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Volume 2

Calculus Volume 2
SENIOR CONTRIBUTING AUTHORS
EDWIN "JED" HERMAN, UNIVERSITY OF WISCONSIN-STEVENS POINT GILBERT STRANG, MASSACHUSETTS INSTITUTE OF TECHNOLOGY

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Table of Contents
Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Chapter 1: Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.1 Approximating Areas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.2 The Definite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 1.3 The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 1.4 Integration Formulas and the Net Change Theorem . . . . . . . . . . . . . . . . . . . . . . 64 1.5 Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 1.6 Integrals Involving Exponential and Logarithmic Functions . . . . . . . . . . . . . . . . . . 93 1.7 Integrals Resulting in Inverse Trigonometric Functions . . . . . . . . . . . . . . . . . . . . 106 Chapter 2: Applications of Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 2.1 Areas between Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 2.2 Determining Volumes by Slicing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 2.3 Volumes of Revolution: Cylindrical Shells . . . . . . . . . . . . . . . . . . . . . . . . . . . 154 2.4 Arc Length of a Curve and Surface Area . . . . . . . . . . . . . . . . . . . . . . . . . . . 169 2.5 Physical Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 2.6 Moments and Centers of Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201 2.7 Integrals, Exponential Functions, and Logarithms . . . . . . . . . . . . . . . . . . . . . . . 219 2.8 Exponential Growth and Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232 2.9 Calculus of the Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243 Chapter 3: Techniques of Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261 3.1 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262 3.2 Trigonometric Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273 3.3 Trigonometric Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285 3.4 Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298 3.5 Other Strategies for Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311 3.6 Numerical Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 316 3.7 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330 Chapter 4: Introduction to Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 351 4.1 Basics of Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352 4.2 Direction Fields and Numerical Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365 4.3 Separable Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 381 4.4 The Logistic Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393 4.5 First-order Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 408 Chapter 5: Sequences and Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 427 5.1 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 427 5.2 Infinite Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 450 5.3 The Divergence and Integral Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 471 5.4 Comparison Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 485 5.5 Alternating Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 496 5.6 Ratio and Root Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 509 Chapter 6: Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 531 6.1 Power Series and Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 532 6.2 Properties of Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 544 6.3 Taylor and Maclaurin Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 561 6.4 Working with Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 581 Chapter 7: Parametric Equations and Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . 605 7.1 Parametric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 606 7.2 Calculus of Parametric Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 625 7.3 Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 642 7.4 Area and Arc Length in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . 662 7.5 Conic Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 671 Appendix A: Table of Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 699 Appendix B: Table of Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 705 Appendix C: Review of Pre-Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 707 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 819

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Preface

1

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 Resources
Customization
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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.
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Errata
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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

2

Preface

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 2 Chapter 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 3 Chapter 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 as modern applications and methods. Derivations and explanations are based on years of classroom experience on the part of 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 areas of physics, chemistry, engineering, and computer science. Student Projects in each chapter give students opportunities to explore interesting sidelights in pure and applied mathematics, from showing that the number e is irrational, to calculating the center of mass of the Grand Canyon Skywalk or the terminal speed of a skydiver. Chapter Opening Applications pose problems that are solved later in the chapter, using the ideas covered in that chapter. Problems include the hydraulic force against the Hoover Dam, and the comparison of the relative intensity of two earthquakes. Definitions, Rules, and Theorems are highlighted throughout the text, including over 60 Proofs of theorems.
Assessments That Reinforce Key Concepts
In-chapter Examples walk students through problems by posing a question, stepping out a solution, and then asking students to practice the skill with a “Check Your Learning” component. The book also includes assessments at the end of each chapter so students can apply what they’ve learned through practice problems. Many exercises are marked with a [T] to indicate they are suitable for solution by technology, including calculators or Computer Algebra Systems (CAS). Answers for selected exercises are available in the Answer Key at the back of the book. The book also includes assessments at the end 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. Exponential and logarithmic functions are presented in Chapter 2. Integration of these functions is covered in Chapters 1 for instructors who want to include them with other types of functions. These discussions, however, are in separate sections that can be skipped for instructors who prefer to wait until the integral definitions are given before teaching the calculus derivations of exponentials and logarithms.
Comprehensive Art Program
Our art program is designed to enhance students’ understanding of concepts through clear and effective illustrations,

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Preface

3

diagrams, and photographs.

Additional Resources
Student and Instructor Resources
We’ve compiled additional resources for both students and instructors, including Getting Started Guides, an instructor solution manual, and PowerPoint slides. Instructor resources require a verified instructor account, which can be requested on 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 students and instructors everywhere. Their tools integrate seamlessly with our OpenStax titles at a low cost. To access the partner resources for your text, visit your book page on openstax.org.
About The Authors
Senior Contributing Authors
Gilbert Strang, Massachusetts Institute of Technology Dr. Strang received his PhD from UCLA in 1959 and has been teaching mathematics at MIT ever since. His Calculus online textbook is one of eleven that he has published and is the basis from which our final product has been derived and updated for today’s student. Strang is a decorated mathematician and past Rhodes Scholar at Oxford University.
Edwin “Jed” Herman, University of Wisconsin-Stevens Point Dr. Herman earned a BS in Mathematics from Harvey Mudd College in 1985, an MA in Mathematics from UCLA in 1987, and a PhD in Mathematics from the University of Oregon in 1997. He is currently a Professor at the University of Wisconsin-Stevens Point. He has more than 20 years of experience teaching college mathematics, is a student research mentor, is experienced in course development/design, and is also an avid board game designer and player.

4
Contributing Authors
Catherine Abbott, Keuka College Nicoleta Virginia Bila, Fayetteville State University Sheri J. Boyd, Rollins College Joyati Debnath, Winona State University Valeree Falduto, Palm Beach State College Joseph Lakey, New Mexico State University Julie Levandosky, Framingham State University David McCune, William Jewell College Michelle Merriweather, Bronxville High School Kirsten R. Messer, Colorado State University - Pueblo Alfred K. Mulzet, Florida State College at Jacksonville William Radulovich (retired), Florida State College at Jacksonville Erica M. Rutter, Arizona State University David Smith, University of the Virgin Islands Elaine A. Terry, Saint Joseph’s University David Torain, Hampton University
Reviewers
Marwan A. Abu-Sawwa, Florida State College at Jacksonville Kenneth J. Bernard, Virginia State University John Beyers, University of Maryland Charles Buehrle, Franklin & Marshall College Matthew Cathey, Wofford College Michael Cohen, Hofstra University William DeSalazar, Broward County School System Murray Eisenberg, University of Massachusetts Amherst Kristyanna Erickson, Cecil College Tiernan Fogarty, Oregon Institute of Technology David French, Tidewater Community College Marilyn Gloyer, Virginia Commonwealth University Shawna Haider, Salt Lake Community College Lance Hemlow, Raritan Valley Community College Jerry Jared, The Blue Ridge School Peter Jipsen, Chapman University David Johnson, Lehigh University M.R. Khadivi, Jackson State University Robert J. Krueger, Concordia University Tor A. Kwembe, Jackson State University Jean-Marie Magnier, Springfield Technical Community College Cheryl Chute Miller, SUNY Potsdam Bagisa Mukherjee, Penn State University, Worthington Scranton Campus Kasso Okoudjou, University of Maryland College Park Peter Olszewski, Penn State Erie, The Behrend College Steven Purtee, Valencia College Alice Ramos, Bethel College Doug Shaw, University of Northern Iowa Hussain Elalaoui-Talibi, Tuskegee University Jeffrey Taub, Maine Maritime Academy William Thistleton, SUNY Polytechnic Institute A. David Trubatch, Montclair State University Carmen Wright, Jackson State University Zhenbu Zhang, Jackson State University
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Preface

Chapter 1 | Integration

5

1 | INTEGRATION

Figure 1.1 Iceboating is a popular winter sport in parts of the northern United States and Europe. (credit: modification of work by 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 move very quickly, and many ice boating enthusiasts are drawn to the sport because of the speed. Top iceboat racers can attain

6

Chapter 1 | Integration

speeds up to five times the wind speed. If we know how fast an iceboat is moving, we can use integration to determine how far 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 wide variety of mechanical and physical applications. In this chapter, we first introduce the theory behind integration and use integrals to calculate areas. From there, we develop the Fundamental Theorem of Calculus, which relates differentiation and integration. 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 by the 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 approximations to the total area. In this process, an area bounded by curves is filled with rectangles, triangles, and shapes with exact area formulas. 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. Taking a 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 general cases.
Sigma (Summation) Notation
As mentioned, we will use shapes of known area to approximate the area of an irregular region bounded by curves. This process often requires adding up long strings of numbers. To make it easier to write down these lengthy sums, we look at some 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 20 without 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
20
∑ i,
i=1
which is much more compact.
Typically, sigma notation is presented in the form
n
∑ ai
i=1
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,
7
beginning with the value when i = 1 and ending with the value when i = n. For example, an expression like ∑ si is i=2

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Chapter 1 | Integration

7

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 we like 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

+

215.

Solution a. Write

5
∑ 3i = 3 + 32 + 33 + 34 + 35
i=1

= 363.

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

5

∑
i=1

1 i2

.

1.1 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.

n

(1.1)

∑ c = nc

i=1

2.

n

n

∑ cai = c ∑ ai

i=1

i=1

(1.2)

3.

n

n

n

∑ ∑ ∑ ⎛⎝ai

+

b

⎞
i⎠

=

ai +

bi

i=1

i=1

i=1

(1.3)

4.

n

n

n

∑ ∑ ∑ ⎛⎝ai

−

b

⎞
i⎠

=

ai −

bi

i=1

i=1

i=1

(1.4)

8

Chapter 1 | Integration

5.

n

m

n

∑ ai = ∑ ai + ∑ ai

i=1

i=1

i=m+1

(1.5)

Proof
We prove properties 2. and 3. here, and leave proof of the other properties to the Exercises.
2. We have
n
∑ cai = ca1 + ca2 + ca3 + ⋯ + can
i=1
= c(a1 + a2 + a3 + ⋯ + an)

3. We have

n
= c ∑ ai. i=1

n

∑ ⎛⎝a

i

+

b

⎞
i⎠

i=1

=

⎛⎝a 1

+

b

⎞
1⎠

+

⎛⎝a 2

+

b

⎞
2⎠

+

⎛⎝a 3

+

b

⎞
3⎠

+

⋯

+

a⎛
⎝

n

+

b

⎞
n⎠

=

(a 1

+

a2

+

a3

+

⋯

+

a n)

+

⎛⎝b 1

+

b2

+

b3

+

⋯

+

b

⎞
n⎠

n

n

= ∑ ai + ∑ bi.

i=1

i=1

□

A few more formulas for frequently found functions simplify the summation process further. These are shown in the next rule, 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

n
∑
i=1

i

=

1

+

2

+

⋯

+

n

=

n(n + 2

1).

2. The sum of consecutive integers squared is given by

n
∑
i=1

i2

=

12

+

22

+

⋯

+

n2

=

n(n

+

1)(2n 6

+

1).

3. The sum of consecutive integers cubed is given by

n
∑
i=1

i3

=

13

+

23

+

⋯

+

n3

=

n2 (n + 4

1) 2 .

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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Chapter 1 | Integration

9

b.

The sum of the terms

⎛⎝i 3

−

i

2⎞ ⎠

for

i

=

1,

2,

3,

4,

5,

6.

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

200

200

∑ (i − 3)2

∑ =

⎛⎝i2 − 6i + 9⎞⎠

i=1

i=1

200

200

200

= ∑ i2 − ∑ 6i + ∑ 9

i=1

i=1

i=1

200

200

200

= ∑ i2 − 6∑ i + ∑ 9

i=1

i=1 i=1

=

200(200

+ 1)(400 6

+

1)

−

6⎡⎣200(2020

+

1)⎤ ⎦

+

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.

6

6

6

∑ ⎛⎝i

3

−

i

2⎞ ⎠

= ∑ i3 − ∑ i2

i=1

i=1

i=1

=

62 (6 + 4

1) 2

−

6(6

+

1)⎛⎝2(6) 6

+

1⎞ ⎠

=

1764 4

−

546 6

= 350

1.2 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

10
∑ i3
i=0

=

(10)2 (10 4

+

1) 2

=

100(121) 4

= 3025.

