MAT 168: Calculus II with Analytic Geometry · MAT 168: Calculus II with Analytic Geometry James V....

MAT 168: Calculus II with Analytic Geometry

James V. Lambers

February 7, 2012

Contents

1 Integrals 5

1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.1.1 Differential Calculus and Quotient Formulas . . . . . . 5

1.1.2 Integral Calculus and Product Formulas . . . . . . . . 9

1.2 Areas and Distances . . . . . . . . . . . . . . . . . . . . . . . 11

1.2.1 The Area Problem . . . . . . . . . . . . . . . . . . . . 12

1.2.2 Sigma Notation . . . . . . . . . . . . . . . . . . . . . . 15

1.2.3 Computing the Exact Area . . . . . . . . . . . . . . . 16

1.2.4 The Distance Problem . . . . . . . . . . . . . . . . . . 18

1.3 The Definite Integral . . . . . . . . . . . . . . . . . . . . . . . 20

1.3.1 Evaluating Integrals . . . . . . . . . . . . . . . . . . . 22

1.3.2 The Midpoint Rule . . . . . . . . . . . . . . . . . . . . 26

1.3.3 Properties of the Definite Integral . . . . . . . . . . . 27

1.4 The Fundamental Theorem of Calculus . . . . . . . . . . . . . 29

1.4.1 Part 1: Differentiation Undoes Integration . . . . . . . 31

1.4.2 Part 2: Integration Undoes Differentiation . . . . . . . 38

1.4.3 Statement of the Fundamental Theorem of Calculus . 38

1.4.4 Differentiation and Integration as Inverse Processes . . 39

1.4.5 Computing Area Using the Fundamental Theorem ofCalculus . . . . . . . . . . . . . . . . . . . . . . . . . . 40

1.4.6 Net Area vs. Total Area . . . . . . . . . . . . . . . . . 44

1.4.7 Displacement vs. Distance . . . . . . . . . . . . . . . . 47

1.4.8 Average Values . . . . . . . . . . . . . . . . . . . . . . 48

1.5 Indefinite Integrals . . . . . . . . . . . . . . . . . . . . . . . . 51

1.5.1 The Net Change Theorem . . . . . . . . . . . . . . . . 56

1.6 The Substitution Rule . . . . . . . . . . . . . . . . . . . . . . 57

1.6.1 Definite Integrals . . . . . . . . . . . . . . . . . . . . . 62

1.6.2 Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . 64

3

4 CONTENTS

2 Techniques of Integration 672.1 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . 672.2 Trigonometric Integrals . . . . . . . . . . . . . . . . . . . . . 722.3 Trigonometric Substitution . . . . . . . . . . . . . . . . . . . 802.4 Integration of Rational Functions . . . . . . . . . . . . . . . . 86

2.4.1 Simple Proper Rational Functions . . . . . . . . . . . 872.4.2 The Method of Partial Fractions . . . . . . . . . . . . 88

2.5 Approximate Integration . . . . . . . . . . . . . . . . . . . . . 962.5.1 Riemann Sums . . . . . . . . . . . . . . . . . . . . . . 972.5.2 Simpson’s Rule and Other Approximation Methods . 98

2.6 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . 1102.6.1 Infinite Intervals . . . . . . . . . . . . . . . . . . . . . 1112.6.2 Discontinuous Integrands . . . . . . . . . . . . . . . . 1142.6.3 A Comparison Test for Improper Integrals . . . . . . . 117

3 Applications of Integration 1193.1 Areas Between Curves . . . . . . . . . . . . . . . . . . . . . . 1193.2 Volume by Slices . . . . . . . . . . . . . . . . . . . . . . . . . 1283.3 Volume by Shells . . . . . . . . . . . . . . . . . . . . . . . . . 1333.4 Arc Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144

Chapter 1

Integrals

1.1 Introduction

In your previous calculus course, you learned about differential calculus,which is the study of the rate of change of one quantity with respect toanother. We briefly review the main ideas of differential calculus beforeintroducing the closely related branch of mathematics known as integralcalculus, which is the subject of this course.

1.1.1 Differential Calculus and Quotient Formulas

Suppose that an object moves in a straight line during some interval in time.How can we compute the velocity of the object at any particular point intime within this interval? If we know that the object’s velocity does notchange over time, then we can use the simple formula

velocity =distance traveled

elapsed time=

final position− initial position

final time− initial time. (1.1)

Since the velocity is known to be constant during the given time interval,we can conclude that this formula yields the velocity of the object at anypoint in time between the initial time and the final time.

What if the object is not necessarily moving at a constant velocity? Inthis case, equation (1.1) is only useful for computing the average velocityof the object between the initial time and the final time. To compute thevelocity at a particular instant t0, we must be more resourceful. We canapproximate this quantity, which is called the instantaneous velocity, byusing equation (1.1) over some small interval in time that contains t0, suchas the interval [t0, t0 + h], where h is some small positive number.

5

6 CHAPTER 1. INTEGRALS

For concreteness, we let s(t) be the function that describes the relation-ship between the position of the object and time. Specifically, for each realnumber t between the initial time and the final time, s(t) is the position ofthe object, at time t, on the straight line along which it is moving. Then,the velocity of the object at time t0 can be approximated by

s(t0 + h)− s(t0)

(t0 + h)− t0=s(t0 + h)− s(t0)

h. (1.2)

From equation (1.1), we can see that this quantity is actually the averagevelocity of the object on the interval [t0, t0 + h], but since h is small, it isreasonable to assume that the object’s velocity cannot vary much duringsuch a short period of time, and we can conclude that the instantaneousvelocity at time t0 can be approximated by this average velocity.

The smaller the value of h, the less the object’s velocity can vary overthe interval [t0, t0 + h]. Therefore, it is reasonable to conclude that as wechoose smaller and smaller values of h, the average velocity over the interval[t0, t0 + h] becomes a better approximation of the instantaneous velocity ofthe object at time t0. We can therefore define the instantaneous velocity attime t0 to be the limit of the average velocity as the width of the interval[t0, t0+h], which is h, approaches zero. Specifically, if we denote this velocityby v(t0), then

v(t0) = limh→0

s(t0 + h)− s(t0)

h. (1.3)

Informally, we are defining the instantaneous velocity at time t0 to be theaverage velocity over an “infinitely small” interval in time containing theinstant t0.

In general, if two quantities x and y are related by the equation y = f(x),where f is a given function, then the instantaneous rate of change of y withrespect to x when x = x0 is given by the derivative of f(x) at x = x0. Thederivative of f(x) at x = x0, which we denote by f ′(x0), is defined by

f ′(x0) = limh→0

f(x0 + h)− f(x0)

h, (1.4)

provided this limit exists. The derivative of a function f(x) is itself a func-tion, denoted by f ′(x), that is defined at every point x0 at which the limitin equation (1.4) exists and whose value at x0 is the quantity f ′(x0) definedin equation (1.4). This notion of the derivative as a function allows us tocompute derivatives of many functions using differentiation rules, which ismuch more efficient than using the definition of the derivative directly.

1.1. INTRODUCTION 7

Example 1 Suppose that a ball is thrown straight up with an initial veloc-ity of 100 ft/s, and that the initial height of the ball is 6 ft. Then, the height,in feet, of the ball t seconds after it has been thrown can be described bythe function s(t), where

s(t) = −16t2 + 100t+ 6, t ≥ 0. (1.5)

Our goal is to compute the velocity of the ball 5 seconds after it has beenthrown.

After 5 seconds, the height of the ball is s(5) = 106 ft. Since the heightof the ball has changed by 106 − 6 = 100 ft in 5 seconds, it is temptingto use the simple formula for velocity in equation (1.1) to conclude thatthe velocity of the ball during this time is 100/5 = 20 ft/s, but since theinitial velocity of the ball is 100 ft/s, we see that the velocity is clearly notconstant, so this formula cannot be used directly to compute the velocity atany particular point in time.

Instead, we can approximate the velocity of the ball after 5 seconds byusing the formula in equation (1.1) to compute the average velocity betweent = 5 and t = 6, which is

s(6)− s(5)

6− 5=

30− 106

1= −76 ft/s. (1.6)

However, we can obtain a more accurate approximation by computing theaverage velocity between t = 5 and t = 5.5, which is

s(5.5)− s(5)

5.5− 5=

72− 106

0.5= −68 ft/s. (1.7)

We can use this approach to obtain the exact velocity at t = 5 using alittle abstraction. If we approximate this velocity by computing the averagevelocity between t = 5 and t = 5 + h, where h is assumed to be some small,positive number, then we have

s(5 + h)− s(5)

(5 + h)− 5=

[−16(5 + h)2 + 100(5 + h) + 6]− 106

h

=−16(25 + 10h+ h2) + 500 + 100h+ 6− 106

h

=−400− 160h− 16h2 + 500 + 100h+ 6− 106

h

=−16h2 − 60h

h= −16h− 60. (1.8)


Taking the limit as h approaches zero, we find that the velocity of the ballafter 5 seconds is −60 ft/s.

In this case, however, the function s(t) that describes the height of theball is a function whose derivative can be computed using differentiationrules. In particular, we can use the Power Rule, the Sum Rule, and theConstant Multiple Rule to obtain

v(t) = s′(t) = −32t+ 100. (1.9)

Since velocity is the rate of change of position, and s(t) describes the posi-tion, or height, of the ball as a function of time, its derivative v(t) = s′(t)describes the velocity of the ball as a function of time. That is, v(t), fort ≥ 0, represents the velocity of the ball t seconds after it has been thrown.Substituting t = 5 into v(t) yields v(5) = −60, so we conclude, as before,that the velocity of the ball after 5 seconds is −60 ft/s. 2

Based on this example, we can summarize the advantage that the deriva-tive provides us with regard to the use of “quotient formulas” such as theformula in equation (1.1).

The derivative, being the limit of a quotient of differences, allowsquotient formulas to be applied to more general problems.

For example, the cost to produce a single unit of a product, known asmarginal cost, can be obtained by the simple quotient formula

marginal cost =cost to produce all units

number of units produced, (1.10)

but this formula is valid only if the cost to produce each unit is the same.This is rarely the case, due to the distinction between fixed costs, suchas the cost of buying manufacturing equipment, that are independent ofthe number of units, and variable costs, such as workers’ wages or coststo maintain equipment, which increase as the number of units increases.Therefore, in order to compute the marginal cost, it is important to recognizethat marginal cost is the rate of change of total production cost with respectto the number of units produced. Therefore, given a function C(x) thatdescribes the total production cost in terms of the number of units, whichis denoted by x, the marginal cost after producing x units is given by thederivative C ′(x). This derivative can be interpreted as the cost of producingone additional unit, given that x units have already been produced.

1.1. INTRODUCTION 9

1.1.2 Integral Calculus and Product Formulas

It is natural to ask whether other types of formulas can be generalized tomore difficult problems using limits. For example, consider the formula forthe area A of a rectangle of width w and height h,

A = wh. (1.11)

How can we use this formula to compute the area of a shape that is not arectangle? We can proceed by approximating the given shape by a numberof rectangles. Then, we can approximate the area of the shape by usingequation (1.11) to compute the area of each rectangle, and then adding allof these areas together.

The accuracy of this approach depends on how well we have approxi-mated the shape using rectangles. If we use more, smaller rectangles insteadof fewer large ones, then the approximate area obtained by adding the areasof these rectangles is likely to be more accurate. Therefore, we can definethe area of the shape to be the limit of this approximate area as the num-ber of rectangles becomes infinite, with each rectangle becoming “infinitelysmall.”

Example 2 Figure 1.1 shows how rectangles can be used to approximatea given shape, so that the area of the shape can easily be approximated bycomputing the areas of all of the rectangles using equation (1.11). Using fourrectangles, we obtain an approximate area of 5.75 square units, whereas witheight rectangles, we obtain an approximate area of 5.1875 square units. Laterin this course, we will learn how to compute the exact area of this shape,which is 4.6 square units. Note that the area obtained using eight rectanglesis much more accurate than that obtained using only four rectangles. 2

As another example, suppose we rearrange the formula for velocity de-scribed in equation (1.1) to obtain the formula

distance traveled = velocity× elapsed time. (1.12)

As with equation (1.1), this formula is valid when the velocity is constant,but what if it is not? Again, we need to be more resourceful. We canapproximate the distance traveled by dividing the interval of time into smallsubintervals, and then using equation (1.12) to approximate the distancetraveled during each of these subintervals. Then, we can add all of thesedistances together to approximate the distance traveled during the entireinterval of time.


Figure 1.1: Approximation the shape bounded by the curve y = x2 + 1 andthe lines y = 0, x = 0 and x = 2, using four rectangles (left plot) and eightrectangles (right plot).

This approach yields a reasonable approximation to the distance trav-eled because, as previously mentioned, the velocity cannot vary much duringvery short periods of time, and therefore equation (1.12) yields an accurateapproximation of the distance traveled during each subinterval. As we dividethe original time interval into more and more subintervals, our approxima-tion of the distance becomes more and more accurate. Therefore, we candefine the exact distance traveled to be the limit of this approximate dis-tance as the number of subintervals becomes infinite, provided that eachsubinterval becomes infinitely small.

In both cases, we wished to compute some quantity z using a “productformula” of the form z = xy. In the first example, z is area, x is width andy is height; in the second, z is distance, x is time and y is velocity. However,in both cases, the formula z = xy is only valid if x and y are constant

1.2. AREAS AND DISTANCES 11

values that are independent of one another. If this is not the case, we canstill use this formula on smaller instances of the same problem, and addthe results together. The process of approximating the quantity z in thismanner, and taking the limit as the number of smaller problems becomesinfinite, is called integration, and the exact value of z is known as a definiteintegral. The study of definite integrals and their computation is calledintegral calculus.

We now relate the definite integral to the derivative by the followingstatement that summarizes the usefulness of the definite integral in the sameway that the usefulness of the derivative was described earlier.

The definite integral, being the limit of a sum of products, allowsproduct formulas to be applied to more general problems.

In this course, we will use the definite integral to expand the usefulness ofa number of product formulas, including not only the formulas for area anddistance that we have discussed, but also formulas for computing quantitiessuch as the volume of a solid object, or the work that is required to movean object a given distance.

As we will see, the concept of the definite integral closely parallels theconcept of the derivative. Both concepts are defined using a limit, which isthe essential ingredient that allows us to obtain an exact value for the quan-tity we wish to compute, instead of a mere approximation. Furthermore,both derivatives and definite integrals can be viewed as functions, whichleads to the development of rules that can be used to compute these exactvalues with much greater efficiency than by computing the appropriate limitdirectly.

The significance of this one advantage cannot be overstated, for it hasserved as one of the most fundamental catalysts for scientific advancementover the past few centuries. This benefit is a direct result of the resourceful-ness and dedication of the mathematicians who developed differential andintegral calculus in order to solve the problems of their day as efficientlyand accurately as possible. We would do well to emulate their virtues in ourefforts to solve the problems of today.

1.2 Areas and Distances

There are many cases in which some quantity is defined to be the productof two other quantities. Some examples are:


• If an object is traveling at constant velocity v, then the distance d thatit travels within a time span of length t is given by the relation d = vt.

• By definition, a rectangle of width w has constant height h. The areaA of the rectangle is given by the formula A = wh.

• The mass m of an object with volume v and density d is given by theformula m = dv.

Unfortunately, in many applications, we cannot necessarily assume that cer-tain quantities such as velocity or density are constant, and therefore wecannot use formulas such as d = vt directly. However, we can use themindirectly in these more difficult cases by employing the most fundamentalconcept of calculus, the limit.

1.2.1 The Area Problem

Suppose we wish to compute the area A of a shape that is not a rectangle. Tosimplify the discussion, we assume that the shape is bounded by the verticallines x = a and x = b, the x-axis y = 0, and the curve defined by somecontinuous function y = f(x), where f(x) ≥ 0 for a ≤ x ≤ b. If the functionf(x) is not constant on the interval [a, b], then the shape is not a rectangle, sowe cannot compute its area by using the formula A = wh directly. However,we can use this formula indirectly. We begin by attempting to approximatethe shape by a rectangle, since three sides of the boundary of the shape doform a portion of a rectangle. How should we choose this rectangle? Weconsider this question in the following example.

Example 3 Suppose that we wish to approximate the shape shown in Fig-ure 1.2 by a rectangle. Certainly, the base of the rectangle should coincidewith the base of the shape, since it is a line segment connecting the points(0, 0) and (2, 0). It follows that the width of the rectangle is 2 units.

Choosing the height, however, is more difficult. Figure 1.3 illustrates theconsequences of two choices for the height. If the height of the rectangle isdetermined by the height of the shape on the left side, then the resultingrectangle has height 1, so the area of the rectangle is 2. As can easily beseen from the figure, this value is too small, since a significant portion ofthe shape lies outside the rectangle. On the other hand, if the height of therectangle is determined by the height of the shape on the right side, then theresulting rectangle has height 5, so the area of the rectangle is 10. Clearly,this value is too large, because the rectangle contains not only the entire


Figure 1.2: Region bounded by the vertical lines x = 0 and x = 2, thehorizontal line y = 0, and the curve y = x2 + 1. In this section we will learnhow to compute the exact area of this region.

shape, but also a significant area outside of the shape. In summary, all weknow so far is that the area of the shape is a number between 2 and 10.

The reason why neither of these estimates are close to the exact area isthat the height of the shape varies significantly between x = 0 and x = 2,whereas the height of a rectangle is constant. However, if we examine anysmall subinterval [c, d] of [0, 2], we find that the function f(x) = x2 +1 tendsnot to vary much within the subinterval. Therefore, a single rectangle couldyield an accurate approximation to the area bounded by y = f(x), x = c,x = d and y = 0. 2

The discussion in the preceding example suggests that we can obtaina better approximation to the area of a non-rectangular shape by approx-imating the shape itself with a number of rectangles, instead of just one


Figure 1.3: Attempts to approximate the shape from Figure 1.2 by a rect-angle. In the left plot, the height of the rectangle is the height of the leftside of the shape, whereas in the right plot, the height of the rectangle isequal to the height of the right side of the shape.

rectangle. Then, we can use the formula A = wh to compute the area ofeach rectangle, and add these areas together.

We begin by dividing the interval [a, b] into n subintervals of equal width∆x = (b − a)/n. These subintervals have endpoints [x0, x1], [x1, x2], . . .,[xn−1, xn], where xi = a + i∆x, for i = 0, 1, 2, . . . , n. Then, we define nrectangles as follows: for each i = 1, 2, . . . , n, the ith rectangle has a basedefined by the line segment connecting the points (xi−1, 0) and (xi, 0). Itfollows that each rectangle has a width of ∆x, since, for i = 1, 2, . . . , n,

xi − xi−1 = (a+ i∆x)− (a+ (i− 1)∆x) = [i+ (i− 1)]∆x = ∆x. (1.13)

We can choose the height of each rectangle using a similar approach as witha single rectangle: setting the height of the rectangle equal to the height of


the shape at some point. Which point should we choose?

We consider the first rectangle, whose base is the line segment connecting(a, 0) to (a + ∆x, 0). The purpose of this rectangle is to approximate thearea of a portion of the shape. This portion is the region bounded by thevertical lines x = a and x = a+∆x, the horizontal line y = 0, and the curvey = f(x). To approximate the area of this portion with the area of the givenrectangle, we can proceed as before, and set the height of the rectangle equalto the height of either the left side or the right side of this portion of theshape. For concreteness, we choose the right side, in which case the heightof the rectangle is f(a + ∆x), or f(x1). Therefore, the area of this firstrectangle is f(x1)∆x.

Proceeding in this fashion with the other rectangles, we choose the heightof the ith rectangle to be f(xi), for i = 1, 2, . . . , n. Then, the area of theith rectangle is f(xi)∆x. From the areas of these n rectangles, we obtainan approximation An for the area of the shape:

A ≈ An = f(x1)∆x+ f(x2)∆x+ · · ·+ f(xn)∆x. (1.14)

Example 4 Figure 1.1 illustrates this approach to approximating the areaof the shape shown in Figure 1.2, using n = 4 and n = 8 rectangles. In thecase where n = 4, the width of each rectangle is (2 − 0)/4 = 1/2, and theheights of the rectangles are given by f(1/2), f(1), f(3/2), and f(2), wheref(x) = x2 + 1. Computing the areas of these four rectangles and addingthem together, we obtain the value 5.75. In the case where n = 8, the widthof each rectangle is (2− 0)/8 = 1/4, and the heights of the eight rectanglesare equal to f(i/4), for i = 1, 2, . . . , 8. Computing the areas of these eightrectangles and adding them together, we obtain the value 5.1875.

As can be seen from Figure 1.1, in both cases, the approximate area thatwe have computed is too large, because the union of the rectangles containsthe entire shape as well as some additional area. However, these estimatesof the area are much more accurate than our previous estimates of 2 and 10that were obtained by approximating the shape with a single rectangle. 2

1.2.2 Sigma Notation

For convenience, we will write summations such as the expression for An inequation (1.14) using sigma notation:

An = f(x1)∆x+ · · ·+ f(xn)∆x =n∑i=1

f(xi)∆x. (1.15)


The Σ symbol indicates that the expression to the right, in this case f(xi)∆x,is to be evaluated for certain values of some index variable, and the resultingvalues are to be added. The index variable is indicated below the Σ; in thiscase, the index variable is i. The starting value of the index variable is alsoindicated below the Σ; in this case, the starting value is 1. The index variable“counts” from the starting value to the final value, which is indicated abovethe Σ; in this case, the final value is n. In summary, the index variable iassumes the values 1, 2, . . . , n. For each of these values of i, the value of theexpression f(xi)∆x is computed, and all of these values are added together.

In general, sigma notation is useful for concisely describing the sum ofconsecutive elements of a sequence am, am+1, am+2, . . . , an, where m and nare integers. Using this notation, the sum of all of these elements can bewritten as

n∑i=m

ai = am + am+1 + am+2 + · · ·+ an. (1.16)

The index variable i specified below the Σ effectively “counts” from thestarting value m to the final value n. For each value of i between m and n,including m and n, the value of ai is included in the sum. It should be notedthat the letter i is not always used as the index variable, though this is themost common choice. It should also be noted that n need not be greaterthan m. If m = n, then the summation includes only one term, which is am.If m > n, then the summation does not include any terms, and thereforethe value of the sum is zero.

1.2.3 Computing the Exact Area

We now return to the problem of computing the area A of a non-rectangularshape. If we approximate the shape by n rectangles using the approachdescribed previously, we obtain the approximate area An given in equation(1.15). As can be seen from Figure 1.1, we can obtain a more accurateapproximation by choosing a larger value for n. Why is this the case? Theanswer lies in the fact that each rectangle is used to approximate the areaof a portion of the original shape. Since each rectangle has the same width∆x = (b − a)/n, it follows that as n increases, each rectangle is associatedwith a smaller portion of the shape. The smaller the portion is, the less thatthe height of the portion can vary. Therefore, the area of the portion canbe approximated more accurately by the area of a rectangle.

This discussion suggests that we can obtain the exact area of the shapeby computing the limit of An, as n becomes infinite, assuming that this limit


exists. This is in fact the case when the function f(x), which defines thetop boundary of the shape, is continuous on the interval [a, b]. This leadsto the following definition.

Definition 1 Let f(x) be a continuous function on an interval [a, b], andassume that f(x) ≥ 0 on [a, b]. The area A of the region bounded by thevertical lines x = a and x = b, the horizonal line y = 0, and the curvey = f(x), is

A = limn→∞

n∑i=1

f(xi)∆x, (1.17)

where ∆x = (b− a)/n and xi = a+ i∆x, for i = 1, 2, . . . , n.

We now illustrate how this definition of area can be used to compute theexact area of a non-rectangular shape.

