deer. Johnson notes:
“A deer population that has plenty to eat and is not hunted by humans or other predators will double everythree years.” (George Johnson, “The Problem of Exploding Deer Populations Has No Attractive Solutions,”January 12, 2001, accessed April 9, 2015, http://www.txtwriter.com/onscience/Articles/deerpops.html.) This
observation corresponds to a rate of increase r = ln(2)


3
= 0.2311, so the approximate growth rate is 23.11%


per year. (This assumes that the population grows exponentially, which is reasonable––at least in the shortterm––with plentiful food supply and no predators.) The KDFWR also reports deer population densities for 32
counties in Kentucky, the average of which is approximately 27 deer per square mile. Suppose this is the deer
density for the whole state (39,732 square miles). The carrying capacity K is 39,732 square miles times 27
deer per square mile, or 1,072,764 deer.


Chapter 4 | Introduction to Differential Equations 395




a. For this application, we have P0 = 900,000, K = 1,072,764, and r = 0.2311. Substitute these values
into Equation 4.8 and form the initial-value problem.


b. Solve the initial-value problem from part a.
c. According to this model, what will be the population in 3 years? Recall that the doubling time predicted


by Johnson for the deer population was 3 years. How do these values compare?
d. Suppose the population managed to reach 1,200,000 deer. What does the logistic equation predict will


happen to the population in this scenario?
Solution


a. The initial value problem is
dP
dt


= 0.2311P

⎝1 −


P
1,072,764



⎠, P(0) = 900,000.


b. The logistic equation is an autonomous differential equation, so we can use the method of separation ofvariables.Step 1: Setting the right-hand side equal to zero gives P = 0 and P = 1,072,764. This means that if the
population starts at zero it will never change, and if it starts at the carrying capacity, it will never change.Step 2: Rewrite the differential equation and multiply both sides by:


dP
dt


= 0.2311P


1,072,764 − P
1,072,764





dP = 0.2311P


1,072,764 − P
1,072,764



⎠dt.


Divide both sides by P(1,072,764 − P):
dP


P(1,072,764 − P)
= 0.2311


1,072,764
dt.


Step 3: Integrate both sides of the equation using partial fraction decomposition:



dP
P(1,072,764 − P)


= ⌠


0.2311
1,072,764


dt


1
1,072,764






1
P


+ 1
1,072,764 − P



⎠dP =


0.2311t
1,072,764


+ C


1
1,072,764



⎝ln|P| − ln|1,072,764 − P|⎞⎠ = 0.2311t1,072,764 + C.


Step 4: Multiply both sides by 1,072,764 and use the quotient rule for logarithms:
ln| P1,072,764 − P | = 0.2311t + C1.


Here C1 = 1,072,764C. Next exponentiate both sides and eliminate the absolute value:


e
ln| P1,072,764 − P | = e0.2311t + C1


| P1,072,764 − P | = C2 e0.2311t
P


1,072,764 − P
= C2 e


0.2311t.


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Here C2 = eC1 but after eliminating the absolute value, it can be negative as well. Now solve for:
P = C2 e


0.2311t (1,072,764 − P).


P = 1,072,764C2 e
0.2311t − C2Pe


0.2311t


P + C2Pe
0.2311t = 1,072,764C2 e


0.2311t


P⎛⎝1 + C2 e
0.2311t⎞


⎠ = 1,072,764C2 e
0.2311t


P(t) =
1,072,764C2 e


0.2311t


1 + C2 e
0.2311t


.


Step 5: To determine the value of C2, it is actually easier to go back a couple of steps to where C2 was
defined. In particular, use the equation


P
1,072,764 − P


= C2 e
0.2311t.


The initial condition is P(0) = 900,000. Replace P with 900,000 and t with zero:
P


1,072,764 − P
= C2 e


0.2311t


900,000
1,072,764 − 900,000


= C2 e
0.2311(0)


900,000
172,764


= C2


C2 =
25,000
4,799


≈ 5.209.


Therefore
P(t) =


1,072,764⎛⎝
25000
4799



⎠e


0.2311t


1 + ⎛⎝
25000
4799



⎠e


0.2311t


= 1,072,764(25000)e
0.2311t


4799 + 25000e0.2311t
.


Dividing the numerator and denominator by 25,000 gives
P(t) = 1,072,764e


0.2311t


0.19196 + e0.2311t
.


Figure 4.21 is a graph of this equation.


Chapter 4 | Introduction to Differential Equations 397




Figure 4.21 Logistic curve for the deer population with aninitial population of 900,000 deer.


c. Using this model we can predict the population in 3 years.
P(3) = 1,072,764e


0.2311(3)


0.19196 + e0.2311(3)
≈ 978,830 deer


This is far short of twice the initial population of 900,000. Remember that the doubling time is based
on the assumption that the growth rate never changes, but the logistic model takes this possibility intoaccount.


d. If the population reached 1,200,000 deer, then the new initial-value problem would be
dP
dt


= 0.2311P

⎝1 −


P
1,072,764



⎠, P(0) = 1,200,000.


The general solution to the differential equation would remain the same.
P(t) =


1,072,764C2 e
0.2311t


1 + C2 e
0.2311t


To determine the value of the constant, return to the equation
P


1,072,764 − P
= C2 e


0.2311t.


Substituting the values t = 0 and P = 1,200,000, you get
C2 e


0.2311(0) = 1,200,000
1,072,764 − 1,200,000


C2 = −
100,000
10,603


≈ − 9.431.


Therefore


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P(t) =
1,072,764C2 e


0.2311t


1 + C2 e
0.2311t


=
1,072,764⎛⎝−


100,000
10,603



⎠e


0.2311t


1 + ⎛⎝−
100,000
10,603



⎠e


0.2311t


= − 107,276,400,000e
0.2311t


100,000e0.2311t − 10,603


≈ 10,117,551e
0.2311t


9.43129e0.2311t − 1
.


This equation is graphed in Figure 4.22.


Figure 4.22 Logistic curve for the deer population with aninitial population of 1,200,000 deer.


Solving the Logistic Differential Equation
The logistic differential equation is an autonomous differential equation, so we can use separation of variables to find thegeneral solution, as we just did in Example 4.14.
Step 1: Setting the right-hand side equal to zero leads to P = 0 and P = K as constant solutions. The first solution
indicates that when there are no organisms present, the population will never grow. The second solution indicates that whenthe population starts at the carrying capacity, it will never change.
Step 2: Rewrite the differential equation in the form


dP
dt


= rP(K − P)
K


.


Then multiply both sides by dt and divide both sides by P(K − P). This leads to
dP


P(K − P)
= r


K
dt.


Multiply both sides of the equation by K and integrate:



K
P(K − P)


dP = ∫ rdt.


Chapter 4 | Introduction to Differential Equations 399




The left-hand side of this equation can be integrated using partial fraction decomposition. We leave it to you to verify that
K


P(K − P)
= 1


P
+ 1
K − P


.


Then the equation becomes
∫ 1P +


1
K − P


dP = ∫ rdt
ln|P| − ln|K − P| = rt + C


ln| PK − P | = rt + C.
Now exponentiate both sides of the equation to eliminate the natural logarithm:


e
ln| PK − P | = ert + C
| PK − P | = eC ert.


We define C1 = ec so that the equation becomes
(4.9)P


K − P
= C1 e


rt.


To solve this equation for P(t), first multiply both sides by K − P and collect the terms containing P on the left-hand
side of the equation:


P = C1 e
rt (K − P)


P = C1Ke
rt − C1Pe


rt


P + C1Pe
rt = C1Ke


rt.


Next, factor P from the left-hand side and divide both sides by the other factor:
(4.10)P⎛⎝1 + C1 ert⎞⎠ = C1Kert


P(t) =
C1Ke


rt


1 + C1 e
rt .


The last step is to determine the value of C1. The easiest way to do this is to substitute t = 0 and P0 in place of P in
Equation 4.9 and solve for C1 :


P
K − P


= C1 e
rt


P0
K − P0


= C1 e
r(0)


C1 =
P0


K − P0
.


Finally, substitute the expression for C1 into Equation 4.10:


P(t) =
C1Ke


rt


1 + C1 e
rt =


P0
K − P0


Kert


1 +
P0


K − P0
ert


Now multiply the numerator and denominator of the right-hand side by ⎛⎝K − P0⎞⎠ and simplify:


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P(t) =


P0
K − P0


Kert


1 +
P0


K − P0
ert


=


P0
K − P0


Kert


1 +
P0


K − P0
ert


·
K − P0
K − P0


=
P0Ke


rt



⎝K − P0



⎠+ P0 e


rt .


We state this result as a theorem.
Theorem 4.2: Solution of the Logistic Differential Equation
Consider the logistic differential equation subject to an initial population of P0 with carrying capacity K and growth
rate r. The solution to the corresponding initial-value problem is given by


(4.11)
P(t) =


P0Ke
rt



⎝K − P0



⎠+ P0 e


rt .


Now that we have the solution to the initial-value problem, we can choose values for P0, r, and K and study the solution
curve. For example, in Example 4.14 we used the values r = 0.2311, K = 1,072,764, and an initial population of
900,000 deer. This leads to the solution


P(t) =
P0Ke


rt



⎝K − P0



⎠+ P0 e


rt


= 900,000(1,072,764)e
0.2311t


(1,072,764 − 900,000) + 900,000e0.2311t


= 900,000(1,072,764)e
0.2311t


172,764 + 900,000e0.2311t
.


Dividing top and bottom by 900,000 gives
P(t) = 1,072,764e


0.2311t


0.19196 + e0.2311t
.


This is the same as the original solution. The graph of this solution is shown again in blue in Figure 4.23, superimposedover the graph of the exponential growth model with initial population 900,000 and growth rate 0.2311 (appearing in
green). The red dashed line represents the carrying capacity, and is a horizontal asymptote for the solution to the logisticequation.


Chapter 4 | Introduction to Differential Equations 401




Figure 4.23 A comparison of exponential versus logisticgrowth for the same initial population of 900,000 organisms
and growth rate of 23.11%.


Working under the assumption that the population grows according to the logistic differential equation, this graph predictsthat approximately 20 years earlier (1984), the growth of the population was very close to exponential. The net growth
rate at that time would have been around 23.1% per year. As time goes on, the two graphs separate. This happens
because the population increases, and the logistic differential equation states that the growth rate decreases as the populationincreases. At the time the population was measured (2004), it was close to carrying capacity, and the population was
starting to level off.
The solution to the logistic differential equation has a point of inflection. To find this point, set the second derivative equalto zero:


P(t) =
P0Ke


rt



⎝K − P0



⎠+ P0 e


rt


P′ (t) =
rP0K



⎝K − P0



⎠ert





⎝K − P0



⎠+ P0 e


rt⎞

2


P″(t) =
r2P0K



⎝K − P0




2 ert − r2P0


2K⎛⎝K − P0

⎠e2rt





⎝K − P0



⎠+ P0 e


rt⎞

3


=
r2P0K



⎝K − P0



⎠ert



⎝K − P0



⎠− P0 e


rt⎞





⎝K − P0



⎠+ P0 e


rt⎞

3


.