10

Chapter 1 | Integration

1.3

20

Evaluate the sum indicated by the notation ∑ (2k + 1).

k=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 small shapes 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 −n a.

We do this by selecting equally spaced points

x0, x1, x2 ,…, xn with x0 = a, xn = b, and

for i = 1, 2, 3,…, n.

xi − xi − 1 = b −n a

We denote the width of each subinterval with the notation Δx, so Δx = b −n a 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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Chapter 1 | Integration

11

We can use this regular partition as the basis of a method for estimating the area under the curve. We next examine two methods: 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.

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

(1.6)

n
= ∑ f (xi − 1)Δx i=1

Figure 1.3 In the left-endpoint approximation of area under a curve, the height of each rectangle is determined by the function value 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 the left-endpoint approximation, but now the heights of the rectangles are determined by the function values at the right of each subinterval.

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

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

(1.7)

n
= ∑ f (xi)Δx. i=1
The notation Rn indicates this is a right-endpoint approximation for A (Figure 1.4).

12

Chapter 1 | Integration

Figure 1.4 In the right-endpoint approximation of area under a curve, the height of each rectangle is determined by the function value at the right of each subinterval. Note that the right-endpoint approximation differs from the left-endpoint approximation 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 area between f (x) and the x-axis. When the left endpoints are used to calculate height, we have a left-endpoint approximation.

Thus,

6
A ≈ L6 = ∑ f (xi − 1)Δx = f (x0)Δx + f (x1)Δx + f (x2)Δx + f (x3)Δx + f (x4)Δx + f (x5)Δx i=1 = 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 endpoints and (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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Chapter 1 | Integration

13

6
A ≈ R6 = ∑ f (xi)Δx = f (x1)Δx + f (x2)Δx + f (x3)Δx + f (x4)Δx + f (x5)Δx + f (x6)Δx i=1 = 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

− 4

0)

=

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.

14

Chapter 1 | Integration

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.

1.4 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, neither the left-endpoint approximation nor the right-endpoint approximation is a particularly accurate estimate of the area under the 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 more precisely.
We can demonstrate the improved approximation obtained through smaller intervals with an example. Let’s explore the idea of 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

− 4

0

=

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 dividing the region from a to b into four equal intervals, the area under the curve is approximately equal to the sum of the areas of the rectangles.
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Chapter 1 | Integration

15

Figure 1.9 shows the same curve divided into eight subintervals. Comparing the graph with four rectangles in Figure 1.8 with 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-endpoint approximation.
The graph in Figure 1.10 shows the same function with 32 rectangles inscribed under the curve. There appears to be little white 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 curve for a left-endpoint approximation.
We can carry out a similar process for the right-endpoint approximation method. A right-endpoint approximation of the same 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.

16

Chapter 1 | Integration

Figure 1.11 Now we divide the area under the curve into four equal 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 a region 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 subintervals for a right-endpoint approximation.
Based on these figures and calculations, it appears we are on the right track; the rectangles appear to approximate the area under the curve better as n gets larger. Furthermore, as n increases, both the left-endpoint and right-endpoint approximations appear 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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Chapter 1 | Integration

17

methods. The idea that the approximations of the area under the curve get better and better as n gets larger and larger is very important, 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 Approximations as 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 been

determined 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 ⎛⎝x*i

⎞ ⎠

as

the

height

of

our

rectangle.

This

gives

us

an

estimate

for

the

area

of

the form

n

∑ A ≈

f ⎛⎝x*i ⎞⎠Δx.

i=1

A sum of this form is called a Riemann sum, named for the 19th-century mathematician Bernhard Riemann, who developed the 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 x*i be any point in [xi − 1, xi]. A Riemann sum is defined for f (x) as

n
∑ f ⎛⎝x*i ⎞⎠Δx.
i=1

Recall that with the left- and right-endpoint approximations, the estimates seem to get better and better as n get larger and larger. The same thing happens with Riemann sums. Riemann sums give better approximations for larger values of n. We are now ready to define the area under a curve in terms of Riemann sums.

Definition

n

∑ Let f (x) be a continuous, nonnegative function on an interval ⎡⎣a, b⎤⎦, and let

f ⎛⎝x*i ⎞⎠Δx be a Riemann sum for

i=1

f (x).

Then, the area under the curve

y

=

f (x)

on

⎡⎣a,

b⎤ ⎦

is given by

n
∑ A = n l→im∞i = 1 f ⎛⎝x*i ⎞⎠Δx.

18

Chapter 1 | Integration

See a graphical demonstration (http://www.openstaxcollege.org/l/20_riemannsums) of the construction 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 limit of 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 used to calculate limits of sums.

Second, we

must consider

what to

do

if the

expression

converges

to

different

limits

for different choices of

⎧⎩⎨x*i

.⎫
⎬ ⎭

Fortunately, this does not happen. Although the proof is beyond the scope of this text, it can be shown that if f (x) is

n
∑ continuous on the closed interval ⎡⎣a, b⎤⎦, then n l→im∞i = 1 f ⎛⎝x*i ⎞⎠Δx exists and is unique (in other words, it does not depend

on the choice of ⎧⎩⎨x*i ⎫⎭⎬).

We look at some examples shortly.

But, before

we do, let’s take

a moment

and

talk about some specific choices

for

⎧⎩⎨x*i

.⎫
⎬ ⎭

Although any choice for

⎧⎩⎨x*i

⎫ ⎬ ⎭

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

⎧
⎩⎨x*i

⎫ ⎬ ⎭

to guarantee one result or the other.

If we want an overestimate,

for example,

we can choose

⎧
⎩⎨x*i

⎫ ⎬ ⎭

such that for

i = 1, 2, 3,…, n,

f ⎛⎝x*i

⎞ ⎠

≥

f (x)

for all

x ∈ [xi − 1,

xi].

In other words, we choose

⎧⎩⎨x*i

⎫ ⎬ ⎭

so that for

i = 1,

2,

3,…, n,

f ⎛⎝x*i

⎞ ⎠

is the maximum function value on

n

∑ the interval

[xi − 1,

xi].

If we select

⎧⎩⎨x*i

⎫ ⎬ ⎭

in this way, then the Riemann sum

f ⎛⎝x*i ⎞⎠Δx is called an upper sum.

i=1

Similarly, if we want an underestimate, we can choose

⎧⎩⎨x*i

⎫ ⎬ ⎭

so that for

i = 1,

2,

3,…, n,

f ⎛⎝x*i

⎞ ⎠

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 = 14. 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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Chapter 1 | Integration

19

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

The Riemann sum is

4

∑ ⎛⎝10 − x2⎞⎠(0.25)

=

0.25⎡⎣10

−

(1.25)

2

+

10

−

(1.5)

2

+

10

−

(1.75)2

+

10

−

(2)

2⎤ ⎦

k=1

= 0.25[8.4375 + 7.75 + 6.9375 + 6]

= 7.28.

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

1.5 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) = sin x

over the interval

⎡⎣a,

b⎤ ⎦

=

⎡⎣0,

π 2

⎤⎦;

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.

20

Chapter 1 | Integration

Figure 1.15

The graph of

y

=

sin x

is divided into six regions:

Δx

=

π/2 6

=

1π2.

The

intervals are

⎡⎣0,

1π2 ⎤⎦,

⎡⎣1π2,

π 6

⎤⎦,

⎡π ⎣6

,

π 4

⎤⎦,

⎡π ⎣4

,

π 3

⎤⎦,

⎡π ⎣3

,

5π 12

⎤⎦,

and

⎡⎣51π2 ,

π 2

⎤⎦.

Note that

f (x) = sin x is

increasing on the interval

⎡⎣0,

π 2

⎤⎦,

so a left-endpoint approximation gives us the lower sum. A left-endpoint

approximation is the Riemann sum

∑ i

5 =

0

sin

xi

⎛⎝1π2 ⎞⎠.

We have

A

≈

sin(0)⎛⎝1π2

⎞ ⎠

+

sin⎛⎝1π2 ⎞⎠⎛⎝1π2

⎞ ⎠

+

sin⎛⎝π6

⎞⎛ π ⎠⎝12

⎞ ⎠

+

sin⎛⎝π4

⎞⎛ π ⎞ ⎠⎝12 ⎠

+

sin⎛⎝π3

⎞⎛ π ⎠⎝12

⎞ ⎠

+

sin⎛⎝51π2

⎞⎛ π ⎞ ⎠⎝12 ⎠

= 0.863.

1.6

Using the function

f (x) =

sin x

over the interval

⎡⎣0,

π 2

⎤⎦,

find an upper sum; let n = 6.

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Chapter 1 | Integration

21

1.1 EXERCISES

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

10

10

a. ∑ i and ∑ k

i=1

k=1

10

15

b. ∑ i and ∑ (i − 5)

i=1

i=6

10

9

∑ ∑ c.

i(i − 1) and

⎛ ⎝

j

+

1⎞ ⎠

j

i=1

j=0

10

10

d. ∑ i(i − 1) and ∑ ⎛⎝k2 − k⎞⎠

i=1

k=1

In the following exercises, use the rules for sums of powers of integers to compute the sums.

10
2. ∑ i i=5

10
3. ∑ i2 i=5

100

100

Suppose that ∑ ai = 15 and ∑ bi = −12. In the

i=1

i=1

following exercises, compute the sums.

100

∑ 4.

⎛⎝a i

+

b

⎞
i⎠

i=1

100

∑ 5.

⎛⎝a i

−

b

⎞
i⎠

i=1

100

∑ 6.

⎛⎝3a

i

−

4b

⎞
i⎠

i=1

100

∑ 7.

⎛⎝5a

i

+

4b

⎞
i⎠

i=1

In the following exercises, use summation properties and formulas to rewrite and evaluate the sums.

20

∑ 8.

100⎛⎝k2 − 5k + 1⎞⎠

k=1

50

∑ 9.

⎛ ⎝

j2

−

2

j⎞⎠

j=1

20

∑ 10.

⎛ ⎝

j

2

−

10

j⎞⎠

j = 11

25

∑ 11.

⎡⎣(2k)2 − 100k⎤⎦

k=1

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 and right sums for the given functions on the indicated interval.

12.

L4 for

f (x)

=

x

1 −1

on [2, 3]

13. R4 for g(x) = cos(πx) on [0, 1]

14.

L6 for

f (x) =

1 x(x − 1)

on

⎡⎣2,

5⎤ ⎦

15.

R6 for

f (x) =

1 x(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 the graph of f.

21. Compute the left and right Riemann sums—L6 and R6, respectively—for f (x) = (3 − |3 − x|) on ⎡⎣0, 6⎤⎦.
Compute their average value and compare it with the area under 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 but do not evaluate them.

22

Chapter 1 | Integration

24. L30 for f (x) = x2 on [1, 2]

25. L10 for f (x) = 4 − x2 on [−2, 2]
26. R20 for f (x) = sin x on [0, π]
27. R100 for ln x on [1, e]
In the following exercises, graph the function then use a calculator or a computer program to evaluate the following left and right endpoint sums. Is the area under the curve between 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]

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 mi 10
to his daily routine each week, what is the total mileage on Joe’s shoes after 25 weeks?

38. The following table gives approximate values of the average annual atmospheric rate of increase in carbon dioxide (CO2) each decade since 1960, in parts per million (ppm). Estimate the total increase in atmospheric CO2 between 1964 and 2013.

Decade

Ppm/y

1964–1973 1.07

1974–1983 1.34

1984–1993 1.40

1994–2003 1.87

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,

π⎤ 4⎦

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

2004–2013 2.07
Table 1.2 Average Annual Atmospheric CO2 Increase, 1964–2013 Source: http://www.esrl.noaa.gov/ gmd/ccgg/trends/.