Example 5 We will use Definition 1 to compute the area A of the shapeshown in Figure 1.2. In this case, we have a = 0, b = 2, and f(x) = x2 + 1.We have

A = limn→∞

n∑i=1

f(xi)∆x

= limn→∞

n∑i=1

(x2i + 1

) 2− 0

n

= limn→∞

n∑i=1

[(0 + i∆x)2 + 1

] 2

n

= limn→∞

n∑i=1

[(i2

n

)2

+ 1

]2

n

= limn→∞

n∑i=1

[4i2

n2+ 1

]2

n

= limn→∞

n∑i=1

(8i2

n3+

2

n

)= lim

n→∞

[(8 · 12

n3+

2

n

)+

(8 · 22

n3+

2

n

)+ · · ·+

(8 · n2

n3+

2

n

)]= lim

n→∞

[8

n3(12 + 22 + · · ·+ n2) +

(2

n+ · · ·+ 2

n

)]. (1.18)


To compute this limit, we first note that 2/n is being added n times, whichyields n(2/n) = 2. Second, we use the formula

12 + 22 + · · ·+ n2 =n(n+ 1)(2n+ 1)

6. (1.19)

Using the Limit Laws

limn→∞

a = a, limn→∞

cf(n) = c limn→∞

f(n), limn→∞

f(n)+g(n) = limn→∞

f(n)+ limn→∞

g(n),

(1.20)where a and c are constants, we obtain

A = limn→∞

[8

n3

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

6+ 2

]= 2 +

8

6limn→∞

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

n3

= 2 +4

3limn→∞

2n2 + 3n+ 1

n2

= 2 +4

3limn→∞

(2 +

3

n+

1

n2

)= 2 +

4

32

= 4.6. (1.21)

2

1.2.4 The Distance Problem

Just as the formula A = wh for the area of a rectangle cannot be useddirectly to compute the area of shape whose height is not constant, theformula d = vt cannot be used directly to compute the distance that anobject travels during a given interval in time if the object’s velocity is notconstant. However, we can use this formula indirectly, just as we usedA = wh indirectly to solve the area problem.

Suppose that an object is traveling at a non-constant velocity. If theobject starts moving at time t = a and continues until t = b, then whatis the total distance traveled? For concreteness, we let v(t) be a functionthat indicates the object’s velocity at time t, where a ≤ t ≤ b. Then, wecan approximate the distance by assuming that in very small intervals oftime, the velocity is nearly constant. This is a reasonable assumption if thefunction v(t) that represents the velocity is a continuous function, which wewill assume in this case.


We then divide the interval [a, b] into n subintervals of width ∆t =(b − a)/n, with endpoints [ti−1, ti] where ti = a + i∆t for i = 0, 1, 2, . . . , n.It follows that between time ti−1 and time ti, for t = 1, 2, . . . , n, the dis-tance traveled is approximately v(ti)∆t, and the total distance d is thereforeapproximated by the sum of these n distances, which we denote by dn:

d ≈ dn = v(t1)∆t+ v(t2)∆t+ · · ·+ v(tn)∆t. (1.22)

Proceeding as in the area problem, we can define the total distance traveledto be the limit of the approximation dn as n approaches infinity:

d = limn→∞

n∑i=1

v(ti)∆t. (1.23)

Example 6 The velocity of a ball thrown straight up with an initial velocityof 100 ft/s is described by the function v(t) = −32t+ 100, where t denotesthe number of seconds that have elapsed since the ball has been thrown, andv(t) is the velocity at time t, measured in ft/s. Using the above definition ofdistance traveled, we can compute the distance that the ball travels in thefirst two seconds after it has been thrown by dividing the interval [0, 2] inton subintervals of width ∆t = (2− 0)/n, and computing the sum

dn =

n∑i=1

v(ti)2

n=

n∑i=1

(−32ti+100)2

n, ti = 0+i∆t = 2i/n, i = 1, 2, . . . , n.

(1.24)It follows that the total distance d that the ball travels in its first two secondsof flight is

d = limn→∞

dn

= limn→∞

n∑i=1

(−32

2i

n+ 100

)2

n

= limn→∞

n∑i=1

(−128i

n2+

200

n

)= lim

n→∞

[(−128 · 1

n2+

200

n

)+

(−128 · 2

n2+

200

n

)+ · · ·+

(−128 · n

n2+

200

n

)]= lim

n→∞

[−128

n2(1 + 2 + · · ·+ n) +

(200

n+ · · ·+ 200

n

)]. (1.25)


To compute this limit, we first note that 200/n is being added n times,which yields n(200/n) = 200. Second, we use the formula

1 + 2 + · · ·+ n =n(n+ 1)

2. (1.26)

Using the Limit Laws

limn→∞

a = a, limn→∞

cf(n) = c limn→∞

f(n), limn→∞

f(n)+g(n) = limn→∞

f(n)+ limn→∞

g(n),

(1.27)where a and c are constants, we obtain

d = limn→∞

[−128

n2

n(n+ 1)

2+ 200

]= 200− 128

2limn→∞

n(n+ 1)

n2

= 200− 64 limn→∞

n+ 1

n

= 200− 64 limn→∞

(1 +

1

n

)= 200− 64

= 136. (1.28)

We conclude that in its first two seconds of flight, the ball travels 136 feet up.Therefore, if the ball was thrown from a height of 6 feet, the altitude of theball after two seconds would be 142 feet. Note that the altitude of the ballat a particular time cannot be determined exclusively from our knowledgeof the velocity; the initial altitude must also be known. This point will beexamined in greater detail in upcoming sections. 2

1.3 The Definite Integral

In the previous section, we approximated the area under the graph of acontinuous, nonnegative function y = f(x) between the vertical lines x = aand x = b by computing the sum of the areas of n rectangles determinedby the division of the interval [a, b] into n subintervals of equal width ∆x =(b− a)/n. Intuitively, we concluded that as n→∞, the approximate area

A ≈ An =

n∑i=1

f(xi)∆x, (1.29)

1.3. THE DEFINITE INTEGRAL 21

where xi = a + i∆x for i = 1, 2, . . . , n, converges to the exact area of thegiven region.

More generally, suppose that for each n = 1, 2, . . ., we define the quantityRn by choosing points a = x0 < x1 < · · · < xn = b, and computing the sum

Rn =n∑i=1

f(x∗i )∆xi, ∆xi = xi − xi−1, xi−1 ≤ x∗i ≤ xi. (1.30)

The sum in equation (1.30) is known as a Riemann sum. Note that theinterval [a, b] need not be divided into subintervals of equal width, and thatf(x) can be evaluated at arbitrary points belonging to each subinterval.

The Riemann sum Rn approximates the area under y = f(x), betweenx = a and x = b, by the sum of the area of n rectangles, where the widthof the ith rectangle is ∆xi, and the height is f(x∗i ), for i = 1, 2, . . . , n. Thisis a generalization of our previous approximations of the area, in which werequired that each rectangle had the same width ∆x = (b − a)/n, and, fori = 1, 2, . . . , n, the point x∗i belonging to the ith subinterval [xi−1, xi] wasalways chosen to be the left endpoint xi−1 or the right endpoint xi.

If f(x) ≥ 0 on [a, b], then, as n→∞, the Riemann sum Rn converges tothe area under the curve y = f(x), provided that the widths of all subinter-vals [xi−1, xi], for i = 1, 2, . . . , n, also approach zero. If f assumes negativevalues on [a, b], then under the same conditions on the widths of the subin-tervals, Rn converges to the net area between the graph of f and the x-axis,where area below the x-axis is counted negatively. This is due to the factthat the height of each rectangle is determined by f(x), so when f(x) isnegative, rectangles will have negative height, and therefore negative area.This will be discussed in greater detail in the next section.

Our definition of the area under the graph of f(x) between x = a andx = b, while precise, is also, unfortunately, rather cumbersome. This can bealleviated by introducing notation that concisely describes the whole processof computing the area in the manner in which we have described. To thatend, we state the following definition.

Definition 2 Let the function f(x) be continuous on the interval [a, b]. Fur-thermore, let {Rn}∞n=1 be a sequence of Riemann sums, as defined in equation(1.30), that has the property that

limn→∞

[max

1≤i≤n∆xi

]= 0. (1.31)


We define the definite integral of f(x) from a to b to be∫ b

af(x) dx = lim

n→∞Rn. (1.32)

The function f(x) is called the integrand, and the values a and b are calledthe lower and upper limits of integration, respectively. The process ofcomputing the value of a definite integral is called integration.

The concept of the definite integral embodies the entire process of ap-proximating the area of a complicated region by reducing the problem toseveral simpler problems of computing areas of simple regions, and then ob-taining the exact area through a limiting process. By representing the resultof such a complex process using such simple notation, we can much moreeasily obtain solutions to a wide variety of problems. This is the primaryadvantage offered by calculus: it provides us with a simple means of workingmathematically with intuitive concepts such as distance and area.

1.3.1 Evaluating Integrals

Now that we have defined the definite integral for the purpose of comput-ing quantities such as areas and distances, we now discuss the problem ofactually evaluating a definite integral of the form∫ b

af(x) dx, (1.33)

for a given function f(x) and limits of integration a and b. Certainly, oneapproach is to use the definition directly. This entails approximating theintegral (1.33) using a Riemann sum of the areas of n rectangles, and thencomputing the limit as n→∞. In the case where f(x) is a polynomial, it ishelpful to use various well-known formulas for evaluating summations, suchas

n∑i=1

i =n(n+ 1)

2(1.34)

orn∑i=1

i2 =n(n+ 1)(2n+ 1)

6. (1.35)

Similar formulas for higher powers of i can be derived by solving systems oflinear equations.


The following summation rules are also useful for reducing the complex-ity of Riemann sums:

n∑i=1

c = nc (1.36)

n∑i=1

cai = c

n∑i=1

ai (1.37)

n∑i=1

ai + bi =

n∑i=1

ai +

n∑i=1

bi (1.38)

n∑i=1

ai − bi =

n∑i=1

ai −n∑i=1

bi (1.39)

We now show how these rules can be helpful for the task of evaluating adefinite integral using the definition.

Example 7 Compute the area of the region bounded by the curve y = f(x),where

f(x) = x2 + 3x+ 2, (1.40)

as well as the horizontal line y = 0, the vertical line x = 0, and the verticalline x = 3.

Solution The area A of this region, which is shown in Figure 1.4, is givenby the value of the definite integral

A =

∫ 3

0f(x) dx =

∫ 3

0x2 + 3x+ 2 dx. (1.41)

To compute this integral, we approximate the region by n rectangles of equalwidth, with height determined by f(x). Specifically, we divide the interval[0, 3] into n subintervals of equal width ∆x = (3 − 0)/n = 3/n. Thesesubintervals have endpoints [x0, x1], [x1, x2], . . ., [xn−1, xn] where xi = i∆x,for i = 0, . . . , n.

These points are used to determine the rectangles that approximatethe original region. The ith rectangle is determined by the ith subinter-val [xi−1, xi] as follows: the width is the length of the subinterval, whichis ∆x = 3/n, and the height is given by f(xi) = x2

i + 3xi + 2. Figure 1.5illustrates these rectangles for the case of n = 4.

To approximate the area of the region, we can compute the areas of then rectangles and add them. The resulting sum is an example of a Riemann


Figure 1.4: Region bounded by the curve y = x2 + 3x + 2 and the linesy = 0, x = 0 and x = 3.

sum. As n, the number of rectangles, increases, we can expect that theapproximate area will approach the exact area. This is in fact the case,which is why the definite integral is defined to be the limit of a Riemannsum of the areas of n rectangles, as n becomes infinite. In other words,

A =

∫ b

af(x) dx = lim

n→∞

n∑i=1

f(xi)∆x, (1.42)

where the sigma notation∑n

i=1 is used to denote the addition of n expres-sions that are indexed by the variable i that “counts” from 1 to n.

We now compute this limit of a Riemann sum to obtain the area of ourregion. Using the laws of sums, we have

A =

∫ 3

0x2 + 3x+ 2 dx


Figure 1.5: Approximating rectangles for the case of n = 4.

= limn→∞

n∑i=1

(x2i + 3xi + 2)∆x

= limn→∞

n∑i=1

[(i∆x)2 + 3(i∆x) + 2]∆x

= limn→∞

n∑i=1

[(3i

n

)2

+ 3

(3i

n

)+ 2

]3

n

= limn→∞

3

n

n∑i=1

[(3i

n

)2

+ 3

(3i

n

)+ 2

]

= limn→∞

3

n

[n∑i=1

(3i

n

)2

+n∑i=1

3

(3i

n

)+

n∑i=1

2

]


= limn→∞

3

n

[n∑i=1

9i2

n2+

n∑i=1

9i

n+

n∑i=1

2

]

= limn→∞

3

n

[9

n2

n∑i=1

i2 +9

n

n∑i=1

i+ 2n∑i=1

1

]

= limn→∞

3

n

[9

n2

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

6+

9

n

n(n+ 1)

2+ 2n

]= lim

n→∞

[27n(n+ 1)(2n+ 1)

6n3+

27n(n+ 1)

2n2+ 6

]= lim

n→∞

[27(2n3 + 3n2 + n)

6n3+

27(n2 + n)

2n2+ 6

]= lim

n→∞

[27

6

(2 +

3

n+

1

n2

)+

27

2

(1 +

1

n

)+ 6

]=

27

6limn→∞

(2 +

3

n+

1

n2

)+

27

2limn→∞

(1 +

1

n

)+ 6

= 9 +27

2+ 6

= 28.5 (1.43)

2

Certainly, this process of evaluating a definite integral using the defini-tion is quite tedious. In the next section, we will see how a diligent study ofthe behavior of the definite integral can lead to much more efficient methodsof evaluation, at least for certain types of integrands such as the polynomialfeatured in the preceding example.

1.3.2 The Midpoint Rule

Suppose that we are computing a Riemann sum by dividing [a, b] into nsubintervals [xi−1, xi], for i = 1, 2, . . . , n, where x0 = a and xn = b. Bythe definition of a Riemann sum in equation (1.30), we can approximate theunderlying definite integral (1.33) by evaluating the integrand f(x) at anypoint x∗i in each subinterval [xi−1, xi] to obtain the height of the correspond-ing rectangle. However, there is a particular advantage to choosing x∗i to bethe midpoint of the subinterval, mi = (xi−1 +xi)/2. In the case where f is alinear function on [xi−1, xi], the area of the corresponding rectangle is equalto the exact area under f(x), since the region bounded by the graph of f ,the x-axis, and the vertical lines x = xi−1 and x = xi is a trapezoid with


average height mi. This technique of choosing x∗i to be the midpoint mi iscalled the Midpoint Rule. For more general functions, the Midpoint Ruletends to yield a more accurate approximation than using either the left orright endpoint of each subinterval.

1.3.3 Properties of the Definite Integral

The properties (1.36)-(1.39) of summations can be used to prove analogousproperties for definite integrals. For example, by (1.36),∫ b

ac dx = c(b− a). (1.44)

Similarly, the rules (1.37)-(1.39) can be used to establish the following addi-tional properties of definite integrals. In stating these properties, we assumethat f(x) and g(x) are continuous on the interval [a, b], and that c is a con-stant. ∫ b

acf(x) dx = c

∫ b

af(x) dx, (1.45)∫ b

af(x) + g(x) dx =

∫ b

af(x) dx+

∫ b

ag(x) dx, (1.46)∫ b

af(x)− g(x) dx =

∫ b

af(x) dx−

∫ b

ag(x) dx. (1.47)

These properties show that the definite integral is a linear function of theintegrand, just as the differentiation operator d/dx is a linear operator,in view of the Sum Rule, Difference Rule and Constant Multiple Rule fordifferentiation.

There are some useful inequalities pertaining to definite integrals. Firstof all, if f(x) ≤ g(x), then∫ b

af(x) dx ≤

∫ b

ag(x) dx. (1.48)

It follows that if m ≤ f(x) ≤M , then

m(b− a) ≤∫ b

af(x) dx ≤M(b− a). (1.49)

This property will be very useful in the next section, when we prove a veryimportant theoretical result concerning definite integrals.


Intuitively, the area under the graph of f from a to b, combined with thearea under the graph of f from b to c, should equal the area under f froma to c. More precisely,∫ b

af(x) dx+

∫ c

bf(x) dx =

∫ c

af(x) dx. (1.50)

A related property is that interchanging the limits of integration reversesthe sign of the integral; i.e.,∫ b

af(x) dx = −

∫ a

bf(x) dx. (1.51)

This is easy to see by examining a Riemann sum, since it includes differencesof x-values that are negated by interchanging the limits. We illustrate thisin the following example.

Example 8 By the definition of the definite integral, and the definition ofa Riemann sum given in equation (1.30), we can define∫ b

af(x) dx = lim

n→∞Rn, (1.52)

where the Riemann sum Rn is defined using equally spaced points a =x0, x1, x2, . . . , xn = b, where xi = a + i∆x, for i = 0, 1, 2, . . . , n. Such aRiemann sum has the form

Rn =n∑i=1

f(x∗i )∆x, (1.53)

where ∆x = (b− a)/n and x∗i is any point in the subinterval [xi−1, xi]. Thisis the case regardless of whether b > a or a > b. In the latter case, ∆x isnegative instead of positive.

For example, if n = 4, a = 1, and b = 3, then we have ∆x = (3− 1)/4 =1/2, and xi = 1+ i/2, for i = 0, 1, 2, 3, 4. It follows that if we choose x∗i = xifor i = 1, 2, 3, 4, then

R4 = f(x1)∆x+ f(x2)∆x+ f(x3)∆x+ f(x4)∆x

= f(3/2)(1/2) + f(2)(1/2) + f(5/2)(1/2) + f(3)(1/2)

=1

2

[f

(3

2

)+ f(2) + f

(3

2

)+ f(3)

]. (1.54)

1.4. THE FUNDAMENTAL THEOREM OF CALCULUS 29

On the other hand, suppose that we interchange the limits of integrationso that a = 3 and b = 1. Then, ∆x = (1 − 3)/4 = −1/2, and xi = 3 − i/2,for i = 0, 1, 2, 3, 4. If we choose each point x∗i in (1.53) so that x∗i = xi−1,for i = 1, 2, 3, 4, then we have

R4 = f(x0)∆x+ f(x1)∆x+ f(x2)∆x+ f(x3)∆x

= f(3)(−1/2) + f(5/2)(−1/2) + f(2)(−1/2) + f(3/2)(−1/2)

= −1

2

[f

(3

2

)+ f(2) + f

(3

2

)+ f(3)

]. (1.55)

We see that the two Riemann sums are identical, except that they are ofopposite sign. As n→∞, this leads to the relation from equation (1.51). 2

One important consequence of the property (1.51) is that for any functionf(x) and any number a, ∫ a

af(x) dx = 0. (1.56)

This can be proved by setting b = a in equation (1.51). We find that thedefinite integral of f(x) from a to a is equal to its own negation, whichimplies that it must be equal to zero.

1.4 The Fundamental Theorem of Calculus

In order to evaluate a definite integral, we must compute the limit of a Rie-mann sum, which, as we have seen, can be very tedious. In this section, wewill discover a much more efficient method for evaluating definite integralsthat will prove to be useful in many cases.

Our discussion will be very similar to the derivation of differentiationrules. By the definition of a derivative, the derivative of a function f(x) ata point x = x0 is a number, denoted by f ′(x0), and that number representsthe instantaneous rate of change of y = f(x) with respect to x, at x = x0.In order to compute the derivative, using the definition, it is necessary tocompute a limit, which can also be rather tedious.

To alleviate this difficulty, we choose to view the derivative of f(x) as afunction in its own right. That is, we can define the function f ′(x) to be thefunction whose value at every point x0 in the domain of f is equal to thederivative of f at x0, assuming that the derivative exists at x0. Specifically,f ′ is defined by

f ′(x) = limh→0

f(x+ h)− f(x)

h, (1.57)


for each x in the domain of f . By allowing x to be a variable, insteadof a fixed number, we are able to use the definition of the derivative toobtain functions that represent the derivatives of certain types of functions,such as polynomials or trigonometric functions, instead of having to usethe definition directly every time we want to compute the derivative of afunction at a specific point.

Example 9 Consider the function f(x) = x2. We can compute the deriva-tive of f(x) at x = 2 by using the definition of the derivative to obtain

f ′(2) = limh→0

f(2 + h)− f(2)

h

= limh→0

(2 + h)2 − 4

h

= limh→0

4 + 4h+ h2 − 4

h= lim

h→04 + h

= 4. (1.58)

Alternatively, we can use the definition of the derivative in a more abstractmanner, allowing x to vary instead of fixing x = 2. This yields

f ′(x) = limh→0

f(x+ h)− f(x)

h

= limh→0

(x+ h)2 − x2

h

= limh→0

x2 + 2hx+ h2 − x2

h= lim

h→02x+ h

= 2x. (1.59)

Then, we can easily compute f ′(x) at any value of x. 2

By being even more abstract, and using the definition of the derivativewith types of functions instead of specific functions, we obtain the variousdifferentiation rules with which we are familiar.

We will use the same approach to simplify the process of computingdefinite integrals. First, we will view the definite integral of a function f(x)over an interval [a, b] as a function in its own right, and try to understandthis function’s behavior. It turns out that this perspective will lead toa remarkable discovery concerning the relationship between integrals andderivatives.


1.4.1 Part 1: Differentiation Undoes Integration

Let f(x) be a function that is continuous for all x ≥ 0. How does thedefinite integral of f change as the limits of integration change? It is ourhope that the answer to this question will provide us with a more efficientmethod for computing definite integrals. To answer this question, we viewthe definite integral as a function of the limits of integration, which meansthat the limits are now variables, instead of fixed numbers.

For simplicity, we will only allow the upper limit to vary, and we willkeep the lower limit fixed at 0. We define

g(x) =

∫ x

0f(t) dt, x > 0. (1.60)

Then, the value of g(x) at any point x0 > 0 is the definite integral of f(x)from 0 to x0.

How does g(x) change as x changes? We can answer this question byattempting to compute g′(x), which is the instantaneous rate of change ofy = g(x) with respect to x. Using the definition of the derivative, and theproperties of definite integrals introduced in Section 1.3, we obtain

g′(x) = limh→0

g(x+ h)− g(x)

h

= limh→0

1

h

[∫ x+h

0f(t) dt−

∫ x

0f(t) dt

]= lim

h→0

1

h

∫ x+h

xf(t) dt. (1.61)

To compute this limit, we note that if h is sufficiently small, then f iscontinuous on the interval [x, x + h]. We may assume that h is sufficientlysmall, because we are computing a limit as h approaches 0. By the ExtremeValue Theorem, f assumes a maximum value and a minimum value on theinterval [x, x+ h].

Let the maximum and minimum values of f on [x, x+ h] be attained atthe points M(h) and m(h) respectively. That is,

f(m(h)) ≤ f(t) ≤ f(M(h)), x ≤ t ≤ x+ h. (1.62)

Then, by the properties of definite integrals,

f(m(h))h ≤∫ x+h

xf(t) dt ≤ f(M(h))h, (1.63)


since h is the width of the interval [x, x+ h].

Because we are computing the limit as h approaches 0, we are not con-cerned with the case of h = 0. Therefore, we can divide through by h andobtain

f(m(h)) ≤ 1

h

∫ x+h

xf(t) dt ≤ f(M(h)). (1.64)

Now, we let h approach 0. Because x ≤ m(h) ≤ x+ h, and x ≤M(h) ≤x+ h, it follows from the Squeeze Theorem that

limh→0

m(h) = x, limh→0

M(h) = x. (1.65)

Because f is continuous on [x, x+ h], it follows from the definition of conti-nuity that

limh→0

f(m(h)) = f(x), limh→0

f(M(h)) = f(x). (1.66)

By the Squeeze Theorem, it follows from equations (1.64) and (1.66) that

limh→0

1

h

∫ x+h

xf(t) dt = f(x). (1.67)

We conclude that

g′(x) = limh→0

1

h

∫ x+h

xf(t) dt = f(x). (1.68)

In words, this equation states that the instantaneous rate of change of thearea of a region with respect to its width is equal to the height of the regionat the point where the width is changing. This is easy to see in the simplecase of a rectangle of width w and height h. Since the area A of the rectangleis given by the formula A = wh, we can easily determine that the rate ofchange of the area with respect to the width is given by

dA

dw=d(wh)

dw= h, (1.69)

since the height of a rectangle is constant. Now, however, we can computethe rate of change of the area of a region with respect to its width, even ifthe height is not constant.