Setting the numerator equal to zero,
r2P0K



⎝K − P0



⎠ert



⎝K − P0



⎠− P0 e


rt⎞
⎠ = 0.


As long as P0 ≠ K, the entire quantity before and including ert is nonzero, so we can divide it out:

⎝K − P0



⎠− P0 e


rt = 0.


Solving for t,


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4.14


P0 e
rt = K − P0


ert =
K − P0
P0


lnert = ln
K − P0
P0


rt = ln
K − P0
P0


t = 1r ln
K − P0
P0


.


Notice that if P0 > K, then this quantity is undefined, and the graph does not have a point of inflection. In the logistic
graph, the point of inflection can be seen as the point where the graph changes from concave up to concave down. This iswhere the “leveling off” starts to occur, because the net growth rate becomes slower as the population starts to approach thecarrying capacity.


A population of rabbits in a meadow is observed to be 200 rabbits at time t = 0. After a month, the
rabbit population is observed to have increased by 4%. Using an initial population of 200 and a growth rate of
0.04, with a carrying capacity of 750 rabbits,


a. Write the logistic differential equation and initial condition for this model.
b. Draw a slope field for this logistic differential equation, and sketch the solution corresponding to aninitial population of 200 rabbits.
c. Solve the initial-value problem for P(t).
d. Use the solution to predict the population after 1 year.


Chapter 4 | Introduction to Differential Equations 403




Student Project: Logistic Equation with a Threshold Population
An improvement to the logistic model includes a threshold population. The threshold population is defined to be theminimum population that is necessary for the species to survive. We use the variable T to represent the threshold
population. A differential equation that incorporates both the threshold population T and carrying capacity K is


(4.12)dP
dt


= −rP⎛⎝1 −
P
K



⎝1 −


P
T



where r represents the growth rate, as before.
1. The threshold population is useful to biologists and can be utilized to determine whether a given species shouldbe placed on the endangered list. A group of Australian researchers say they have determined the thresholdpopulation for any species to survive: 5000 adults. (Catherine Clabby, “AMagic Number,” American Scientist


98(1): 24, doi:10.1511/2010.82.24. accessed April 9, 2015, http://www.americanscientist.org/issues/pub/a-magic-number). Therefore we use T = 5000 as the threshold population in this project. Suppose that the
environmental carrying capacity in Montana for elk is 25,000. Set up Equation 4.12 using the carrying
capacity of 25,000 and threshold population of 5000. Assume an annual net growth rate of 18%.


2. Draw the direction field for the differential equation from step 1, along with several solutions for different
initial populations. What are the constant solutions of the differential equation? What do these solutionscorrespond to in the original population model (i.e., in a biological context)?


3. What is the limiting population for each initial population you chose in step 2? (Hint: use the slope field to
see what happens for various initial populations, i.e., look for the horizontal asymptotes of your solutions.)


4. This equation can be solved using the method of separation of variables. However, it is very difficult to get thesolution as an explicit function of t. Using an initial population of 18,000 elk, solve the initial-value problem
and express the solution as an implicit function of t, or solve the general initial-value problem, finding a
solution in terms of r, K, T , and P0.


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4.4 EXERCISES
For the following problems, consider the logistic equation
in the form P′ = CP − P2. Draw the directional field and
find the stability of the equilibria.
168. C = 3
169. C = 0
170. C = −3
171. Solve the logistic equation for C = 10 and an initial
condition of P(0) = 2.
172. Solve the logistic equation for C = −10 and an
initial condition of P(0) = 2.
173. A population of deer inside a park has a carryingcapacity of 200 and a growth rate of 2%. If the initial
population is 50 deer, what is the population of deer at any
given time?
174. A population of frogs in a pond has a growth rateof 5%. If the initial population is 1000 frogs and the
carrying capacity is 6000, what is the population of frogs
at any given time?
175. [T] Bacteria grow at a rate of 20% per hour in a
petri dish. If there is initially one bacterium and a carryingcapacity of 1 million cells, how long does it take to reach
500,000 cells?
176. [T] Rabbits in a park have an initial population of
10 and grow at a rate of 4% per year. If the carrying
capacity is 500, at what time does the population reach
100 rabbits?
177. [T] Two monkeys are placed on an island. After 5
years, there are 8 monkeys, and the estimated carrying
capacity is 25 monkeys. When does the population of
monkeys reach 16 monkeys?
178. [T] A butterfly sanctuary is built that can hold 2000
butterflies, and 400 butterflies are initially moved in. If
after 2 months there are now 800 butterflies, when does
the population get to 1500 butterflies?
The following problems consider the logistic equation withan added term for depletion, either through death oremigration.


179. [T] The population of trout in a pond is given by
P′ = 0.4P⎛⎝1 −


P
10000



⎠− 400, where 400 trout are


caught per year. Use your calculator or computer softwareto draw a directional field and draw a few sample solutions.What do you expect for the behavior?
180. In the preceding problem, what are the stabilities ofthe equilibria 0 < P1 < P2?
181. [T] For the preceding problem, use software togenerate a directional field for the value f = 400. What
are the stabilities of the equilibria?
182. [T] For the preceding problems, use software togenerate a directional field for the value f = 600. What
are the stabilities of the equilibria?
183. [T] For the preceding problems, consider the casewhere a certain number of fish are added to the pond, or
f = −200. What are the nonnegative equilibria and their
stabilities?
It is more likely that the amount of fishing is governed bythe current number of fish present, so instead of a constantnumber of fish being caught, the rate is proportional tothe current number of fish present, with proportionalityconstant k, as
P′ = 0.4P⎛⎝1 −


P
10000



⎠− kP.


184. [T] For the previous fishing problem, draw adirectional field assuming k = 0.1. Draw some solutions
that exhibit this behavior. What are the equilibria and whatare their stabilities?
185. [T] Use software or a calculator to draw directionalfields for k = 0.4. What are the nonnegative equilibria and
their stabilities?
186. [T] Use software or a calculator to draw directionalfields for k = 0.6. What are the equilibria and their
stabilities?
187. Solve this equation, assuming a value of k = 0.05
and an initial condition of 2000 fish.
188. Solve this equation, assuming a value of k = 0.05
and an initial condition of 5000 fish.
The following problems add in a minimal threshold valuefor the species to survive, T , which changes the
differential equation to P′(t) = rP⎛⎝1 − PK ⎞⎠⎛⎝1 − TP⎞⎠.


Chapter 4 | Introduction to Differential Equations 405




189. Draw the directional field of the threshold logisticequation, assuming K = 10, r = 0.1, T = 2. When does
the population survive? When does it go extinct?
190. For the preceding problem, solve the logisticthreshold equation, assuming the initial condition
P(0) = P0.


191. Bengal tigers in a conservation park have a carryingcapacity of 100 and need a minimum of 10 to survive.
If they grow in population at a rate of 1% per year, with
an initial population of 15 tigers, solve for the number of
tigers present.
192. A forest containing ring-tailed lemurs in Madagascarhas the potential to support 5000 individuals, and the
lemur population grows at a rate of 5% per year. A
minimum of 500 individuals is needed for the lemurs to
survive. Given an initial population of 600 lemurs, solve
for the population of lemurs.
193. The population of mountain lions in NorthernArizona has an estimated carrying capacity of 250 and
grows at a rate of 0.25% per year and there must be 25
for the population to survive. With an initial population of
30 mountain lions, how many years will it take to get
the mountain lions off the endangered species list (at least
100)?


The following questions consider the Gompertz equation,a modification for logistic growth, which is often used formodeling cancer growth, specifically the number of tumorcells.
194. The Gompertz equation is given by
P(t)′ = α ln




K
P(t)

⎠P(t). Draw the directional fields for this


equation assuming all parameters are positive, and giventhat K = 1.
195. Assume that for a population, K = 1000 and
α = 0.05. Draw the directional field associated with this
differential equation and draw a few solutions. What is thebehavior of the population?
196. Solve the Gompertz equation for generic α and K
and P(0) = P0.
197. [T] The Gompertz equation has been used to modeltumor growth in the human body. Starting from one tumorcell on day 1 and assuming α = 0.1 and a carrying
capacity of 10 million cells, how long does it take to reach
“detection” stage at 5 million cells?


198. [T] It is estimated that the world human populationreached 3 billion people in 1959 and 6 billion in 1999.
Assuming a carrying capacity of 16 billion humans, write
and solve the differential equation for logistic growth, anddetermine what year the population reached 7 billion.
199. [T] It is estimated that the world human populationreached 3 billion people in 1959 and 6 billion in 1999.
Assuming a carrying capacity of 16 billion humans, write
and solve the differential equation for Gompertz growth,and determine what year the population reached 7 billion.
Was logistic growth or Gompertz growth more accurate,considering world population reached 7 billion on October
31, 2011?


200. Show that the population grows fastest when itreaches half the carrying capacity for the logistic equation
P′ = rP⎛⎝1 −


P
K

⎠.


201. When does population increase the fastest in the
threshold logistic equation P′(t) = rP⎛⎝1 − PK ⎞⎠⎛⎝1 − TP⎞⎠?
202. When does population increase the fastest for the
Gompertz equation P(t)′ = α ln⎛⎝ KP(t)⎞⎠P(t)?
Below is a table of the populations of whooping cranes inthe wild from 1940 to 2000. The population rebounded
from near extinction after conservation efforts began. Thefollowing problems consider applying population modelsto fit the data. Assume a carrying capacity of 10,000
cranes. Fit the data assuming years since 1940 (so your
initial population at time 0 would be 22 cranes).


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Year (years sinceconservation began)
WhoopingCranePopulation


1940(0) 22


1950(10) 31


1960(20) 36


1970(30) 57


1980(40) 91


1990(50) 159


2000(60) 256


Source: https://www.savingcranes.org/images/stories/site_images/conservation/whooping_crane/pdfs/historic_wc_numbers.pdf
203. Find the equation and parameter r that best fit the
data for the logistic equation.
204. Find the equation and parameters r and T that best
fit the data for the threshold logistic equation.
205. Find the equation and parameter α that best fit the
data for the Gompertz equation.
206. Graph all three solutions and the data on the samegraph. Which model appears to be most accurate?
207. Using the three equations found in the previousproblems, estimate the population in 2010 (year 70 after
conservation). The real population measured at that timewas 437. Which model is most accurate?


Chapter 4 | Introduction to Differential Equations 407




4.5 | First-order Linear Equations
Learning Objectives


4.5.1 Write a first-order linear differential equation in standard form.
4.5.2 Find an integrating factor and use it to solve a first-order linear differential equation.
4.5.3 Solve applied problems involving first-order linear differential equations.


Earlier, we studied an application of a first-order differential equation that involved solving for the velocity of an object.In particular, if a ball is thrown upward with an initial velocity of v0 ft/s, then an initial-value problem that describes the
velocity of the ball after t seconds is given by


dv
dt


= −32, v(0) = v0.