34. Let tj denote the time that it took Tejay van Garteren to ride the jth stage of the Tour de France in 2014. If there
21
were a total of 21 stages, interpret ∑ t j. j=1
35. Let r j denote the total rainfall in Portland on the jth
31
day of the year in 2009. Interpret ∑ r j. j=1
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
365
d1 + ∑ δ j. j=2

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Chapter 1 | Integration

23

39. The following table gives the approximate increase in sea level in inches over 20 years starting in the given year. Estimate the net change in mean sea level from 1870 to 2010.
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 Sea Level Increases, 1870–1990 Source: http://link.springer.com/article/ 10.1007%2Fs10712-011-9119-1

40. The following table gives the approximate increase in dollars in the average price of a gallon of gas per decade since 1950. If the average price of a gallon of gas in 2010 was $2.60, what was the average price of a gallon of gas in 1950?
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 Gas Price Increases, 1950–2000 Source: http://epb.lbl.gov/homepages/ Rick_Diamond/docs/ lbnl55011-trends.pdf.

24

Chapter 1 | Integration

41. The following table gives the percent growth of the 43. U.S. population beginning in July of the year indicated. If the U.S. population was 281,421,906 in July 2000, estimate the U.S. population in July 2010.
Year % Change/Year
2000 1.12

2001 0.99

2002 0.93 44.
2003 0.86

2004 0.93

2005 0.93

2006 0.97

2007 0.96

2008 0.95

45.

2009 0.88

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

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

47. [T] Use a computer algebra system to compute the Riemann sum, LN, for N = 10, 30, 50 for

f (x) =

1 1 + x2

on

[−1,

1].

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Chapter 1 | Integration

25

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 computer program 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 computer program 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 = ln x on the interval [1, 2], which has an exact area of 2 ln(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

RN − LN

= (b − a) ×

f (b) − N

f (a).

general,

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)

− N

f (a).

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 squares enclosed 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.

26

59. A unit circle is made up of n wedges equivalent to the inner 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 = B sin⎛⎝2nπ ⎞⎠. Use this information to argue that the area

of a unit circle is equal to π.

Chapter 1 | Integration

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Chapter 1 | Integration

27

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:
n
∑ A = n l→im∞i = 1 f ⎛⎝x*i ⎞⎠Δx.
However, this definition came with restrictions. We required f (x) to be continuous and nonnegative. Unfortunately, realworld problems don’t always meet these restrictions. In this section, we look at how to apply the concept of the area under the 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

b

n

∫ ∑ a f (x)dx = n l→im∞i = 1 f ⎛⎝x*i ⎞⎠Δx,

(1.8)

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 on Applications of Derivatives (http://cnx.org/content/m53602/latest/) , where we used the indefinite integral symbol (without the a and b above and below) to represent an antiderivative. Although the notation for indefinite integrals may look similar to the notation for a definite integral, they are not the same. A definite integral is a number. An indefinite integral is a family of functions. Later in this chapter we examine how these concepts are related. However, close attention should 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, who is 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 the interval, ⎡⎣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 the computation of the integral. We could use any variable we like as the variable of integration:

b

b

b

∫ f (x)dx = ∫ f (t)dt = ∫ f (u)du

a

a

a

28

Chapter 1 | Integration

n
∑ Previously, we discussed the fact that if f (x) is continuous on ⎡⎣a, b⎤⎦, then the limit n l→im∞i = 1 f ⎛⎝x*i ⎞⎠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 not strictly necessary. Any partition can be used to form a Riemann sum. However, if a nonregular partition is used to define the definite integral, it is not sufficient to take the limit as the number of subintervals goes to infinity. Instead, we must take the limit as the width of the largest subinterval goes to zero. This introduces a little more complex notation in our limits and makes the calculations more difficult without really gaining much additional insight, so we stick with regular partitions for the Riemann sums.

Example 1.7

Evaluating an Integral Using the Definition

Use the definition of the definite integral to evaluate ∫ 2x2 dx. Use a right-endpoint approximation to generate 0
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 −n a = 2n.

Since we are using a right-endpoint approximation to generate Riemann sums, for each i, we need to calculate the 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⎡⎣2n

⎤ ⎦

=

2ni.

Thus, the function value at the right endpoint of the interval is

f (xi)

=

x

2 i

=

⎛⎝2ni ⎞⎠2

=

4i 2 n2

.

Then the Riemann sum takes the form

n
∑
i=1

f (xi)Δx

=

n
∑
i=1

⎛⎝4ni22⎞⎠2n

=

n
∑
i=1

8i 2 n3

=

n

8 n3

∑
i=1

i 2.

n
Using the summation formula for ∑ i2, we have i=1

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Chapter 1 | Integration

29

n

n

∑ f (xi)Δx
i=1

= n83i∑= 1 i2

=

n83⎡⎣n(n

+

1)(2n 6

+

1)⎤ ⎦

=

n83⎡⎣2n3

+

3n 2 6

+

n⎤ ⎦

=

16n 3

+ 24n2 6n 3

+

n

=

8 3

+

4n

+

1 6n

2

.

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

∫

2
x

2

dx

0

n
= ∑ n l→im∞i = 1 f (xi)Δx

=

n l→im∞⎛⎝83

+

4n

+

1 6n

⎞ 2⎠

=

n l→im∞⎛⎝83⎞⎠

+

n l→im∞⎛⎝4n⎞⎠

+

n

l→im∞⎛⎝61n

2

⎞ ⎠

=

8 3

+

0

+

0

=

8 3

.

1.7

Use the definition of

the definite

integral to

evaluate

∫

3
(2x − 1)dx.

Use

a

right-endpoint

approximation

0

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 chapter we develop techniques for evaluating definite integrals without taking limits of Riemann sums. However, for now, we can rely on the fact that definite integrals represent the area under the curve, and we can evaluate definite integrals by using geometric formulas to calculate that area. We do this to confirm that definite integrals do, indeed, represent areas, so we can then 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

6
∫

9 − (x − 3)2dx.

3

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

30

Chapter 1 | Integration

∫ 6 9 − (x − 3)2dx, 3
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,

∫ 6 9 − (x − 3)2 3

=

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.

1.8

Use the

formula for the area of a trapezoid to evaluate

∫

4
(2x + 3)dx.

2

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

⎧
⎩⎨x*i

}

as the left endpoint of each interval. Construct a rectangle on each

subinterval of height

f ⎛⎝x*i

⎞ ⎠

and

width

Δx.

When

f ⎛⎝x*i

⎞ ⎠

is

positive,

the

product

f ⎛⎝x*i ⎞⎠Δx

represents the area of the

rectangle, as before. When

f ⎛⎝x*i

⎞ ⎠

is

negative,

however,

the

product

f ⎛⎝x*i ⎞⎠Δx

represents the negative of the area of the

rectangle. The Riemann sum then becomes

8
∑ f ⎛⎝x*i ⎞⎠Δx = ⎛⎝Area of rectangles above the x-axis⎞⎠ − ⎛⎝Area of rectangles below the x-axis⎞⎠
i=1

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Chapter 1 | Integration

31

Figure 1.17 For a function that is partly negative, the Riemann sum is the area of the rectangles above the x-axis less the 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,

2
∫ f (x)dx 0

n
= ∑ n l→im∞i = 1 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 A1 less 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 is positive. If the area below the x-axis is larger, the net signed area is negative. If the areas above and below the x-axis are equal, the net signed area is zero.
Example 1.9
Finding the Net Signed Area

32

Chapter 1 | Integration

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 = 12bh, the area of triangle A1, above the axis, is
A1 = 123(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

∫ 2xdx −3

=

A1

−

A2

=

9

−

9

=

0.

Figure 1.19 The area above the curve and below the x-axis equals 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.
1.9 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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Chapter 1 | Integration

33

velocity of an object as a function of time, then the area under the curve tells us how far the object is from its original position. 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 car travels 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 its original position (Figure 1.20). Using integral notation, we have
2
∫ 75dt = 150. 0

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 a speed 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 a speed of 40 mph for 3 hours, it will be back at it starting position (Figure 1.21). Again, using integral notation, we have

2

5

∫ 60dt + ∫ −40dt = 120 − 120

0

2

= 0.

In this case the displacement is zero.

34

Chapter 1 | Integration

Figure 1.21 The area above the axis and the area below the axis are 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 area between 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 use the absolute value function. Thus, the total distance traveled by the car is

2

5

2

5

∫ |60|dt + ∫ |−40|dt = ∫ 60dt + ∫ 40dt

0

2

0

2

= 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
b
∫a f (x)dx = A1 − A2.
The total area between f (x) and the x-axis is given by
b
∫ a| f (x)|dx = A1 + A2.

Example 1.10

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Chapter 1 | Integration

35

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-axis

over

⎡⎣0,

6⎤ ⎦

is A2 plus A1.

We have

6

∫ 0

|(x

−

2)|dx

=

A2

+

A 1.

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

The total area, then, is

A2

=

12bh

=

1 2

·

2·2

=

2

A1

=

12bh

=

1 2

·4·

4

=

8.

A1 + A2 = 8 + 2 = 10.

1.10 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 to the limits of integration. These properties, along with the rules of integration that we examine later in this chapter, help us manipulate expressions to evaluate definite integrals.

Rule: Properties of the Definite Integral

1.

a
∫ f (x)dx = 0

(1.9)

a

36

Chapter 1 | Integration

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

a

b

∫ b

f (x)dx

=

−∫a

f (x)dx

(1.10)

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

b

b

b

∫

∫ ⎡
⎣

f

(x)

+

g(x)⎤⎦dx

=

f (x)dx + ∫

g(x)dx

a

a

a

(1.11)

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

⌠ ∫ b

b

⎡ ⎣

f

(x)

−

g(x)⎤⎦dx

=

⌠

f (x)dx −

⌡a

⌡a

b
g(x)dx
a

(1.12)

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

b

b

∫a c f (x)dx = c∫a f (x)

(1.13)

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

6.

b

c

b

∫ f (x)dx = ∫ f (x)dx + ∫ f (x)dx

a

a

c

(1.14)

Although this formula normally applies when c is between a and b, the formula holds for all values of a, b, and c, 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 ∫ 1 ⎛⎝−3x3 + 2x + 2⎞⎠dx. We apply properties 3. and 5. to get −2

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Chapter 1 | Integration

37

∫ 1 ⎛⎝−3x3 + 2x + 2⎞⎠dx = ∫ 1 −3x3 dx + ∫ 1 2xdx + ∫ 1 2dx

−2

−2

−2

−2

= −3∫ 1 x3 dx + 2∫ 1 xdx + ∫ 1 2dx.

−2

−2

−2

1.11 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

8

5

8

If it is known that ∫ f (x)dx = 10 and ∫ f (x)dx = 5, find the value of ∫ f (x)dx.

0

0

5

Solution By property 6.,
Thus,

b

c

b

∫ f (x)dx = ∫ f (x)dx + ∫ f (x)dx.

a

a

c

8

5

8

∫ f (x)dx = ∫ f (x)dx + ∫ f (x)dx

0

0

5

8
10 = 5 + ∫ f (x)dx 5

8
5 = ∫ f (x)dx. 5

1.12

5

5

If it is known that ∫ f (x)dx = −3 and ∫ f (x)dx = 4,

find the value of

∫

2
f (x)dx.

1

2

1

Comparison Properties of Integrals
A picture can sometimes tell us more about a function than the results of computations. Comparing functions by their graphs as well as by their algebraic expressions can often give new insight into the process of integration. Intuitively, we might say that 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.

38

Chapter 1 | Integration

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

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

b
∫ f (x)dx ≥ 0. a

b

b

∫a f (x)dx ≥ ∫a g(x)dx.

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

b
m(b − a) ≤ ∫ f (x)dx a ≤ 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

∫

1
g(x)dx

≥

∫

1
f (x)dx

(Figure

1.24). The thin, red-shaded area shows

just

0

0

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

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Chapter 1 | Integration

39

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 more clearly.

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 following test scores in your algebra class: 89, 90, 56, 78, 100, and 69. Your semester grade is your average of test scores and you want 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 6

+

100

+

69

=

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 x*i

in each subinterval and calculate

f ⎛⎝x*i

⎞ ⎠

for

i = 1, 2,…, n.

In other words, consider each

f ⎛⎝x*i

⎞ ⎠

as a sampling

of the

function

over

each

subinterval.