More generally, suppose that we define the function F (x) by

F (x) =

∫ x

af(t) dt, (1.70)


where f(t) is continuous on [a, b]. Then, Part 1 of the Fundamental Theoremof Calculus states that for any x in (a, b), F is continuous and differentiableat x, and F ′(x) = f(x). The function F (x) is called an antiderivative off(x). Later in this section, we will see how antiderivatives can be used toeasily evaluate definite integrals.

Example 10 Differentiate the function

F (x) =

∫ x

3t2 + 1 dt. (1.71)

Solution Recall Part 1 of the Fundamental Theorem of Calculus, whichstates that if f(x) is continuous, and the function F (x) is defined by theintegral

F (x) =

∫ x

af(t) dt, (1.72)

then F ′(x) = f(x). Applying this result to the given function F (x), we have

F ′(x) =d

dx

[∫ x

3t2 + 1 dt

]= x2 + 1. (1.73)

2


F (x) =

∫ 0

xt2 + 1 dt. (1.74)

Solution We use the following property of definite integrals:∫ b

af(x) dx = −

∫ a

bf(x) dx. (1.75)

It follows that

F ′(x) =d

dx

[∫ 0

xt2 + 1 dt

]=

d

dx

[−∫ x

0t2 + 1 dt

]= − d

dx

[∫ x

0t2 + 1 dt

]= −(x2 + 1). (1.76)


It should be noted that except for the minus sign, the derivative is thesame as in the previous example. In other words, the integral from 0 to xhas the same derivative as integral from 3 to x. This makes sense becausethe rate of change of the area under sin(t2 + 1), with respect to the widthx, should only depend on the height of the region, at the point where theregion is changing, which is at t = x. 2


F (x) =

∫ x

3sin(t2 + 1) dt. (1.77)

Solution Recall Part 1 of the Fundamental Theorem of Calculus, whichstates that if f(x) is continuous, and the function F (x) is defined by theintegral

F (x) =

∫ x

af(t) dt, (1.78)

then F ′(x) = f(x). Note that the value of F (x) at any x is the area underthe curve y = f(t), between t = a and t = x.

Applying this result to the given function F (x), we have

F ′(x) =d

dx

[∫ x

3sin(t2 + 1) dt

]= sin(x2 + 1). (1.79)

2


F (x) =

∫ x2

3sin(t2 + 1) dt. (1.80)

Solution Let g(x) be the function defined by

g(x) =

∫ x

3sin(t2 + 1) dt. (1.81)

From the previous example, we know that g′(x) = sin(x2 + 1). The functionF (x) in this example can be written as g(x2), since F (x) and g(x) areidentical, except that the upper limit of integration for g(x) is x and theupper limit for F (x) is x2. Therefore, F (x) can be obtained from g(x) simplyby replacing x with x2 in the definition of g(x).


We can differentiate F (x) = g(x2) using the Chain Rule, which yields

F ′(x) =d

dx[g(x2)] = g′(x2)(2x) = sin((x2)2 + 1)(2x) = 2x sin(x4 + 1).

(1.82)2


F (x) =

∫ x2

1/xsin(t2 + 1) dt. (1.83)

Solution Let g(x) be defined by∫ x

0sin(t2 + 1) dt. (1.84)

Then, as in the previous example, g′(x) = sin(x2 + 1). Using the propertiesof definite integrals, we can write F (x) as

F (x) =

∫ x2

1/xsin(t2 + 1) dt

=

∫ 0

1/xsin(t2 + 1) dt+

∫ x2

0sin(t2 + 1) dt

= −∫ 1/x

0sin(t2 + 1) dt+

∫ x2

0sin(t2 + 1) dt

= −g(1/x) + g(x2). (1.85)

Therefore, we can use the Chain Rule to obtain

F ′(x) =d

dx[g(x2)]− d

dx[g(1/x)]

= g′(x2)(2x)− g′(1/x)(−1/x2)

= sin((x2)2 + 1)(2x) + sin((1/x)2 + 1)(1/x2)

= 2x sin(x4 + 1) +sin(

1x2

+ 1)

x2. (1.86)

2


F (x) =

∫ sinx

3t2 + 1 dt. (1.87)


Solution If we define

g(x) =

∫ x

3t2 + 1 dt, (1.88)

then it follows that F (x) = g(sinx). From Example 10, we know that

g′(x) = x2 + 1. (1.89)

Using the Chain Rule, we obtain

F ′(x) =d

dx[g(sinx)]

= g′(sinx) cosx

= [(sinx)2 + 1] cosx. (1.90)

2


F (x) =

∫ sinx

x2t2 + 1 dt. (1.91)

Solution We use the following property of definite integrals,∫ b

af(x) dx =

∫ c

af(x) dx+

∫ b

cf(x) dx, (1.92)

with a = x2, b = sinx, and c = 0. Furthermore, we define

g(x) =

∫ x

0t2 + 1 dt, (1.93)

and note that by Part 1 of the Fundamental Theorem of Calculus,

g′(x) = x2 + 1. (1.94)

It follows that

F ′(x) =d

dx

[∫ sinx

x2t2 + 1 dt

]=

d

dx

[∫ sinx

0t2 + 1 dt+

∫ 0

x2t2 + 1 dt

]=

d

dx

[∫ sinx

0t2 + 1 dt−

∫ x2

0t2 + 1 dt

]


=d

dx

[∫ sinx

0t2 + 1 dt

]− d

dx

[∫ x2

0t2 + 1 dt

]

=d

dx[g(sinx)]− d

dx

[g(x2)

]= g′(sinx) cosx− g′(x2)(2x)

= [(sinx)2 + 1] cosx− [(x2)2 + 1](2x)

= (sin2 x+ 1) cosx− 2x(x4 + 1). (1.95)

2


F (x) =

∫ g(x)

h(x)f(t) dt. (1.96)

Solution We define the function G(x) by

G(x) =

∫ x

0f(t) dt. (1.97)

By Part 1 of the Fundamental Theorem of Calculus, G′(x) = f(x). Usingthe same approach as in the previous example, we rewrite F (x) using theproperties of definite integrals and obtain

F (x) =

∫ 0

h(x)f(t) dt+

∫ g(x)

0f(t) dt

= −∫ h(x)

0f(t) dt+

∫ g(x)

0f(t) dt

= −G(h(x)) +G(g(x)). (1.98)

Using the Chain Rule, we obtain

F ′(x) =d

dx[G(g(x))]− d

dx[G(h(x))]

= G′(g(x))g′(x)−G′(h(x))h′(x)

= f(g(x))g′(x)− f(h(x))h′(x). (1.99)

2


1.4.2 Part 2: Integration Undoes Differentiation

Previously, we learned that the derivative of the integral of a function f(x)with respect to the integral’s upper limit was equal to f(x). Specifically, if

g(x) =

∫ x

af(t) dt, (1.100)

then, by Part 1 of the Fundamental Theorem of Calculus, g′(x) = f(x).This result, however, is only one part of this theorem. The second part,which follows from the first part described in equation (1.100), tells us howwe can evaluate the definite integral∫ b

af(t) dt. (1.101)

Suppose that F (x) is any antiderivative of f ; that is, F ′(x) = f(x). Thenthe antiderivative g(x) defined in equation (1.100) satisfies g(x) = F (x) +Cfor some constant C, since g′(x)− F ′(x) ≡ 0 implies g(x)− F (x) must be aconstant. By the properties of definite integrals and the definition of g(x),we have g(a) = 0. It follows that∫ b

af(t) dt = g(b)

= g(b)− g(a)

= F (b) + C − [F (a) + C]

= F (b)− F (a). (1.102)

In summary, we can compute a definite integral of f simply by finding anantiderivative of f , and then evaluating the antiderivative at the limits ofintegration.

1.4.3 Statement of the Fundamental Theorem of Calculus

We are now ready to summarize these results and formally state the mostimportant theoretical result in all of differential and integral calculus. Inthe statement of the following theorem, note that we assume that functionsare differentiable on a closed interval. A function f(x) is differentiable on aclosed interval [a, b] if it is differentiable on (a, b) and f is differentiable fromright at a and differentiable from the left at b; that is, the one-sided limits

limx→a+

f(x)− f(a)

x− aand lim

x→b−f(x)− f(b)

x− b(1.103)

both exist. This is similar to the definition of continuity on a closed interval.


Theorem 1 (The Fundamental Theorem of Calculus) Let f(x) be continu-ous on [a, b].

1. If the function g(x) is defined by the definite integral

g(x) =

∫ x

af(t) dt, (1.104)

then g(x) is continuous on [a, b] and differentiable on [a, b], and

g′(x) = f(x), a ≤ x ≤ b. (1.105)

2. If F (x) is any antiderivative of f(x) on [a, b]; that is, if

F ′(x) = f(x), a ≤ x ≤ b, (1.106)

then ∫ b

af(x) dx = F (b)− F (a). (1.107)

Example 18 Let f(x) = x2. Then, if we define

g(x) =

∫ x

0f(t) dt =

∫ x

0t2 dt, (1.108)

then, by Part 1 of the Fundamental Theorem of Calculus, g′(x) = x2. Fur-thermore, if F (x) = x3/3, then F (x) is an antiderivative of f(x), sinceF ′(x) = x2. It follows from Part 2 of the Fundamental Theorem of Calculusthat ∫ 4

0x2 dx = F (4)− F (0) =

43

3− 03

3=

64

3. (1.109)

This implies that the area under the graph of x2 from x = 0 to x = 4 isequal to 64/3. 2

1.4.4 Differentiation and Integration as Inverse Processes

In words, the Fundamental Theorem of Calculus states that the derivativeof an integral of a function is the function itself, and that the integral of aderivative of a function is the function, up to an additive constant. In otherwords, differentiation and integration are inverse processes, just as divisionand multiplication are inverse operations. This parallel is even closer thanit would appear on the surface. Consider the following:


• When speed is constant over time, distance is the product of speedand time. In general, distance is the integral of speed over time.

• If speed is constant over time, then it is equal to the ratio of distanceto time. In general, the speed at any given time is the derivative, orrate of change, of distance traveled with respect to time.

1.4.5 Computing Area Using the Fundamental Theorem ofCalculus

Example 19 Let h and b be positive constants. Use the FundamentalTheorem of Calculus to compute the area of the region bounded by thecurve y = f(x), where f(x) = hx/b, the horizontal line y = 0, and thevertical lines x = 0 and x = b.

Solution The region is shown in Figure 1.6. It is a triangle with height h

Figure 1.6: Region bounded by the curve y = f(x) = hx/b, the horizontalline y = 0, and the vertical lines x = 0 and x = b.


and a base of length b. By the definition of the definite integral, the area Aof this region is given by

A =

∫ b

0f(x) dx =

∫ b

0

hx

bdx. (1.110)

We use Part 2 of the Fundamental Theorem of Calculus, which statesthat for any function f that is continuous on [a, b],∫ b

af(x) dx = F (b)− F (a), (1.111)

where F (x) is any antiderivative of f(x); that is, F ′(x) = f(x). We thereforeneed to find an antiderivative of f(x) = hx/b.

To accomplish this, we use two anti-differentiation rules: first, if F is anantiderivative of f , then, for any constant c, cF is an antiderivative of cf .Second, the antiderivative of xn, for any integer n 6= −1, is xn/(n + 1). Itfollows that if f(x) = hx/b, then

F (x) =h

b

x1+1

1 + 1=hx2

2b(1.112)

is an antiderivative of f(x). This can be verified by differentiating F (x),which yields

F ′(x) =d

dx

[hx2

2b

]=

h

2b

d

dx[x2] =

h

2b(2x) =

hx

b= f(x). (1.113)

Now that we have found an antiderivative of f(x), we apply the Funda-mental Theorem of Calculus and obtain∫ b

0f(x) dx = F (b)− F (0) =

h(b2)

2b− h(02)

2b=hb2

2b=

1

2bh, (1.114)

which is the familiar formula for the area of a triangle of height h and baseof length b. 2

Example Compute the area of the region bounded by the graph of

y = f(x) =y2 − y1

hx+ y1, (1.115)

the line y = 0, and the lines x = 0 and x = h, where y1, y2, and h are allconstants.


Solution This region is a trapezoid of width h and heights y1 and y2. Thefunction f(x) is a linear function, defining a line that passes through thepoints (0, y1) and (h, y2). To compute the area of this region, we again usePart 2 of the Fundamental Theorem of Calculus.

Using the rule that an antiderivative of xn is xn+1/(n + 1), providedn 6= −1, we can determine that the function F (x), defined by

F (x) =y2 − y1

h

x2

2+ y1x, (1.116)

is an antiderivative of f(x). This can be verified by differentiating F (x), asin the previous example.

From Part 2 of the Fundamental Theorem of Calculus, we find that thearea A of the given region is

A =

∫ h

0f(x) dx

=

∫ h

0

y2 − y1

hx+ y1 dx

= F (h)− F (0)

=

[y2 − y1

h

h2

2+ y1h

]−[y2 − y1

h

02

2+ y1(0)

]=

y2 − y1

h

h2

2+ y1h

= (y2 − y1)h

2+ y1h

=h

2y2 −

h

2y1 + hy1

=h

2y2 +

h

2y1

= hy1 + y2

2, (1.117)

which is the well-known formula for the area of a trapezoid of width h andheights y1 and y2. 2

Example 20 Compute the area of the region bounded by the curve y =sinx, where 0 ≤ x ≤ π, and the line y = 0.

Solution The region is shown in Figure 1.7. To compute the area A of thisregion, we need to evaluate the definite integral

A =

∫ π

0sinx dx. (1.118)


Figure 1.7: Region bounded by y = sinx and y = 0

To accomplish this, we first obtain an antiderivative of f(x) = sinx.Since

d

dx[cosx] = − sinx, (1.119)

it follows that F (x) = − cosx is an antiderivative of sinx, as can be verifiedby noting that

F ′(x) =d

dx[− cosx] = − d

dx[cosx] = −(− sinx) = sinx. (1.120)

Therefore, by Part 2 of the Fundamental Theorem of Calculus, the areaA of the region is given by

A =

∫ π

0sinx dx

= F (π)− F (0)

= − cosπ − (− cos 0)


= −(−1)− (−1)

= 1 + 1

= 2. (1.121)

2

1.4.6 Net Area vs. Total Area

We know that the definite integral∫ b

af(x) dx

can be used to compute the area of the region bounded by the vertical linesx = a and x = b, the horizontal line y = 0, and the curve y = f(x), whenf(x) ≥ 0 on [a, b]. If f(x) < 0 on [a, b], then the integral represents thenegative of the area of the region, because it is the limit of a sum of areas ofrectangles whose heights are negative. We now consider the interpretationof the integral when the integrand f(x) can be either positive or negativeon [a, b].

Example 21 Evaluate the definite integral∫ 2π

0sinx dx. (1.122)

What is the geometric interpretation of this integral?

Solution Using the Fundamental Theorem of Calculus, and the fact thatan antiderivative of f(x) = sinx is F (x) = − cosx, we obtain∫ 2π

0sinx dx = F (2π)− F (0)

= − cos(2π)− (− cos(0))

= −1− (−1)

= −1 + 1

= 0. (1.123)

The graph of f(x) = sinx on the interval [0, 2π] is shown in Figure 1.8.What is the geometric significance of the integral being equal to 0? Notethat in the previous example, we determined that the integral of the same


Figure 1.8: Graph of y = sinx, 0 ≤ x ≤ 2π

function from x = 0 to x = π is equal to 2, and that is the area of the regionbounded by the curve y = sinx and the lines y = 0, x = 0 and x = π. FromFigure 1.8, we can visually determine that the area of the region boundedby the curve y = sinx and the lines y = 0, x = π and x = 2π is also equalto 2. Given this information, why is the integral from x = 0 to x = 2π equalto 0? We can answer this question by computing∫ 2π

πsinx dx = F (2π)− F (π)

= − cos(2π)− (− cos(π))

= −1− (−(−1))

= −1− 1

= −2. (1.124)

This example illustrates that∫ ba f(x) dx is not always equal to the area of

the region bounded by the curve y = f(x) and the lines y = 0, x = a and


x = b. Rather, it is the net area above the x-axis; that is, area above the x-axis is counted positively, while area below the x-axis is counted negatively.Therefore, the area between x = 0 and x = π, which is positive, cancelswith the area between x = π and x = 2π, which is negative, thus explainingwhy the integral is equal to zero. 2

In general, to compute the gross or total area, rather than the net area,of the region bounded by y = f(x), y = 0, x = a and x = b, it is necessaryto evaluate the definite integral∫ b

a|f(x)| dx. (1.125)

This is best accomplished by dividing the interval [a, b] into subintervalsbased on where f(x) is positive or negative, and then integrating over eachof these subintervals, using the fact that |x| = −x when x < 0 in orderto eliminate the absolute value from the problem. Then, the integral fromx = a to x = b is the sum of the integrals over the subintervals.

Example 22 To compute the total area under y = sinx from x = 0 tox = 2π, we need to compute

A =

∫ 2π

0| sinx| dx. (1.126)

We can compute this integral as follows: we note that sinx ≥ 0 on [0, π],while sinx ≤ 0 on [π, 2π]. It follows that | sinx| = sinx on [0, π], while| sinx| = − sinx on [π, 2π]. We then have

A =

∫ 2π

0| sinx| dx

=

∫ π

0| sinx| dx+

∫ 2π

π| sinx| dx

=

∫ π

0sinx dx+

∫ 2π

π− sinx dx

=

∫ π

0sinx dx−

∫ 2π

πsinx dx

= − cosx|π0 − [− cosx]|2ππ= [− cos(π)− (− cos(0))]− [− cos(2π)− (− cos(π))]

= [−(−1)− (−1)]− [−1− (−(−1))]


= [1 + 1]− [−1− 1]

= 2− (−2)

= 2 + 2

= 4. (1.127)

This value is the total area of the region enclosed by the curve y = sinxand the lines y = 0, x = 0 and x = 2π. Note that in this case we used thenotation

f(x)|ba = f(b)− f(a). (1.128)

2

1.4.7 Displacement vs. Distance

We now revisit the problem of computing the distance that an object hastraveled along a straight line, given its velocity v(t), for a ≤ t ≤ b. If the ve-locity changes sign from positive, then it follows that the object has changeddirection from right to left, and “positive distance” previously traveled tothe right is offset by “negative distance” traveled to the left.

Just as the definite integral of a function f(x) represents the net areaunder the curve y = f(x), the definite integral of a function v(t) representsthe displacement, or “net distance”, traveled by an object along a straightline with velocity v(t). In order to compute the total distance traveled bythe object, it is necessary to integrate |v(t)| instead.

Example 23 A falling object has a constant acceleration of a(t) = −32 ft/s2.Since acceleration is the derivative of the velocity v(t), it follows that v(t)is an antiderivative of a(t). From Part 2 of the Fundamental Theorem ofCalculus, we have ∫ t

0a(s) ds = v(t)− v(0).

Therefore, if the initial velocity is v0 = v(0) ft/s, then the velocity is givenby rearranging the above equation:

v(t) = v(0) +

∫ t

0a(s) ds = v0 +

∫ t

0(−32) ds = v0 + (−32s)|t0 = v0 − 32t.

To compute the displacement d(t), whose derivative is the velocity v(t), weagain apply Part 2 of the Fundamental Theorem of Calculus to obtain∫ t

0v(s) ds = d(t)− d(0),


and if we denote the initial position by d0 = d(0) ft, we have

d(t) = d0 +

∫ t

0v(s) ds

= d0 +

∫ t

0(v0 − 32s) ds

= d0 +(v0s− 16s2

)∣∣t0

= d0 + v0t− 16t2.

Now, suppose that d0 = 6 ft and v0 = 128 ft/s. Then, to compute thedisplacement between t = 0 s and t = 5 s, we use the fact that displacementis an antiderivative of velocity to obtain∫ 5

0v(s) ds = d(5)− d(0)

= [6 + 128(5)− 16(5)2]− [6 + 128(0)− 16(0)2]

= 246− 6 = 240 ft.

On the other hand, to compute the distance traveled by the object betweent = 0 s and t = 5 s, we must compute∫ 5

0|v(s)| ds =

∫ 5

0|128− 32s| ds

=

∫ 4

0(128− 32s) ds+

∫ 5

4(32s− 128) ds

=(128s− 16s2

)∣∣40

+(16s2 − 128s

)∣∣54

= 256 + (−240− (−256))

= 272 ft.

The integral from s = 0 to s = 5 is divided into two integrals at s = 4,because v(s) ≥ 0 on [0, 4], while v(s) ≤ 0 on [4, 5]. By integrating v(s) froms = 0 to s = 4, and −v(s) from s = 4 to s = 5, we obtain the integral of|v(s)| from s = 0 to s = 5, which represents the distance traveled betweent = 0 s and t = 5 s. 2

1.4.8 Average Values

Suppose that we wish to compute the average value of that a function f(x)assumes over the interval [a, b]. If f(x) is a constant function f(x) = c, then


we know that the average value is simply c. For a more general function f ,we can approximate the average by dividing [a, b] into subintervals of width∆x = (b − a)/n, and assuming that f is approximately constant over eachsubinterval. Taking the average over the ith subinterval to be f(x∗i ), wherex∗i is the midpoint of the ith subinterval, we obtain the approximate average

Avg f ≈ 1

n

n∑i=1

f(x∗i ) (1.129)

which can be rewritten as

Avg f ≈ 1

b− a

n∑i=1

f(x∗i )∆x. (1.130)

Letting n→∞, we obtain the exact average as a definite integral

Avg f =1

b− a

∫ b

af(x) dx. (1.131)

Let f be differentiable over the interval [a, b]. Recall from differentialcalculus that by the Mean Value Theorem for derivatives, there exists somepoint c in [a, b] such that f ′(c) is equal to the average of f ′(x) over [a, b]. Itis natural to ask if there is some point c in [a, b] such that the total changein f from a to b is equal to f(c)(b − a), the area of the rectangle that hasheight f(c) and width (b−a). The following theorem answers this question.

Theorem 2 (Mean Value Theorem for Integrals) If f is continuous on[a, b], then ∫ b

af(x) dx = f(c)(b− a) (1.132)

for some c in (a, b).

Proof Since f is continuous on [a, b], it follows from Part 1 of the Funda-mental Theorem of Calculus that the function F (x) defined by

F (x) =

∫ x

af(t) dt (1.133)

is continuous on [a, b] and differentiable on (a, b), and F ′(x) = f(x) on (a, b).By the Mean Value Theorem for derivatives,

F (b)− F (a)

b− a= F ′(c) (1.134)


for some c in (a, b). However, by Part 2 of the Fundamental Theorem ofCalculus,

F (b)− F (a) =

∫ b

af(x) dx. (1.135)

Substituting the right side of equation (1.135) into equation (1.134) yields

1

b− a

∫ b

af(x) dx = F ′(c), (1.136)

and since F ′(x) = f(x) for any x in (a, b), it follows that

1

b− a

∫ b

af(x) dx = f(c), (1.137)

and multiplying both sides by b − a yields the conclusion of the theorem,stated in equation (1.132). 2

Remark From equation (1.137), we obtain an intuitive interpretation of theMean Value Theorem for Integrals: a function f that is continuous on [a, b]assumes its average value at some point in (a, b). 2

Example 24 Find b such that the average value of f(x) = x2 + b on theinterval [0, 2] is equal to 8.

Solution The average value of f(x) = x2 + b on [0, 2] is given by

1

2− 0

∫ 2

0x2 + b dx =

1

2

(x3

3+ bx

)∣∣∣∣20

=1

2

(8

3+ 2b

)=

4

3+ b. (1.138)

It follows that in order for the average value of f(x) on [0, 2] to be equal to8, we must have

b = 8− 4

3=

20

3. (1.139)

2

Example 25 Compute the average value of f(x) = x2 on the interval [0, 2].

Solution We use the formula for the average value of f(x) on an interval[a, b],

Avg f =1

b− a

∫ b

af(x) dx. (1.140)

This yields

Avg f =1

2− 0

∫ 2

0x2 dx =

1

2

x3

3

∣∣∣∣20

=1

2

8

3=

4

3. (1.141)

2

1.5. INDEFINITE INTEGRALS 51

Example 26 Find the value of b so that the average value of f(x) = x3 onthe interval [0, b] is equal to 4.