This model assumes that the only force acting on the ball is gravity. Now we add to the problem by allowing for thepossibility of air resistance acting on the ball.
Air resistance always acts in the direction opposite to motion. Therefore if an object is rising, air resistance acts in adownward direction. If the object is falling, air resistance acts in an upward direction (Figure 4.24). There is no exactrelationship between the velocity of an object and the air resistance acting on it. For very small objects, air resistance isproportional to velocity; that is, the force due to air resistance is numerically equal to some constant k times v. For larger
(e.g., baseball-sized) objects, depending on the shape, air resistance can be approximately proportional to the square of the
velocity. In fact, air resistance may be proportional to v1.5, or v0.9, or some other power of v.


Figure 4.24 Forces acting on a moving baseball: gravity actsin a downward direction and air resistance acts in a directionopposite to the direction of motion.


We will work with the linear approximation for air resistance. If we assume k > 0, then the expression for the force FA
due to air resistance is given by FA = −kv. Therefore the sum of the forces acting on the object is equal to the sum of
the gravitational force and the force due to air resistance. This, in turn, is equal to the mass of the object multiplied by itsacceleration at time t (Newton’s second law). This gives us the differential equation


mdv
dt


= −kv − mg.


Finally, we impose an initial condition v(0) = v0, where v0 is the initial velocity measured in meters per second. This
makes g = 9.8 m/s2. The initial-value problem becomes


(4.13)mdv
dt


= −kv − mg, v(0) = v0.


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The differential equation in this initial-value problem is an example of a first-order linear differential equation. (Recall thata differential equation is first-order if the highest-order derivative that appears in the equation is 1.) In this section, we
study first-order linear equations and examine a method for finding a general solution to these types of equations, as well assolving initial-value problems involving them.
Definition
A first-order differential equation is linear if it can be written in the form


(4.14)a(x)y′ + b(x)y = c(x),
where a(x), b(x), and c(x) are arbitrary functions of x.


Remember that the unknown function y depends on the variable x; that is, x is the independent variable and y is the
dependent variable. Some examples of first-order linear differential equations are



⎝3x


2 − 4⎞⎠y′ + (x − 3)y = sinx


(sinx)y′ − (cosx)y = cotx


4xy′ + (3lnx)y = x3 − 4x.


Examples of first-order nonlinear differential equations include

⎝y′⎞⎠4 − ⎛⎝y′⎞⎠3 = (3x − 2)⎛⎝y + 4⎞⎠
4y′ + 3y3 = 4x − 5



⎝y′⎞⎠2 = siny + cosx.


These equations are nonlinear because of terms like ⎛⎝y′⎞⎠4, y3, etc. Due to these terms, it is impossible to put these
equations into the same form as Equation 4.14.
Standard Form
Consider the differential equation



⎝3x


2 − 4⎞⎠y′ + (x − 3)y = sinx.


Our main goal in this section is to derive a solution method for equations of this form. It is useful to have the coefficient of
y′ be equal to 1. To make this happen, we divide both sides by 3x2 − 4.


y′ +



x − 3
3x2 − 4



⎠y =


sinx
3x2 − 4


This is called the standard form of the differential equation. We will use it later when finding the solution to a generalfirst-order linear differential equation. Returning to Equation 4.14, we can divide both sides of the equation by a(x). This
leads to the equation


(4.15)y′ + b(x)
a(x)


y = c(x)
a(x)


.


Now define p(x) = b(x)
a(x)


and q(x) = c(x)
a(x)


. Then Equation 4.14 becomes
(4.16)y′ + p(x)y = q(x).


We can write any first-order linear differential equation in this form, and this is referred to as the standard form for a first-order linear differential equation.


Chapter 4 | Introduction to Differential Equations 409




(4.17)


4.15


Example 4.15
Writing First-Order Linear Equations in Standard Form
Put each of the following first-order linear differential equations into standard form. Identify p(x) and q(x) for
each equation.


a. y′ = 3x − 4y
b. 3xy′


4y − 3
= 2 (here x > 0)


c. y = 3y′ − 4x2 + 5
Solution


a. Add 4y to both sides:
y′ + 4y = 3x.


In this equation, p(x) = 4 and q(x) = 3x.
b. Multiply both sides by 4y − 3, then subtract 8y from each side:


3xy′
4y − 3


= 2


3xy′ = 2⎛⎝4y − 3⎞⎠
3xy′ = 8y − 6


3xy′ − 8y = −6.


Finally, divide both sides by 3x to make the coefficient of y′ equal to 1:
y′ − 8


3x
y = − 2


3x
.


This is allowable because in the original statement of this problem we assumed that x > 0. (If x = 0
then the original equation becomes 0 = 2, which is clearly a false statement.)
In this equation, p(x) = − 8


3x
and q(x) = − 2


3x
.


c. Subtract y from each side and add 4x2 − 5:
3y′ − y = 4x2 − 5.


Next divide both sides by 3:
y′ − 1


3
y = 4


3
x2 − 5


3
.


In this equation, p(x) = − 1
3
and q(x) = 4


3
x2 − 5


3
.


Put the equation (x + 3)y′
2x − 3y − 4


= 5 into standard form and identify p(x) and q(x).


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Integrating Factors
We now develop a solution technique for any first-order linear differential equation. We start with the standard form of afirst-order linear differential equation:


(4.18)y′ + p(x)y = q(x).
The first term on the left-hand side of Equation 4.15 is the derivative of the unknown function, and the second termis the product of a known function with the unknown function. This is somewhat reminiscent of the power rule from theDifferentiation Rules (http://cnx.org/content/m53575/latest/) section. If we multiply Equation 4.16 by a yet-to-be-determined function µ(x), then the equation becomes


(4.19)µ(x)y′ + µ(x)p(x)y = µ(x)q(x).
The left-hand side Equation 4.18 can be matched perfectly to the product rule:


d
dx

⎣ f (x)g(x)⎤⎦ = f ′ (x)g(x) + f (x)g′ (x).


Matching term by term gives y = f (x), g(x) = µ(x), and g′ (x) = µ(x)p(x). Taking the derivative of g(x) = µ(x) and
setting it equal to the right-hand side of g′ (x) = µ(x)p(x) leads to


µ′ (x) = µ(x)p(x).


This is a first-order, separable differential equation for µ(x). We know p(x) because it appears in the differential equation
we are solving. Separating variables and integrating yields


µ′ (x)
µ(x)


= p(x)




µ′ (x)
µ(x)


dx = ∫ p(x)dx


ln|µ(x)| = ∫ p(x)dx + C


e
ln|µ(x)| = e


∫ p(x)dx + C


|µ(x)| = C1 e
∫ p(x)dx


µ(x) = C2 e
∫ p(x)dx


.


Here C2 can be an arbitrary (positive or negative) constant. This leads to a general method for solving a first-order linear
differential equation. We first multiply both sides of Equation 4.16 by the integrating factor µ(x). This gives


(4.20)µ(x)y′ + µ(x)p(x)y = µ(x)q(x).
The left-hand side of Equation 4.19 can be rewritten as d


dx

⎝µ(x)y⎞⎠.


(4.21)d
dx

⎝µ(x)y⎞⎠ = µ(x)q(x).


Next integrate both sides of Equation 4.20 with respect to x.
(4.22)⌠



d
dx

⎝µ(x)y⎞⎠dx = ∫ µ(x)q(x)dx


µ(x)y = ∫ µ(x)q(x)dx.


Divide both sides of Equation 4.21 by µ(x):
(4.23)y = 1


µ(x)

⎣∫ µ(x)q(x)dx + C



⎦.


Since µ(x) was previously calculated, we are now finished. An important note about the integrating constant C: It may


Chapter 4 | Introduction to Differential Equations 411




seem that we are inconsistent in the usage of the integrating constant. However, the integral involving p(x) is necessary in
order to find an integrating factor for Equation 4.15. Only one integrating factor is needed in order to solve the equation;therefore, it is safe to assign a value for C for this integral. We chose C = 0. When calculating the integral inside the
brackets in Equation 4.21, it is necessary to keep our options open for the value of the integrating constant, because ourgoal is to find a general family of solutions to Equation 4.15. This integrating factor guarantees just that.
Problem-Solving Strategy: Solving a First-order Linear Differential Equation


1. Put the equation into standard form and identify p(x) and q(x).
2. Calculate the integrating factor µ(x) = e∫ p(x)dx.
3. Multiply both sides of the differential equation by µ(x).
4. Integrate both sides of the equation obtained in step 3, and divide both sides by µ(x).
5. If there is an initial condition, determine the value of C.


Example 4.16
Solving a First-order Linear Equation
Find a general solution for the differential equation xy′ + 3y = 4x2 − 3x. Assume x > 0.
Solution


1. To put this differential equation into standard form, divide both sides by x:
y′ + 3xy = 4x − 3.


Therefore p(x) = 3x and q(x) = 4x − 3.
2. The integrating factor is µ(x) = e∫ (3/x)dx = e3lnx = x3.
3. Multiplying both sides of the differential equation by µ(x) gives us


x3 y′ + x3 ⎛⎝
3
x

⎠y = x


3 (4x − 3)


x3 y′ + 3x2 y = 4x4 − 3x3


d
dx

⎝x


3 y⎞⎠ = 4x
4 − 3x3.


4. Integrate both sides of the equation.


d
dx

⎝x


3 y⎞⎠dx = ∫ 4x4 − 3x3dx


x3 y = 4x
5


5
− 3x


4


4
+ C


y = 4x
2


5
− 3x


4
+ Cx−3.


5. There is no initial value, so the problem is complete.
Analysis


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4.16


You may have noticed the condition that was imposed on the differential equation; namely, x > 0. For any
nonzero value of C, the general solution is not defined at x = 0. Furthermore, when x < 0, the integrating
factor changes. The integrating factor is given by Equation 4.19 as f (x) = e∫ p(x)dx. For this p(x) we get


e
∫ p(x)dx =


e
∫ (3/x)dx


= e3ln|x| = |x|3,


since x < 0. The behavior of the general solution changes at x = 0 largely due to the fact that p(x) is not
defined there.


Find the general solution to the differential equation (x − 2)y′ + y = 3x2 + 2x. Assume x > 2.


Now we use the same strategy to find the solution to an initial-value problem.
Example 4.17
A First-order Linear Initial-Value Problem
Solve the initial-value problem


y′ + 3y = 2x − 1, y(0) = 3.


Solution
1. This differential equation is already in standard form with p(x) = 3 and q(x) = 2x − 1.
2. The integrating factor is µ(x) = e∫ 3dx = e3x.
3. Multiplying both sides of the differential equation by µ(x) gives


e3x y′ + 3e3x y = (2x − 1)e3x


d
dx

⎣ye


3x⎤
⎦ = (2x − 1)e


3x.


Integrate both sides of the equation:


d
dx

⎣ye


3x⎤
⎦dx = ∫ (2x − 1)e3x dx


ye3x = e
3x


3
(2x − 1) − ⌠



2
3
e3x dx


ye3x = e
3x (2x − 1)


3
− 2e


3x


9
+ C


y = 2x − 1
3


− 2
9
+ Ce−3x


y = 2x
3


− 5
9
+ Ce−3x.