The

average

value

of the function

may then

be

approximated as

f ⎛⎝x*1

⎞ ⎠

+

f

⎛⎝x*2

⎞ ⎠
n

+

⋯

+

f ⎛⎝x*n

⎞
⎠,

which is basically the same expression used to calculate the average of discrete values.

But we know Δx = b −n a,

so

n

=

b−a Δx

,

and we get

f ⎛⎝x*1

⎞ ⎠

+

f

⎛⎝x*2

⎞ ⎠
n

+

⋯

+

f ⎛⎝x*n

⎞
⎠=

f ⎛⎝x*1

⎞ ⎠

+

f ⎛⎝x*2

⎞ ⎠

+

⋯

+

(b − a)

f ⎛⎝x*n

⎞
⎠.

Δx

n
Following through with the algebra, the numerator is a sum that is represented as ∑ f ⎛⎝x*i ⎞⎠, and we are dividing by a i=1
fraction. To divide by a fraction, invert the denominator and multiply. Thus, an approximate value for the average value of the function is given by

40

Chapter 1 | Integration

n

∑ f ⎛⎝x*i

⎞ ⎠

i=1

(b − a)

Δx

n

∑ =

⎛⎝bΔ−xa⎞⎠i = 1

f ⎛⎝x*i

⎞ ⎠

n

∑ =

⎛
⎝b

1 −

⎞
a ⎠i

=

1

f ⎛⎝x*i

⎞⎠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 of a function is given by

b

1 −

n
∑ an l→im∞i = 1

f (xi)Δx

=

b

1 −

b
a∫a 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

f ave

=

b

1 −

b
a∫a

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 = 12h(a + b), where h represents height, and a and b represent the two parallel sides. Then,
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Chapter 1 | Integration

41

5
∫ x + 1dx 0

= 12h(a + b)

=

1 2

·

5 · (1

+

6)

= 325.

Thus the average value of the function is

5

1 −

5
0∫0 x

+

1dx

=

1 5

·

35 2

=

7 2

.

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

42

1.2 EXERCISES

In the following exercises, express the limits as integrals.

71.

n
∑ 60. n l→im∞i = 1 ⎛⎝x*i ⎞⎠Δx over [1, 3]

n

∑ 61.

n l→im∞i = 1 ⎛⎝5⎛⎝x*i

⎞2
⎠

−

3⎛⎝x*i

⎞⎠3⎞⎠Δx

over

[0,

2]

n
∑ 62. n l→im∞i = 1 sin2 ⎛⎝2πx*i ⎞⎠Δx over [0, 1]

n
∑ 63. n l→im∞i = 1 cos2 ⎛⎝2πx*i ⎞⎠Δx over [0, 1]

72.

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

n
64. Ln = 1n ∑ i −n 1 i=1

n

65.

Rn

=

1n ∑

i n

i=1

n

73.

66.

Ln

=

2n

∑

⎛⎝1 + 2i

−n

1

⎞ ⎠

i=1

n

67.

Rn

=

3n

∑

⎛⎝3

+

3

i n

⎞ ⎠

i=1

n

68.

Ln

=

2nπ

∑

2π i

−n

1

cos⎛⎝2π

i

−n

1

⎞ ⎠

i=1

69.

Rn

=

n
1n ∑ i=1

⎛⎝1

+

ni ⎞⎠log⎛⎝⎛⎝1

+

i ⎞2⎞ n⎠ ⎠

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

70.

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Chapter 1 | Integration

Chapter 1 | Integration

43

74.

75.

In the following exercises, evaluate the integral using area formulas.

3
76. ∫ (3 − x)dx 0

3
77. ∫ (3 − x)dx 2

78.

3
∫ (3 − |x|)dx

−3

6
79. ∫ (3 − |x − 3|)dx 0

80.

2
∫

4 − x2dx

−2

81. ∫ 5 4 − (x − 3)2dx 1

82.

12
∫

36 − (x − 6)2dx

0

3
83. ∫ (3 − |x|)dx −2
In the following exercises, use averages of values at the left (L) and right (R) endpoints to compute the integrals of the piecewise linear functions with graphs that pass through the given 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]

4

2

Suppose that ∫ f (x)dx = 5 and ∫ f (x)dx = −3, and

0

0

4
∫ g(x)dx = −1

and

2
∫ g(x)dx = 2.

In the following

0

0

exercises, compute the integrals.

4

∫ 88.

⎛ ⎝

f

(x)

+

g(x)⎞⎠d

x

0

4

∫ 89.

⎛ ⎝

f

(x)

+

g(x)⎞⎠d

x

2

2

∫ 90.

⎛ ⎝

f

(x)

−

g(x)⎞⎠d

x

0

4

∫ 91.

⎛ ⎝

f

(x)

−

g(x)⎞⎠d

x

2

2

∫ 92.

3⎛
⎝

f

(x)

−

4g(x)⎞⎠dx

0

4

∫ 93.

4⎛
⎝

f

(x)

−

3g(x)⎞⎠dx

2

In the following exercises, use the identity

A

0

A

∫ f (x)dx = ∫ f (x)dx + ∫ f (x)dx to compute the

−A

−A

0

integrals.

π

94.

⌠ ⌡−π

sin t 1 + t2

dt

(Hint: sin(−t) = −sin(t))

44

Chapter 1 | Integration

95.

∫

−

π π

1

+

t cos

t

dt

3
96. ∫ (2 − x)dx (Hint: Look at the graph of f.) 1

97.

∫

4
(x

−

3)

3

dx

(Hint: Look at the graph of f.)

2

In the following exercises,

1
∫ xdx 0

=

1 2

,

∫ 1x2 dx 0

=

13,

and

compute the integrals.

given that

∫

1
x

3

0

dx

=

14,

∫ 98.

1⎛⎝1 + x + x2 + x3⎞⎠dx

0

∫ 99.

1⎛⎝1 − x + x2 − x3⎞⎠dx

0

100.

∫

1
(1

−

x)2

dx

0

101.

∫

1
(1

−

2x)3

dx

0

102.

⌠⌡01⎛⎝6x

−

4 3

x 2⎞⎠d x

∫ 103.

1⎛⎝7 − 5x3⎞⎠dx

0

In the following exercises, use the comparison theorem.

104. Show that ∫ 3⎛⎝x2 − 6x + 9⎞⎠dx ≥ 0. 0

3
105. Show that ∫ (x − 3)(x + 2)dx ≤ 0. −2

106.

Show that

1
∫

1 + x3dx ≤ ∫ 1

1 + x2dx.

0

0

107. Show that ∫ 2 1 + xdx ≤ ∫ 2 1 + x2dx.

1

1

108.

Show that

π/2
∫ sintdt 0

≥

π4 .

(Hint: sint ≥ 2πt over

⎡⎣0,

π 2

⎤⎦)

π/4
109. Show that ∫ costdt ≥ π 2/4. −π/4
In the following exercises, find the average value fave of f between 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) = sin x, a = 0, b = 2π

115. f (x) = cos x, a = 0, b = 2π

In the following exercises, approximate the average value using Riemann sums L100 and R100. How does your answer compare 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 = tan x

over the interval

⎡⎣0,

π 4

⎤⎦;

the

exact

solution is 2 lnπ(2).

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 using the 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 − 83.

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

solution is

1 4

⎛⎝e

4

−

1⎞⎠.

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Chapter 1 | Integration

45

122.

[T]

y

=

⎛⎝12

⎞x ⎠

over the interval

[0, 4];

the exact

solution is 641ln5(2).

123.

[T]

y

=

x

sin⎛⎝x

2⎞ ⎠

over the interval

[−π, 0];

the

exact solution is

cos⎛⎝π

2⎞ ⎠

2π

−

1

.

124.

Suppose that

A

=

∫

2π
sin

2

tdt

and

0

B

=

∫

2π
cos

2

tdt.

Show that

A + B = 2π

and

A = B.

0

125. Suppose that A = ∫ π/4 sec2 tdt = π and −π/4

B = ∫ π/4 tan2 tdt. −π/4

Show that

A−B=

π2 .

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

b

b

p(b) = ℓ(b), and that ∫ p(t)dt > ∫ ℓ(t)dt. Explain

a

a

d

d

why ∫c p(t) > ∫c 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 ∫ B⎛⎝ax2 + bx + c⎞⎠dx as large as A

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

=

∫ ai

−

1

f

(t)dt.

b
a. Explain why ∫a f (t)dt = A1 + A2 + ⋯ + AN.

| |b

b

b. Then, explain why ∫a f (t)dt ≤ ∫a | f (t)|dt.

131. Suppose f and g are continuous functions such that

d

d

∫

f (t)dt ≤ ∫

g(t)dt

for every subinterval

⎡⎣c,

d⎤ ⎦

of

c

c

⎡⎣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
∫ f (t)dt i−1

=

i.

Show that

N
∫ f (t)dt 0

=

N(N + 2

1).

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

has

i
∫

f (t)dt = i2.

Show

that

i−1

N
∫ 0

f (t)dt

=

N(N

+

1)(2N 6

+

1).

136. [T] Compute the left and right Riemann sums L10

and R10 and their average

L10 + R10 2

for

f (t) = t2

over

[0, 1].

Given that

∫

1
t

2

dt

=

0.3–3,

to how many

0

decimal places is L10 + R10 accurate? 2

46

Chapter 1 | Integration

137. [T] Compute the left and right Riemann sums, L10

and R10, and their average

L10 + R10 2

for

f

(t)

=

⎛⎝4

−

t

2⎞ ⎠

over [1, 2]. Given that ∫ 2⎛⎝4 − t2⎞⎠dt = 1.6–6, to how 1

many decimal places is L10 + R10 accurate? 2

138.

If ∫ 5 1 + t4dt = 41.7133...,

1

∫ 5 1 + u4du ? 1

what is

1
139. Estimate ∫ tdt using the left and right endpoint 0
sums, each with a single rectangle. How does the average of these left and right endpoint sums compare with the
1
actual value ∫ tdt ? 0

140.

Estimate

1
∫ tdt

by comparison with the area of a

0

single rectangle with height equal to the value of t at the

midpoint

t

=

1 2

.

How does this midpoint estimate compare

1
with the actual value ∫ tdt ? 0

141. From the graph of sin(2πx) shown:

a.

Explain why

∫

1
sin(2πt)dt

=

0.

0

a+1
b. Explain why, in general, ∫ sin(2πt)dt = 0 for a any value of a.

142. If f is 1-periodic integrable over [0, 1],
1
∫ f (t)dt = 0 ? 0

⎛ ⎝

f

(t

+

1)

=

f (t)⎞⎠,

is it always

odd, and true that

1
143. If f is 1-periodic and ∫ f (t)dt = A, is it 0

1+a

necessarily true that ∫

f (t)dt = A for all A?

a

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Chapter 1 | Integration

47

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 area formulas and limits of Riemann sums, and both approaches are extremely cumbersome. In this section we look at some more powerful and useful techniques for evaluating definite integrals.
These new techniques rely on the relationship between differentiation and integration. This relationship was discovered and explored 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 this section. 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 relationships he discovered, codified as Newton’s laws and the law of universal gravitation, are still taught as foundational material in physics today, and his calculus has spawned entire fields of mathematics. To learn more, read a brief biography (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 for Integrals, 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 at

the 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

This formula can also be stated as

f

(c)

=

b

1 −

b
a∫a

f

(x)dx.

b
∫a f (x)dx = f (c)(b − a).

(1.15)

Proof
Since f (x) is continuous on ⎡⎣a, b⎤⎦, by the extreme value theorem (see Maxima 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

48

Chapter 1 | Integration

Dividing by b − a gives us

b
m(b − a) ≤ ∫a f (x)dx ≤ M(b − a).

m

≤

b

1 −

b
a∫a f

(x)dx

≤

M.

Since

b

1 −

a

∫

b a

f

(x)d

x

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

and the proof is complete. □

f

(c)

=

b

1 −

b
a∫a

f

(x)dx,

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

4

1 −

0∫

4
(8
0

−

2x)d

x.

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 = 12(4)(8) = 16.

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

14(16) = 4. Set the average value equal to f (c) and solve for c.