Solution The average value of f(x) on [0, b] is given by

1

b− 0

∫ b

0x3 dx =

1

b

x4

4

∣∣∣∣b0

=1

b

b4

4=b3

4. (1.142)

Therefore, in order for the average value to be equal to 4, we must haveb3/4 = 4. It follows that b3 = 16, which yields the solution b = 3

√16 ≈

2.5198. 2

1.5 Indefinite Integrals

As evaluating integrals can be accomplished by computing antiderivatives, itis desirable to know the antiderivatives of commonly used functions. For easeof exposition, the notation of an integral sign without limits of integrationis used to denote the set of all antiderivatives of a function. We now definethis notation precisely.

Definition 3 (Indefinite Integral) Let f(x) be a continuous function. Theindefinite integral of f(x), denoted by the symbol∫

f(x) dx, (1.143)

is the set of all antiderivatives of f(x).

It is important to note that the indefinite integral of f(x) does not repre-sent a specific function. Instead, it represents a family of infinitely manyfunctions, all of which have the common property that their derivatives areequal to f(x).

Fortunately, this family of functions is very easy to describe. If we adda constant to a function, its derivative does not change, since the derivativeof a constant is zero. Furthermore, by the Mean Value Theorem, any twofunctions that have the same derivative differ from one another by a con-stant. It follows that if F (x) is any antiderivative of f(x), then a functionG(x) is also an antiderivative of f(x) if and only if

G(x) = F (x) + C, (1.144)


where C is a constant. Since all antiderivatives of f(x) have the form F (x)+C, we write

∫f(x) dx = F (x) + C, (1.145)

where F (x) is any antiderivative of f(x), and C represents an arbitrary con-stant. This notation is used in order to specify the “most general antideriva-tive” of f(x), because this generality may be needed if the antiderivative isused in another problem that may specify a value for C, as in the followingexample.

Example 27 We can determine that an antiderivative of the function f(t) =−32 is F (t) = −32t, since the derivative of −32t with respect to t is −32.However, the most general antiderivative of this function is F (t) = −32t+C,where C is an arbitrary constant. If we interpret f(t) as the acceleration ofan ball thrown in to the air, then the velocity of the ball is a function of theform F (t) = −32t + C, where the constant C is the initial velocity of theball. Therefore, the antiderivative −32t, without the constant C, is of littleuse, because it is not general enough to describe the velocity. 2

Several indefinite integrals can be obtained easily from differentiationrules, as shown in the following table. We assume that F (x) and G(x) areantiderivatives of f(x) and g(x), respectively; that is, F ′(x) = f(x) andG′(x) = g(x). With each indefinite integral, we add the arbitrary constantC in order to specify the most general indefinite integral possible.


Differentiation Rule Indefinite Integralddx [f(x) + g(x)] = f ′(x) + g′(x)

∫f(x) + g(x) dx = F (x) +G(x) + C

ddx [f(x)− g(x)] = f ′(x)− g′(x)

∫f(x) + g(x) dx = F (x)−G(x) + C

ddx [cf(x)] = cf ′(x)

∫cf(x) dx = cF (x) + C

ddx [xn] = nxn−1

∫xn dx = xn+1

n+1 + C, n 6= −1ddx [sinx] = cosx

∫cosx dx = sinx+ C

ddx [cosx] = − sinx

∫sinx dx = − cosx+ C

ddx [tanx] = sec2 x

∫sec2 x dx = tanx+ C

ddx [cotx] = − csc2 x

∫csc2 x dx = − cotx+ C

ddx [secx] = secx tanx

∫secx tanx dx = secx+ C

ddx [cscx] = − cscx cotx

∫cscx cotx dx = − cscx+ C

ddx [sin−1 x] = 1√

1−x2∫

1√1−x2 dx = sin−1 x+ C

ddx [tan−1 x] = 1

1+x2

∫1

1+x2dx = tan−1 x+ C

ddx [ln |x|] = 1

x

∫1x dx = ln |x|+ C

ddx [ex] = ex

∫ex dx = ex + C

ddx [ax] = ax ln a

∫ax dx = ax

ln a + Cddx [f(g(x))] = f ′(g(x))g′(x)

∫f(g(x))g′(x) dx = F (g(x)) + C

ddx [f(x)g(x)] = g(x)f ′(x) + f(x)g′(x)

∫F (x)g(x) dx = F (x)G(x)−

∫G(x)f(x) dx

In the last indefinite integral, which is actually a statement of the techniqueof integration by parts, the arbitrary constant C is not included because anindefinite integral appears on both sides of the equation, and therefore it isnot necessary to include C in order to specify the most general indefiniteintegral.

Example 28 Compute ∫x3 − 6x2 + 5x− 4 dx. (1.146)

Solution We use the following anti-differentiation rules, which are namedaccording to the differentiation rules from which they are obtained. In de-scribing these rules, we assume that F ′(x) = f(x) and G′(x) = g(x).∫

f(x) + g(x) dx = F (x) +G(x) + C (Sum Rule) (1.147)∫f(x)− g(x) dx = F (x)−G(x) + C (Difference Rule) (1.148)∫

cf(x) dx = cF (x) + C (Constant Multiple Rule)(1.149)


∫xn dx =

xn+1

n+ 1, n 6= −1 (Power Rule) (1.150)

We proceed as follows:∫x3 − 6x2 + 5x− 4 dx =

∫x3 dx−

∫6x2 dx+

∫5x dx−

∫4 dx

=

∫x3 dx− 6

∫x2 dx+ 5

∫x dx− 4

∫1 dx

=x4

4− 6

x3

3+ 5

x2

2− 4x+ C

=x4

4− 2x3 +

5x2

2− 4x+ C. (1.151)

Note that even though we evaluated each term as a separate indefinite in-tegral, it is only necessary to include one term that is an arbitrary constantC, because it can be equal to any constant. 2

Example 29 Compute ∫secx tanx− cosx dx. (1.152)

Solution Using the Difference Rule as well as the rules∫secx tanx dx = secx+ C, (1.153)∫

cosx dx = sinx+ C, (1.154)

we obtain∫secx tanx− cosx dx =

∫secx tanx dx−

∫cosx dx = secx− sinx+ C.

(1.155)2

Example 30 Compute ∫8x7 + sec2 x dx. (1.156)

Solution Using the Sum Rule, the Constant Multiple Rule, the Power Ruleand the rule ∫

sec2 x dx = tanx+ C, (1.157)


we obtain∫8x7+sec2 x dx = 8

∫x7 dx+

∫sec2 x dx = 8

x8

8+tanx+C = x8+tanx+C.

(1.158)2

Example 31 Given that the acceleration a(t) of an object at time t is givenby

a(t) = 7 cos t+ 5, (1.159)

and its initial velocity is v(0) = −4 and its initial position is s(0) = 10,compute the object’s velocity function v(t) and the position function s(t).

Solution Since v′(t) = a(t), it follows that v(t) must be an antiderivativeof a(t). Therefore, v(t) has the form

v(t) = F (t) + C, (1.160)

where F (t) is any antiderivative of a(t) and the constant C is to be deter-mined so that v(0) = −4.

To compute an antiderivative of a(t), we use the Sum Rule, the ConstantMultiple Rule, and the rule∫

cosx dx = sinx+ C (1.161)

and obtain

v(t) =

∫a(t) dt = 7

∫cos t dt+ 5

∫1 dt = 7 sin t+ 5t+ C. (1.162)

We now specify the constant C so that v(0) = −4. Substituting t = 0 intothe equation v(t) = 7 sin t+ 5t+ C yields

v(0) = 7 sin 0 + 5(0) + C, (1.163)

which simplifies to −4 = C since v(0) = −4 and sin 0 = 0. Therefore, thevelocity of the object is given by

v(t) = 7 sin t+ 5t− 4. (1.164)


To obtain the position function, we use the fact that s′(t) = v(t); thatis, s(t) is an antiderivative of v(t). We have∫

v(t) dt =

∫7 sin t+ 5t− 4 dt

= 7

∫sin t dt+ 5

∫t dt− 4

∫1 dt

= −7 cos t+5t2

2− 4t+ C. (1.165)

We need to determine the value of C so that s(0) = 10. Substituting t = 0into the equation

s(t) = −7 cos t+5t2

2− 4t+ C (1.166)

yields

s(0) = −7 cos 0 + C, (1.167)

but since cos 0 = 1 and s(0) = 10, we have 10 = −7 + C, which impliesC = 17. We conclude that

s(t) = −7 cos t+5t2

2− 4t+ 17 (1.168)

is the position of the object at time t. 2

1.5.1 The Net Change Theorem

Recall the second part of the Fundamental Theorem of Calculus,

F ′(x) = f(x) =⇒∫ b

af(x) dx = F (b)− F (a). (1.169)

By rewriting this relation as∫ b

aF ′(x) dx = F (b)− F (a), (1.170)

we obtain the interpretation that the integral of a rate of change of a quantityfrom a to b represents the net change in that quantity from a to b. Thisresult is known as the Net Change Theorem.

Example 32 Let v(t) represent the velocity of an object that is moving ina straight line, and s(t) represent its position. Then s′(t) = v(t), and, by the

1.6. THE SUBSTITUTION RULE 57

Net Change Theorem, the total displacement of the object from the timet = 0 until the time t = 5 is given by

s(5)− s(0) =

∫ 5

0s(t) dt =

∫ 5

0v(t) dt. (1.171)

Alternatively, if the position s(0) is known, and the velocity v(t) is known,then the position at any time t = b is given by the following rearrangementof the Net Change Theorem,

s(b) = s(0) +

∫ b

0v(t) dt. (1.172)

2

1.6 The Substitution Rule

In the last section, it was mentioned that indefinite integrals can be obtainedfrom differentiation rules. This includes the Chain Rule for differentiation,which states that

d

dx[f(g(x))] = f ′(g(x))g′(x). (1.173)

Reversing this process, suppose that F is an antiderivative of a continuousfunction f , and that g is a differentiable function. It follows from (1.173)that

d

dx[F (g(x))] = f(g(x))g′(x) (1.174)

which yields ∫f(g(x))g′(x) dx = F (g(x)) + C, (1.175)

where C is an arbitrary constant.If we make a change of variable u = g(x) on the right side of equation

(1.175), then from the relation∫f(x) dx = F (x) + C (1.176)

we obtain the Substitution Rule∫f(g(x))g′(x) dx =

∫f(u) du. (1.177)

Note that the constant C is dropped because both sides of the equation(1.177) are indefinite integrals.


The Substitution Rule is used to compute indefinite integrals by attempt-ing to recognize whether the integrand is of the form f(g(x))g′(x) for somechoice of f and g. This illustrates the primary asymmetry of differentiationand integration: differentiation tends to be a much more straightforwardprocess, involving the application of a sequence of rules, where the sequenceis easily determined. Integration, on the other hand, requires the determi-nation of whether the integrand can be viewed as the result of some differen-tiation rule, and such determination consists of a difficult pattern-matchingprocess. This contrast shows in the fact that differentiation is much easierto automate by a computer program than integration.

Example 33 Evaluate the indefinite integral∫x sin(x2) dx. (1.178)

Solution Consider the portion of the integrand sin(x2). This is a com-position of two functions, with f(u) = sinu being the outer function andg(x) = x2 being the inner function. Since the derivative of the inner func-tion g(x) = x2 is g′(x) = 2x, and since this derivative appears as a factor inthe integrand, we can recognize that the integrand can be written as

x sin(x2) =1

22x sin(x2) =

1

2g′(x)f(g(x)). (1.179)

Therefore, we can apply the Substitution Rule,∫f(g(x))g′(x) dx =

∫f(u) du, (1.180)

where the substitution is u = g(x).To apply this rule, we first note that the indefinite integral, or most

general antiderivative, of f(x) = sinx is∫sinx dx = − cosx+ C. (1.181)

Using this result, we obtain∫x sin(x2) dx =

∫1

22x sin(x2) dx

=1

2

∫2x sin(x2) dx


=1

2

∫g′(x) sin(g(x)) dx

=1

2

∫g′(x)f(g(x)) dx

=1

2

∫f(u) du

=1

2[− cosu+ C]

=1

2[− cos(g(x)) + C]

=1

2[− cos(x2) + C]

= −1

2cos(x2) + C, (1.182)

since 1/2 multiplied by an arbitrary constant is still an arbitrary constant.Note that the Substitution Rule only applies on integrands that involve a

composition of functions that has the form f(g(x)), where g′(x) appears as afactor in the integrand. Therefore, when evaluating an integral, it is wise tobegin by examining the integrand to determine whether it has such a form,by trying to recognize the outer function f(x) and the inner function g(x).If such a composition can be recognized, then it is necessary to computeg′(x) and determine whether it also appears in the integrand. 2

Example 34 Compute ∫tanx dx. (1.183)

Solution Writing tanx = sinx/ cosx and letting u = cosx, we have du =− sinx dx. It follows that∫

tanx dx =

∫sinx

cosxdx

= −∫du

u

= − ln |u|+ C

= − ln | cosx|+ C

= ln | cosx|−1 + C

= ln

∣∣∣∣ 1

cosx

∣∣∣∣+ C

= ln | secx|+ C. (1.184)


As will be seen in later sections, similar integration rules can be obtainedfor cotx, secx, and cscx. 2

Example 35 Evaluate ∫e3x dx. (1.185)

Solution Using the substitution u = 3x, we have du = 3 dx, or dx = du/3,which yields ∫

e3x dx =1

3

∫eu du =

1

3eu + C =

1

3e3x + C. (1.186)

2

Example 36 Evaluate ∫ex cos(ex) dx. (1.187)

Solution Using the substitution u = ex, we have du = ex dx which yields∫ex cos(ex) dx =

∫cosu du = sinu+ C = sin(ex) + C. (1.188)

Example 37 Evaluate ∫ex

1 + exdx. (1.189)

Solution Using the substitution u = 1+ex, we have du = ex dx which yields∫ex

1 + exdx =

∫1

udu = lnu+ C = ln |1 + ex|+ C. (1.190)

2

Example 38 Evaluate ∫(sinx)ecosx dx. (1.191)

Solution Using the substitution u = cosx, we have du = − sinx dx whichyields ∫

(sinx)ecosx dx = −∫eu du = −eu + C = −ecosx + C. (1.192)

2


Example 39 Evaluate ∫2x dx√1− x4

. (1.193)

Solution Using the substitution u = x2, we have du = 2x dx and therefore∫2x dx√1− x4

=

∫du√

1− u2= sin−1 u+ C = sin−1 x2 + C. (1.194)

2

Example 40 Evaluate the indefinite integral∫x7(1− x4)1/3 dx. (1.195)

Solution This integrand includes a factor (1− x4)1/3, which can be viewedas a composition of two functions. That is, (1 − x4)1/3 = f(g(x)) wheref(u) = u1/3 and g(x) = 1− x4. We have g′(x) = −4x3, which appears as afactor in the integrand; that is, we can write

x7(1− x4)1/3 = x4(x3)(1− x4)1/3 = −1

4x4(4x3)(1− x4)1/3. (1.196)

Because of the additional factor of x4, however, it can be difficult to applythe Substitution Rule directly. We therefore use an explicit substitutionu = 1− x4, which yields du = −4x3, and obtain∫

x7(1− x4)1/3 dx =

∫−1

4x4(4x3)(1− x4)1/3 dx = −1

4

∫(1− u)u1/3 du,

(1.197)since u = 1−x4 implies x4 = 1−u. We can now evaluate this integral usingthe Power Rule. We have

−1

4

∫(1− u)u1/3 du = −1

4

∫u1/3 − u4/3 du

= −1

4

[u4/3

4/3− u7/3

7/3

]+ C

= − 3

16u4/3 +

3

28u7/3 + C. (1.198)

However, we need to compute the indefinite integral of x7(1 − x4)1/3. Inorder to accomplish this, we note that by the substitution u = g(x), wehave transformed the original problem of integrating∫

f(g(x))g′(x) dx = F (g(x)) + C, (1.199)


where F is any antiderivative of f , to the simpler problem∫f(u) du = F (u) + C. (1.200)

In other words, we have computed the function

F (u) = − 3

16u4/3 +

3

28u7/3. (1.201)

Therefore, the solution to our original problem is∫x7(1− x4)1/3 dx = F (g(x)) + C

= − 3

16[g(x)]4/3 +

3

28[g(x)]7/3 + C

= − 3

16(1− x4)4/3 +

3

28(1− x4)7/3 + C (1.202)

In summary, we can apply the Substitution Rule by choosing an appropriatesubstitution u = g(x) (where the integrand is a composition f(g(x)), with gbeing the inner function) to obtain a new integral with respect to u, wherethe integrand is simply f(u). After evaluating that integral, we use thesubstitution u = g(x) to obtain our final answer as a function of x. 2

1.6.1 Definite Integrals

We now consider the evaluation of definite integrals using substitution. Sup-pose that g is continuously differentiable on [a, b] and that F is an antideriva-tive of a function f that is continuous on the image of [a, b] under g. Fromthe second part of the Fundamental Theorem of Calculus and the relation(1.174), it follows that∫ b

af(g(x))g′(x) dx = F (g(b))− F (g(a))

=

∫ g(b)

g(a)F ′(u) du

=

∫ g(b)

g(a)f(u) du. (1.203)

We see that the interval a ≤ x ≤ b is mapped by the substitution u = g(x)to the interval g(a) ≤ u ≤ g(b).


Example 41 Evaluate the definite integral∫ 2π

02x sin(x2) dx. (1.204)

Solution We will discuss two approaches to solving this problem. In aprevious example, we noted that the integrand can be written in the formf(g(x))g′(x), where f(u) = sinu and g(x) = x2. Because an antiderivativeof f(u) = sinu is F (u) = − cosu, it follows that∫

2x sin(x2) dx = − cos(x2) + C, (1.205)

in view of the Substitution Rule∫f(g(x))g′(x) dx = F (g(x)) + C. (1.206)

We can then use the Fundamental Theorem of Calculus to obtain∫ 2π

02x sin(x2) dx = F (g(2π))− F (g(0))

= − cos((2π)2)− (− cos(02))

= − cos(4π2)− (− cos(02))

≈ −(−0.2070)− (−1)

≈ 1 + 0.2070

≈ 1.2070. (1.207)

Alternatively, we can use the substitution u = x2, since x2 is the innerfunction of the composition appearing in the integrand. This allows us torewrite the integral as∫ 2π

02x sin(x2) dx =

∫ 4π2

0sinu du, (1.208)

since u = x2 implies du = 2x dx. The new limits u = 0 and u = 2π2 areobtained by applying the substitution u = x2 to the original limits x = 0and x = 2π. We can now evaluate a simpler definite integral and obtain∫ 2π

02x sin(x2) dx =

∫ 4π2

0sinu du

= − cosu|4π2

0


= − cos(4π2)− (− cos(0))

≈ −(−0.2070)− (−1)

≈ 1 + 0.2070

≈ 1.2070. (1.209)

2

Example 42 Evaluate the integral∫ 2

0

1

1 + 4x2dx. (1.210)

Solution Using the substitution u = 2x, we have du = 2 dx and therefore∫ 2

0

1

1 + 4x2dx =

1

2

∫ 4

0

1

1 + u2du (1.211)

since u = 0 when x = 0 and u = 2(2) = 4 when x = 2. We then have

1

2

∫ 4

0

1

1 + u2du =

1

2tan−1 u

∣∣40

=1

2(tan−1 4−tan−1 0) =

1

2tan−1 4 ≈ 0.6629.

(1.212)2

1.6.2 Symmetry

The Substitution Rule can be used to easily evaluate certain integrals by wayof the simple substitution u = −x. Specifically, suppose that f is continuouson the interval [−a, a]. If f is an even function (i.e., f(−x) = f(x)), thenthe area under f from 0 to a is exactly the same as the area under f from−a to 0, and it follows that∫ a

−af(x) dx = 2

∫ a

0f(x) dx. (1.213)

On the other hand, if f is an odd function (i.e., f(−x) = −f(x)), then thenet area under f from 0 to a is equal in magnitude to the net area under ffrom −a to 0, but of opposite sign. Therefore∫ a

−af(x) dx = 0. (1.214)

These rules are particularly useful for trigonometric functions, as sinx is anodd function and cosx is an even function.


Example 43 Since sinx is an odd function, we can immediately conclude,without using any anti-differentiation rules, that∫ π

−πsinx dx = 0. (1.215)

2

Example 44 The function f(x) = x2 is even, since f(−x) = (−x)2 = x2 =f(x). It follows that ∫ 4

−4x2 dx = 2

∫ 4

0x2 dx

= 2x3

3

∣∣∣∣40

= 243

3

= 264

3

=128

3. (1.216)

2

Chapter 2

Techniques of Integration

2.1 Integration by Parts

Recall the Product Rule for differentiation,

d

dx[f(x)g(x)] = f(x)g′(x) + g(x)f ′(x). (2.1)

Like other differentiation rules, this rule can be reversed to obtain an in-tegration rule. Specifically, the product f(x)g(x) is an antiderivative off(x)g′(x) + g(x)f ′(x); that is,∫

f(x)g′(x) + g(x)f ′(x) dx = f(x)g(x) + C. (2.2)

Rearranging algebraically, we obtain the equation∫f(x)g′(x) dx = f(x)g(x)−

∫g(x)f ′(x) dx, (2.3)

where the arbitrary constant C can be dropped because an indefinite integralappears on both sides of the equation. If we let u = f(x) and v = g(x), thenwe have du = f ′(x) dx and dv = g′(x) dx. It follows that equation (2.3) canbe rewritten as ∫

u dv = uv −∫v du. (2.4)

This rule is known as the rule of integration by parts.To apply this rule properly to an indefinite integral of the form∫

h(x) dx, (2.5)

67

68 CHAPTER 2. TECHNIQUES OF INTEGRATION

it is necessary to identify the functions that will play the role of u and vin the above rule. In other words, we must be able to write h(x) dx in theform u dv, where u and v are chosen so that

∫v du can be evaluated more

easily than the original integral. Typically, one chooses u = u(x) to be someportion of h(x), and then sets dv = h(x)/u(x) dx. Then, v and du can easilybe determined.

Example 45 Evaluate ∫lnx dx. (2.6)

Solution We use integration by parts with

u = lnx, du =1

xdx, v = x, dv = dx (2.7)

to obtain∫lnx dx = x lnx−

∫x

1

xdx = x lnx−

∫1 dx = x lnx− x+ C. (2.8)

2

Example 46 Evaluate ∫tan−1 x dx. (2.9)


u = tan−1 x, du =1

1 + x2dx, v = x, dv = dx (2.10)

as well as a substitution u = x2 + 1 to obtain∫tan−1 x dx = x tan−1 x−

∫x

1 + x2dx

= x tan−1 x− 1

2

∫1

udu

= x tan−1 x− 1

2ln |u|+ C

= x tan−1 x− 1

2ln |x2 + 1|+ C. (2.11)

2

2.1. INTEGRATION BY PARTS 69

Example 47 Evaluate ∫ex sinx dx. (2.12)


u = ex, du = ex dx, v = − cosx, dv = sinx dx (2.13)

to obtain ∫ex sinx dx = −ex cosx dx+

∫ex cosx dx. (2.14)

We then use integration by parts again with

u = ex, du = ex dx, v = sinx, dv = cosx dx (2.15)

to obtain∫ex sinx dx = −ex cosx dx+ ex sinx−

∫ex sinx dx. (2.16)

We have obtained the same integral that we started with, but with theopposite sign. We can therefore add this integral to both sides to obtain

2

∫ex sinx dx = ex sinx− ex cosx+ C (2.17)

which yields ∫ex sinx dx =

ex

2(sinx− cosx) + C. (2.18)

2

We now examine how integration by parts can be used to evaluate adefinite integral. If we integrate both sides of the Product Rule for differen-tiation from a to b, we have∫ b

a

d

dx[f(x)g(x)] dx =

∫ b

af(x)g′(x) + g(x)f ′(x) dx. (2.19)

It follows from Part 2 of the Fundamental Theorem of Calculus that

f(x)g(x)|ba =

∫ b

af(x)g′(x) + g(x)f ′(x) dx, (2.20)

or, rearranging,∫ b

af(x)g′(x) dx = f(x)g(x)|ba −

∫ b

ag(x)f ′(x) dx. (2.21)


Therefore, to apply integration parts to a definite integral of the form∫ b

ah(x) dx, (2.22)

we identify the functions u and v as in the case of the indefinite integral.Then, we can simply compute u(b)v(b)−u(a)v(a), and evaluate the integralof v(x)u′(x) from a to b.