4. Now substitute x = 0 and y = 3 into the general solution and solve for C:


Chapter 4 | Introduction to Differential Equations 413




4.17


y = 2
3
x − 5


9
+ Ce−3x


3 = 2
3
(0) − 5


9
+ Ce−3(0)


3 = −5
9
+ C


C = 32
9
.


Therefore the solution to the initial-value problem is
y = 2


3
x − 5


9
+ 32


9
e−3x.


Solve the initial-value problem y′ − 2y = 4x + 3 y(0) = −2.


Applications of First-order Linear Differential Equations
We look at two different applications of first-order linear differential equations. The first involves air resistance as it relatesto objects that are rising or falling; the second involves an electrical circuit. Other applications are numerous, but most aresolved in a similar fashion.
Free fall with air resistance
We discussed air resistance at the beginning of this section. The next example shows how to apply this concept for a ball invertical motion. Other factors can affect the force of air resistance, such as the size and shape of the object, but we ignorethem here.
Example 4.18
A Ball with Air Resistance
A racquetball is hit straight upward with an initial velocity of 2 m/s. The mass of a racquetball is approximately
0.0427 kg. Air resistance acts on the ball with a force numerically equal to 0.5v, where v represents the
velocity of the ball at time t.


a. Find the velocity of the ball as a function of time.
b. How long does it take for the ball to reach its maximum height?
c. If the ball is hit from an initial height of 1 meter, how high will it reach?


Solution
a. The mass m = 0.0427 kg, k = 0.5, and g = 9.8 m/s2. The initial velocity is v0 = 2 m/s. Therefore


the initial-value problem is
0.0427dv


dt
= −0.5v − 0.0427(9.8), v0 = 2.


Dividing the differential equation by 0.0427 gives


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dv
dt


= −11.7096v − 9.8, v0 = 2.


The differential equation is linear. Using the problem-solving strategy for linear differential equations:
Step 1. Rewrite the differential equation as dv


dt
+ 11.7096v = −9.8. This gives p(t) = 11.7096 and


q(t) = −9.8


Step 2. The integrating factor is µ(t) = e∫ 11.7096dt = e11.7096t.
Step 3. Multiply the differential equation by µ(t):


e11.7096t dv
dt


+ 11.7096ve11.7096t = −9.8e11.7096t


d
dt

⎣ve


11.7096t⎤
⎦ = −9.8e


11.7096t.


Step 4. Integrate both sides:


d
dt

⎣ve


11.7096t⎤
⎦dt = ∫ −9.8e11.7096t dt


ve11.7096t = −9.8
11.7096


e11.7096t + C


v(t) = −0.8369 + Ce−11.7096t.


Step 5. Solve for C using the initial condition v0 = v(0) = 2:
v(t) = −0.8369 + Ce−11.7096t


v(0) = −0.8369 + Ce−11.7096(0)


2 = −0.8369 + C
C = 2.8369.


Therefore the solution to the initial-value problem is v(t) = 2.8369e−11.7096t − 0.8369.
b. The ball reaches its maximum height when the velocity is equal to zero. The reason is that when thevelocity is positive, it is rising, and when it is negative, it is falling. Therefore when it is zero, it is neitherrising nor falling, and is at its maximum height:


2.8369e−11.7096t − 0.8369 = 0


2.8369e−11.7096t = 0.8369


e−11.7096t = 0.8369
2.8369


≈ 0.295


lne−11.7096t = ln0.295 ≈ − 1.221


−11.7096t = −1.221
t ≈ 0.104.


Therefore it takes approximately 0.104 second to reach maximum height.
c. To find the height of the ball as a function of time, use the fact that the derivative of position is velocity,i.e., if h(t) represents the height at time t, then h′ (t) = v(t). Because we know v(t) and the initial


height, we can form an initial-value problem:
h′ (t) = 2.8369e−11.7096t − 0.8369, h(0) = 1.


Chapter 4 | Introduction to Differential Equations 415




4.18


Integrating both sides of the differential equation with respect to t gives
∫ h′ (t)dt = ∫ 2.8369e−11.7096t − 0.8369dt


h(t) = − 2.8369
11.7096


e−11.7096t − 0.8369t + C


h(t) = −0.2423e−11.7096t − 0.8369t + C.


Solve for C by using the initial condition:
h(t) = −0.2423e−11.7096t − 0.8369t + C


h(0) = −0.2423e−11.7096(0) − 0.8369(0) + C
1 = −0.2423 + C
C = 1.2423.


Therefore
h(t) = −0.2423e−11.7096t − 0.8369t + 1.2423.


After 0.104 second, the height is given by
h(0.2) = −0.2423e−11.7096t − 0.8369t + 1.2423 ≈ 1.0836 meter.


The weight of a penny is 2.5 grams (United States Mint, “Coin Specifications,” accessed April 9, 2015,
http://www.usmint.gov/about_the_mint/?action=coin_specifications), and the upper observation deck of theEmpire State Building is 369 meters above the street. Since the penny is a small and relatively smooth object,
air resistance acting on the penny is actually quite small. We assume the air resistance is numerically equal to
0.0025v. Furthermore, the penny is dropped with no initial velocity imparted to it.


a. Set up an initial-value problem that represents the falling penny.
b. Solve the problem for v(t).
c. What is the terminal velocity of the penny (i.e., calculate the limit of the velocity as t approaches


infinity)?


Electrical Circuits
A source of electromotive force (e.g., a battery or generator) produces a flow of current in a closed circuit, and this currentproduces a voltage drop across each resistor, inductor, and capacitor in the circuit. Kirchhoff’s Loop Rule states that the sumof the voltage drops across resistors, inductors, and capacitors is equal to the total electromotive force in a closed circuit.We have the following three results:


1. The voltage drop across a resistor is given by
ER = Ri,


where R is a constant of proportionality called the resistance, and i is the current.
2. The voltage drop across an inductor is given by


EL = Li′,


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where L is a constant of proportionality called the inductance, and i again denotes the current.
3. The voltage drop across a capacitor is given by


EC =
1
C
q,


where C is a constant of proportionality called the capacitance, and q is the instantaneous charge on the capacitor. The
relationship between i and q is i = q′.
We use units of volts (V) to measure voltage E, amperes (A) to measure current i, coulombs (C) to measure charge
q, ohms (Ω) to measure resistance R, henrys (H) to measure inductance L, and farads (F) to measure capacitance
C. Consider the circuit in Figure 4.25.


Figure 4.25 A typical electric circuit, containing a voltage
generator ⎛⎝VS⎞⎠, capacitor (C), inductor (L), and resistor
(R).


Applying Kirchhoff’s Loop Rule to this circuit, we let E denote the electromotive force supplied by the voltage generator.
Then


EL + ER + EC = E.


Substituting the expressions for EL, ER, and EC into this equation, we obtain
(4.24)Li′ + Ri + 1


C
q = E.


If there is no capacitor in the circuit, then the equation becomes
(4.25)Li′ + Ri = E.


This is a first-order differential equation in i. The circuit is referred to as an LR circuit.
Next, suppose there is no inductor in the circuit, but there is a capacitor and a resistor, so L = 0, R ≠ 0, and C ≠ 0. Then
Equation 4.23 can be rewritten as


(4.26)Rq′ + 1
C
q = E,


which is a first-order linear differential equation. This is referred to as an RC circuit. In either case, we can set up and solvean initial-value problem.
Example 4.19
Finding Current in an RL Electric Circuit
A circuit has in series an electromotive force given by E = 50sin20t V, a resistor of 5Ω, and an inductor of
0.4 H. If the initial current is 0, find the current at time t > 0.


Chapter 4 | Introduction to Differential Equations 417




Solution
We have a resistor and an inductor in the circuit, so we use Equation 4.24. The voltage drop across the resistor isgiven by ER = Ri = 5i. The voltage drop across the inductor is given by EL = Li′ = 0.4i′. The electromotive
force becomes the right-hand side of Equation 4.24. Therefore Equation 4.24 becomes


0.4i′ + 5i = 50sin20t.


Dividing both sides by 0.4 gives the equation
i′ + 12.5i = 125sin20t.


Since the initial current is 0, this result gives an initial condition of i(0) = 0. We can solve this initial-value
problem using the five-step strategy for solving first-order differential equations.
Step 1. Rewrite the differential equation as i′ + 12.5i = 125sin20t. This gives p(t) = 12.5 and
q(t) = 125sin20t.


Step 2. The integrating factor is µ(t) = e∫ 12.5dt = e12.5t.
Step 3. Multiply the differential equation by µ(t):


e12.5t i′ + 12.5e12.5t i = 125e12.5t sin20t
d
dt

⎣ie


12.5t⎤
⎦ = 125e


12.5t sin20t.


Step 4. Integrate both sides:


d
dt

⎣ie


12.5t⎤
⎦dt = ∫ 125e12.5t sin20t dt


ie12.5t = ⎛⎝
250sin20t − 400cos20t


89

⎠e


12.5t + C


i(t) = 250sin20t − 400cos20t
89


+ Ce−12.5t.


Step 5. Solve for C using the initial condition v(0) = 2:
i(t) = 250sin20t − 400cos20t


89
+ Ce−12.5t


i(0) = 250sin20(0) − 400cos20(0)
89


+ Ce−12.5(0)


0 = −400
89


+ C


C = 400
89


.


Therefore the solution to the initial-value problem is
i(t) = 250sin20t − 400cos20t + 400e


−12.5t


89
= 250sin20t − 400cos20t


89
+ 400e


−12.5t


89
.


The first term can be rewritten as a single cosine function. First, multiply and divide by 2502 + 4002 = 50 89:
250sin20t − 400cos20t


89
= 50 89


89


250sin20t − 400cos20t


50 89



= − 50 89
89


8cos20t


89
− 5sin20t


89

⎠.


Next, define φ to be an acute angle such that cosφ = 8
89


. Then sinφ = 5
89


and


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4.19


−50 89
89


8cos20t


89
− 5sin20t


89

⎠ = −


50 89
89



⎝cosφcos20t − sinφsin20t⎞⎠


= − 50 89
89


cos⎛⎝20t + φ⎞⎠.


Therefore the solution can be written as
i(t) = − 50 89


89
cos ⎛⎝20t + φ⎞⎠+ 400e


−12.5t


89
.


The second term is called the attenuation term, because it disappears rapidly as t grows larger. The phase shift is
given by φ, and the amplitude of the steady-state current is given by 50 89


89
. The graph of this solution appears


in Figure 4.26:


Figure 4.26


A circuit has in series an electromotive force given by E = 20sin5t V, a capacitor with capacitance
0.02 F, and a resistor of 8 Ω. If the initial charge is 4 C, find the charge at time t > 0.