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

8 − 2c = 4 c=2

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Chapter 1 | Integration

49

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

1.14

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

Solution We are looking for the value of c such that

Replacing f (c) with c2, we have

f

(c)

=

3

1 −

0

∫

3
x
0

2

dx

=

1 3

(9)

=

3.

c2 = 3 c = ± 3.
Since − 3 is outside the interval, take only the positive value. Thus, c = 3 (Figure 1.27).

50

Chapter 1 | Integration

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

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

Fundamental Theorem of Calculus Part 1: Integrals and Antiderivatives
As mentioned earlier, the Fundamental Theorem of Calculus is an extremely powerful theorem that establishes the relationship between differentiation and integration, and gives us a way to evaluate definite integrals without using Riemann sums or calculating areas. The theorem is comprised of two parts, the first of which, the Fundamental Theorem of Calculus, 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
x
F(x) = ∫ f (t)dt, a
then F′ (x) = f (x) over ⎡⎣a, b⎤⎦.

(1.16)

Before we delve into the proof, a couple of subtleties are worth mentioning here. First, a comment on the notation. Note that we 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 like it’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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Chapter 1 | Integration

51

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

F′ (x)

=

lim
h→0

F(x

+

h) h

−

F(x)

=

hli→m0

1 h

⎡x
⎣⎢∫a

+

h

f

(t)dt

−

x
∫a

f

⎤
(t)dt⎥
⎦

=

lim
h→0

1 h

⎡x
⎣⎢∫a

+

h

f

(t)dt

+

a
∫ x

f

⎤
(t)dt⎥
⎦

=

hli→m0

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)d

x

=

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 f (c)
h→0

=

cli→mx f (c)

=

f (x).

Putting all these pieces together, we have

and the proof is complete. □

F′ (x)

=

hli→m0

1 h

∫

x x

+

h

f

(x)dx

= lim f (c)
h→0

= f (x),

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

x

g(x)

=

⌠ ⌡1 t 3

1 +

dt. 1

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

g′ (x)

=

1 x3 +

. 1

52

Chapter 1 | Integration

1.16

Use the Fundamental Theorem of Calculus, Part 1 to find the derivative of

r
g(r) = ∫

x2 + 4dx.

0

Example 1.18
Using the Fundamental Theorem and the Chain Rule to Calculate Derivatives

x
Let F(x) = ∫ sintdt. Find F′ (x). 1

Solution
u(x)
Letting u(x) = x, we have F(x) = ∫ sin tdt. Thus, by the Fundamental Theorem of Calculus and the chain 1
rule,

F′ (x)

=

sin⎛⎝u(x)⎞⎠

du dx

= sin(u(x)) · ⎛⎝12x−1/2⎞⎠

=

sin 2

x x

.

1.17

x3

Let F(x) = ∫ costdt. Find F′ (x).

1

Example 1.19
Using the Fundamental Theorem of Calculus with Two Variable Limits of Integration

Let

F(x)

=

∫

2x
t

3

dt.

Find

F′ (x).

x

Solution

We have

F(x)

=

∫

2x
t
x

3

dt.

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

get

F(x)

=

∫ 2x
t

3

x

dt

=

∫

0
t

3

dt

+

∫

2x
t

3

dt

x

0

=

−∫

x
t

3

dt

+

∫

2x
t

3

dt.

0

0

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Chapter 1 | Integration

53

Differentiating the first term, we obtain

d dx

⎡⎣−∫

x
t
0

3

dt⎤⎦

=

−x

3.

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

Thus,

∫ ∫ d

⎡ ⎢

dx⎣

2x
t

3

⎤
dt⎥

0

⎦

=

d dx

⎡ ⎢ ⎣

u(x) ⎤

t3 dt⎥

0

⎦

=

(u(x))3

du dx

= (2x)3 · 2

= 16x3.

F′ (x)

=

d dx

⎡⎣−∫

x
t
0

3

dt⎤⎦

+

∫ d
dx

⎡ ⎢ ⎣

2x
t

3

⎤
dt⎥

0

⎦

= −x3 + 16x3

= 15x3.

1.18

x2

Let F(x) = ∫x costdt. 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 efforts by mathematicians for approximately 500 years, new techniques emerged that provided scientists with the necessary tools to explain many phenomena. Using calculus, astronomers could finally determine distances in space and map planetary orbits. Everyday financial problems such as calculating marginal costs or predicting total profit could now be handled with simplicity and accuracy. Engineers could calculate the bending strength of materials or the three-dimensional motion of objects. 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 by comparison. It almost seems too simple that the area of an entire curved region can be calculated by just evaluating an antiderivative 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

b
∫a f (x)dx = F(b) − F(a).

(1.17)

We often see the notation

F(x)|

b a

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

54

Chapter 1 | Integration

antiderivative for the integrand, then we can evaluate the definite integral by evaluating the antiderivative at the endpoints of 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)⎤⎦

n

∑ =

⎡⎣F(xi) − F(xi − 1)⎤⎦.

i=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
n
F(b) − F(a) = ∑ f (ci)Δx. i=1
Taking the limit of both sides as n → ∞, we obtain

□
Example 1.20

n
F(b) − F(a) = ∑ n l→im∞i = 1 f (ci)Δx
b
= ∫ f (x)dx. a

Evaluating an Integral with the Fundamental Theorem of Calculus

Use The Fundamental Theorem of Calculus, Part 2 to evaluate

∫ 2 ⎛⎝t2 − 4⎞⎠dt. −2

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

If

y

=

x n,

∫

xn dx

=

xn + 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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Chapter 1 | Integration

55

∫ 2 ⎛⎝t2 − 4⎞⎠dt −2

=

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

= − 332.

Analysis

Notice that we did not include the “+ C” term when we wrote the antiderivative. The reason is that, according to the Fundamental Theorem of Calculus, Part 2, any antiderivative works. So, for convenience, we chose the antiderivative 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 curve and the x-axis is all below the x-axis. Area is always positive, but a definite integral can still produce a negative number (a net signed area). For example, if this were a profit function, a negative number indicates the company is operating at a loss over the given interval.

Figure 1.28 The evaluation of a definite integral can produce a 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:

∫

9 1

x

− x

1

d

x.

Solution

56

Chapter 1 | Integration

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

9

⌠ ⌡1

x−1 x 1/2

d

x

=

⌠9⎛ x ⌡1⎝x 1/2

−

x11/2⎞⎠dx.

Use the properties of exponents to simplify:

∫ ⌠9⎛ x
⌡1⎝x 1/2

−

x11/2⎞⎠dx

=

9⎛⎝x1/2 − x−1/2⎞⎠dx.
1

Now, integrate using the power rule:

See Figure 1.29.

| ∫ 9⎛⎝x1/2 − x−1/2⎞⎠dx 1

=

⎛⎜x ⎝

3/2 3 2

−

x 112/2 ⎞⎠⎟

9 1

= ⎡⎣⎢(9)323/2 − (9)121/2⎤⎦⎥ − ⎡⎣⎢(1)323/2 − (1)121/2⎤⎦⎥

= ⎡⎣23(27) − 2(3)⎤⎦ − ⎡⎣23(1) − 2(1)⎤⎦

=

18

−

6

−

2 3

+

2

= 430.

Figure 1.29 The area under the curve from x = 1 to x = 9 can be calculated by evaluating a definite integral.
1.19 Use The Fundamental Theorem of Calculus, Part 2 to evaluate ∫ 2x−4 dx. 1
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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Chapter 1 | Integration

57

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⎛⎝π2t⎞⎠ 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

5
∫ (5 + 2t)dt. 0

Using the power rule, we have

| ∫ 5 (5 + 2t)dt 0

=

⎛⎝5t

+

t

2⎞ ⎠

5 0

= (25 + 25) = 50.

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

5
∫ 10 0

+

cos⎛⎝π2

t⎞⎠dt.

We know sin t is an antiderivative of cos t, so it is reasonable to expect that an antiderivative of cos⎛⎝π2t⎞⎠ would

involve sin⎛⎝π2t⎞⎠. However, when we differentiate sin⎛⎝π2t⎞⎠,

we get

π 2

cos⎛⎝π2

t⎞⎠

as a result of the chain rule, so we

have to account for this additional coefficient when we integrate. We obtain

| 5
∫ 10 0

+

cos⎛⎝π2

t⎞⎠dt

=

⎛⎝10t

+

π2

sin⎛⎝π2 t⎞⎠⎞⎠

5 0

=

⎛⎝50

+

π2

⎞ ⎠

−

⎛⎝0

−

π2sin 0⎞⎠

≈ 50.6.

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

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

58
A Parachutist in Free Fall

Chapter 1 | Integration

Figure 1.30 Skydivers can adjust the velocity of their dive by changing the position of their body during the free 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 adjustments to her body position in the air to control how fast she falls. If she arches her back and points her belly toward the ground, she reaches a terminal velocity of approximately 120 mph (176 ft/sec). If, instead, she orients her body with her 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 simplify our calculations. Julie executes her jumps from an altitude of 12,500 ft. After she exits the aircraft, she immediately starts 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 ripcord and 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 the
distance 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 the ground.
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Chapter 1 | Integration

59

On Julie’s second jump of the day, she decides she wants to fall a little faster and orients herself in the “head down” position. Her terminal velocity in this position is 220 ft/sec. Answer these questions based on this velocity:
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 as fast when her parachute opens. If she begins this maneuver at an altitude of 4000 ft, how long does she spend in a free fall before beginning the reorientation? Some jumpers wear “ wingsuits” (see Figure 1.31). These suits have fabric panels between the arms and legs and allow the wearer to glide around in a free fall, much like a flying squirrel. (Indeed, the suits are sometimes called “flying squirrel suits.”) When wearing these suits, terminal velocity can be reduced to about 30 mph (44 ft/sec), allowing the wearers a much longer time in the air. Wingsuit flyers still use parachutes to land; although the vertical velocities are within the margin of safety, horizontal velocities can exceed 70 mph, much too fast to land safely.

Figure 1.31 The fabric panels on the arms and legs of a wingsuit work to reduce the vertical velocity of a skydiver’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, how long does she get to spend gliding around in the air?

60

Chapter 1 | Integration

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 must be going the same speed at some point.
145. Two mountain climbers start their climb at base camp, taking two different routes, one steeper than the other, and arrive at the peak at exactly the same time. Is it necessarily true that, at some point, both climbers increased in altitude at the same rate?
146. To get on a certain toll road a driver has to take a card that lists the mile entrance point. The card also has a timestamp. When going to pay the toll at the exit, the driver is surprised to receive a speeding ticket along with the toll. Explain how this can happen.
x
147. Set F(x) = ∫ (1 − t)dt. Find F′ (2) and the 1
average value of F ′ over [1, 2].

In the following exercises, use the Fundamental Theorem of Calculus, Part 1, to find each derivative.

148.

d dx

∫

x
e
1

−t

2

dt

149.

d dx

∫

x
e
1

cos

t

dt

150.

d dx

∫

x 3

9 − y2dy

x
151. d ⌠ ds dx⌡4 16 − s2

152.

d dx

∫

2x
tdt
x

153.

d dx

∫

0

x
tdt

154.

∫ d sin x
dx 0

1 − t2dt

155.

∫ d 1
dx cos x

1 − t2dt

156.

d dx

x
⌠ ⌡1 1

t2 +t

4

dt

157.

d

x2 ⌠

d x ⌡1

1

t +

t

dt

158.

d dx

ln
∫ 0

x
e

t

dt

159.

d dx

e2
∫ lnu 1

2

du

x
160. The graph of y = ∫ f (t)dt, where f is a piecewise 0
constant function, is shown here.

a. Over which intervals is f positive? Over which intervals is it negative? Over which intervals, if any, is it equal to zero?
b. What are the maximum and minimum values of f? c. What is the average value of f?
x
161. The graph of y = ∫ f (t)dt, where f is a piecewise 0
constant function, is shown here.

a. Over which intervals is f positive? Over which intervals is it negative? Over which intervals, if any, 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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Chapter 1 | Integration

61

x
162. The graph of y = ∫ ℓ(t)dt, where ℓ is a piecewise 0
linear function, is shown here.

a. Over which intervals is ℓ positive? Over which intervals is it negative? Over which, if any, is it zero?
b. Over which intervals is ℓ increasing? Over which is it decreasing? Over which, if any, is it constant?
c. What is the average value of ℓ?
x
163. The graph of y = ∫ ℓ(t)dt, where ℓ is a piecewise 0
linear function, is shown here.

a. Over which intervals is ℓ positive? Over which intervals is it negative? Over which, if any, is it zero?
b. Over which intervals is ℓ increasing? Over which is it decreasing? Over which intervals, if any, is it constant?
c. What is the average value of ℓ?
In the following exercises, use a calculator to estimate the area under the curve by computing T10, the average of the left- and right-endpoint Riemann sums using N = 10 rectangles. Then, using the Fundamental Theorem of Calculus, 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] ∫ (cos x − sin x)dx over [0, π]

169.