Example 48 Compute ∫ π/2

0x cosx dx. (2.23)


u = x, du = dx, v = sinx, dv = cosx dx (2.24)

to obtain ∫ π/2

0x cosx dx = x sinx|π/20 −

∫ π/2

0sinx dx

= x sinx+ cosx|π/20

=π

2sin

π

2+ cos

π

2− cos 0

=π

2− 1. (2.25)

2

A common usage of integration by parts is to evaluate integrals of theform ∫

xmf(x) dx, (2.26)

where the function f(x) can easily be anti-differentiated repeatedly, as is thecase with various exponential or trigonometric functions. For such an inte-grand, one can choose u = xm, which yields du = mxm−1 dx. By applyingintegration by parts in this way m times, the power of x in the integrandcan be reduced all the way to zero, hopefully yielding a simpler integrand.

Example 49 Evaluate ∫x2ex dx. (2.27)

2.1. INTEGRATION BY PARTS 71


u = x2, du = 2x dx, v = ex, dv = ex dx (2.28)

to obtain ∫x2ex dx = x2ex − 2

∫xex dx. (2.29)

We then use integration by parts again with

u = x, du = dx, v = ex, dv = ex dx (2.30)

to obtain ∫x2ex dx = x2ex − 2

∫xex dx

= x2ex − 2

[xex −

∫ex dx

]= x2ex − 2 [xex − ex + C]

= x2ex − 2xex + 2ex + C

= (x2 − 2x+ 2)ex + C. (2.31)

2

Example 50 Derive a formula for∫sinn x dx (2.32)

where n is an integer, and n ≥ 2.


u = sinn−1 x, du = (n− 1) sinn−2 x cosx dx, v = − cosx, dv = sinx dx(2.33)

to obtain∫sinn x dx = − sinn−1 x cosx+ (n− 1)

∫sinn−2 x cos2 x dx. (2.34)

Writing cos2 x = 1− sin2 x yields∫sinn x dx = − sinn−1 x cosx+ (n− 1)

∫sinn−2 x cos2 x dx

= − sinn−1 x cosx+ (n− 1)

∫sinn−2 x(1− sin2 x) dx


∫sinn−2 x− sinn x dx


∫sinn−2 x dx− (n− 1)

∫sinn x dx.(2.35)


Moving the last integral on the right side to the left side of the equationyields

n

∫sinn x dx = − sinn−1 x cosx+ (n− 1)

∫sinn−2 x dx (2.36)

or ∫sinn x dx = − 1

nsinn−1 x cosx+

n− 1

n

∫sinn−2 x dx. (2.37)

2

2.2 Trigonometric Integrals

So far we have obtained several integration rules by reversing various differ-entiation rules. Additional integration rules can be obtained by employingtrigonometric identities to convert integrands involving trigonometric func-tions into a form in which other integration rules can be applied. For suchintegrands, the resulting rules can then be used directly, instead of contin-uing to rely on the trigonometric identities used to establish them.

For example, suppose we wish to evaluate an integral of the form∫sinm x cosn x dx. (2.38)

If n is odd, then we can use the identity sin2 x+ cos2 x = 1 to express n− 1of the powers of cosine in terms of sine. Then, we can use the substitutionu = sinx. Because du = cosx dx, the resulting integral with respect to uwill be simple to evaluate, since the integrand will consist of several termsof the form um+2k, for k = 0, . . . , (n− 1)/2, which can easily be integratedusing the power rule. An integral of the form (2.38) where m is odd can behandled similarly, using the same identity.

Example 51 Evaluate ∫sin3 x cos2 x dx. (2.39)

Solution Since the power of sine is odd, we write all but one power of sinein terms of cosines:∫

sin3 x cos2 x dx =

∫sinx(1−cos2 x) cos2 x dx =

∫sinx cos2 x dx−

∫sinx cos4 x dx.

(2.40)

2.2. TRIGONOMETRIC INTEGRALS 73

Using the substitution u = cosx, with du = − sinx dx, yields∫sinx cos2 x dx−

∫sinx cos4 x dx = −

∫u2 du+

∫u4 du =

u5

5−u

3

3+C =

cos5 x

5−cos3 x

3+C.

(2.41)2

If both m and n are even, then the half-angle formulas

sin2 x =1

2(1− cos 2x), cos2 x =

1

2(1 + cos 2x) (2.42)

can be used to reduce the powers of sine and cosine until every term in thetransformed integrand either has an odd power of sine or cosine, or until asimpler integration rule can be used. If a term of the form sinx cosx arises,this can be handled using the substitution u = cosx or u = sinx, but onemay find it easier to use the double-angle formula

sinx cosx =1

2sin 2x. (2.43)

Example 52 Evaluate ∫sin2 x cos2 x dx. (2.44)

Solution Using the double-angle formula

sin 2x = 2 sinx cosx (2.45)

yields ∫sin2 x cos2 x dx =

∫ (sin 2x

2

)2

dx =1

4

∫sin2 2x dx. (2.46)

We can then use a half-angle formula

sin2 x =1− cos 2x

2(2.47)

to obtain ∫sin2 x cos2 x dx =

1

4

∫1− cos 4x

2dx

=1

8

∫1− cos 4x dx

=1

8

(x− sin 4x

4

)=

x

8− sin 4x

32+ C. (2.48)

2


Integrals of the form ∫tanm x secn x dx (2.49)

can be handled using a similar strategy. In this case, the useful identity issec2 x = 1 + tan2 x. If the power of secant is even, then all but two of thepowers of secant can be expressed in terms of tanx using this identity, andthen the substitution u = tanx can be used to obtain a simpler integrand.If, on the other hand, the power of tangent is odd and n > 0, then all butone of the powers of tangent can be expressed in terms of secx, and then thesubstitution u = secx can be used, since we would have du = secx tanx.

Example 53 Evaluate ∫sec4 x tan3 x dx. (2.50)

Solution We write all but two powers of secant as powers of tangents, usingthe identity

sec2 x = tan2 x+ 1. (2.51)

This yields∫sec4 x tan3 x dx =

∫sec2 x(tan2 x+1) tan3 x dx =

∫sec2 x tan5 x dx+

∫sec2 x tan3 x dx.

(2.52)Using the substitution u = tanx, with du = sec2 x dx, yields∫

sec4 x tan3 x dx =


∫sec2 x tan3 x dx

=

∫u5 du+

∫u3 du

=u6

6+u4

4+ C

=tan6 x

6+

tan4 x

4+ C. (2.53)

An alternative approach is to write all but one power of tangent as powersof secants, using the identity

tan2 x = sec2 x− 1. (2.54)


This yields∫sec4 x tan3 x dx =

∫sec4 x(sec2 x− 1) tanx dx

=

∫sec6 x tanx dx−

∫sec4 x tanx dx. (2.55)

Using the substitution u = secx, with du = secx tanx dx, yields∫sec4 x tan3 x dx =

∫sec6 x tanx dx−

∫sec4 x tanx dx

=

∫sec5 x secx tanx dx−

∫sec3 x secx tanx dx

=

∫u5 du−

∫u3 du

=u6

6− u4

4+ C

=sec6 x

6− sec4 x

4+ C. (2.56)

2

Other cases are not as straightforward. It may be necessary to use othertrigonometric identities, integration by parts, or the rules for integratingtanx and secx: ∫

tanx dx = ln | secx|+ C, (2.57)∫secx dx = ln | secx+ tanx|+ C. (2.58)

Example 54 Evaluate ∫sec3 x tan2 x dx. (2.59)

Solution Since the power of tangent is even and the power of secant is odd,it is not possible to rewrite powers of one in terms of powers of the otherand use a simple substitution as in the previous example. Instead, we canproceed using integration by parts with

u = secx tan2 x, dv = sec2 x, (2.60)

which implies

du = (2 sec3 x tanx+ secx tan3 x) dx, v = tanx. (2.61)


Before applying integration by parts, we use the identity tan2 x = sec2 x− 1to write

du = (2 sec3 x tanx+ secx(sec2 x− 1) tanx) dx

= (2 sec3 x tanx+ sec2 x secx tanx− secx tanx) dx

= (3 sec3 x tanx− secx tanx) dx. (2.62)

Integrating by parts, we obtain∫sec3 x tan2 x dx = secx tan3 x− 3


∫secx tan2 x dx.

(2.63)Rearranging algebraically yields

4

∫sec3 x tan2 x dx = secx tan3 x+

∫secx tan2 x dx (2.64)

or ∫sec3 x tan2 x dx =

1

4secx tan3 x+

1

4

∫secx tan2 x dx. (2.65)

We now focus on the remaining integral∫secx tan2 x dx. (2.66)

Rewriting the powers of tangent as powers of secant yields∫secx tan2 x dx =

∫secx(sec2 x− 1) dx =

∫sec3 x− secx dx. (2.67)

To integrate secx, we proceed as follows, using the substitution u = secx+tanx with du = (secx tanx+ sec2 x) dx:∫

secx dx =

∫secx

secx+ tanx

secx+ tanxdx

=

∫sec2 x+ secx tanx

secx+ tanxdx

=

∫du

u

= ln |u|+ C

= ln | secx+ tanx|+ C. (2.68)

To integrate sec3 x, we use integration by parts with

u = secx, dv = sec2 x dx (2.69)


which yieldsdu = secx tanx dx, v = tanx. (2.70)

It follows that ∫sec3 x dx = secx tanx−

∫secx tan2 x dx. (2.71)

Using the identity tan2 x = sec2 x− 1 yields∫sec3 x dx = secx tanx−

∫secx(sec2 x−1) dx = secx tanx−

∫sec3 x dx+

∫secx dx.

(2.72)Rearranging algebraically yields

2

∫sec3 x dx = secx tanx+

∫secx dx (2.73)

or ∫sec3 x dx =

1

2secx tanx+

1

2

∫secx dx. (2.74)

Applying our earlier result of integrating secx, we obtain∫sec3 x dx =

1

2secx tanx+

1

2ln | secx+ tanx|. (2.75)

Finally, we put all of the pieces together and obtain∫sec3 x tan2 x dx =

1

4secx tan3 x+

1

4

∫secx tan2 x dx

=1

4secx tan3 x+

1

4

[∫sec3 x dx−

∫secx dx

]=

1

4secx tan3 x+

1

4

[1

2secx tanx+

1

2ln | secx+ tanx| − ln | secx+ tanx|

]=

1

4secx tan3 x+

1

4

[1

2secx tanx− 1

2ln | secx+ tanx|

]=

1

4secx tan3 x+

1

8secx tanx− 1

8ln | secx+ tanx|. (2.76)

2

Trigonometric identities can also be helpful for integrals of the forms∫sinmx cosnx dx, (2.77)

78 CHAPTER 2. TECHNIQUES OF INTEGRATION∫sinmx sinnx dx, (2.78)

or ∫cosmx cosnx dx. (2.79)

The identities

sinA cosB =1

2[sin(A−B) + sin(A+B)], (2.80)

sinA sinB =1

2[cos(A−B)− cos(A+B)], (2.81)

cosA cosB =1

2[cos(A−B) + cos(A+B)], (2.82)

which can be derived from the identities for cosines and sines of sums anddifferences, can be used to convert these integrands into much simpler onesinvolving only a sine or cosine, in which case the rules for integrating thesefunctions can be employed.

Example 55 Prove that for any nonnegative integers m and n,∫ π

−πsinmx cosnx dx = 0. (2.83)

Solution Using the identity

sinA cosB =1

2[sin(A−B) + sin(A+B)], (2.84)

we obtain∫ π

−πsinmx cosnx dx =

1

2

∫ π

−πsin(m− n)x+ sin(m+ n)x dx. (2.85)

Since sinx is an odd function, it follows that sin kx is odd for any integer k.Since the integral of any odd function over an interval of the form [−a, a] isequal to 0, we must have∫ π

−πsin(m− n)x dx = 0,

∫ π

−πsin(m+ n)x dx = 0, (2.86)

and therefore we must have∫ π

−πsinmx cosnx dx = 0. (2.87)

2


By now it should be clear that there are a great many integrals thatcannot be evaluated using the Fundamental Theorem of Calculus withoutsome ingenuity. Although we have greatly expanded the range of integrandsfor which integration rules can be developed, there are still many integrandsfor which no such rules exist, and therefore the definition of the definiteintegral must be used instead.

Example 56 Evaluate ∫ π

−πsin 3x cos 7x dx. (2.88)

Solution We use the product sum identity

sinA cosB =1

2[sin(A+B) + sin(A−B)] (2.89)

with A = 3x and B = 7x. We then have∫ π

−πsin 3x cos 7x dx =

∫ π

−π

1

2[sin(3x+ 7x) + sin(3x− 7x)] dx

=1

2

∫ π

−πsin 10x+ sin(−4x) dx

=1

2

(− cos 10x

10+− cos(−4x)

−4

)∣∣∣∣π−π

=

(− cos 10x

20+

cos(−4x)

8

)∣∣∣∣π−π

=

(− cos 10π

20+

cos(−4π)

8

)−(− cos(−10π)

20+

cos(4π)

8

)=

(−1

20+

1

8

)−(−1

20+

1

8

)=−1

20+

1

8− −1

20− 1

8= 0. (2.90)

The antiderivative in the third step can be obtained using separate substi-tutions u = 10x and u = 4x in each term, or by noting that in general,∫

sin ax dx = −cos ax

a+ C. (2.91)

2


2.3 Trigonometric Substitution

Recall the substitution rule,∫f(g(x))g′(x) dx =

∫f(u) du. (2.92)

In applying this rule, one makes the substitution u = g(x), which yieldsdu = g′(x) dx. Using this relation, the original integrand f(g(x))g′(x) isthen rewritten as f(u).

In rewriting the integrand in this fashion, one can write the substitutionin such a way that x is expressed as a function of u, instead of the otherway around: x = g−1(u). Then, we have dx = du/g′(g−1(u)). This is notalways the easiest approach to applying the substitution rule, but can bevery effective when the substitution to be employed is more complicated,or if the function g−1 is simpler to work with than g. This approach tosubstitution is known as inverse substitution, but it should be recognizedthat it is nothing more than an alternative approach to u-substitution.

This is precisely the case when evaluating an integral of the form∫ √a2 − x2 dx. (2.93)

Suppose that we use the substitution x = a sin θ. Clearly, it is preferable towrite the substitution in this manner than expressing u as a function of x.From the relation dx = a cos θ dθ, we obtain the new integral∫ √

a2 − a2 sin2 θ(a cos θ) dθ =

∫a2 cos2 θ dθ, (2.94)

which can easily be evaluated using a half-angle formula. For more compli-cated integrands involving the expression

√a2 − x2, the same substitution

can sometimes be used.

Example 57 Evaluate ∫ 2

0x√

4− x2 dx. (2.95)

Solution One approach is to use the substitution u = 4−x2, which impliesdu = −2x dx. The new limits can be obtained by substituting the limitsx = 0 and x = 2 into the relation u = 4− x2, which yields u = 4 and u = 0.

2.3. TRIGONOMETRIC SUBSTITUTION 81

We then have ∫ 2

0x√

4− x2 dx = −1

2

∫ 0

4u1/2 du

=1

2

∫ 4

0u1/2 du

=1

2

u3/2

3/2

∣∣∣∣∣4

0

=1

2

2

3u3/2

∣∣∣∣40

=1

343/2

=1

3(41/2)3

=1

323

=8

3. (2.96)

Alternatively, we can use an inverse substitution x = 2 sin θ, which impliesdx = 2 cos θ dθ. We then have√

4− x2 =√

4− (2 sin θ)2 =√

4− 4 sin2 θ = 2√

1− sin2 θ = 2√

cos2 θ = 2 cos θ.(2.97)

It follows that∫ 2

0x√

4− x2 dx =

∫ π/2

0(2 sin θ)(2 cos θ)(2 cos θ) dθ

= 8

∫ π/2

0sin θ cos2 θ dθ

= −8

∫ 0

1u2 du, u = cos θ, du = − sin θ dθ

= 8

∫ 1

0u2 du

= 8u3

3

∣∣∣∣10

=8

3. (2.98)

2


Similar substitutions can be used for integrands containing expressions ofthe form

√a2 + x2 or

√x2 − a2. In the first case, the substitution x = a tan θ

can be used in conjunction with the identity 1 + tan2 θ = sec2 θ. In thesecond case, the substitution x = a sec θ can be used in conjunction withthe identity sec2 θ − 1 = tan2 θ.

Example 58 Evaluate ∫dx√x2 + 4

. (2.99)

Solution We use the inverse substitution x = 2 tan θ, which implies dx =2 sec2 θ dθ. Furthermore,√x2 + 4 =

√(2 tan θ)2 + 4 =

√4 tan2 θ + 4 = 2

√tan2 θ + 1 = 2

√sec2 θ = 2 sec θ.

(2.100)Therefore ∫

dx√x2 + 4

=

∫2 sec2 θ dθ

2 sec θ

=

∫sec θ dθ

= ln | sec θ + tan θ|+ C

= ln

∣∣∣∣∣√x2 + 4

2+x

2

∣∣∣∣∣+ C

= ln1

2|√x2 + 4 + x|+ C

= ln |√x2 + 4 + x|+ ln

1

2+ C

= ln |√x2 + 4 + x| − ln 2 + C

= ln |√x2 + 4 + x|+ C (2.101)

since C represents an arbitrary constant. 2

Often, these substitutions must be used in conjunction with other inte-gration techniques. We illustrate this with a simple example. Suppose wewish to evaluate ∫ √

a2 + x2 dx. (2.102)

From the discussion above, we should use the substitution x = a tan θ. Thisyields ∫ √

a2 + x2 dx =

∫a2 sec3 θ dθ. (2.103)


However, in order to evaluate the new integral, we must resort to inte-gration by parts, which yields∫

sec3 θ dθ = sec θ tan θ −∫

sec θ tan2 θ dθ. (2.104)

Using the identity tan2 θ = sec2 θ − 1, we obtain∫sec3 θ dθ = sec θ tan θ −

∫sec3 θ dθ +

∫sec θ dθ. (2.105)

Rearranging algebraically and using the integration rule for sec θ, we finallyobtain∫ √

a2 + x2 dx =

∫a2 sec3 θ dθ

=a2

2[sec θ tan θ + ln | sec θ + tan θ|] + C

=x

2

√x2 + a2 +

a2

2ln(√

x2 + a2 + x)

+ C.(2.106)

The key lesson to be learned from this example is that one should beprepared to use any integration technique at his or her disposal, since, formany integrands, some combination of such techniques is necessary to eval-uate the given integral. Unlike the more straightforward process of differ-entiation, anti-differentiation often requires substantial ingenuity and open-mindedness.

Example 59 Evaluate ∫ √1− x− x2 dx. (2.107)

Solution Before we can use a trigonometric substitution, we need to writethe expression 1− x− x2 as a sum or difference of two squares. Completingthe square, we have

x2 + x− 1 =

(x+

1

2

)2

− 1− 1

4=

(x+

1

2

)2

− 5

4(2.108)

and therefore ∫ √1− x− x2 dx =

∫ √5

4−(x+

1

2

)2

dx. (2.109)


Since we now have a difference of two squares under the radical sign, andsince the first term is a constant, we try to write the radical in the form√a2 − a2 sin2 θ for some choice of a and some substitution involving a sin θ.

Matching a2 − a2 sin2 θ to 5/4 − (x + 1/2)2, we find that we must havea =

√5/4 =

√5/2, and also

x+1

2=

√5

2sin θ. (2.110)

Using this inverse substitution, we have

dx =

√5

2cos θ dθ (2.111)

and√5

4−(x+

1

2

)2

=

√√√√5

4−

(√5

2sin θ

)2

=

√5

4−√

54 sin2 θ =

√5

2

√1− sin2 θ =

√5

2cos θ.

(2.112)It follows that∫ √

1− x− x2 dx =

∫ (√5

2cos θ

)2

dθ =5

4

∫cos2 θ dθ. (2.113)

Using the half-angle formula

cos2 θ =1 + cos 2θ

2(2.114)

yields ∫cos2 θ dθ =

∫1 + cos 2θ

2dθ =

1

2

(θ +

sin 2θ

2

)+ C (2.115)

and therefore ∫ √1− x− x2 dx =

5

8θ +

5

16sin 2θ + C. (2.116)

From our original substitution, we obtain

θ = sin−1

(2√5

(x+

1

2

)), (2.117)


and using the double-angle formula

sin 2θ = 2 sin θ cos θ (2.118)

along with the relations

x+1

2=

√5

2sin θ (2.119)

and √5

4−(x+

1

2

)2

=

√5

2cos θ (2.120)

yields

sin 2θ = 2 sin θ cos θ

= 2

(2√5

(x+

1

2

)) 2√5

√5

4−(x+

1

2

)2

=8

5

(x+

1

2

)√1− x− x2. (2.121)

Now we can assemble our final answer,∫ √1− x− x2 dx =

5

8sin−1

(2√5

(x+

1

2

))+

1

2

(x+

1

2

)√1− x− x2+C.

(2.122)2

Example 60 Evaluate ∫x2√x2 − a2 dx (2.123)

where a is a positive constant.

Solution We use the inverse substitution

x = a sec θ (2.124)

which impliesdx = a sec θ tan θ dθ (2.125)

and√x2 − a2 =

√(a sec θ)2 − a2 =

√a2 sec2 θ − a2 = a

√sec2 θ − 1 = a

√tan2 θ = a tan θ.

(2.126)


Therefore,∫x2√x2 − a2 dx =

∫(a sec θ)2(a tan θ)(a sec θ tan θ) dθ = a4

∫sec3 θ tan2 θ dθ.

(2.127)This integral was worked out in the previous section. Using that result, weobtain∫x2√x2 − a2 dx = a4

[1

4sec θ tan3 θ +

1

8sec θ tan θ − 1

8ln | sec θ + tan θ|

]+C.

(2.128)Using the relations

x = a sec θ (2.129)

and √x2 − a2 = a tan θ, (2.130)

we obtain∫x2√x2 − a2 dx =

1

4x(x2 − a2)3/2 +

a2

8x√x2 − a2 − a4

8ln

∣∣∣∣∣xa +

√x2 − a2

a

∣∣∣∣∣+ C

=1

4x(x2 − a2)3/2 +

a2

8x√x2 − a2 − a4

8ln

1

a

∣∣∣x+√x2 − a2

∣∣∣+ C

=1

4x(x2 − a2)3/2 +

a2

8x√x2 − a2 − a4

8ln∣∣∣x+

√x2 − a2

∣∣∣− a4

8ln

1

a+ C

=1

4x(x2 − a2)3/2 +

a2

8x√x2 − a2 − a4

8ln∣∣∣x+

√x2 − a2

∣∣∣+ C (2.131)

since a is a constant and C is an arbitrary constant. 2

2.4 Integration of Rational Functions

In this section we consider the integration of rational functions. A functionf(x) is said to be rational if it can be written as

f(x) =P (x)

Q(x), (2.132)

where P (x) and Q(x) are polynomials. If the degree of P is less than that ofQ, then f is said to be a proper rational function; otherwise it is improper.We shall only consider proper rational functions, because if f is improper,

2.4. INTEGRATION OF RATIONAL FUNCTIONS 87

then we can divide P by Q using long division of polynomials to obtain therepresentation

f(x) = S(x) +R(x)

Q(x), (2.133)

where S(x) and R(x) are both polynomials, and the rational function g(x) =R(x)/Q(x) is proper. Therefore we can integrate f(x) by integrating eachterm of S(x) using the power rule and then handling g(x) separately.