Chapter 4 | Introduction to Differential Equations 419




4.5 EXERCISES
Are the following differential equations linear? Explainyour reasoning.
208. dy


dx
= x2 y + sinx


209. dy
dt


= ty


210. dy
dt


+ y2 = x


211. y′ = x3 + ex
212. y′ = y + ey
Write the following first-order differential equations instandard form.
213. y′ = x3 y + sinx
214. y′ + 3y − lnx = 0
215. −xy′ = (3x + 2)y + xex


216. dy
dt


= 4y + ty + tan t


217. dy
dt


= yx(x + 1)


What are the integrating factors for the followingdifferential equations?
218. y′ = xy + 3
219. y′ + ex y = sinx
220. y′ = x ln(x)y + 3x
221. dy


dx
= tanh(x)y + 1


222. dy
dt


+ 3ty = et y


Solve the following differential equations by usingintegrating factors.
223. y′ = 3y + 2
224. y′ = 2y − x2
225. xy′ = 3y − 6x2


226. (x + 2)y′ = 3x + y
227. y′ = 3x + xy
228. xy′ = x + y
229. sin(x)y′ = y + 2x
230. y′ = y + ex
231. xy′ = 3y + x2


232. y′ + lnx = yx
Solve the following differential equations. Use yourcalculator to draw a family of solutions. Are there certaininitial conditions that change the behavior of the solution?
233. [T] (x + 2)y′ = 2y − 1
234. [T] y′ = 3et/3 − 2y
235. [T] xy′ + y


2
= sin(3t)


236. [T] xy′ = 2cosxx − 3y
237. [T] (x + 1)y′ = 3y + x2 + 2x + 1
238. [T] sin(x)y′ + cos(x)y = 2x
239. [T] x2 + 1y′ = y + 2
240. [T] x3 y′ + 2x2 y = x + 1
Solve the following initial-value problems by usingintegrating factors.
241. y′ + y = x, y(0) = 3
242. y′ = y + 2x2, y(0) = 0
243. xy′ = y − 3x3, y(1) = 0
244. x2 y′ = xy − lnx, y(1) = 1
245. ⎛⎝1 + x2⎞⎠y′ = y − 1, y(0) = 0
246. xy′ = y + 2x lnx, y(1) = 5


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247. (2 + x)y′ = y + 2 + x, y(0) = 0
248. y′ = xy + 2xex, y(0) = 2
249. xy′ = y + 2x, y(0) = 1
250. y′ = 2y + xex, y(0) = −1
251. A falling object of mass m can reach terminal
velocity when the drag force is proportional to its velocity,with proportionality constant k. Set up the differential
equation and solve for the velocity given an initial velocityof 0.
252. Using your expression from the preceding problem,what is the terminal velocity? (Hint: Examine the limitingbehavior; does the velocity approach a value?)
253. [T] Using your equation for terminal velocity, solvefor the distance fallen. How long does it take to fall 5000
meters if the mass is 100 kilograms, the acceleration due
to gravity is 9.8 m/s2 and the proportionality constant is
4?


254. A more accurate way to describe terminal velocity isthat the drag force is proportional to the square of velocity,with a proportionality constant k. Set up the differential
equation and solve for the velocity.
255. Using your expression from the preceding problem,what is the terminal velocity? (Hint: Examine the limitingbehavior: Does the velocity approach a value?)
256. [T] Using your equation for terminal velocity, solvefor the distance fallen. How long does it take to fall 5000
meters if the mass is 100 kilograms, the acceleration due
to gravity is 9.8 m/s2 and the proportionality constant
is 4? Does it take more or less time than your initial
estimate?
For the following problems, determine how parameter a
affects the solution.
257. Solve the generic equation y′ = ax + y. How does
varying a change the behavior?
258. Solve the generic equation y′ = ax + y. How does
varying a change the behavior?
259. Solve the generic equation y′ = ax + xy. How does
varying a change the behavior?
260. Solve the generic equation y′ = x + axy. How does
varying a change the behavior?


261. Solve y′ − y = ekt with the initial condition
y(0) = 0. As k approaches 1, what happens to your
formula?


Chapter 4 | Introduction to Differential Equations 421




asymptotically semi-stable solution
asymptotically stable solution


asymptotically unstable solution


autonomous differential equation
carrying capacity
differential equation
direction field (slope field)


equilibrium solution
Euler’s Method
general solution (or family of solutions)
growth rate
initial population
initial value(s)
initial velocity
initial-value problem
integrating factor
linear
logistic differential equation
order of a differential equation
particular solution
phase line
separable differential equation
separation of variables
solution curve
solution to a differential equation


CHAPTER 4 REVIEW
KEY TERMS


y = k if it is neither asymptotically stable nor asymptotically unstable
y = k if there exists ε > 0 such that for any value c ∈ (k − ε, k + ε) the solution


to the initial-value problem y′ = f (x, y), y(x0) = c approaches k as x approaches infinity
y = k if there exists ε > 0 such that for any value c ∈ (k − ε, k + ε) the


solution to the initial-value problem y′ = f (x, y), y(x0) = c never approaches k as x approaches infinity
an equation in which the right-hand side is a function of y alone


the maximum population of an organism that the environment can sustain indefinitely
an equation involving a function y = y(x) and one or more of its derivatives


a mathematical object used to graphically represent solutions to a first-order differentialequation; at each point in a direction field, a line segment appears whose slope is equal to the slope of a solution tothe differential equation passing through that point
any solution to the differential equation of the form y = c, where c is a constant


a numerical technique used to approximate solutions to an initial-value problem
the entire set of solutions to a given differential equation


the constant r > 0 in the exponential growth function P(t) = P0 ert
the population at time t = 0


a value or set of values that a solution of a differential equation satisfies for a fixed value of theindependent variable
the velocity at time t = 0


a differential equation together with an initial value or values
any function f (x) that is multiplied on both sides of a differential equation to make the side


involving the unknown function equal to the derivative of a product of two functions
description of a first-order differential equation that can be written in the form a(x)y′ + b(x)y = c(x)


a differential equation that incorporates the carrying capacity K and growth rate r into
a population model


the highest order of any derivative of the unknown function that appears in theequation
member of a family of solutions to a differential equation that satisfies a particular initial condition


a visual representation of the behavior of solutions to an autonomous differential equation subject to variousinitial conditions
any equation that can be written in the form y′ = f (x)g(y)


a method used to solve a separable differential equation
a curve graphed in a direction field that corresponds to the solution to the initial-value problem passingthrough a given point in the direction field


a function y = f (x) that satisfies a given differential equation


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standard form


step size
threshold population


the form of a first-order linear differential equation obtained by writing the differential equation in theform y′ + p(x)y = q(x)
the increment h that is added to the x value at each step in Euler’s Method


the minimum population that is necessary for a species to survive
KEY EQUATIONS


• Euler’s Method
xn = x0 + nh


yn = yn − 1 + h f (xn − 1, yn − 1), where h is the step size


• Separable differential equation
y′ = f (x)g(y)


• Solution concentration
du
dt


= INFLOW RATE − OUTFLOW RATE


• Newton’s law of cooling
dT
dt


= k(T − Ts)


• Logistic differential equation and initial-value problem
dP
dt


= rP⎛⎝1 −
P
K

⎠, P(0) = P0


• Solution to the logistic differential equation/initial-value problem
P(t) =


P0Ke
rt



⎝K − P0



⎠+ P0 e


rt


• Threshold population model
dP
dt


= −rP⎛⎝1 −
P
K



⎝1 −


P
T



• standard form
y′ + p(x)y = q(x)


• integrating factor
µ(x) = e


∫ p(x)dx


KEY CONCEPTS
4.1 Basics of Differential Equations


• A differential equation is an equation involving a function y = f (x) and one or more of its derivatives. A solution
is a function y = f (x) that satisfies the differential equation when f and its derivatives are substituted into the
equation.


• The order of a differential equation is the highest order of any derivative of the unknown function that appears inthe equation.
• A differential equation coupled with an initial value is called an initial-value problem. To solve an initial-valueproblem, first find the general solution to the differential equation, then determine the value of the constant. Initial-value problems have many applications in science and engineering.


4.2 Direction Fields and Numerical Methods
• A direction field is a mathematical object used to graphically represent solutions to a first-order differential


Chapter 4 | Introduction to Differential Equations 423




equation.
• Euler’s Method is a numerical technique that can be used to approximate solutions to a differential equation.


4.3 Separable Equations
• A separable differential equation is any equation that can be written in the form y′ = f (x)g(y).
• The method of separation of variables is used to find the general solution to a separable differential equation.


4.4 The Logistic Equation
• When studying population functions, different assumptions—such as exponential growth, logistic growth, orthreshold population—lead to different rates of growth.
• The logistic differential equation incorporates the concept of a carrying capacity. This value is a limiting value onthe population for any given environment.
• The logistic differential equation can be solved for any positive growth rate, initial population, and carryingcapacity.


4.5 First-order Linear Equations
• Any first-order linear differential equation can be written in the form y′ + p(x)y = q(x).
• We can use a five-step problem-solving strategy for solving a first-order linear differential equation that may or maynot include an initial value.
• Applications of first-order linear differential equations include determining motion of a rising or falling object withair resistance and finding current in an electrical circuit.


CHAPTER 4 REVIEW EXERCISES
True or False? Justify your answer with a proof or acounterexample.
262. The differential equation y′ = 3x2 y − cos(x)y″ is
linear.
263. The differential equation y′ = x − y is separable.
264. You can explicitly solve all first-order differentialequations by separation or by the method of integratingfactors.
265. You can determine the behavior of all first-orderdifferential equations using directional fields or Euler’smethod.
For the following problems, find the general solution to thedifferential equations.
266. y′ = x2 + 3ex − 2x


267. y′ = 2x + cos−1 x


268. y′ = y⎛⎝x2 + 1⎞⎠


269. y′ = e−y sinx
270. y′ = 3x − 2y
271. y′ = y lny
For the following problems, find the solution to the initialvalue problem.
272. y′ = 8x − lnx − 3x4, y(1) = 5
273. y′ = 3x − cosx + 2, y(0) = 4
274. xy′ = y(x − 2), y(1) = 3


275. y′ = 3y2 (x + cosx), y(0) = −2
276. (x − 1)y′ = y − 2, y(0) = 0


277. y′ = 3y − x + 6x2, y(0) = −1
For the following problems, draw the directional field


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associated with the differential equation, then solve thedifferential equation. Draw a sample solution on thedirectional field.
278. y′ = 2y − y2


279. y′ = 1x + lnx − y, for x > 0
For the following problems, use Euler’s Method with
n = 5 steps over the interval t = [0, 1]. Then solve the
initial-value problem exactly. How close is your Euler’sMethod estimate?
280. y′ = −4yx, y(0) = 1
281. y′ = 3x − 2y, y(0) = 0
For the following problems, set up and solve the differentialequations.
282. A car drives along a freeway, accelerating accordingto a = 5sin(πt), where t represents time in minutes.
Find the velocity at any time t, assuming the car starts
with an initial speed of 60 mph.
283. You throw a ball of mass 2 kilograms into the air
with an upward velocity of 8 m/s. Find exactly the time the
ball will remain in the air, assuming that gravity is given by
g = 9.8 m/s2.