[T]

⌠4 ⌡x 2

d

x

over

[1,

4]

In the following exercises, evaluate each definite integral using the Fundamental Theorem of Calculus, Part 2.

∫ 170.

2 ⎛⎝x2 − 3x⎞⎠dx

−1

∫ 171.

3 ⎛⎝x2 + 3x − 5⎞⎠dx

−2

3
172. ∫ (t + 2)(t − 3)dt −2

∫ 173.

3⎛⎝t2 − 9⎞⎠⎛⎝4 − t2⎞⎠dt

2

174.

∫

2
x

9

dx

1

175.

∫

1
x

99

dx

0

∫ 176.

8⎛⎝4t5/2 − 3t3/2⎞⎠dt

4

177.

⌠⌡14/4⎛⎝x 2

−

1 x2

⎞⎠d

x

2

178.

⌠ ⌡1

2 x3

dx

179.

4 ⌠ ⌡1

21xdx

4

180.

⌠ 2− ⌡1 t2

tdt

16
181. ⌠ dt ⌡1 t1/4

2π
182. ∫ cosθdθ 0

π/2
183. ∫ sinθdθ 0

62

Chapter 1 | Integration

184. ∫ π/4sec2 θdθ 0

π/4
185. ∫ secθ tanθ 0

π/4
186. ∫ cscθ cot θdθ π/3

187. ∫ π/2csc2 θdθ π/4

188.

⌠⌡12⎛⎝t12

−

1 t3

⎞⎠dt

189.

⌠⌡−−21⎛⎝t12

−

1 t3

⎞⎠dt

In the following exercises, use the evaluation theorem to express the integral as a function F(x).

190.

∫

x
t
a

2

dt

191. ∫ xet dt 1

x
192. ∫ costdt 0

x

193.

∫

sin
−x

tdt

In the following exercises, identify the roots of the integrand to remove absolute values, then evaluate using the Fundamental Theorem of Calculus, Part 2.

194.

3
∫ |x|dx

−2

| | 195. ∫ 4 t2 − 2t − 3 dt −2 π
196. ∫ |cost|dt 0

π/2
197. ∫ |sint|dt −π/2

198. Suppose that the number of hours of daylight on

a given day in Seattle is modeled by the function

−3.75

cos⎛⎝π6t

⎞ ⎠

+

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 a

year?

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 of

daylight hours in Seattle between t1 and t2.

d. Compute the mean hours of daylight in Seattle

between 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 day

length.

199. Suppose the rate of gasoline consumption in the

United States can be modeled by a sinusoidal function of

the form

⎛⎝11.21

−

cos⎛⎝π6t

⎞⎞ ⎠⎠

×

10

9

gal/mo.

a. What is the average monthly consumption, and for

which values of t is the rate at time t equal to the

average rate?

b. What is the number of gallons of gasoline

consumed in the United States in a year?

c. Write an integral that expresses the average

monthly U.S. gas consumption during the part of

the 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)

=

b

1 −

b
a∫a 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)

=

b

1 −

b
a∫a f

(t)dt

and at least one point

∫ m

∈

⎡⎣a,

b⎤ ⎦

such that

f (m)

<

b

1 −

a

b
f (t)dt.
a

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63

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

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

203. A point on an ellipse with major axis length 2a and minor axis length 2b has the coordinates
(acosθ, bsinθ), 0 ≤ θ ≤ 2π. a. Show that the distance from this point to the focus at (−c, 0) is d(θ) = a + c cosθ, 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. The perihelion for Earth’s orbit around the Sun is 147,098,290 km and the aphelion is 152,098,232 km.
a. By placing the major axis along the x-axis, find the average 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 value was computed as the average of the perihelion and aphelion 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 its orbit. Assuming that M, m, and the ellipse parameters a

and b (half-lengths of the major and minor axes) are given, set up—but do not evaluate—an integral that expresses in

terms of G, m, M, a, b the average gravitational force

between the Sun and the planet.

64

Chapter 1 | Integration

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 is important to note that these formulas are presented in terms of indefinite integrals. Although definite and indefinite integrals are closely related, there are some key differences to keep in mind. A definite integral is either a number (when the limits of integration are constants) or a single function (when one or both of the limits of integration are variables). An indefinite integral 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 correct approach 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 proper notation based on your choice.
Basic Integration Formulas
Recall the integration formulas given in the table in Antiderivatives (http://cnx.org/content/m53621/latest/#fsid1165043092431) and the rule on properties of definite integrals. Let’s look at a few examples of how to apply these rules.
Example 1.23
Integrating a Function Using the Power Rule

4
Use the power rule to integrate the function ∫ t(1 + t)dt. 1

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

4
∫ t(1 + t)dt

=

∫

4
t

1/2(1

+

t)dt

1

1

∫ = 4⎛⎝t1/2 + t3/2⎞⎠dt. 1

Now apply the power rule:

∫ 4⎛⎝t1/2 + t3/2⎞⎠dt 1

| =

⎛⎝23 t 3/2

+

2 5

t

5/2⎞ ⎠

4 1

=

⎡⎣23 (4) 3/2

+

25(4)

5/2⎤ ⎦

−

⎡⎣23 (1) 3/2

+

2 5

(1)

5/2⎤ ⎦

= 21556.

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

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Chapter 1 | Integration

65

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 value equals 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:
b
F(b) = F(a) + ∫a F '(x)dx
or
b
∫a F '(x)dx = F(b) − F(a).

(1.18)

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, to name only a few applications. Net change accounts for negative quantities automatically without having to write more than one 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 this motion as shown in Figure 1.32.

Figure 1.32 The graph shows speed versus time for the given motion 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

5
∫ v(t)dt 2

=

4
∫ 40dt 2

+

5
⌠ −30dt ⌡4

= 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

66

Chapter 1 | Integration

5
∫ |v(t)|dt 2

=

4
∫ 40dt 2

+

5
⌠ 30dt ⌡4

= 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 function accounts for both forward distance and backward distance. To find net displacement, integrate the velocity function over the 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

| 3
∫ (3t − 5)dt 0

=

3t 2 2

−

5t

3 0

=

⎡3(3) ⎣2

2

−

5(3)⎤⎦

−

0

=

27 2

−

15

=

27 2

−

30 2

= − 32.

The net displacement is

−

3 2

m (Figure 1.33).

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Chapter 1 | Integration

67

Figure 1.33 The graph shows velocity versus time for a particle 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 the absolute 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,

5 3

⎤⎦,

we have |v(t)| = −v(t) over that interval.

Over ⎡⎣53, 3⎤⎦, the function is positive, so |v(t)| = v(t). Thus, we have

68

Chapter 1 | Integration

3
∫ |v(t)|dt 0

=

5/3
⌠ −v(t)dt ⌡0

+

3
∫ v(t)dt 5/3

5/3

3

= ∫ 5 − 3tdt + ∫ 3t − 5dt

0

5/3

| | =

⎛⎝5t

−

3t 2 2

⎞ ⎠

5/3 0

+

⎛⎝32t 2

−

5t⎞⎠

3 5/3

=

⎡⎣5⎛⎝53

⎞ ⎠

−

3(5/3) 2

2

⎤ ⎦

−

0

+

⎡⎣227

−

15⎤⎦

−

⎡3(5/3) ⎣2

2

−

25 3

⎤ ⎦

=

25 3

−

25 6

+

27 2

−

15

−

25 6

+

25 3

= 461.

So, the total distance traveled is 14 m. 6

1.22 Find the net displacement and total distance traveled in meters given the velocity function f (t) = 12et − 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

| ∫ 2⎛⎝5 − t3⎞⎠dt 0

=

⎛⎝5t

−

t4⎞ 4⎠

2 0

=

⎡⎣5(2)

−

(2) 4

4⎤ ⎦

−

0

=

10

−

16 4

= 6.

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Chapter 1 | Integration

69

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 the wind 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. As Andrew gets his iceboat set up, though, the wind begins to pick up. During his first half hour of iceboating, the wind 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 ⎨ ⎩15

for for

0

≤

t

≤

1 2

1 2

≤

t

≤

1.

Recalling that Andrew’s iceboat travels at twice the wind speed, and assuming he moves in a straight line away from 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

1
= ∫ 2v(t)dt.

0

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

70

Chapter 1 | Integration

∫

1
2v(t)dt

0

=

1/2
⌠ 2v(t)dt ⌡0

+

∫ 1 2v(t)dt 1/2

=

1/2
⌠ 2(20t ⌡0

+

5)dt

+

1
∫ 2(15)dt 1/3

=

1/2
⌠ (40t ⌡0

+

10)dt

+

1
∫ 30dt 1/2

=

⎡⎣20t 2

+

10t⎤⎦|

1/2 0

+

[30t]|11/2

= ⎛⎝240 + 5⎞⎠ − 0 + (30 − 15)

= 25.

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

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

⎧20t + 5 v(t) = ⎨
⎩−10t + 15

for for

0

≤

t

≤

1 2

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 in which 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 are symmetric 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

=

a
2∫ 0

f

(x)dx.

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

a
∫−a f (x)dx = 0.

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Chapter 1 | Integration

71

Example 1.28
Integrating an Even Function

Integrate the even function ∫ 2 ⎛⎝3x8 − 2⎞⎠dx and verify that the integration formula for even functions holds. −2

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

| ∫ 2 ⎛⎝3x8 − 2⎞⎠dx −2

=

⎛x9 ⎝3

−

2x⎞⎠

2 −2

=

⎡ ⎢ ⎣

(2)9 3

−

⎤
2(2)⎥
⎦

−

⎡ ⎢ ⎣

(−2)9 3

−

⎤
2(−2)⎥
⎦

=

⎛⎝5132

−

4⎞⎠

−

⎛⎝−

512 3

+

4⎞⎠

= 10300.

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

| ∫ 2⎛⎝3x8 − 2⎞⎠dx 0

=

⎛x9 ⎝3

−

2x⎞⎠

2 0

=

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 area between the curve and the x-axis. Both views show the symmetry about the y-axis.

72

Chapter 1 | Integration

Example 1.29
Integrating an Odd Function

Evaluate the definite integral of the odd function −5 sin x 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 the x-axis over [−π, 0], and the negative area below the x-axis over [0, π]. We have

π

∫

−5
−π

sin

xdx

= −5(−cos x)|−π π

= 5 cos x|−π π

=

[5cosπ]

−

5⎡
⎣

cos(−π

)⎤ ⎦

= −5 − (−5)

= 0.

Figure 1.36 The graph shows areas between a curve and the x-axis for an odd function.
1.24 Integrate the function ∫ 2 x4 dx. −2
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Chapter 1 | Integration

73

1.4 EXERCISES

Use basic integration formulas to compute the following antiderivatives.

207.

∫

⎛ ⎝

x−

1 x

⎞⎠dx

208.

⌠⌡⎛⎝e 2x

−

1 2

e

x/2⎞⎠dx

209. ⌠dx ⌡2x

210.

⌠x ⌡

− x2

1

d

x

π
211. ∫ (sin x − cos x)dx 0

π/2
212. ∫ (x − sin x)dx 0

213. Write an integral that expresses the increase in the perimeter 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 that have equal length s, such as a pentagon or hexagon) has perimeter Ns. Write an integral that expresses the increase in perimeter of a regular N-gon when the length of each side increases 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 outer

sides of length 920 ft. Write an integral to express the area

of the roof of the Pentagon according to these dimensions

and evaluate this area.