Example 61 Evaluate ∫2x3 + 3x2 + 7x+ 4

2x+ 1dx. (2.134)

Solution Dividing the numerator and denominator, we have

2x3+3x2+7x+4 = x2(2x+1)+2x2+7x+4 = (x2+x)(2x+1)+6x+4 = (x2+x+3)(2x+1)+1(2.135)

and therefore

2x3 + 3x2 + 7x+ 4

2x+ 1= x2 + x+ 3 +

1

2x+ 1, (2.136)

which yields, by the substitution u = 2x+ 1,∫2x3 + 3x2 + 7x+ 4

2x+ 1dx =

∫x2 + x+ 3 +

1

2x+ 1dx

=x3

3+x2

2+ 3x+

1

2

∫1

udu

=x3

3+x2

2+ 3x+

1

2ln |2x+ 1|+ C.(2.137)

2

2.4.1 Simple Proper Rational Functions

We know how to integrate some proper rational functions using techniqueswe have already learned. To review:∫

1

x+ adx = ln |x+ a|+ C, (2.138)

∫1

x2 + a2dx =

1

atan−1

(xa

)+ C. (2.139)


Some slightly more complex proper rational functions can be handled usingthese rules in conjunction with other techniques. For example, using thesubstitution u = x2, we have∫

x

x2 + adx =

1

2

∫du

u+ a=

1

2ln |u+ a|+ C =

1

2ln |x2 + a|+ C. (2.140)

To integrate a more general rational function, we can rewrite it as a sum ofsimpler rational functions that can be integrated using simpler rules and/ortechniques such as those illustrated above. Such a sum can be obtainedusing the method of partial fractions.

2.4.2 The Method of Partial Fractions

The method of partial fractions involves the decomposition of a generalproper rational function

f(x) =R(x)

Q(x)(2.141)

into a sum of simpler rational functions that can be integrated using knownintegration techniques such as the rules above. The method is based on thefact that Q(x) can be factored into a product of linear factors of the form(ax+ b) or an irreducible quadratic factor of the form (ax2 + bx+ c), whereb2 − 4ac < 0. This is simply a consequence of the Fundamental Theorem ofAlgebra, which states that any polynomial of degree n has exactly n roots,which can be real numbers or complex numbers.

The first step in the method of partial fractions is to factor Q(x) intothese linear and quadratic factors. The second step is to write f(x) as asum of proper rational functions that have these factors as denominators.This sum is called the partial fraction decomposition of f(x). Finally, each ofterm in this decomposition, which has a much simpler form than f(x), canbe integrated individually. Depending on the factorization of Q(x), thereare four distinct scenarios:

1. Q(x) is a product of linear factors of the form (ax+b), where all of thefactors have distinct roots (in other words, the ratio b/a is different inall factors). Then, since we can write Q(x) as

Q(x) = (a1x+ b1) · · · (akx+ bk) =k∏j=1

(ajx+ bj), (2.142)


we can then write f(x) as

f(x) =R(x)

Q(x)=

A1

a1x+ b1+ · · ·+ Ak

akx+ bk. (2.143)

The constants A1, . . . , Ak can be determined by solving the equation

R(x) =

[A1

a1x+ b1+ · · ·+ Ak

a1x+ bk

]Q(x). (2.144)

This can be accomplished by cancelling the denominators with factorsof Q(x) where possible, resulting in a polynomial on the right side.Then, by matching powers of x on both sides of the equation, weobtain a system of k linear equations for the constants A1, . . . , Ak.Then, we can integrate each term individually as follows:∫

Ajajx+ bj

dx =Ajaj

ln

∣∣∣∣xj +bjaj

∣∣∣∣+ C, j = 1, . . . , k. (2.145)

Example 62 Evaluate ∫A

x+ adx. (2.146)

Solution Using the substitution u = x+ a, we have du = dx and∫A

x+ adx = A

∫1

udu = A ln |u|+ C = A ln |x+ a|+ C. (2.147)

2

Example 63 Evaluate ∫x4 + 2

x2 − 1dx. (2.148)

Solution Dividing the numerator by the denominator yields

x4 + 2 = x2(x2 − 1) + x2 + 2 = (x2 + 1)(x2 − 1) + 3 (2.149)

and thereforex4 + 2

x2 − 1= x2 + 1 +

3

x2 − 1. (2.150)


To evaluate the integral, we need to compute the partial fraction de-composition of

1

x2 − 1=

1

(x+ 1)(x− 1)=

A

x+ 1+

B

x− 1. (2.151)

Multiplying through this equation by x2 − 1 yields

1 =

[A

x+ 1+

B

x− 1

](x+1)(x−1) =

A(x+ 1)(x− 1)

x+ 1+B(x+ 1)(x− 1)

x− 1= A(x−1)+B(x+1).

(2.152)Collecting terms with like powers of x yields

1 = Ax−A+Bx+B = (A+B)x+B −A. (2.153)

Matching like powers of x on both sides of the equation yields thesystem of equations

A+B = 0,

B −A = 1. (2.154)

From the first equation, A = −B. Substituting this relation into thesecond equation yields −A − A = −2A = 1, and therefore we musthave A = −1/2 and B = 1/2. It follows that∫

x4 + 2

x2 − 1dx =

∫x2 + 1 +

3

x2 − 1dx

=x3

3+ x+ 3

∫−1/2

x+ 1+

1/2

x− 1dx

=x3

3+ x− 3

3ln |x+ 1|+ 3

2ln |x− 1|+ C.(2.155)

2

Example 64 Evaluate ∫x− 7

(x− 2)(x+ 3)dx. (2.156)

Solution We need to compute the partial fraction decomposition

x− 7

(x− 2)(x+ 3)=

A

x− 2+

B

x+ 3, (2.157)


where the constants A and B are to be determined. Multiplyingthrough this equation by (x− 2)(x+ 3) yields

x−7 =

[A

x− 2+

B

x+ 3

](x−2)(x+3) = A(x+3)+B(x−2). (2.158)

Collecting terms with like powers of x yields

x− 7 = Ax+ 3A+Bx− 2B = (A+B)x+ 3A− 2B. (2.159)

Matching like powers of x on both sides of the equation yields thesystem of equations

A+B = 1

3A− 2B = −7 (2.160)

for the constants A and B. From the first equation, A = 1 − B.Substituting this relation into the second equation yields 3(1 − B) −2B = 3 − 5B = −7, which yields B = 2 and therefore we must haveA = −1. It follows that∫

x− 7

(x− 2)(x+ 3)dx =

∫−1

x− 2+

2

x+ 3dx

= − ln |x− 2|+ 2 ln |x+ 3|+ C.(2.161)

2

Example 65 Evaluate ∫x+ 3

(x− 2)(x2 − 1)dx. (2.162)

Solution The denominator factors as

(x− 2)(x2 − 1) = (x− 2)(x+ 1)(x− 1) (2.163)

and therefore we must compute a partial fraction decomposition of theform

x+ 3

(x− 2)(x2 − 1)=

A

x− 2+

B

x+ 1+

C

x− 1, (2.164)


where the constants A, B, and C are to be determined. Multiplyingthrough this equation by (x− 2)(x2 − 1) yields

x+ 3 =A(x− 2)(x2 − 1)

x− 2+B(x− 2)(x2 − 1)

x+ 1+C(x− 2)(x2 − 1)

x− 1

= A(x2 − 1) +B(x− 2)(x− 1) + C(x− 2)(x+ 1). (2.165)

To determine A, B and C we use the method of creating zeros. Sub-stituting x = 2 into the above equation yields

5 = A((2)2 − 1) +B(2− 2)(2− 1) + C(2− 2)(2 + 1) = 3A (2.166)

and therefore A = 5/3. Similarly, substituting x = 1 yields

4 = A((1)2−1)+B(1−2)(1−1)+C(1−2)(1+1) = C(−1)(2) = −2C(2.167)

and therefore C = −2. Finally, substituting x = −1 yields

2 = A((−1)2−1)+B(−1−2)(−1−1)+C(−1−2)(−1+1) = B(−3)(−2) = 6B(2.168)

and therefore B = 1/3. It follows that∫x+ 3

(x− 2)(x2 − 1)dx =

∫5/3

x− 2+

1/3

x+ 1− 2

x− 1dx

=5

3ln |x− 2|+ 1

3ln |x+ 1| − 2 ln |x− 1|+ C.(2.169)

2

2. Q(x) is a product of linear factors, where some of the factors repeat.In this case, we can write Q(x) as

Q(x) = (a1x+ b1)m1 · · · (akx+ bk)mk =

k∏j=1

(ajx+ bj)mj , (2.170)

where k is the number of distinct linear factors, and, for each j =1, . . . , k, mj is called the multiplicity of the factor ajx + bj . Since atleast one factor repeats in this scenario, this means mj > 1 for at leastone j.

For each distinct factor ajx + bj of Q(x), the partial fraction decom-position of f(x) includes the terms

A1j

ajx+ bj+

A2j

(ajx+ bj)2+ · · ·+

Amjj

(ajx+ bj)mj. (2.171)


Then, we can solve for the constants {Aij} as in the previous scenario,and integrate each term separately. The terms with denominators ofthe form (ax + b)j , where j > 1, can be integrated using the powerrule.

It follows that f(x) can be written as

f(x) =R(x)

Q(x)=

k∑j=1

mj∑i=1

Aij(ajx+ bj)i

, (2.172)

which yields∫f(x) dx =

k∑j=1

mj∑i=1

Aij

∫dx

(ajx+ bj)i

=k∑j=1

mj∑i=1

Aijaij

∫dx

(x+ bj/aj)i

=k∑j=1

[A1j

aj

∫dx

(x+ bj/aj)+

mj∑i=2

Aijaij

∫dx

(x+ bj/aj)i

]

=k∑j=1

[A1j

ajln

∣∣∣∣xj +bjaj

∣∣∣∣− mj∑i=2

Aijaj(ajx+ bj)i−1

]+ C.(2.173)

Example 66 Evaluate ∫3

(x+ 4)8dx. (2.174)

Solution Using the substitution u = x + 4, we have du = dx, so bythe Power Rule,∫

3

(x+ 4)8dx = 3

∫1

u8du = 3

∫u−8 du = 3

u−7

−7+C = − 3

7(x+ 4)7+C.

(2.175)2

Example 67 Evaluate ∫3x2 − 2x− 3

(x+ 1)x2dx. (2.176)


Solution We must compute a partial fraction decomposition of theform

3x2 − x− 3

(x+ 1)x2=

A

x+ 1+B

x+C

x2(2.177)

due to the repeated linear factor of x in the denominator. Multiplyingthrough this equation by (x+ 1)x2 yields

3x2 − 2x− 3 =

[A

x+ 1+B

x+C

x2

](x+ 1)x2

=A(x+ 1)x2

x+ 1+B(x+ 1)x2

x+C(x+ 1)x2

x2

= Ax2 +B(x+ 1)x+ C(x+ 1)

= Ax2 +Bx2 +Bx+ Cx+ C

= (A+B)x2 + (B + C)x+ C. (2.178)

Matching like powers of x on both sides of the equation, we obtain thesystem of equations

A+B = 3,

B + C = −2,

C = −3. (2.179)

Since C = −3, it follows that B = −2− (−3) = 1. Therefore we musthave A = 3−B = 3− 1 = 2. We then have∫

3x2 − 2x− 3

(x+ 1)x2dx =

∫2

x+ 1+

1

x− 3

x2dx

= 2 ln |x+ 1|+ ln |x| − 3x−1

−1

= 2 ln |x+ 1|+ ln |x|+ 3

x. (2.180)

2

3. Q(x) is a product of linear and irreducible quadratic factors, wherenone of the quadratic factors repeat. Then, for each quadratic fac-tor (ax2 + bx + c) of Q(x), where b2 − 4ac < 0, the partial fractiondecomposition of f(x) includes a term of the form

A(2ax+ b) +B

ax2 + bx+ c, (2.181)


where the constants A and B are to be determined as in the previousscenarios. This term can then be integrated independently of the otherterms in the decomposition by completing the square, which yields∫A(2ax+ b) +B

ax2 + bx+ cdx = A

∫2ax+ b

ax2 + bx+ cdx+

B

a

∫1(

x+ b2a

)2+ 4ac−b2

4a2

dx

= A

∫du

u+B

a

∫1(

x+ b2a

)2+ 4ac−b2

4a2

dx, u = ax2 + bx+ c

= A ln∣∣ax2 + bx+ c

∣∣+B

a

√4a2

4ac− b2tan−1

x+ b2a√

4ac−b24a2

+ C

= A ln∣∣ax2 + bx+ c

∣∣+2B√

4ac− b2tan−1

(2ax+ b√4ac− b2

)+ C.(2.182)

Example 68 Evaluate ∫B

x2 + z2dx. (2.183)

Solution Using the substitution u = x/z, we have du = dx/z and∫B

x2 + z2= B

∫1

x2 + z2dx

= zB

∫1

(zu)2 + z2dx

=zB

z2

∫1

u2 + 1dx

=B

ztan−1 u+ C

=B

ztan−1

(xz

)+ C. (2.184)

The same answer can be obtained immediately by using equation(2.182), with A = 0, b = 0, a = 1 and c = z2. 2

4. Q(x) is a product of linear and irreducible quadratic factors whichcan repeat. In this case, for each factor of Q(x) that has the form(ax2 + bx + c)j , the partial fraction decomposition of f(x) can be


written in such a way that it includes terms of the form

A1(2ax+ b) +B1

ax2 + bx+ c+

j∑i=2

Ci(ax2 + bx+ c)i−1

+(2ax+ b)[Di − (i− 1)Cix]

(ax2 + bx+ c)i.

(2.185)where the constants are to be determined as in the previous scenarios.The first term can be handled as described above. The other termscan be integrated as follows:∫

Di(2ax+ b)

(ax2 + bx+ c)idx = Di

∫du

ui= − Di

(i− 1)(ax2 + bx+ c)i−1+ C,

(2.186)where u = ax2 + bx+ c, and

Ci

∫ [1

(ax2 + bx+ c)i−1− (i− 1)(2ax+ b)x

(ax2 + bx+ c)i

]dx =

Cix

(ax2 + bx+ c)i−1+C.

(2.187)This rule is obtained by applying the Quotient Rule for differentiationin reverse.

It should be noted that the partial fraction decompositions used in the caseof quadratic factors, whether repeated or not, differ substantially from thedecompositions presented in the textbook. The reason for this discrepancy isthat the decompositions presented here, while somewhat more complicated,make the actual integration process far simpler, since terms correspondingto each constant in the numerator can be integrated independently withouthaving to employ algebraic tricks that are difficult to conceive and applycorrectly.

2.5 Approximate Integration

The Fundamental Theorem of Calculus provides a simple method for eval-uating many definite integrals of the form∫ b

af(x) dx. (2.188)

However, the theorem can only be used under the following conditions:

• The integrand f(x) must be a known continuous function. This is notalways the case, since knowledge of f(x) may be limited to a small setof values at specific sampling points in the interval [a, b], where suchvalues are typically obtained from measurements.

2.5. APPROXIMATE INTEGRATION 97

• Even if a formula for f(x) is known, we must still be able to anti-differentiate f(x). If no antiderivative is known, then we cannot applythe Fundamental Theorem of Calculus.

In real applications, integrals for which the Fundamental Theorem of Cal-culus is of no use arise quite frequently. Such integrals cannot be evaluatedexactly, and therefore we must approximate their values instead. This leadsto the study of numerical integration, which is also known as numericalquadrature.

2.5.1 Riemann Sums

Since the definite integral is defined to be the limit of a sequence of Riemannsums, certainly any valid Riemann sum can be used to approximate the valueof the integral. we have already used a few particular types of Riemann sumsto obtain such an approximation:

• We can divide the interval [a, b] into n subintervals of equal width∆x = (b − a)/n, and approximate the area under the graph of f byadding the areas of n rectangles of width ∆x and height determined bythe value of f(x) at the left endpoint of each subinterval. The resultingRiemann sum is∫ b

af(x) dx ≈

n∑i=1

f(xi−1)∆x, xi = a+ i∆x. (2.189)

• Similarly, we can determine the height of each rectangle using the rightendpoint of each subinterval, resulting in the approximation∫ b

af(x) dx ≈

n∑i=1

f(xi)∆x. (2.190)

• We have seen that a more accurate approximation can be obtained bysetting the height of each rectangle equal to the value of f(x) at themidpoint of the corresponding subinterval. The resulting approxima-tion is known as the Midpoint Rule:∫ b

af(x) dx ≈

n∑i=1

f(xi−1/2)∆x, xi−1/2 =xi−1 + xi

2. (2.191)


In addition, we can approximate the integral by averaging the approxima-tions obtained by using left endpoints and right endpoints. The resultingrule, called the Trapezoidal Rule, effectively approximates the area underthe graph of f by n trapezoids of width ∆x and heights f(xi−1) and f(xi),for i = 1, . . . , n.

Intuitively, it makes sense that an approximation using trapezoids ismore accurate than using rectangles. This is true in that the TrapezoidalRule is far more accurate than using left or right endpoints. In fact, not onlyis it more accurate, but the approximation obtained using the TrapezoidalRule converges to the exact value of the integral at a much faster rate thanthe approximation obtained using left or right endpoints. Specifically, theerror in the approximation by left or right endpoints decreases by a factorof 2 when the number of rectangles is doubled. In other words, the error isproportional to ∆x, and therefore we say that these approximations convergelinearly to the exact value as n, the number of rectangles, tends to ∞.

The error in the Trapezoidal Rule, however, decreases by a factor of4 when the number of rectangles is doubled. It follows that the error isproportional to ∆x2, and therefore we say that the approximation obtainedusing the Trapezoidal Rule converges quadratically to the exact value as n,the number of trapezoids, tends to ∞.

Surprisingly, it turns out that the Midpoint Rule, even though it approx-imates the area using rectangles, is even more accurate than the TrapezoidalRule. Its approximations converge quadratically to the exact value, and theerror with n rectangles is approximately half of the error of the TrapezoidalRule using n trapezoids. The reason for this greater accuracy can be seenby examining a particular subinterval [xi−1, xi] of [a, b]. Let xM be the mid-point of this subinterval. Then the area of the rectangle of height f(xM ) andwidth ∆x is equal to the area of the trapezoid whose upper side is tangent tothe graph of f at xM , since such a trapezoid, by the equation of the tangentline, would have heights f(xM ) − f ′(xM )∆x/2 and f(xM ) + f ′(xM )∆x/2.The fact that the upper side of this trapezoid is tangent to the graph of fallows this side to more closely approximate the graph of f on this subin-terval than the line passing through f(xi−1) and f(xi), which is the upperside of the trapezoid used by the Trapezoidal Rule.

2.5.2 Simpson’s Rule and Other Approximation Methods

The left and right endpoint approximations to the definite integral approxi-mate the integrand f(x) by a constant function on each subinterval of [a, b] todetermine the height of each rectangle. The Midpoint Rule and Trapezoidal


Rule approximate f(x) by a linear function on each subinterval, which helpsexplain why they are more accurate. It is natural to ask if even more accu-racy can be obtained by approximating f(x) by a higher-degree polynomialon each subinterval.

The answer to this question is yes, and this fact is the basis of Simpson’sRule. This rule approximates f(x) by a quadratic function on each consecu-tive pair of subintervals, where the number of subintervals is assumed to beeven. For each pair of subintervals [xi−1, xi] and [xi, xi+1], Simpson’s Ruleimplicitly constructs the unique quadratic function that passes through thethree endpoints xi−1, xi, and xi+1. This quadratic function is then inte-grated exactly on the interval [xi−1, xi+1] using the Fundamental Theoremof Calculus. This integral turns out to be equal to

∆x

3[f(xi−1) + 4f(xi) + f(xi+1)], (2.192)

where ∆x = (b − a)/n is the width of each subinterval. Adding the resultsfrom all pairs of subintervals, we obtain the approximation∫ b

af(x) dx ≈ ∆x

3[f(x0)+4f(x1)+2f(x2)+4f(x3)+· · ·+2f(xn−2)+4f(xn−1)+f(xn)].

(2.193)This rule exhibits fourth-order convergence; that is, doubling the number ofsubintervals reduces the error by a factor of 16.

All of the rules we have discussed are examples of interpolatory quadra-ture rules. These rules have the property that they approximate the in-tegrand by polynomials on subintervals of [a, b], and then integrate eachpolynomial exactly over the corresponding subinterval. Effectively, theserules compute the exact integral of a function that interpolates the valuesof the integrand f(x) on [a, b] based on the values of f at the endpoints ofeach subinterval.

Interpolatory quadrature rules that use k equally spaced sampling pointsfor each polynomial approximation have the property that they obtain theexact result if the integrand is a polynomial of degree k or less if k is odd,and degree k − 1 or less if k is even. For example, Simpson’s Rule is ex-act for cubic polynomials, since it uses three nodes in each approximation.However, much higher accuracy can be obtained if we do not require thesampling points to be equally spaced. This is the basic idea behind Gaus-sian quadrature, in which the sampling points are chosen so as to obtain asmuch accuracy as possible. Using this approach with k sampling points foreach polynomial approximation yields exact results for an integrand that is


a polynomial of degree 2k − 1 or less–nearly twice the degree that can beobtained using equally spaced sampling points. It is for this reason thatGaussian quadrature is often the preferred approach for approximating in-tegrals numerically.

For more information on numerical methods for approximating integralsas well as derivatives, see http://www.math.uci.edu/∼jlambers/math105a/.

Example 69 The following table describes a function r(t) that indicatesthe rate of snowfall at time t, where t is measured in hours since some initialtime.

t 0 1/2 1 3/2 2 5/2 3

r(t) 0.7 1.2 2.3 1.7 1.1 0.5 0.2

Estimate the total snowfall over the three-hour period for which data hasbeen gathered.

Solution We will estimate the total snowfall, which is given by the definiteintegral ∫ 3

0r(t) dt, (2.194)

using five different methods: left endpoints, right endpoints, the MidpointRule, the Trapezoidal Rule, and Simpson’s Rule.

1. Left endpoints: the integral can be interpreted not only as total snow-fall, but also the area under the curve y = r(t) from t = 0 to t = 3. Thisarea can be estimated by approximating the region under the curveby six rectangles that are determined as follows: the interval [0, 3] isdivided into six subintervals of equal with ∆t = (3−0)/6 = 1/2. Eachsubinterval has endpoints [ti−1, ti] where ti = i∆t for i = 0, . . . , 6.Then, we have ∫ 3

0r(t) dt =

6∑i=1

∫ ti

ti−1

r(t) dt, (2.195)

and each integral in the above sum can be approximated as follows:∫ ti

ti−1

r(t) dt ≈ r(ti−1)(ti−1 − ti) = r(ti−1)∆t. (2.196)

In other words, the integral of the subinterval [ti−1, ti] is approximatedby computing the area of a rectangle with width ∆t and height equalto the value of r(t) at the left endpoint of the subinterval. Since there


are 6 subintervals, each of width 1/2, it follows that the approximationL6 to the total snowfall obtained using left endpoints is given by

L6 =6∑i=1

r(ti−1)∆t

=6∑i=1

r((i− 1)∆t)∆t

=6∑i=1

r((i− 1)/2)/2

=1

2[r(0) + r(1/2) + r(1) + r(3/2) + r(2) + r(3/2)]

=1

2[0.7 + 1.2 + 2.3 + 1.7 + 1.1 + 0.5]

=1

2(7.5)

= 3.75. (2.197)

The approximating rectangles are shown in Figure 2.1.

2. Right endpoints: we proceed exactly as with left endpoints, but thistime we approximate the integral over each subinterval by the area of arectangle whose height is obtained by evaluating r(t) at the right end-point, ti, of each subinterval [ti−1, ti]. The approximation R6 obtainedusing these right endpoints is

R6 =6∑i=1

r(ti)∆t

=6∑i=1

r(i∆t)∆t

=6∑i=1

r(i/2)/2

=1

2[r(1/2) + r(1) + r(3/2) + r(2) + r(3/2) + r(3)]

=1

2[1.2 + 2.3 + 1.7 + 1.1 + 0.5 + 0.2]

=1

2(7)

= 3.5. (2.198)


Figure 2.1: Approximating rectangles from using left endpoints

The approximating rectangles are shown in Figure 2.2.