284. You drop a ball with a mass of 5 kilograms out an
airplane window at a height of 5000 m. How long does it
take for the ball to reach the ground?
285. You drop the same ball of mass 5 kilograms out
of the same airplane window at the same height, exceptthis time you assume a drag force proportional to the ball’svelocity, using a proportionality constant of 3 and the ball
reaches terminal velocity. Solve for the distance fallen as afunction of time. How long does it take the ball to reach theground?
286. A drug is administered to a patient every 24 hours
and is cleared at a rate proportional to the amount of drugleft in the body, with proportionality constant 0.2. If the
patient needs a baseline level of 5 mg to be in the
bloodstream at all times, how large should the dose be?
287. A 1000 -liter tank contains pure water and a solution
of 0.2 kg salt/L is pumped into the tank at a rate of 1 L/
min and is drained at the same rate. Solve for total amountof salt in the tank at time t.


288. You boil water to make tea. When you pour thewater into your teapot, the temperature is 100°C. After 5
minutes in your 15°C room, the temperature of the tea is
85°C. Solve the equation to determine the temperatures of
the tea at time t. How long must you wait until the tea is at
a drinkable temperature (72°C)?
289. The human population (in thousands) of Nevada in
1950 was roughly 160. If the carrying capacity is
estimated at 10 million individuals, and assuming a
growth rate of 2% per year, develop a logistic growth
model and solve for the population in Nevada at any time(use 1950 as time = 0). What population does your model
predict for 2000? How close is your prediction to the true
value of 1,998,257?
290. Repeat the previous problem but use Gompertzgrowth model. Which is more accurate?


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5 | SEQUENCES ANDSERIES


Figure 5.1 The Koch snowflake is constructed by using an iterative process. Starting with an equilateral triangle, at each stepof the process the middle third of each line segment is removed and replaced with an equilateral triangle pointing outward.


Chapter Outline
5.1 Sequences
5.2 Infinite Series
5.3 The Divergence and Integral Tests
5.4 Comparison Tests
5.5 Alternating Series
5.6 Ratio and Root Tests


Introduction
The Koch snowflake is constructed from an infinite number of nonoverlapping equilateral triangles. Consequently, we canexpress its area as a sum of infinitely many terms. How do we add an infinite number of terms? Can a sum of an infinitenumber of terms be finite? To answer these questions, we need to introduce the concept of an infinite series, a sum withinfinitely many terms. Having defined the necessary tools, we will be able to calculate the area of the Koch snowflake (seeExample 5.8).
The topic of infinite series may seem unrelated to differential and integral calculus. In fact, an infinite series whose termsinvolve powers of a variable is a powerful tool that we can use to express functions as “infinite polynomials.” We canuse infinite series to evaluate complicated functions, approximate definite integrals, and create new functions. In addition,infinite series are used to solve differential equations that model physical behavior, from tiny electronic circuits to Earth-orbiting satellites.
5.1 | Sequences


Learning Objectives
5.1.1 Find the formula for the general term of a sequence.
5.1.2 Calculate the limit of a sequence if it exists.
5.1.3 Determine the convergence or divergence of a given sequence.


Chapter 5 | Sequences and Series 427




In this section, we introduce sequences and define what it means for a sequence to converge or diverge. We show how tofind limits of sequences that converge, often by using the properties of limits for functions discussed earlier. We close thissection with the Monotone Convergence Theorem, a tool we can use to prove that certain types of sequences converge.
Terminology of Sequences
To work with this new topic, we need some new terms and definitions. First, an infinite sequence is an ordered list ofnumbers of the form


a1, a2, a3 ,…, an ,… .


Each of the numbers in the sequence is called a term. The symbol n is called the index variable for the sequence. We use
the notation


{an}n = 1
∞ , or simply





⎨an





⎬,


to denote this sequence. A similar notation is used for sets, but a sequence is an ordered list, whereas a set is not ordered.Because a particular number an exists for each positive integer n, we can also define a sequence as a function whose
domain is the set of positive integers.
Let’s consider the infinite, ordered list


2, 4, 8, 16, 32,… .


This is a sequence in which the first, second, and third terms are given by a1 = 2, a2 = 4, and a3 = 8. You can
probably see that the terms in this sequence have the following pattern:


a1 = 2
1, a2 = 2


2, a3 = 2
3, a4 = 2


4, and a5 = 2
5.


Assuming this pattern continues, we can write the nth term in the sequence by the explicit formula an = 2n. Using this
notation, we can write this sequence as


{2n}n = 1
∞ or





⎨2n





⎬.


Alternatively, we can describe this sequence in a different way. Since each term is twice the previous term, this sequencecan be defined recursively by expressing the nth term an in terms of the previous term an − 1. In particular, we can
define this sequence as the sequence {an} where a1 = 2 and for all n ≥ 2, each term an is defined by the recurrence
relation an = 2an − 1.


Definition
An infinite sequence {an} is an ordered list of numbers of the form


a1, a2 ,…, an ,… .


The subscript n is called the index variable of the sequence. Each number an is a term of the sequence. Sometimes
sequences are defined by explicit formulas, in which case an = f (n) for some function f (n) defined over the
positive integers. In other cases, sequences are defined by using a recurrence relation. In a recurrence relation, oneterm (or more) of the sequence is given explicitly, and subsequent terms are defined in terms of earlier terms in thesequence.


Note that the index does not have to start at n = 1 but could start with other integers. For example, a sequence given by
the explicit formula an = f (n) could start at n = 0, in which case the sequence would be


a0, a1, a2 ,… .


Similarly, for a sequence defined by a recurrence relation, the term a0 may be given explicitly, and the terms an for n ≥ 1
may be defined in terms of an − 1. Since a sequence {an} has exactly one value for each positive integer n, it can be
described as a function whose domain is the set of positive integers. As a result, it makes sense to discuss the graph of a


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sequence. The graph of a sequence {an} consists of all points (n, an) for all positive integers n. Figure 5.2 shows the
graph of {2n}.


Figure 5.2 The plotted points are a graph of the sequence
{2n}.


Two types of sequences occur often and are given special names: arithmetic sequences and geometric sequences. In anarithmetic sequence, the difference between every pair of consecutive terms is the same. For example, consider thesequence
3, 7, 11, 15, 19,… .


You can see that the difference between every consecutive pair of terms is 4. Assuming that this pattern continues, this
sequence is an arithmetic sequence. It can be described by using the recurrence relation






a1 = 3


an = an − 1 + 4 for n ≥ 2.


Note that
a2 = 3 + 4


a3 = 3 + 4 + 4 = 3 + 2 · 4


a4 = 3 + 4 + 4 + 4 = 3 + 3 · 4.


Thus the sequence can also be described using the explicit formula
an = 3 + 4(n − 1)


= 4n − 1.


In general, an arithmetic sequence is any sequence of the form an = cn + b.
In a geometric sequence, the ratio of every pair of consecutive terms is the same. For example, consider the sequence


2, − 2
3
, 2
9
, − 2


27
, 2
81


,… .


We see that the ratio of any term to the preceding term is −1
3
. Assuming this pattern continues, this sequence is a geometric


sequence. It can be defined recursively as
a1 = 2


an = − 13
· an − 1 for n ≥ 2.


Alternatively, since


Chapter 5 | Sequences and Series 429




5.1


a2 = −
1
3
· 2


a3 =

⎝−


1
3



⎝−


1
3

⎠(2) =



⎝−


1
3



2
· 2


a4 =

⎝−


1
3



⎝−


1
3



⎝−


1
3

⎠(2) =



⎝−


1
3



3
· 2,


we see that the sequence can be described by using the explicit formula
an = 2



⎝−


1
3



n − 1
.


The sequence {2n} that we discussed earlier is a geometric sequence, where the ratio of any term to the previous term is
2. In general, a geometric sequence is any sequence of the form an = crn.
Example 5.1
Finding Explicit Formulas
For each of the following sequences, find an explicit formula for the nth term of the sequence.


a. −1
2
, 2
3
, − 3


4
, 4
5
, − 5


6
,…


b. 3
4
, 9
7
, 27
10


, 81
13


, 243
16


,…


Solution
a. First, note that the sequence is alternating from negative to positive. The odd terms in the sequence arenegative, and the even terms are positive. Therefore, the nth term includes a factor of (−1)n. Next,


consider the sequence of numerators {1, 2, 3,…} and the sequence of denominators {2, 3, 4,…}.
We can see that both of these sequences are arithmetic sequences. The nth term in the sequence of
numerators is n, and the nth term in the sequence of denominators is n + 1. Therefore, the sequence
can be described by the explicit formula


an =
(−1)nn
n + 1


.


b. The sequence of numerators 3, 9, 27, 81, 243,… is a geometric sequence. The numerator of the
nth term is 3n The sequence of denominators 4, 7, 10, 13, 16,… is an arithmetic sequence. The
denominator of the nth term is 4 + 3(n − 1) = 3n + 1. Therefore, we can describe the sequence by the
explicit formula an = 3n3n + 1.


Find an explicit formula for the nth term of the sequence ⎧

⎨1
5
, − 1


7
, 1
9
, − 1


11
,…



⎬.


Example 5.2


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5.2


Defined by Recurrence Relations
For each of the following recursively defined sequences, find an explicit formula for the sequence.


a. a1 = 2, an = −3an − 1 for n ≥ 2
b. a1 = 12, an = an − 1 + ⎛⎝12⎞⎠


n for n ≥ 2


Solution
a. Writing out the first few terms, we have


a1 = 2


a2 = −3a1 = −3(2)


a3 = −3a2 = (−3)
22


a4 = −3a3 = (−3)
32.


In general,
an = 2(−3)


n − 1.


b. Write out the first few terms:
a1 =


1
2


a2 = a1 +


1
2



2
= 1


2
+ 1


4
= 3


4


a3 = a2 +


1
2



3
= 3


4
+ 1


8
= 7


8


a4 = a3 +


1
2



4
= 7


8
+ 1


16
= 15


16
.


From this pattern, we derive the explicit formula
an = 2


n − 1
2n


= 1 − 1
2n


.


Find an explicit formula for the sequence defined recursively such that a1 = −4 and an = an − 1 + 6.


Limit of a Sequence
A fundamental question that arises regarding infinite sequences is the behavior of the terms as n gets larger. Since a
sequence is a function defined on the positive integers, it makes sense to discuss the limit of the terms as n → ∞. For
example, consider the following four sequences and their different behaviors as n → ∞ (see Figure 5.3):


a. {1 + 3n} = {4, 7, 10, 13,…}. The terms 1 + 3n become arbitrarily large as n → ∞. In this case, we say that
1 + 3n → ∞ as n → ∞.


b. ⎧

⎨1 − ⎛⎝


1
2



n⎫



⎬ =





⎨1
2
, 3
4
, 7
8
, 15
16


,…



⎬. The terms 1 − ⎛⎝12⎞⎠


n
→ 1 as n → ∞.


c. ⎧⎩⎨(−1)n⎫⎭⎬ = ⎧⎩⎨−1, 1, −1, 1,…⎫⎭⎬. The terms alternate but do not approach one single value as n → ∞.