217. A dodecahedron is a Platonic solid with a surface that consists of 12 pentagons, each of equal area. By how much does the surface area of a dodecahedron increase as the side length of each pentagon doubles from 1 unit to 2 units?
218. An icosahedron is a Platonic solid with a surface that consists of 20 equilateral triangles. By how much does the surface area of an icosahedron increase as the side length of each triangle doubles from a unit to 2a units?
219. Write an integral that quantifies the change in the area of the surface of a cube when its side length doubles from s unit to 2s units and evaluate the integral.

220. Write an integral that quantifies the increase in the volume of a cube when the side length doubles from s unit to 2s units and evaluate the integral.
221. Write an integral that quantifies the increase in the surface area of a sphere as its radius doubles from R unit to 2R units and evaluate the integral.
222. Write an integral that quantifies the increase in the volume of a sphere as its radius doubles from R unit to 2R units and evaluate the integral.
223. Suppose that a particle moves along a straight line with velocity v(t) = 4 − 2t, where 0 ≤ t ≤ 2 (in meters per second). Find the displacement at time t and the total distance 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 line
with 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 line with 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 at an initial speed of 40 m/sec. Acceleration resulting from gravity is −9.8 m/sec2. Neglecting air resistance, solve for the 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 at an initial speed of 60 m/sec. Acceleration resulting from gravity is −9.8 m/sec2. Neglecting air resistance, solve for the 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 units between times t = 2 and t = 3, find the net change in the radius during that time.

74

Chapter 1 | Integration

230. A spherical balloon is being inflated at a constant rate. If the volume of the balloon changes from 36π in.3 to 288π 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 up to height x. If 3
water flows into the tank at a rate of 1 m3/min, find the height of water in the tank after 5 min. Find the change in height between 5 min and 10 min.

232. A horizontal cylindrical tank has cross-sectional area

A(x) = 4⎛⎝6x − x2⎞⎠m2 at height x meters above the bottom

when x ≤ 3.

a. The volume V between heights a and b is

b
∫a A(x)dx. Find the volume at heights between 2
m and 3 m. b. Suppose that oil is being pumped into the tank
at 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 in terms of x, when the height is at x meters? c. How long does it take to fill the tank to 3 m starting from a fill level of 2 m?

233. The following table lists the electrical power in gigawatts—the rate at which energy is consumed—used in a certain city for different hours of the day, in a typical 24-hour period, with hour 1 corresponding to midnight to 1 a.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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Chapter 1 | Integration

75

234. The average residential electrical power use (in hundreds 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

235. The data in the following table are used to estimate the average power output produced by Peter Sagan for each of 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 that 1W = 1 J/s, and the average power output by Sagan during this time interval.

a. Compute the average total energy used in a day in kilowatt-hours (kWh).
b. If a ton of coal generates 1842 kWh, how long does it take for an average residence to burn a ton of coal?
c. Explain why the data might fit a plot of the form
p(t) = 11.5 − 7.5 sin⎛⎝1π2t ⎞⎠.

76

Chapter 1 | Integration

236. The data in the following table are used to estimate the average power output produced by Peter Sagan for each 15-min interval of Stage 1 of the 2012 Tour de France.
Minutes Watts Minutes Watts

15

200

165

170

30

180

180

220

237. The distribution of incomes as of 2012 in the United States in $5000 increments is given in the following table. The kth row denotes the percentage of households with
incomes 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.

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.

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Chapter 1 | Integration

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 Income Distributions Source: http://www.census.gov/ prod/2013pubs/p60-245.pdf

77

19 1.8 40 2.3

20 2.1 41

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

a. Estimate the percentage of U.S. households in 2012 with incomes less than $55,000.
b. What percentage of households had incomes exceeding $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 gravitational force exerted by an object of mass M and one of mass m with centers that are separated by a distance r is

F

=

G

mM r2

,

with G an empirical constant

G = 6.67x10−11 m3 /⎛⎝kg · s2⎞⎠. The work done by a

variable force over an interval

⎡⎣a,

b⎤ ⎦

is defined as

W

b
= ∫a F(x)dx.

If

Earth has mass

5.97219 × 1024

and

radius 6371 km, compute the amount of work to elevate

a polar weather satellite of mass 1400 kg to its orbiting

altitude of 850 km above Earth.

239. For a given motor vehicle, the maximum achievable deceleration from braking is approximately 7 m/sec2 on dry concrete. On wet asphalt, it is approximately 2.5 m/sec2. Given that 1 mph corresponds to 0.447 m/sec, find the total distance that a car travels in meters on dry concrete after the brakes are applied until it comes to a complete stop if the initial velocity is 67 mph (30 m/sec) or if the initial braking velocity is 56 mph (25 m/sec). Find the corresponding distances if the surface is slippery wet asphalt.

240. John is a 25-year old man who weighs 160 lb. He burns 500 − 50t calories/hr while riding his bike for t hours. If an oatmeal cookie has 55 cal and John eats 4t cookies during the tth hour, how many net calories has he lost after 3 hours riding his bike?

241. Sandra is a 25-year old woman who weighs 120 lb. She burns 300 − 50t cal/hr while walking on her treadmill. Her caloric intake from drinking Gatorade is 100t calories during the tth hour. What is her net decrease in calories after walking for 3 hours?

78

Chapter 1 | Integration

242. A motor vehicle has a maximum efficiency of 33 mpg at a cruising speed of 40 mph. The efficiency drops at a rate of 0.1 mpg/mph between 40 mph and 50 mph, and at a rate of 0.4 mpg/mph between 50 mph and 80 mph. What is the efficiency in miles per gallon if the car is cruising at 50 mph? What is the efficiency in miles per gallon if the car is cruising at 80 mph? If gasoline costs $3.50/gal, what is the cost of fuel to drive 50 mi at 40 mph, at 50 mph, and at 80 mph?

243. Although some engines are more efficient at given a horsepower than others, on average, fuel efficiency decreases with horsepower at a rate of 1/25 mpg/ horsepower. If a typical 50-horsepower engine has an average fuel efficiency of 32 mpg, what is the average fuel efficiency of an engine with the following horsepower: 150, 300, 450?

244. [T] The following table lists the 2013 schedule of federal income tax versus taxable income.

Taxable Income Range

The Tax Is …

… Of the Amount Over

$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

245. [T] The following table provides hypothetical data regarding the level of service for a certain highway.

Highway Speed Range (mph)

Vehicles per Hour per Lane

Density Range (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 and highway speed on the y-axis.
b. Compute the average decrease in speed (in miles per hour) per unit increase in congestion (vehicles per hour per lane) as the latter increases from 600 to 1000, from 1000 to 1500, and from 1500 to 2100. Does the decrease in miles per hour depend linearly on the increase in vehicles per hour per lane?
c. Plot minutes per mile (60 times the reciprocal of miles per hour) as a function of vehicles per hour per lane. Is this function linear?
For the next two exercises use the data in the following table, which displays bald eagle populations from 1963 to 2000 in the continental United States.

$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 Taxable Income Source: http://www.irs.gov/pub/irs-prior/ i1040tt--2013.pdf.
Suppose that Steve just received a $10,000 raise. How much of this raise is left after federal taxes if Steve’s salary before receiving the raise was $40,000? If it was $90,000? If it was $385,000?

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Chapter 1 | Integration

79

Year 1963

Population of Breeding Pairs of Bald Eagles
487

1974 791

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 bald eagles per year present for the 37 years by computing the average value of p over [0, 37].

1981 1188

1986 1875

1992 3749

1996 5094

2000 6471
Table 1.11 Population of Breeding Bald Eagle Pairs 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 year present for the 37 years by computing the average value of p over [0, 37].

248. [T] Suppose you go on a road trip and record your speed at every half hour, as compiled in the following table. 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 over the 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 constant rate; rather, the acceleration is variable. For the following exercises, use the following table, which contains the acceleration measured at every second as a driver merges onto a freeway.

80

Chapter 1 | Integration

Time (sec) 1

Acceleration (mph/sec) 11.2

2

10.6

3

8.1

4

5.4

252. [T] The number of hamburgers sold at a restaurant throughout the day is given in the following table, with the accompanying 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

5

0

12

28

15

20

249. [T] The accompanying graph plots the best quadratic

fit, a(t) = −0.70t2 + 1.44t + 10.44, to the data from the

18

30

preceding table. Compute the average value of a(t) to

estimate the average acceleration between t = 0 and

21

45

t = 5.

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

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Chapter 1 | Integration

81

253. [T] An athlete runs by a motion detector, which records her speed, as displayed in the following table. The best linear fit to this data, ℓ(t) = −0.068t + 5.14, is shown in the accompanying graph. Use the average value of 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

82

Chapter 1 | Integration

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 drawback of this method, though, is that we must be able to find an antiderivative, and this is not always easy. In this section we examine a technique, called integration by substitution, to help us find antiderivatives. Specifically, this method helps us find 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—that is, 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 ⌠⌡⎛⎝x2 − 3⎞⎠3 2xdx, we have f (x) = x3, g(x) = x2 − 3, and g '(x) = 2x. Then,
f ⎡⎣g(x)⎤⎦g′ (x) = ⎛⎝x2 − 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 integrand with du. It is also referred to as change of variables because we are changing variables to obtain an expression that is easier to 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,

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

(1.19)

Proof
Let f, g, u, and F be as specified in the theorem. Then ddxF(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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Chapter 1 | Integration

83

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

3

⎫ ⎬ ⎭

⌠ ∫ ⎛⎝x2 − 3⎞⎠

(2xdx) =
⏟

u3 du.

⌡u

du

Using the power rule for integrals, we have

⌠u3 du ⌡

=

u4 4

+

C.

Substitute the original expression for x back into the solution:

u4 4

+

C

=

⎛⎝x 2

− 4

3⎞⎠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 to go 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⎛⎝3x2 + 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⎛⎝3x2 + 4⎞⎠4 dx = ∫ u4 du.
Remember that du is the derivative of the expression chosen for u, regardless of what is inside the integrand. Now we can evaluate the integral with respect to u:

84

Chapter 1 | Integration

∫ u4 du

=

u5 5

+

C

=

⎛⎝3x2 + 5

4⎞⎠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′ = ⎛⎝15⎞⎠5⎛⎝3x2 + 4⎞⎠4 6x = 6x⎛⎝3x2 + 4⎞⎠4.
This is exactly the expression we started with inside the integrand.

1.25 Use substitution to find the antiderivative of ⌠⌡3x2 ⎛⎝x3 − 3⎞⎠2 dx.
Sometimes we need to adjust the constants in our integral if they don’t match up exactly with the expressions we are substituting.
Example 1.31
Using Substitution with Alteration
Use substitution to find the antiderivative of ∫ z z2 − 5dz.
Solution Rewrite the integral as ⌠⌡z⎛⎝z2 − 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 12. we can solve this problem. Thus,
u = z2 − 5 du = 2z dz 12du = 12(2z)dz = z dz.

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Chapter 1 | Integration

85

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

∫ ⌠⌡z⎛⎝z2

−

5⎞⎠1/2

dz

=

1 2

u1/2 du.

Integrate the expression in u:

∫1
2

u1/2 du

=

⎛⎝12

⎞u
⎠

3/2 3

+

C

2

= ⎛⎝12⎞⎠⎛⎝23⎞⎠u3/2 + C

= 13u3/2 + C

=

1 3

⎛⎝z

2

−

5⎞⎠3/2

+

C.

1.26

Use

substitution

to

find

the

antiderivative

of

⌠x ⌡

2

⎛⎝x

3

+

5⎞⎠9

dx.

Example 1.32
Using Substitution with Integrals of Trigonometric Functions

Use substitution to evaluate the integral

⌠ sint ⌡cos3

t

dt.

Solution
We know the derivative of cos t is −sin t, integral, we have

so we set u = cos t. Then du = −sin tdt. Substituting into the

Evaluating the integral, we get

⌠ sint ⌡cos3 t

dt

=

−⌠⌡duu3 .