3. Midpoint Rule: the area under y = r(t) is once again approximatedby computing the areas of rectangles of equal width, but the height ofeach rectangle is determined by evaluating r(t) at the midpoint of eachsubinterval. Because we only know the values of r(t) at select points,we can use at most 3 subintervals of width ∆t = (3 − 0)/3 = 1. Oursubintervals are [0, 1], [1, 2], and [2, 3], with midpoints 1/2, 3/2, and5/2, respectively. The approximation M3 is given by

M3 =3∑i=1

r(ti−1/2)∆t

=

3∑i=1

r((i− 1/2)∆t)∆t


Figure 2.2: Approximating rectangles from using right endpoints

=

3∑i=1

r(i− 1/2)

= r(1/2) + r(3/2) + r(5/2)

= 1.2 + 1.7 + 0.5

= 3.4. (2.199)

The approximating rectangles are shown in Figure 2.3. Alternatively,the Midpoint Rule can be viewed as approximating the region underr(t) on each subinterval [ti−1, ti] by a trapezoid of width ∆t and heightsr(tM ) + (∆t/2)r′(tM ) and r(tM )− (∆t/2)r′(tM ), where tM = (ti−1 +ti)/2 is the midpoint of the subinterval. Of course, we do not know


Figure 2.3: Approximating rectangles from Midpoint Rule

r′(tM ), but we can approximate it using the finite difference

r′(tM ) ≈ r(ti)− r(ti−1)

∆t. (2.200)

The approximating trapezoids are shown in Figure 2.4.

4. Trapezoidal Rule: The interval [0, 3] is once again divided into sixsubintervals of equal width ∆t = 1/2. On each subinterval [ti−1, ti],the region under y = r(t) is approximated by a trapezoid of width ∆tand heights determined by the value of r(t) at the left endpoint ti−1

and at the right endpoint ti. The approximation T6 obtained usingthis approach is

T6 =6∑i=1

∆tr(ti−1) + r(ti)

2


Figure 2.4: Approximating trapezoids from Midpoint Rule

=∆t

2

6∑i=1

r(ti−1) + r(ti)

=∆t

2[(r(t0) + r(t1)) + (r(t1) + r(t2)) + (r(t2) + r(t3))+

(r(t3) + r(t4)) + (r(t4) + r(t5)) + (r(t5) + r(t6))]

=∆t

2[r(t0) + 2r(t1) + 2r(t2) + 2r(t3) + 2r(t4) + 2r(t5) + r(t6)]

=1

4[0.7 + 2(1.2) + 2(2.3) + 2(1.7) + 2(1.1) + 2(0.5) + 0.2]

=1

4[0.7 + 2.4 + 4.6 + 3.4 + 2.2 + 1 + 0.2]

=1

4(14.5)


= 3.625. (2.201)

The approximating trapezoids are shown in Figure 2.5.

Figure 2.5: Trapezoidal Rule

5. Simpson’s Rule: whereas the approach of using left endpoints or rightendpoints approximates the integrand r(t) by a constant function oneach subinterval, and the Midpoint Rule and Trapezoidal Rule ap-proximate r(t) by a linear function on each subinterval, Simpson’sRule approximates r(t) by a quadratic function on each pair of adja-cent subintervals and integrates each quadratic function exactly usingthe Fundamental Theorem of Calculus. The resulting approximationS6 is

S6 =∆x

3[r(t0) + 4r(t1) + 2r(t2) + 4r(t3) + 2r(t4) + 4r(t5) + 2r(t6) + r(t7)]


=1/2

3[0.7 + 4(1.2) + 2(2.3) + 4(1.7) + 2(1.1) + 4(0.5) + 0.2]

=1

6[0.7 + 4.8 + 4.6 + 6.8 + 2.2 + 2 + 0.2]

=1

6(21.3)

= 3.55. (2.202)

The approximating quadratic functions are shown in Figure 2.6. Notethat on the third and fourth subintervals, the quadratic approximationis actually a linear function, since the points (1, 2.3), (3/2, 1.7), and(2, 1.1) all lie on the same line.

Figure 2.6: Approximating quadratic functions from Simpson’s Rule

2


Example 70 Approximate the integral∫ 3

0x2 dx (2.203)

using the following methods:

1. Left endpoints with 3 and 6 subintervals,

2. Right endpoints with 3 and 6 subintervals,

3. The Midpoint Rule with 3 and 6 subintervals,

4. The Trapezoidal Rule with 3 and 6 subintervals,

5. Simpson’s Rule with 2 and 4 subintervals.

Compare to the exact answer∫ 3

0x2 dx =

x3

3

∣∣∣∣30

=27

3= 9. (2.204)

Solution In the case of 3 subintervals, we have width ∆x = (3− 0)/3 = 1,and for 6 subintervals, we have ∆x = (3− 0)/2 = 1/2.

1. Left endpoints: We have

L3 = ∆x[x2

0 + x21 + x2

2

]= 1(02 + 12 + 22) = (0 + 1 + 4) = 5 (2.205)

and

L6 = ∆x[x2

0 + x21 + x2

2 + x23 + x2

4 + x25

]=

1

2

[01 + (1/2)2 + 12 + (3/2)2 + 22 + (5/2)2

]=

1

2[0 + 1/4 + 1 + 9/4 + 4 + 25/4]

=1

2

55

4= 6.875. (2.206)

2. Right endpoints: We have

R3 = ∆x[x2

1 + x22 + x2

3

]= 1(12 + 22 + 32) = (1 + 4 + 9) = 14 (2.207)


and

R6 = ∆x[x2

1 + x22 + x2

3 + x24 + x2

5 + x26

]=

1

2

[(1/2)2 + 12 + (3/2)2 + 22 + (5/2)2 + 32

]=

1

2[1/4 + 1 + 9/4 + 4 + 25/4 + 9]

=1

2

91

4= 11.375. (2.208)

3. The Midpoint Rule: We have

M3 = ∆x[x2

1/2 + x23/2 + x2

5/2

]= 1

[(0.5)2 + (1.5)2 + (2.5)2

]= 8.75

(2.209)and

M6 = ∆x[x2

1/2 + x23/24 +2

5/2 +x27/2 + x2

9/2 + x211/2

]=

1

2

[(0.25)2 + (0.75)2 + (1.25)2 + (1.75)2 + (2.25)2 + (2.75)2

]=

1

217.875

= 8.9375. (2.210)

4. The Trapezoidal Rule: We have

T3 =∆x

2

[x2

0 + 2x21 + 2x2

2 + x23

]=

1

2

[02 + 2(12) + 2(22) + 32

]=

1

2(0+2+8+9) = 9.5

(2.211)and

T6 =∆x

2

[x2

0 + 2x21 + 2x2

2 + 2x23 + 2x2

4 + 2x25 + x2

6

]=

1/2

2

[02 + 2(0.52) + 2(12) + 2(1.52) + 2(22) + 2(2.52) + 32

]=

1

436.5

= 9.125. (2.212)

5. Simpson’s Rule: We have

S2 =∆x

3

[x2

0 + 4x21 + x2

2

]=

1.5

3

[02 + 4(1.52) + 32

]=

1

2[0 + 9 + 9] = 9

(2.213)


and

S4 =∆x

3

[x2

0 + 4x21 + 2x2

2 + 4x23 + x2

4

]=

3/4

3

[02 + 4(0.752) + 2(1.52) + 4(2.252) + 32

]= 9. (2.214)

Because Simpson’s Rule approximates the integrand by a quadratic functionon each pair of subintervals, and the integrand is a quadratic function, itsapproximation is exact no matter how few subintervals are used.

The following table summarizes the error of the other approximations.In all cases the error is obtained by taking the absolute value of the differencebetween the approximation and the exact answer of 9.

Rule Error, n = 3 Error, n = 6

Left endpoints 4 2.125

Right endpoints 5 2.375

Midpoint Rule 0.25 0.0625

Trapezoidal Rule 0.5 0.125

We make the following observations:

• The Midpoint Rule is the most accurate, followed by the TrapezoidalRule. Both of these rules are much more accurate than using eitherleft or right endpoints.

• For the Midpoint Rule and the Trapezoidal Rule, the error is reducedby a factor of 4 when the number of subintervals is doubled.

• When using the left or right endpoints, the error is reduced by a factorof only 2 when the number of subintervals is doubled.

2

2.6 Improper Integrals

So far, we have only considered integrals of the form∫ b

af(x) dx (2.215)

in which the interval [a, b] is finite, and the integrand f(x) is either continu-ous on [a, b], or, if it is discontinuous anywhere in [a, b], the discontinuity is

2.6. IMPROPER INTEGRALS 111

not infinite. In this section, we consider integrals that do not fall into thiscategory, either because the interval [a, b] is infinite, or the integrand be-comes infinite at some point in [a, b]. Such integrals are known as improperintegrals.

2.6.1 Infinite Intervals

In many applications such as in statistics or the study of differential equa-tions on unbounded domains, it is necessary to compute integrals over aninterval [a, b] where either endpoint a or b is infinite. In this case, we candefine such integrals in terms of integrals over finite intervals.

Definition 4 If∫ ta f(x) dx exists for t ≥ a, then we define∫ ∞

af(x) dx = lim

t→∞

∫ t

af(x) dx, (2.216)

provided this limit exists and is finite. Similarly, if∫ bt f(x) dx exists for all

t ≤ b, then we define ∫ b

−∞f(x) dx = lim

t→∞

∫ b

tf(x) dx, (2.217)

provided the limit exists and is finite. If an integral over an infinite intervalexists, we say that the integral is convergent; otherwise, we say that it isdivergent.

Example 71 Evaluate ∫ ∞1

1

xdx. (2.218)

Solution We have ∫ ∞1

1

xdx = lim

b→∞

∫ b

1

1

xdx

= limb→∞

lnx|b1= lim

b→∞ln b− ln 1

= limb→∞

ln b

= ∞ (2.219)

and therefore the integral is divergent. 2



1

x2dx. (2.220)


1

xdx = lim

b→∞

∫ b

1

1

x2dx

= limb→∞

−1

x

∣∣∣∣b1

= limb→∞

−1

b− (−1)

= limb→∞

1− 1

b= 1. (2.221)

2

Example 73 Determine the values of p for which∫ ∞1

1

xpdx (2.222)

is convergent. Assume p > 0.

Solution We know that the integral diverges when p = 1. If p 6= 1, then wehave, by the Power Rule,∫ ∞

1

1

xpdx = lim

b→∞

∫ b

1

1

xpdx

= limb→∞

x−p+1

−p+ 1

∣∣∣∣b1

= limb→∞

1

1− p1

xp−1

∣∣∣∣b1

= limb→∞

1

1− p

(1

bp−1− 1

1p−1

)=

1

p− 1− 1

p− 1limb→∞

1

bp−1. (2.223)

In order for 1/bp−1 to tend to zero as b → ∞, the exponent p − 1 must bepositive, and therefore the integral converges if p > 1. In this case, its valueis 1/(p− 1). 2


Example 74 Determine the volume of the solid of revolution obtained byrevolving the area under 1/x, for x ≥ 1, around the x-axis.

Solution This solid has circular cross-sections of radius 1/x, for all x ≥ 1.It follows that the cross-sectional area is A(x) = π(1/x)2 and therefore thevolume is given by

V =

∫ ∞1

π

(1

x

)2

dx = π

∫ ∞1

1

x2dx = π, (2.224)

based on the result from the previous example. 2


cosx dx. (2.225)


cosx dx = limb→∞

∫ b

0cosx dx

= limb→∞

sinx|b0= lim

b→∞sin b− sin 0

= limb→∞

sin b. (2.226)

However, limb→∞ sin b does not exist because sine oscillates between −1 and1. Therefore the integral is divergent. 2

We can use the integrals defined above to define the integral of a func-tion f(x) over the entire real line. If both

∫∞a f(x) dx and

∫ a−∞ f(x) dx are

convergent, then we define∫ ∞−∞

f(x) dx =

∫ a

−∞f(x) dx+

∫ ∞a

f(x) dx. (2.227)

The process of computing an integral over an infinite interval is suggestedby the definition. For example, to compute

∫∞a f(x) dx, one can compute∫ b

a f(x) dx where the upper limit b is a variable instead of a fixed constant,and then compute the limit of the integral as b→∞.

Example 76 Evaluate ∫ ∞−∞

e−|x| dx. (2.228)


Solution We have∫ ∞−∞

e−|x| dx =

∫ 0

−∞e−|x| dx+

∫ ∞0

e−|x| dx

= lima→−∞

∫ 0

ae−|x| dx+ lim

b→∞

∫ b

0e−|x| dx

= lima→−∞

∫ 0

aex dx+ lim

b→∞

∫ b

0e−x dx

= lima→−∞

ex|0a + limb→∞

−e−x∣∣b0

= lima→−∞

e0 − ea + limb→∞

−e−b − (−e0)

= e0 + e0 − lima→−∞

ea − limb→∞

e−b

= 2. (2.229)

2

2.6.2 Discontinuous Integrands

An integral can be improper even if it is over a finite interval [a, b]. If theintegrand f(x) has a vertical asymptote at some point in [a, b], then the usualdefinition of a definite integral cannot apply, we can define the integral as alimit of integrals computed over an interval on which f(x) is continuous.

Definition 5 Suppose that f(x) has a vertical asymptote at x = a. Thenwe define ∫ b

af(x) = lim

t→a+

∫ b

tf(x) dx, (2.230)

provided the limit exists and is finite. Similarly, if f(x) has a vertical asymp-tote at x = b, then we define∫ b

af(x) = lim

t→b−

∫ t

af(x) dx, (2.231)

provided the limit exists and is finite. Finally, if f(x) has a vertical asymp-tote at x = c, where a < c < b, then we define∫ b

af(x) dx =

∫ c

af(x) dx+

∫ b

cf(x) dx, (2.232)


provided that the two (improper) integrals on the right side of the equationexist. We say that an integral over [a, b] where f(x) has a vertical asymp-tote in [a, b] is convergent if it exists and is finite, and we say that it isdivergent otherwise.

As with integrals over infinite intervals, computing an integral wherethe integrand has a vertical asymptote on [a, b] can proceed using a method

suggested by the definition. For example, to compute∫ ba f(x) dx where f(x)

has a vertical asymptote at x = b, one can compute∫ ca f(x) dx where c is a

variable upper limit assumed to be between a and b, and then take the limitas c→ b from the left.

Example 77 Evaluate ∫ √2

0

dx√2− x2

. (2.233)

Solution The integrand has an infinite discontinuity at x =√

2. Therefore,we have, by the substitution u = x/

√2,∫ √2

0

dx√2− x2

= limb→√

2

∫ b

0

dx√2− x2

= limb→√

2

1√2

∫ b

0

dx√1− x2/2

= limb→√

2

1√2

∫ b

0

dx√1− (x/

√2)2

= limb→1

√2√2

∫ b

0

du√1− u2

= limb→1

∫ b

0

du√1− u2

= limb→1

sin−1 u∣∣b0

= limb→1

sin−1 b− sin−1 0

= limb→1

sin−1 b

=π

2. (2.234)

2


Example 78 Assume p > 0. Determine the values of p for which∫ 1

0

1

xpdx (2.235)

is convergent, and, when it is convergent, determine the integral’s value.

Solution We have∫ 1

0

1

xpdx = lim

a→0+

∫ 1

a

1

xpdx

= lima→0+

x−p+1

−p+ 1

∣∣∣∣1a

= lima→0+

1

1− p1

xp−1

∣∣∣∣1a

= lima→0+

1

1− p

(1

1p−1− 1

ap−1

). (2.236)

For 1/ap−1 to tend to zero as a→ 0, the exponent p− 1 must be negative,and therefore we must have 0 < p < 1. In this case, the integral is convergentto the value 1/(1− p). 2

Example 79 Determine whether the integral∫ 3

1

dx

x2 − 2(2.237)

is convergent.

Solution This integrand has a discontinuity at√

2 ≈ 1.41421356237, whichlies in the interval [1, 3]. Therefore we have∫ 3

1

dx

x2 − 2=

∫ √2

1

dx

x2 − 2+

∫ 3

√2

dx

x2 − 2. (2.238)

However, because 1/(x2 − 2) has the partial fraction decomposition

1

x2 − 2=

A

x+√

2+

B

x−√

2(2.239)

for some constants A and B, it follows that the antiderivative is given by∫dx

x2 − 2= A ln |x+

√2|+B ln |x−

√2|+ C. (2.240)

Since limx→0 lnx = −∞, it follows that the integral from 1 to√

2 is diver-gent, as is the integral from

√2 to 3. Therefore the integral from 1 to 3 does

not exist. 2


2.6.3 A Comparison Test for Improper Integrals

If one only wants to know whether an improper integral over an infiniteinterval is convergent or divergent, without necessarily knowing its exactvalue, then the Comparison Theorem can be useful.

Theorem 3 (Comparison Test) If f(x) ≥ g(x) ≥ 0 for x ≥ a, then thefollowing statements are true:

• If∫∞a f(x) dx is convergent, then

∫∞a g(x) dx is convergent.

• If∫∞a g(x) dx is divergent, then

∫∞a f(x) dx is divergent.

This theorem not easy to prove, but, intuitively, it can be seen that it istrue by comparing the area under the graphs of f and g for x ≥ a, since theregion under g is entirely contained within the region under f .

Example 80 Use the Comparison Theorem to determine whether∫ ∞1

dx

x3 + 7x2 + 2x+ 1(2.241)

is convergent.

Solution For x > 0, we have

x3 + 7x2 + 2x+ 1 > x3 (2.242)

and therefore1

x3 + 7x2 + 2x+ 1<

1

x3. (2.243)

Since ∫ ∞1

1

x3dx = lim

b→∞

∫ b

1

1

x3dx =

1

2, (2.244)

it follows from the Comparison Theorem that∫ ∞1

dx

x3 + 7x2 + 2x+ 1(2.245)

is convergent. 2

Chapter 3

Applications of Integration

In the next few sections, we begin to expand the applicability of the definiteintegral. In particular, we will learn how the definite integral can be usedto compute areas of more general regions than in previous discussion, andeven volumes of certain three-dimensional solids.

3.1 Areas Between Curves

We have already learned that the definite integral∫ b

af(x) dx (3.1)

can be used to compute the area of the region bounded by the curve y =f(x), the horizontal line y = 0, and the vertical lines x = a and x = b. Now,suppose we need to compute the area of a somewhat more complicatedregion: one that is bounded above by the curve y = f(x) and boundedbelow by the curve y = g(x), between the vertical lines x = a and x = b.We assume that f(x) and g(x) are continuous on [a, b], and that f(x) ≥ g(x)on [a, b].

As before, we can approximate this region using rectangles. We dividethe interval [a, b] into n subintervals of width ∆x = (b−a)/n. These subinter-vals are [x0, x1], [x1, x2], . . ., [xn−1, xn], where xi = a+ i∆x for i = 0, . . . , n.We then approximate the region by n rectangles of width ∆x and heightf(x∗i ) − g(x∗i ), where x∗i is any point in the ith subinterval [xi−1, xi]. Notethat the height the ith rectangle happens to be the vertical distance betweenthe curves y = f(x) and y = g(x) at the point x∗i .

119

120 CHAPTER 3. APPLICATIONS OF INTEGRATION

Using these rectangles, we can approximate the area A of the regionusing the Riemann sum

A ≈ Rn =n∑i=1

[f(x∗i )− g(x∗i )]∆xi. (3.2)

As the number of rectangles, n, approaches infinity, we obtain the exact areaof the region between the curves y = f(x) and y = g(x), which is given thedefinite integral

A = limn→∞

Rn =

∫ b

af(x)− g(x) dx. (3.3)

We can therefore use the definite integral to compute the area of the regionbounded above and below by any two curves, not just regions bounded aboveby the curve y = f(x) and bounded below by the horizontal line y = 0.

A similar approach can be used to compute the area of the regionbounded on the right by a curve of the form x = f(y), on the left by acurve x = g(y), below by the horizontal line y = c and above by the hori-zontal line y = d. If f(y) ≥ g(y) on the interval [c, d], then the area A ofsuch a region is given by the definite integral

A =

∫ d

cf(y)− g(y) dy. (3.4)

On a standard graph, x increases to the right, so in order to ensure that thearea has the correct sign, the function whose graph is the left boundary ofthe region, in this case g(y), is subtracted from the function whose graph isthe right boundary, which is f(y) in this case. For this reason, it is helpful toremember “right − left” when integrating with respect to y to compute thearea between to curves. Similarly, it is helpful to remember “top − bottom”when integrating with respect to x, since the integral represents the correctarea when the function whose graph is the bottom boundary is subtractedfrom the function whose graph is the top boundary.

The assumption that f(x) ≥ g(x), or that f(y) ≥ g(y), should not beignored. In general, the above integrals represent the net area of regionsbetween two curves. If g(x) ≥ f(x) on any subinterval of [a, b], then on thatsubinterval, the area between the curves is counted negatively toward thevalue of the integral of f(x) − g(x) from a to b. In general, the area A ofthe region between the graphs of f(x) and g(x) from x = a to x = b is

A =

∫ b

a|f(x)− g(x)| dx, (3.5)

3.1. AREAS BETWEEN CURVES 121

where the absolute value ensures that the area between the two curves isalways counted positively. A similar integral can be used to compute thearea between the curves x = f(y) and x = g(y) from y = c to y = d.

Example 81 Use calculus to compute the area of the triangle with vertices(−4, 2), (1, 7) and (5,−3).

Solution The triangle is displayed in Figure 3.1. We will compute its area

Figure 3.1: Triangle with vertices (−4, 2), (1, 7) and (5,−3)

by computing the area enclosed by the three lines that define the edges ofthe triangle. We begin by computing the equations of these lines. First, weconsider the line that passes through the vertices (−4, 2) and (1, 7). Theslope is given by

7− 2

1− (−4)=

5

5= 1, (3.6)

which yields the equation

y − 7 = 1(x− 1) (3.7)


ory = x+ 6. (3.8)

Proceeding in the same manner with the other two lines, we obtain slopes

−3− 7

5− 1=−10

4= −5

2,−3− 2

5− (−4)= −−5

9, (3.9)

and the corresponding equations

y − 7 = −5

2(x− 1), y − 2 = −5

9(x− (−4)) (3.10)

which simplify to

y = −5

2x+

19

2, y = −5

9x− 2

9. (3.11)

We are now ready to compute the area of the triangle using definiteintegrals. We divide the triangle along the dashed line shown in Figure 3.1.This yields two smaller triangles, the area of which can be computed using asingle definite integral in each case. The left triangle is the region boundedby the lines x = 1, y = x + 6, and y = −5x/9 − 2/9. The line y = x + 6defines the top of the triangle, while the line y = −5x/9 − 2/9 defines thebottom. The limits of integration are dictated by the right boundary x = 1and the fact that the two lines y = x+ 6 and y = −5x/9− 2/9 intersect atx = −4. It follows that its area, which we denote by A1, is

A1 =

∫ 1

−4(x+ 6)−

(−5

9x− 2

9

)dx

=

∫ 1

−4x+ 6 +

5

9x+

2

9dx

=

∫ 1

−4

9

9x+

54

9+

5

9x+

2

9dx

=

∫ 1

−4

14

9x+

56

9dx

=14

9

∫ 1

−4x+ 4 dx

=14

9

[x2

2+ 4x

]∣∣∣∣1−4

=14

9

{[1

2+ 4

]−[

(−4)2

2+ 4(−4)

]}


=14

9

{9

2− [8− 16]

}=

14

9

(9

2+ 8

)=

14

9

(25

2

)=

350

18

=175

9. (3.12)

We use the same approach to compute the area of the triangle to theright of the dashed line. This triangle is the region bounded by the linesx = 1, y = −5x/2 + 19/2, and y = −5x/9−2/9. The line y = −5x/2 + 19/2defines the top of the triangle, while the line y = −5x/9 − 2/9 defines thebottom. The limits of integration are dictated by the right boundary x = 1and the fact that the two lines y = −5x/2 + 19/2 and y = −5x/9 − 2/9intersect at x = 5. It follows that its area, which we denote by A2, is

A2 =

∫ 5

1

(−5

2x+

19

2

)−(−5

9x− 2

9

)dx

=

∫ 5

1−5

2x+

19

2+

5

9x+

2

9dx

=

∫ 5

1−45

18x+

171

18+

10

18x+

4

18dx

=

∫ 5

1−35

18x+

175

18dx

=1

18

∫ 5

1−35x+ 175 dx

=1

18

[−35

x2

2+ 175x

]∣∣∣∣51

=1

18

{[−35

52

2+ 175(5)

]−[−35

2+ 175

]}=

1

18

{−875

2+ 875 +

35

2− 175

}=

1

18

{−840

2+ 700

}=

1

18(−420 + 700)


=280

18

=140

9. (3.13)

We conclude that the area A of the entire triangle is

A = A1 +A2 =175

9+

140

9=

315

9= 35. (3.14)

2

Example 82 Compute the area of the region bounded by the curves y =sinx and y = cosx, as well as the lines x = 0 and x = 2π.