Chapter 5 | Sequences and Series 431




d. ⎧

⎨(−1)


n


n



⎬ =





⎨−1, 1


2
, − 1


3
, 1
4
,…



⎬. The terms alternate for this sequence as well, but (−1)nn → 0 as n → ∞.


Figure 5.3 (a) The terms in the sequence become arbitrarily large as n → ∞. (b) The terms
in the sequence approach 1 as n → ∞. (c) The terms in the sequence alternate between 1
and −1 as n → ∞. (d) The terms in the sequence alternate between positive and negative
values but approach 0 as n → ∞.


From these examples, we see several possibilities for the behavior of the terms of a sequence as n → ∞. In two of the
sequences, the terms approach a finite number as n → ∞. In the other two sequences, the terms do not. If the terms of a
sequence approach a finite number L as n → ∞, we say that the sequence is a convergent sequence and the real number
L is the limit of the sequence. We can give an informal definition here.


Definition
Given a sequence {an}, if the terms an become arbitrarily close to a finite number L as n becomes sufficiently
large, we say {an} is a convergent sequence and L is the limit of the sequence. In this case, we write


lim
n → ∞


an = L.


If a sequence {an} is not convergent, we say it is a divergent sequence.


From Figure 5.3, we see that the terms in the sequence ⎧

⎨1 − ⎛⎝


1
2



n⎫



⎬ are becoming arbitrarily close to 1 as n becomes


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very large. We conclude that ⎧

⎨1 − ⎛⎝


1
2



n⎫



⎬ is a convergent sequence and its limit is 1. In contrast, from Figure 5.3, we see


that the terms in the sequence 1 + 3n are not approaching a finite number as n becomes larger. We say that {1 + 3n} is
a divergent sequence.
In the informal definition for the limit of a sequence, we used the terms “arbitrarily close” and “sufficiently large.” Althoughthese phrases help illustrate the meaning of a converging sequence, they are somewhat vague. To be more precise, we nowpresent the more formal definition of limit for a sequence and show these ideas graphically in Figure 5.4.
Definition
A sequence {an} converges to a real number L if for all ε > 0, there exists an integer N such that |an − L| < ε
if n ≥ N. The number L is the limit of the sequence and we write


lim
n → ∞


an = L or an → L.


In this case, we say the sequence {an} is a convergent sequence. If a sequence does not converge, it is a divergent
sequence, and we say the limit does not exist.


We remark that the convergence or divergence of a sequence {an} depends only on what happens to the terms an as
n → ∞. Therefore, if a finite number of terms b1, b2 ,…, bN are placed before a1 to create a new sequence


b1, b2 ,…, bN, a1, a2 ,…,


this new sequence will converge if {an} converges and diverge if {an} diverges. Further, if the sequence {an} converges
to L, this new sequence will also converge to L.


Figure 5.4 As n increases, the terms an become closer to L. For values of n ≥ N, the
distance between each point (n, an) and the line y = L is less than ε.


As defined above, if a sequence does not converge, it is said to be a divergent sequence. For example, the sequences
{1 + 3n} and ⎧⎩⎨(−1)n⎫⎭⎬ shown in Figure 5.4 diverge. However, different sequences can diverge in different ways. The
sequence ⎧⎩⎨(−1)n⎫⎭⎬ diverges because the terms alternate between 1 and −1, but do not approach one value as n → ∞.
On the other hand, the sequence {1 + 3n} diverges because the terms 1 + 3n → ∞ as n → ∞. We say the sequence
{1 + 3n} diverges to infinity and write lim


n → ∞
(1 + 3n) = ∞. It is important to recognize that this notation does not imply


the limit of the sequence {1 + 3n} exists. The sequence is, in fact, divergent. Writing that the limit is infinity is intended


Chapter 5 | Sequences and Series 433




only to provide more information about why the sequence is divergent. A sequence can also diverge to negative infinity. Forexample, the sequence {−5n + 2} diverges to negative infinity because −5n + 2 → −∞ as n → −∞. We write this as
lim


n → ∞
(−5n + 2) = → −∞.


Because a sequence is a function whose domain is the set of positive integers, we can use properties of limits of functionsto determine whether a sequence converges. For example, consider a sequence {an} and a related function f defined
on all positive real numbers such that f (n) = an for all integers n ≥ 1. Since the domain of the sequence is a subset
of the domain of f , if lim


x → ∞
f (x) exists, then the sequence converges and has the same limit. For example, consider


the sequence ⎧

⎨1n



⎬ and the related function f (x) = 1x . Since the function f defined on all real numbers x > 0 satisfies


f (x) = 1x → 0 as x → ∞, the sequence ⎧⎩⎨1n⎫⎭⎬ must satisfy 1n → 0 as n → ∞.


Theorem 5.1: Limit of a Sequence Defined by a Function
Consider a sequence {an} such that an = f (n) for all n ≥ 1. If there exists a real number L such that


lim
x → ∞


f (x) = L,


then {an} converges and
lim


n → ∞
an = L.


We can use this theorem to evaluate lim
n → ∞


rn for 0 ≤ r ≤ 1. For example, consider the sequence ⎧⎩⎨(1/2)n⎫⎭⎬ and the related
exponential function f (x) = (1/2)x. Since lim


x → ∞
(1/2)x = 0, we conclude that the sequence ⎧⎩⎨(1/2)n⎫⎭⎬ converges and its


limit is 0. Similarly, for any real number r such that 0 ≤ r < 1, lim
x → ∞


r x = 0, and therefore the sequence {rn}
converges. On the other hand, if r = 1, then lim


x → ∞
r x = 1, and therefore the limit of the sequence {1n} is 1. If r > 1,


lim
x → ∞


r x = ∞, and therefore we cannot apply this theorem. However, in this case, just as the function r x grows without
bound as n → ∞, the terms rn in the sequence become arbitrarily large as n → ∞, and we conclude that the sequence
{rn} diverges to infinity if r > 1.
We summarize these results regarding the geometric sequence {rn}:


rn → 0 if 0 < r < 1


rn → 1 if r = 1


rn → ∞ if r > 1.


Later in this section we consider the case when r < 0.
We now consider slightly more complicated sequences. For example, consider the sequence ⎧⎩⎨(2/3)n + (1/4)n⎫⎭⎬. The terms
in this sequence are more complicated than other sequences we have discussed, but luckily the limit of this sequence isdetermined by the limits of the two sequences ⎧⎩⎨(2/3)n⎫⎭⎬ and ⎧⎩⎨(1/4)n⎫⎭⎬. As we describe in the following algebraic limit laws,
since ⎧⎩⎨(2/3)n⎫⎭⎬ and {1/4)n⎫⎭⎬ both converge to 0, the sequence ⎧⎩⎨(2/3)n + (1/4)n⎫⎭⎬ converges to 0 + 0 = 0. Just as we were
able to evaluate a limit involving an algebraic combination of functions f and g by looking at the limits of f and g (see
Introduction to Limits (http://cnx.org/content/m53483/latest/) ), we are able to evaluate the limit of a sequencewhose terms are algebraic combinations of an and bn by evaluating the limits of {an} and ⎧⎩⎨bn⎫⎭⎬.


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Theorem 5.2: Algebraic Limit Laws
Given sequences {an} and ⎧⎩⎨bn⎫⎭⎬ and any real number c, if there exist constants A and B such that limn → ∞an = A
and lim


n → ∞
bn = B, then


i. lim
n → ∞


c = c


ii. lim
n → ∞


can = c limn → ∞an = cA


iii. lim
n → ∞



⎝an ± bn



⎠ = limn → ∞an ± limn → ∞bn = A ± B


iv. lim
n → ∞



⎝an · bn



⎠ = ⎛⎝ limn → ∞an



⎠ ·

⎝ limn → ∞bn



⎠ = A ·B


v. lim
n → ∞


an
bn

⎠ =


lim
n → ∞


an
lim


n → ∞
bn


= A
B
, provided B ≠ 0 and each bn ≠ 0.


Proof
We prove part iii.
Let ϵ > 0. Since lim


n → ∞
an = A, there exists a constant positive integer N1 such that for all n ≥ N1. Since


lim
n → ∞


bn = B, there exists a constant N2 such that |bn − B| < ε/2 for all n ≥ N2. Let N be the largest of N1 and
N2. Therefore, for all n ≥ N,
|(an + bn)−(A + B)| ≤ |an − A| + |bn − B| < ε2


+ ε
2
= ε.



The algebraic limit laws allow us to evaluate limits for many sequences. For example, consider the sequence ⎧



⎨ 1
n2



⎬. As


shown earlier, lim
n → ∞


1/n = 0. Similarly, for any positive integer k, we can conclude that
lim


n → ∞
1
nk


= 0.


In the next example, we make use of this fact along with the limit laws to evaluate limits for other sequences.
Example 5.3
Determining Convergence and Finding Limits
For each of the following sequences, determine whether or not the sequence converges. If it converges, find itslimit.


a. ⎧

⎨5 − 3


n2





b. ⎧

⎨3n


4 − 7n2 + 5
6 − 4n4







c. ⎧

⎨2


n


n2





Chapter 5 | Sequences and Series 435




d. ⎧

⎨⎛⎝1 +


4
n



n⎫





Solution
a. We know that 1/n → 0. Using this fact, we conclude that


lim
n → ∞


1
n2


= lim
n → ∞




1
n

⎠. limn → ∞




1
n

⎠ = 0.


Therefore,
lim


n → ∞

⎝5 −


3
n2

⎠ = limn → ∞5 − 3 limn → ∞


1
n2


= 5 − 3.0 = 5.


The sequence converges and its limit is 5.
b. By factoring n4 out of the numerator and denominator and using the limit laws above, we have


lim
n → ∞


3n4 − 7n2 + 5
6 − 4n4


= lim
n → ∞


3 − 7
n2


+ 5
n4


6


n4
− 4


=
lim


n → ∞

⎝3 −


7


n2
+ 5


n4



lim
n → ∞



6


n4
− 4



=



⎝ limn → ∞(3)− limn → ∞


7


n2
+ lim


n → ∞
5


n4




⎝ limn → ∞


6


n4
− lim


n → ∞
(4)



=



⎝ limn → ∞(3)−7 · limn → ∞


1


n2
+ 5 · lim


n → ∞
1


n4




⎝6 · limn → ∞


1


n4
− lim


n → ∞
(4)



= 3 − 7 · 0 + 5 · 0
6 · 0 − 4


= − 3
4
.


The sequence converges and its limit is −3/4.
c. Consider the related function f (x) = 2x /x2 defined on all real numbers x > 0. Since 2x → ∞ and


x2 → ∞ as x → ∞, apply L’Hôpital’s rule and write
lim


x → ∞
2x


x2
= lim


x → ∞
2x ln2
2x


Take the derivatives of the numerator and denominator.


= lim
x → ∞


2x (ln2)2


2
Take the derivatives again.


= ∞.