−⌠⌡duu3 = −∫ u−3 du

Putting the answer back in terms of t, we get

=

−⎛⎝−

1 2

⎞⎠u

−2

+

C.

⌠ sint ⌡cos3 t

dt

=

1 2u 2

+

C

=

1 2cos2 t

+

C.

86

Chapter 1 | Integration

1.27

Use substitution to evaluate the integral

⌠ cost ⌡sin2 t

dt.

Sometimes we need to manipulate an integral in ways that are more complicated than just multiplying or dividing by a constant. We need to eliminate all the expressions within the integrand that are in terms of the original variable. When we are done, u should be the only variable in the integrand. In some cases, this means solving for the original variable in terms of 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 −

1dx.

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 dx x−1

=

∫

u

+ u

1du

= ∫ 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 = 23u3/2 + 2u1/2 + C

= 23(x − 1)3/2 + 2(x − 1)1/2 + C

= (x − 1)1/2 ⎡⎣23(x − 1) + 2⎤⎦ + C

=

(x

−

1) 1/2

⎛⎝23 x

−

2 3

+

63 ⎞⎠

=

(x

−

1) 1/2

⎛⎝23 x

+

4 3

⎞ ⎠

= 23(x − 1)1/2 (x + 2) + C.

1.28 Use substitution to evaluate the indefinite integral ∫ cos3 t sint dt.
Substitution for Definite Integrals
Substitution can be used with definite integrals, too. However, using substitution to evaluate a definite integral requires a change to the limits of integration. If we change variables in the integrand, the limits of integration change as well.

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Chapter 1 | Integration

87

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,

∫ b
⌠ f ⎛⎝g(x)⎞⎠g′ (x)dx = ⌡a

g(b)
f (u)du.
g(a)

Although we will not formally prove this theorem, we justify it with some calculations here. From the substitution rule for indefinite integrals, if F(x) is an antiderivative of f (x), we have

Then

∫ f ⎛⎝g(x)⎞⎠g′ (x)dx = F⎛⎝g(x)⎞⎠ + C.

b
∫ f ⎡⎣g(x)⎤⎦g′ (x)dx a

| =

F⎛⎝g(x)⎞⎠

x x

==

b a

= F⎛⎝g(b)⎞⎠ − F⎛⎝g(a)⎞⎠

=

F(u)|

u u

= =

g(b) g(a)

(1.20)

and we have the desired result.
Example 1.34

g(b)
= ∫ f (u)du, g(a)

Using Substitution to Evaluate a Definite Integral

Use

substitution

to evaluate

1
⌠x ⌡0

2

⎛⎝1

+

2x

3⎞⎠5d

x.

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

To adjust the limits of integration, x = 1, u = 1 + 2(1) = 3. Then

du = 6x2 dx 16du = x2 dx.
note that when

x = 0, u = 1 + 2(0) = 1,

and when

Evaluating this expression, we get

∫ 1

⌠x ⌡0

2

⎛⎝1

+

2x

3⎞5 ⎠

dx

=

1 6

3
u

5

du.

1

88

Chapter 1 | Integration

| 16∫13u5 du

=

⎛⎝16

⎞⎛u 6 ⎠⎝ 6

⎞ ⎠

3 1

=

1 36

⎡⎣(3)

6

−

(1)

6⎤ ⎦

=

182 9

.

1.29

Use substitution to evaluate the definite integral

0
⌠⌡−1y⎛⎝2y 2

−

3⎞⎠5

d

y.

Example 1.35
Using Substitution with an Exponential Function

Use substitution to evaluate

∫

1
xe

4x

2

+

3

dx.

0

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

∫1
xe

4x

2

+

3

dx

0

= 18∫37eu du

| =

18 e u

7 3

=

e7

− 8

e3

≈ 134.568.

1.30

Use substitution

to

evaluate

1
⌠⌡0x

2

cos⎛⎝π2

x

3⎞⎠dx.

Substitution may be only one of the techniques needed to evaluate a definite integral. All of the properties and rules of integration apply independently, and trigonometric functions may need to be rewritten using a trigonometric identity before we 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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Chapter 1 | Integration

89

Using Substitution to Evaluate a Trigonometric Integral

Use substitution to evaluate

∫

π/2
cos

2

θ

dθ.

0

Solution

Let us first use a trigonometric identity to rewrite the integral. The trig identity

cos2 θ

=

1

+

cos 2θ 2

allows us

to rewrite the integral as

∫ π/2cos2 0

θdθ

=

π/2 ⌠

1

⌡0

+

cos 2

2θ dθ.

Then,

⌠⌡0π/2⎛⎝1

+

cos 2

2θ

⎞⎠dθ

= ⌠⌡0π/2⎛⎝12 + 12cos 2θ⎞⎠dθ

=

1 2

∫

π/2
dθ
0

+

π/2
∫ cos2θdθ. 0

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 12du = dθ. Also, when θ = 0, u = 0, and when θ = π/2, u = π. Expressing the second integral in terms of u, we have

π/2

1⌠ 2⌡0

dθ

+

12∫

π/2
cos
0

2θdθ

=

π/2

1⌠ 2 ⌡0

dθ

+

1 2

⎛⎝12

⎞⎠∫

π
cos
0

udu

| | =

θ 2

θ θ

= =

π/2 0

+

1 4

sin

u

u u

= =

θ 0

=

⎛π ⎝4

−

0⎞⎠

+

(0

−

0)

=

π4 .

90

Chapter 1 | Integration

1.5 EXERCISES
254. Why is u-substitution referred to as change of variable?
255. 2. If f = g ∘ h, when reversing the chain rule, ddx(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 using differentiation. Then, using the indicated u-substitution,
identify f such that the integral takes the form ∫ f (u)du.

256.
∫ x x + 1dx = 125(x + 1)3/2 (3x − 2) + C; u = x + 1

257.

⌠ ⌡

x2 x−

1dx(x

>

1)

=

2 15

x − 1⎛⎝3x2 + 4x + 8⎞⎠ + C; u = x − 1

258.

⌠x ⌡

4x2 + 9dx =

1 12

⎛⎝4x

2

+

9⎞⎠3/2

+

C;

u

=

4x2

+

9

259.

⌠ ⌡

x 4x2 +

dx 9

=

1 4

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

260.

⌠ ⌡(4x2

x +

9)2dx

=

−

1 8(4x 2

+

; 9)

u

=

4x2

+

9

In the following exercises, find the antiderivative using the indicated substitution.

261. ∫ (x + 1)4 dx; u = x + 1

262. ∫ (x − 1)5 dx; u = x − 1

263. ∫ (2x − 3)−7 dx; u = 2x − 3

264. ∫ (3x − 2)−11 dx; u = 3x − 2

265.

⌠ ⌡

x

x 2+

dx; 1

u

=

x2 + 1

266.

⌠ ⌡

x 1−

x2dx;

u

=

1−

x2

267. ⌠⌡(x − 1)⎛⎝x2 − 2x⎞⎠3 dx; u = x2 − 2x 268. ⌠⌡⎛⎝x2 − 2x⎞⎠⎛⎝x3 − 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 of variables to determine the indefinite integral.
271. ∫ x(1 − x)99 dx

272.

⌠⌡t⎛⎝1

−

t

2⎞10 ⎠

dt

273. ∫ (11x − 7)−3 dx

274. ∫ (7x − 11)4 dx

275. ∫ cos3 θ sinθdθ

276. ∫ sin7 θ cosθdθ

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

3

x2 −

3⎞⎠2

dx

282. ⌠ x3 dx ⌡ 1 − x2

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Chapter 1 | Integration

91

283.

⌠⎮ ⌡⎛⎝1

y5

−

y

3⎞3/2 ⎠

dy

99
284. ∫ cosθ(1 − cosθ) sinθdθ

285. ∫ ⎛⎝1 − cos3 θ⎞⎠10 cos2 θ sin θdθ

286.

⌠(cos ⌡

θ

−

1)⎛⎝cos

2

θ

−

2

cos

θ⎞⎠3

sin

θdθ

287. ⌠⌡⎛⎝sin2 θ − 2 sin θ⎞⎠⎛⎝sin3 θ − 3sin2 θ⎞⎠3 cos θdθ
In the following exercises, use a calculator to estimate the area under the curve using left Riemann sums with 50 terms, then use substitution to solve for the exact answer.
288. [T] y = 3(1 − x)2 over [0, 2]

289.

[T]

y

=

x⎛⎝1

−

x

2⎞3 ⎠

over

[−1,

2]

290. [T] y = sin x(1 − cos x)2 over [0, π]

291.

[T]

y=

x ⎛⎝x2 + 1⎞⎠2

over

[−1,

1]

In the following exercises, use a change of variables to evaluate the definite integral.

292.

∫

1
x

1 − x2dx

0

1
293. ⌠ x dx ⌡0 1 + x2

2
294. ⌠ t dt ⌡0 5 + t2

1
295. ⌠ t dt ⌡0 1 + t3

296. ∫ π/4sec2 θ tanθdθ 0

π/4

297.

⌠ ⌡0

sin θ cos4 θ

dθ

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 the indicated interval. If possible, estimate a value of C that would need to be added to the antiderivative to make it
x
equal to the definite integral F(x) = ∫ f (t)dt, with a the a
left endpoint of the given interval.

298. [T] ∫ (2x + 1)ex2 + x − 6 dx over [−3, 2]

∫ 299. [T]

cos⎛⎝ln(2x)⎞⎠ x

d

x

on

[0,

2]

300.

[T]

⌠ ⌡

3x2 + 2x + 1 dx x3 + x2 + x + 4

over

[−1,

2]

301.

[T]

⌠ sin x ⌡cos3 x

dx

over

⎡⎣−

π 3

,

π⎤ 3⎦

∫ 302. [T]

(x + 2)e−x2 − 4x + 3 dx

over

⎡⎣−5,

1⎤ ⎦

303. [T] ∫ 3x2 2x3 + 1dx over [0, 1]

b
∫ 304. If h(a) = h(b) in g '⎛⎝h(x)⎞⎠h(x)dx, what can you a
say about the value of the integral?

305. Is the substitution u = 1 − x2 in the definite integral

2

⌠ ⌡0 1

x −

x

2dx

okay? If not, why not?

In the following exercises, use a change of variables to show that each definite integral is equal to zero.
306. ∫ πcos2 (2θ)sin(2θ)dθ 0

∫ 307.

π
t

cos⎛⎝t

2⎞⎠sin⎛⎝t

2⎞⎠dt

0

1
308. ∫ (1 − 2t)dt 0

92

Chapter 1 | Integration

1

309.

⌠⎮⎮ ⌡0

⎛⎝1

1 − 2t

+

⎛⎝t

−

1 2

⎞2⎞dt ⎠⎠

310.

⌠⌡0πsin⎛⎝⎜⎛⎝t

−

π 2

⎞⎠3⎞⎠⎟cos⎛⎝t

−

π 2

⎞⎠dt

2
311. ∫ (1 − t)cos(πt)dt 0

312. ∫ 3π/4sin2 t cos tdt π/4

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

⎡a ⎣c

,

b⎤ c⎦

for

c

>

0.

314.

Find the area under the graph of

f

(t)

=

⎛⎝1

t

+

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)

=

⎛⎝1

t

−

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 − x2dx.

Use the substitution

x = cost

to

−1

express the area of a semicircle as the integral of a

trigonometric function. You do not need to compute the

integral.

317. The area of the top half of an ellipse with a major

axis 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

a

as

⌠b ⌡−a

1 − ax22dx.

Use the substitution

x = acost

to

express this area in terms of an integral of a trigonometric

function. You do not need to compute the integral.

318. [T] The following graph is of a function of the form f (t) = a sin(nt) + b sin(mt). Estimate the coefficients a and b, and the frequency parameters n and m. Use these
π
estimates to approximate ∫ f (t)dt. 0
319. [T] The following graph is of a function of the form f (x) = a cos(nt) + b cos(mt). Estimate the coefficients a and b and the frequency parameters n and m. Use these
π
estimates to approximate ∫ f (t)dt. 0

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