We conclude that the sequence diverges.
d. Consider the function f (x) = ⎛⎝1 + 4x⎞⎠


x defined on all real numbers x > 0. This function has the
indeterminate form 1∞ as x → ∞. Let


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5.3


y = lim
x → ∞



⎝1 +


4
x



x
.


Now taking the natural logarithm of both sides of the equation, we obtain
ln(y) = ln



⎣ limx → ∞



⎝1 +


4
x



x⎤
⎦.


Since the function f (x) = lnx is continuous on its domain, we can interchange the limit and the natural
logarithm. Therefore,


ln(y) = lim
x → ∞

⎣ln

⎝1 +


4
x



x⎤
⎦.


Using properties of logarithms, we write
lim


x → ∞

⎣ln

⎝1 +


4
x



x⎤
⎦ = limx → ∞x ln



⎝1 +


4
x

⎠.


Since the right-hand side of this equation has the indeterminate form ∞· 0, rewrite it as a fraction to
apply L’Hôpital’s rule. Write


lim
x → ∞


x ln⎛⎝1 +
4
x

⎠ = limx → ∞


ln(1 + 4/x)
1/x


.


Since the right-hand side is now in the indeterminate form 0/0, we are able to apply L’Hôpital’s rule.
We conclude that


lim
x → ∞


ln(1 + 4/x)
1/x


= lim
x → ∞


4
1 + 4/x


= 4.


Therefore, ln(y) = 4 and y = e4. Therefore, since lim
x → ∞



⎝1 +


4
x



x
= e4, we can conclude that the


sequence ⎧

⎨⎛⎝1 +


4
n



n⎫



⎬ converges to e4.


Consider the sequence ⎧



⎝5n


2 + 1⎞⎠/e
n⎫

⎬. Determine whether or not the sequence converges. If it converges,


find its limit.
Recall that if f is a continuous function at a value L, then f (x) → f (L) as x → L. This idea applies to sequences
as well. Suppose a sequence an → L, and a function f is continuous at L. Then f (an) → f (L). This property often
enables us to find limits for complicated sequences. For example, consider the sequence 5 − 3


n2
. From Example 5.3a.


we know the sequence 5 − 3
n2


→ 5. Since x is a continuous function at x = 5,


lim
n → ∞


5 − 3
n2


= lim
n → ∞

⎝5 −


3
n2

⎠ = 5.


Chapter 5 | Sequences and Series 437




5.4


Theorem 5.3: Continuous Functions Defined on Convergent Sequences
Consider a sequence {an} and suppose there exists a real number L such that the sequence {an} converges to L.
Suppose f is a continuous function at L. Then there exists an integer N such that f is defined at all values an for
n ≥ N, and the sequence ⎧⎩⎨ f (an)⎫⎭⎬ converges to f (L) (Figure 5.5).


Proof
Let ϵ > 0. Since f is continuous at L, there exists δ > 0 such that | f (x) − f (L)| < ε if |x − L| < δ. Since the
sequence {an} converges to L, there exists N such that |an − L| < δ for all n ≥ N. Therefore, for all n ≥ N,
|an − L| < δ, which implies | f (an)− f (L)| < ε. We conclude that the sequence ⎧⎩⎨ f (an)⎫⎭⎬ converges to f (L).


Figure 5.5 Because f is a continuous function as the inputs
a1, a2, a3 ,… approach L, the outputs
f (a1), f (a2), f (a3),… approach f (L).


Example 5.4
Limits Involving Continuous Functions Defined on Convergent Sequences
Determine whether the sequence ⎧



⎨cos⎛⎝3/n


2⎞




⎬ converges. If it converges, find its limit.


Solution
Since the sequence ⎧



⎨3/n2





⎬ converges to 0 and cosx is continuous at x = 0, we can conclude that the sequence





⎨cos⎛⎝3/n


2⎞




⎬ converges and


lim
n → ∞


cos


3
n2

⎠ = cos(0) = 1.


Determine if the sequence ⎧

⎨ 2n + 1


3n + 5



⎬ converges. If it converges, find its limit.


Another theorem involving limits of sequences is an extension of the Squeeze Theorem for limits discussed inIntroduction to Limits (http://cnx.org/content/m53483/latest/) .


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Theorem 5.4: Squeeze Theorem for Sequences
Consider sequences {an}, ⎧⎩⎨bn⎫⎭⎬, and {cn}. Suppose there exists an integer N such that


an ≤ bn ≤ cn for all n ≥ N.


If there exists a real number L such that
lim


n → ∞
an = L = limn → ∞cn,


then ⎧⎩⎨bn⎫⎭⎬ converges and limn → ∞bn = L (Figure 5.6).


Proof
Let ε > 0. Since the sequence {an} converges to L, there exists an integer N1 such that |an − L| < ε for all n ≥ N1.
Similarly, since {cn} converges to L, there exists an integer N2 such that |cn − L| < ε for all n ≥ N2. By assumption,
there exists an integer N such that an ≤ bn ≤ cn for all n ≥ N. Let M be the largest of N1, N2, and N. We must
show that |bn − L| < ε for all n ≥ M. For all n ≥ M,


−ε < −|an − L| ≤ an − L ≤ bn − L ≤ cn − L ≤ |cn − L| < ε.


Therefore, −ε < bn − L < ε, and we conclude that |bn − L| < ε for all n ≥ M, and we conclude that the sequence ⎧⎩⎨bn⎫⎭⎬
converges to L.


Figure 5.6 Each term bn satisfies an ≤ bn ≤ cn and the
sequences {an} and {cn} converge to the same limit, so the
sequence ⎧⎩⎨bn⎫⎭⎬ must converge to the same limit as well.


Example 5.5
Using the Squeeze Theorem
Use the Squeeze Theorem to find the limit of each of the following sequences.


a. ⎧

⎨cosn
n2





b. ⎧

⎨⎛⎝−


1
2



n⎫





Chapter 5 | Sequences and Series 439




5.5


Solution
a. Since −1 ≤ cosn ≤ 1 for all integers n, we have


− 1
n2


≤ cosn
n2


≤ 1
n2


.


Since −1/n2 → 0 and 1/n2 → 0, we conclude that cosn/n2 → 0 as well.
b. Since


− 1
2n


≤ ⎛⎝−
1
2



n
≤ 1


2n


for all positive integers n, −1/2n → 0 and 1/2n → 0, we can conclude that (−1/2)n → 0.


Find lim
n → ∞


2n − sinn
n .


Using the idea from Example 5.5b. we conclude that rn → 0 for any real number r such that −1 < r < 0. If r < −1,
the sequence {rn} diverges because the terms oscillate and become arbitrarily large in magnitude. If r = −1, the
sequence {rn} = ⎧⎩⎨(−1)n⎫⎭⎬ diverges, as discussed earlier. Here is a summary of the properties for geometric sequences.


(5.1)rn → 0 if |r| < 1
(5.2)rn → 1 if r = 1
(5.3)rn → ∞ if r > 1
(5.4)⎧



⎨rn⎫⎭⎬ diverges if r ≤ −1Bounded Sequences


We now turn our attention to one of the most important theorems involving sequences: the Monotone ConvergenceTheorem. Before stating the theorem, we need to introduce some terminology and motivation. We begin by defining whatit means for a sequence to be bounded.
Definition
A sequence {an} is bounded above if there exists a real number M such that


an ≤ M


for all positive integers n.
A sequence {an} is bounded below if there exists a real number M such that


M ≤ an


for all positive integers n.
A sequence {an} is a bounded sequence if it is bounded above and bounded below.
If a sequence is not bounded, it is an unbounded sequence.


For example, the sequence {1/n} is bounded above because 1/n ≤ 1 for all positive integers n. It is also bounded below
because 1/n ≥ 0 for all positive integers n. Therefore, {1/n} is a bounded sequence. On the other hand, consider the


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sequence {2n}. Because 2n ≥ 2 for all n ≥ 1, the sequence is bounded below. However, the sequence is not bounded
above. Therefore, {2n} is an unbounded sequence.
We now discuss the relationship between boundedness and convergence. Suppose a sequence {an} is unbounded. Then it is
not bounded above, or not bounded below, or both. In either case, there are terms an that are arbitrarily large in magnitude
as n gets larger. As a result, the sequence {an} cannot converge. Therefore, being bounded is a necessary condition for a
sequence to converge.
Theorem 5.5: Convergent Sequences Are Bounded
If a sequence {an} converges, then it is bounded.


Note that a sequence being bounded is not a sufficient condition for a sequence to converge. For example, the sequence



⎨(−1)n⎫⎭⎬ is bounded, but the sequence diverges because the sequence oscillates between 1 and −1 and never approaches a
finite number. We now discuss a sufficient (but not necessary) condition for a bounded sequence to converge.
Consider a bounded sequence {an}. Suppose the sequence {an} is increasing. That is, a1 ≤ a2 ≤ a3…. Since the
sequence is increasing, the terms are not oscillating. Therefore, there are two possibilities. The sequence could diverge toinfinity, or it could converge. However, since the sequence is bounded, it is bounded above and the sequence cannot divergeto infinity. We conclude that {an} converges. For example, consider the sequence





⎨1
2
, 2
3
, 3
4
, 4
5
,…



⎬.


Since this sequence is increasing and bounded above, it converges. Next, consider the sequence



⎨2, 0, 3, 0, 4, 0, 1, − 1


2
, − 1


3
, − 1


4
,…



⎬.


Even though the sequence is not increasing for all values of n, we see that −1/2 < −1/3 < −1/4 < ⋯. Therefore,
starting with the eighth term, a8 = −1/2, the sequence is increasing. In this case, we say the sequence is eventually
increasing. Since the sequence is bounded above, it converges. It is also true that if a sequence is decreasing (or eventuallydecreasing) and bounded below, it also converges.
Definition
A sequence {an} is increasing for all n ≥ n0 if


an ≤ an + 1 for all n ≥ n0.


A sequence {an} is decreasing for all n ≥ n0 if
an ≥ an + 1 for all n ≥ n0.


A sequence {an} is a monotone sequence for all n ≥ n0 if it is increasing for all n ≥ n0 or decreasing for all
n ≥ n0.


We now have the necessary definitions to state the Monotone Convergence Theorem, which gives a sufficient condition forconvergence of a sequence.
Theorem 5.6: Monotone Convergence Theorem
If {an} is a bounded sequence and there exists a positive integer n0 such that {an} is monotone for all n ≥ n0,


Chapter 5 | Sequences and Series 441




then {an} converges.


The proof of this theorem is beyond the scope of this text. Instead, we provide a graph to show intuitively why this theoremmakes sense (Figure 5.7).


Figure 5.7 Since the sequence {an} is increasing and
bounded above, it must converge.


In the following example, we show how the Monotone Convergence Theorem can be used to prove convergence of asequence.
Example 5.6
Using the Monotone Convergence Theorem
For each of the following sequences, use the Monotone Convergence Theorem to show the sequence convergesand find its limit.


a. ⎧

⎨4


n


n!





b. {an} defined recursively such that
a1 = 2 and an + 1 =


an
2


+ 1
2an


for all n ≥ 2.


Solution
a. Writing out the first few terms, we see