Solution The area that is to be computed is shown in Figure 3.2. The

Figure 3.2: Region bounded by y = sinx, y = cosx, x = 0 and x = 2π,shaded

region whose area we will compute is shaded in the figure. We can see thatthis region can be divided into three regions:


• the region bounded above by y = cosx, below by y = sinx, to the leftby x = 0, and to the right by x = π/4, which is where y = cosx andy = sinx intersect

• the region bounded above by y = sinx, below by y = cosx, to the leftby x = π/4, and to the right by x = 5π/4, which is another point atwhich y = cosx and y = sinx intersect

• the region bounded above by y = cosx, below by y = sinx, to the leftby x = 3π/4, and to the right by x = 2π.

The area of these regions can be obtained by evaluating the definite integrals

A1 =

∫ π/4

0cosx−sinx dx, A2 =

∫ 5π/4

π/4sinx−cosx dx, A3 =

∫ 2π

5π/4cosx−sinx dx.

(3.15)Using anti-differentiation rules, we obtain∫

cosx− sinx dx = sinx+ cosx+ C. (3.16)

It follows from the Fundamental Theorem of Calculus that

A1 =

∫ π/4

0cosx− sinx dx

= sinx+ cosx|π/40

= [sin(π/4) + cos(π/4)]− [sin 0 + cos 0]

=

[√2

2+

√2

2

]− [0 + 1]

=√

2− 1, (3.17)

A2 =

∫ 5π/4

π/4sinx− cosx dx

= − cosx− sinx|5π/4π/4

= [− cos(5π/4)− sin(5π/4)]− [− cos(π/4)− sin(π/4)]

=

[−

(−√

2

2

)−

(−√

2

2

)]−

[−√

2

2−√

2

2

]

=

[√2

2+

√2

2

]+

[√2

2+

√2

2

]= 2

√2, (3.18)


and

A3 =

∫ 2π

5π/4cosx− sinx dx

= sinx+ cosx|2π5π/4= [sin(2π) + cos(2π)]− [sin(5π/4) + cos(5π/4)]

= [0 + 1]−

[−√

2

2−√

2

2

]= 1 +

√2. (3.19)

We conclude that the area A of the entire region between the curves is givenby

A = A1 +A2 +A3 = (√

2− 1) + 2√

2 + (1 +√

2) = 4√

2. (3.20)

It should be noted that this area is given by the definite integral∫ 2π

0| cosx− sinx| dx, (3.21)

which we computed by dividing the interval [0, 2π] into subintervals on whichcosx− sinx is either positive or negative, but not both.

Example 83 Compute the area of the region bounded by the line y = xand the parabola x = y2 − 2.

Solution This region is shown in Figure 3.3. The curve x = y2 − 2 cannotbe described by an equation of the form y = f(x), because it does not passthe vertical line test. Therefore, it is difficult to compute the area of theshaded region in the figure by integrating with respect to x. Instead, we willintegrate with respect to y, with our integrand being the distance betweenthe right curve and the left curve, as a function of y. This distance is givenby y − (y2 − 2), since, for each y-value in the shaded region, the line x = yis to the right of the curve x = y2 − 2.

The limits of integration are given by the y-values at which these twocurves intersect, since we are integrating with respect to y. These y-valuessatisfy the equation y = y2− 2, or y2− y− 2 = 0. The polynomial y2− y− 2factors into (y − 2)(y + 1), so the curves intersect at y = −1 and y = 2.Therefore, the area A of the shaded region is given by the definite integral

A =

∫ 2

−1y − (y2 − 2) dy. (3.22)


Figure 3.3: Area between x = y and x = y2 − 2, shaded

We have

A =

∫ 2

−1y − (y2 − 2) dy

=

∫ 2

−1y − y2 + 2 dy

=y2

2− y3

3+ 2y

∣∣∣∣2−1

=

[22

2− 23

3+ 2(2)

]−[

(−1)2

2− (−1)3

3+ 2(−1)

]=

[2− 8

3+ 4

]−[

1

2+

1

3− 2

]= 2− 8

3+ 4− 1

2− 1

3+ 2


= 8− 9

3− 1

2

=9

2. (3.23)

2

3.2 Volume by Slices

Suppose that S is a solid that lies between the planes x = a and x =b. If A(x) represents the area of the cross-section, or slice, obtained byintersecting S with the plane that crosses the x-axis at x and is perpendicularto the x-axis, then the volume of S is

V =

∫ b

aA(x) dx. (3.24)

This formula is a generalization of the formula for the volume of a cylinderof radius r and height (b − a) placed parallel to the x-axis, which is V =πr2(b − a). In this case, the solid has a constant radius, independent of x,and therefore each cross-section has constant area A(x) ≡ πr2.

The more general formula (3.24) can be obtained by dividing [a, b] intosubintervals [x0, x1], . . ., [xn−1, xn] where x0 = a and xn = b. On eachsubinterval [xi−1, xi], we then approximate the portion of S between theplanes x = xi−1 and x = xi by a cylinder of height xi − xi−1 and radiusA(xi). Just as the definite integral allows us to compute the area of a regionwith non-constant height, it allows us to compute the volume of a solidwhose cross-sectional area is non-constant.

Some solids whose volumes can be computed using the formula (3.24)are known as solids of revolution, as they can be obtained by revolving thearea of some two-dimensional region around a line. For such solids, thecross-sectional area A(x) is typically very easy to determine. To illustrate,we discuss some examples:

• Suppose that the region in question is the area below the curve y =f(x), where f(x) ≥ 0, from x = a to x = b. Furthermore, suppose thatthis region is revolved around the x-axis to obtain the solid S. Then,for each x in [a, b], the cross-section at x is a disc of radius f(x), whichimplies A(x) = π[f(x)]2. The volume can then be computed by usingequation (3.24) with this choice of A(x). This approach to computingthe volume of such a solid is called the disc method.

3.2. VOLUME BY SLICES 129

• Suppose that the region is the area between two curves y = f(x) andy = g(x), where f(x) ≥ g(x) ≥ 0, from x = a to x = b. As before,the region is revolved around the x-axis. Then, for each x betweena and b, the cross-section of the solid at x is the region between twoconcentric circles. The function f(x) determines the outer radius ofthe cross-section, and g(x) determines the inner radius. It follows thatA(x) = π[f(x)2 − g(x)2]; that is, A(x) is the difference of the areas oftwo concentric circles. Because each cross-section resembles a washer,the method of computing the volume of such a solid by integratingA(x) is called the washer method.

Many solids, however, are not solids of revolution. In such cases, thecross-sectional area A(x) may still be easy to determine. For example, thecross-section may be a triangle or a rectangle. If the cross-section is not ashape whose area is easy to compute for each x, then it may be necessaryto intersect the solid with planes that are perpendicular to the y-axis or thez-axis and determine the cross-sectional area as a function of y or z insteadof as a function of x.

It is important to recognize the parallel between computing areas andcomputing volumes using definite integrals. In both cases, an object isapproximated by n simpler objects, such as rectangles or cylinders, so thatthe desired quantity be obtained by computing it easily for each of thesimpler objects, and adding the results to obtain an approximation Rn,which is a Riemann sum. Then, by determining the overall result as afunction of n and computing limn→∞Rn, the exact result is obtained. Itis a interesting exercise to inquire as to what other simple formulas can begeneralized to more difficult problems using this approach.

Example 84 We will compute the volume of conical “beam,” a cross-section of which is shown in Figure 3.4. For each x between 0 and 5, thecross-section of the beam at x can be viewed as a rectangle of length 3x andheight 2x, with a half-disc of radius x removed from each side. It followsthat the cross-sectional area is given by A(x) = 6x2−πx2. Using the formulain equation (3.24), we can compute volume of this solid as follows:∫ 5

06x2 − πx2 dx = 2x3 − π

3x3∣∣∣50

= 125(

2− π

3

). (3.25)

2

Example 85 Suppose that we form a solid by revolving the region undery = x2, from x = 0 to x = 1, around the x-axis. The resulting solid is


Figure 3.4: Cross-section of conical beam for some x between 0 and 5

a curved funnel with cross-sectional area A(x) = π(x2)2 = πx4. It followsthat its volume is ∫ 1

0πx4 dx = π

x5

5

∣∣∣∣10

=π

5. (3.26)

2

Example 86 Suppose that we form a solid by revolving the same region asin Example 85, except that in this case, instead of revolving it around thex-axis, we revolve it around line y = −2. Then, the cross-section of x is adisc with radius x2 +2, instead of x2. It follows that the volume of this solidis∫ 1

0π(x2+2)2 dx = π

∫ 1

0x4+4x2+4 dx = π

(x5

5+

4x3

3+ 4x

)∣∣∣∣10

= π

(1

5+

4

3+ 4

)=

83π

15.

(3.27)2

3.2. VOLUME BY SLICES 131

Example 87 Suppose that we form a solid by revolving the region betweeny = x and y = x2 around x-axis. Then, for each x between 0 and 1 (wherethe curves y = x and y = x2 intersect), the cross-section of the solid at x isa washer with outer radius x and inner radius x2, since x ≥ x2 on [0, 1]. Itfollows that the volume of the solid, obtained by the washer method, is∫ 1

0πx2 − πx4 dx = π

(x3

3− x5

5

)∣∣∣∣10

= π

(1

3− 1

5

)=

2π

15. (3.28)

2

Example 88 Suppose that we form a solid by revolving the region undery = 2−

√x, from x = 0 to x = 1, around line y = 2. This region is illustrated

in Figure 3.5. For each x between 0 and 1, the cross-section of the solid atx is a washer. The outer radius of the washer is 2, the distance between theline y = 2 and the line y = 0. The inner radius is the distance between thecurve y = 2−

√x and the line y = 2, which is 2− (2−

√x) =

√x. It follows

that the volume of the solid is∫ 1

0π(22 −

√x

2) dx = π

∫ 1

04− x dx = π

(4x− x2

2

)∣∣∣∣10

= π

(4− 1

2

)=

7π

2.

2

Example 89 Suppose that we form a solid of revolution using the sameregion as in Example 88, (shown in Figure 3.5) except that this time, werevolve the region around the y-axis, which is the vertical line x = 0. Inthis case, we can still use the formula in equation (3.24), except that thecross-sectional area is now a function of y. For each y between 0 and 2, thecross-section at y is a disc.

For 0 ≤ y ≤ 1, the radius of the disc is equal to 1, and for 1 ≤ y ≤ 2,the radius is equal to the horizontal distance between the line x = 0 and thecurve y = 2 −

√x. Rewriting the equation of the curve as a function of y,

we obtain x = (2− y)2, and it follows that the radius of the disc is (2− y)2,for 1 ≤ y ≤ 2.

Because two functions are needed to describe the cross-sectional areaA(y), one for 0 ≤ y ≤ 1 and one for 1 ≤ y ≤ 2, it is necessary to breakup the integral of A(y) from 0 to 2 into two integrals, one for each of thesesubintervals. It follows that the volume V of the solid is

V =

∫ 1

0π(12) dy +

∫ 2

1π[(2− y)2]2 dy


Figure 3.5: The region bounded by y = 2−√x, y = 0, x = 0, and x = 1 is

to be revolved around the line y = 2.

=

∫ 1

0π dy +

∫ 2

1π(2− y)4 dy

= πy|10 − π∫ 0

1u4 du

= π + π

∫ 1

0u4 du

= π + πu5

5

∣∣∣∣10

=6π

5

In the third step, we used the substitution u = 2−y to simplify the integrandin the second integral. In the next section, we will compute the volume ofthis same solid using another method, called the shell method, in which we

3.3. VOLUME BY SHELLS 133

will not need to break up the integral. 2

3.3 Volume by Shells

In the previous section we learned how we could compute the volumes of asolid S that lay between the planes x = a and x = b by integrating its cross-sectional A(x) area over the interval [a, b]. Unfortunately, this techniqueof computing volume “by slices”, where each slice is a cylinder of infinitelysmall height, is only practical if the cross-sectional area is relatively easy todetermine.

If this is not the case, then we can instead try to determine whetherthe solid can be approximated by several concentric cylindrical shells. Forexample, suppose that b > a ≥ 0, and we have a solid that can be obtainedby rotating the region bounded by y = f(x) (where f(x) ≥ 0), x = a, x = b,and y = 0. Then, the volume of resulting solid can be approximated by firstdividing the interval [a, b] into n subintervals of equal width ∆x = (b−a)/n,with the ith subinterval having endpoints xi−1 and xi, where xi = i∆x fori = 0, . . . , n.

For each subinterval, we denote the midpoint x∗i by (xi−1 +xi)/2. Then,we compute the volumes of n shells of thickness ∆x = (b − a)/n, heightf(x∗i ), inner radius xi−1, and outer radius xi, for i = 1, . . . , n. The volumeof the ith shell is Vi = 2πx∗i f(x∗i )∆x, and therefore the volume V of theentire solid is approximated by

V ≈n∑i=1

Vi =

n∑i=1

2πx∗i f(x∗i )∆x. (3.29)

Letting n→∞, this approximation converges to the exact volume, with thesummation converging to the definite integral

V =

∫ b

a2πxf(x) dx. (3.30)

More generally, if the solid can be viewed as a collection of concentriccylindrical shells of radius r(x) and height f(x), for a ≤ x ≤ b, the volumeof the solid is given by

V =

∫ b

a2πr(x)f(x) dx. (3.31)

If the solid is formed by rotating a region around the y-axis, then r(x) = x.However, a different r(x) must be used if the solid is obtained by rotating


a region around a different line. For example, if the line around which theregion is rotated is x = c, where c must be outside the interval [a, b], thenr(x) = x− c if c ≤ a, while r(x) = c− x if c ≥ b.

Example 90 In the previous section, we used the disc method to computethe volume of the solid obtained by revolving the region bounded by y =2 −√x, y = 0, x = 0 and x = 1 around the y-axis. Now, we will compute

the volume of the same solid using the shell method. The average radius ofthe shell at x is equal to x, and the height of this shell is equal to the heightof the region at x, which is 2−

√x. It follows that the volume of the solid is∫ 1

02πx(2−

√x) dx = 2π

∫ 1

02x−x3/2 dx = 2π

(x2 − 2x5/2

5

)∣∣∣∣∣1

0

= 2π

(1− 2

5

)=

6π

5.

(3.32)2

Example 91 Let f(x) = x2. Consider the region bounded by the curvey = f(x), the horizontal line y = 0, and the vertical lines x = 1 and x = 2.Compute the volume of the solid obtained by revolving this region aroundthe y-axis.

Solution We use the method of cylindrical shells. The given region is shownin Figure 3.6, while the solid obtained by revolving the region around the y-axis is shown in Figure 3.7. This solid can be approximated by a collectionof concentric cylindrical shells. The approximation proceeds as follows: first,we divide the interval [1, 2] into n subintervals of equal width ∆x = 1/n.These intervals have endpoints [x0, x1], [x1, x2], . . ., [xn−1, xn] where xi =1 + i∆x, for i = 0, 1, 2, . . . , n. Then, we approximate the region belowy = f(x) by rectangles of width ∆x and height f(x∗i ), where x∗i is any pointin the interval [xi−1, xi] for i = 1, . . . , n.

Then, by revolving each rectangle around the y-axis, we obtain n cylin-drical shells that approximate the solid, just as the rectangles approximatethe region. This process of revolving each rectangle around the y-axis toobtain a shell is illustrated in Figure 3.8.

For each i = 1, . . . , n, the ith shell has thickness ∆x, height f(x∗i ), innerradius xi−1 and outer radius xi. The volume Vi of this shell is given by

Vi = 2π

(xi−1 + xi

2

)f(x∗i )∆x. (3.33)

By adding the volume of all of these shells, we obtain a Riemann sum thatyields an approximation to the volume of the solid. As the number of subin-


Figure 3.6: Region bounded by y = x2, y = 0, x = 1 and x = 2, shaded

tervals, n, becomes infinite, this approximation converges to the exact vol-ume. The volume V of the solid is therefore given by

V = limn→∞

n∑i=1

Vi

= limn→∞

n∑i=1

2π

(xi−1 + xi

2

)f(x∗i )∆x

= limn→∞

n∑i=1

2π

(xi −

∆x

2

)f(x∗i )∆x

=

∫ 2

12πxf(x) dx. (3.34)


Figure 3.7: Solid obtained by revolving region shown in Figure 1.4 aroundthe y-axis

Evaluating the resulting definite integral yields

V =

∫ 2

12πx(x2) dx

= 2π

∫ 2

1x3 dx

= 2πx4

4

∣∣∣∣21

= 2π

(24

4− 14

4

)= 2π

(4− 1

4

)


Figure 3.8: Cylindrical shell obtained by revolving rectangle around the y-axis. The rectangle is one of several that is used to approximate the regionbounded by y = x2, y = 0, x = 1, x = 2, whose outline is shown.

=15π

2. (3.35)

2

Example 92 Let f(x) = x2. Consider the region from the previous ex-ample, that is bounded by the curve y = f(x), the horizontal line y = 0,and the vertical lines x = 1 and x = 2. Compute the volume of the solidobtained by revolving this region around the vertical line x = −2.

Solution In this case, we can approximate the solid by cylindrical shells asbefore, but the center and radii of the shells is different. Because the centerof the solid is the line x = −2, the inner radii of the ith shell, corresponding


to the subinterval [xi−1, xi], is xi−1 + 2, since that is the distance betweenthe inner boundary of the shell and the center. Similarly, the outer radius ofthe ith shell is xi+2. Proceeding in the previous example, we can determinethat the volume V of the solid is given by

V = limn→∞

n∑i=1

Vi

= limn→∞

n∑i=1

2π

((xi−1 + 2) + (xi + 2)

2

)f(x∗i )∆x

= limn→∞

n∑i=1

2π

(xi + 2− ∆x

2

)f(x∗i )∆x

=

∫ 2

12π(x+ 2)f(x) dx. (3.36)


V =

∫ 2

12π(x+ 2)x2 dx

= 2π

∫ 2

1x3 + 2x2 dx

= 2π

(x4

4+

2x3

3

)∣∣∣∣21

= 2π

(24

4+

2 · 23

3− 14

4− 2 · 13

3

)= 2π

(4 +

16

3− 1

4− 2

3

)= 2π

(15

4+

14

3

)=

101π

6. (3.37)

2

Example 93 Let f(x) = x2. Consider the region from the previous ex-ample, that is bounded by the curve y = f(x), the horizontal line y = 0,and the vertical lines x = 1 and x = 2. Compute the volume of the solidobtained by revolving this region around the horizontal line y = −1.

Solution We will compute the volume of this solid using both the washermethod (that is, volume by slices) and the shell method. Using the washer


method, we see that for each x in the interval [1, 2], the corresponding washerhas inner radius 1 and outer radius x2 + 1. It follows that the volume V ofthe solid is given by

V =

∫ 2

1π(x2 + 1)2 − π12 dx

= π

∫ 2

1x4 + 2x2 + 1− 1 dx

= π

∫ 2

1x4 + 2x2 dx

= π

(x5

5+

2x3

3

)∣∣∣∣21

= π

(25

5+

2 · 23

3− 15

5− 2 · 13

3

)= π

(32

5+

16

3− 1

5− 2

3

)= π

(31

5+

14

3

)=

163π

15. (3.38)

A sample washer is illustrated in Figure 3.9.

Using the shell method, we integrate with respect to y because we arerevolving the region around a horizontal line. For each y in the interval[1, 4], we have a cylindrical shell centered at the line y = −1 with thicknessdy, average radius y + 1, and height 2 − √y, since that is the horizontaldistance between the line x = 2 and the curve y = x2, or, equivalently,x =√y. Furthermore, for each y in the interval [0, 1], we have a cylindrical

shell centered at the line y = −1 with thickness dy, average radius y + 1,and height 1. It follows that the volume V is given by

V =

∫ 1

02π(y + 1)(1) dy +

∫ 4

12π(y + 1)(2−√y) dy

= 2π

[∫ 1

0y + 1 dy +

∫ 4

12y − y3/2 − y1/2 + 2 dy

]

= 2π

(y2

2+ y

)∣∣∣∣10

+

(y2 − y5/2

5/2− y3/2

3/2+ 2y

)∣∣∣∣∣4

1


Figure 3.9: The region bounded by y = x2, y = 0, x = 1 and x = 2 is to berevolved around the line y = −1. The washer corresponding to x = 1.4 isshown.

= 2π

[(12

2+ 1

)+

(42 − 2 · 45/2

5− 2 · 43/2

3+ 2 · 4

)−

(12 − 2 · 15/2

5− 2 · 13/2

3+ 2 · 1

)]

= 2π

[(1

2+ 1

)+

(16− 2 · 32

5− 2 · 8

3+ 8

)−(

1− 2

5− 2

3+ 2

)]= 2π

[3

2+

(24− 64

5− 16

3

)−(

3− 2

5− 2

3

)]= 2π

[3

2+ 21− 64

5− 16

3+

2

5+

2

3

]= 2π

[45

30+

630

30− 384

30− 160

30+

12

30+

20

30

]


= 2π

[163

30

]=

163π

15. (3.39)

A sample shell is illustrated in Figure 3.10. 2

Figure 3.10: The region bounded by y = x2, y = 0, x = 1 and x = 2 is tobe revolved around the line y = −1. The shell corresponding to y = 1.32 isshown.

Example 94 Let f(x) = x2. Consider the region that is bounded by thecurve y = f(x), the horizontal line y = 0, and the vertical lines x = 1 andx = 2. This region in shown in Figure 3.6. Compute the volume of the solidobtained by revolving this region around the vertical line x = 3.


Solution In this case, we can approximate the solid by cylindrical shells.We can divide the interval [1, 2] into n subintervals of width ∆x = 1/n.These subintervals have endpoints [xi−1, xi], where xi = 1 + i∆x. Becausethe center of the solid is the line x = 3, the inner radius of the ith shell,corresponding to the subinterval [xi−1, xi], is 3−xi, since that is the distancebetween the inner boundary of the shell and the center. Similarly, the outerradius of the ith shell is 3 − xi−1. It follows that the volume Vi of the ithshell is

Vi = 2π

((3− xi−1) + (3− xi)

2

)f(x) dx. (3.40)

We can determine that the volume V of the solid is given by

V = limn→∞

n∑i=1

Vi

= limn→∞

n∑i=1

2π

((3− xi−1) + (3− xi)

2

)f(x∗i )∆x

= limn→∞

n∑i=1

2π

(3− xi +

∆x

2

)f(x∗i )∆x

=

∫ 2

12π(3− x)f(x) dx. (3.41)


V =

∫ 2

12π(3− x)x2 dx

= 2π

∫ 2

13x2 − x3 dx

= 2π

(x3 − x4

4

)∣∣∣∣21

= 2π

(23 − 24

4− 13 +

14

4

)= 2π

(7− 4 +

1

4

)= 2π

(3 +

1

4

)=

13π

2. (3.42)


In general, if the axis of revolution is the line x = k, then the average radiusof the shell corresponding to each x is |x− k|. 2

Example 95 Let f(x) = x2. Consider the region from the previous ex-ample, that is bounded by the curve y = f(x), the horizontal line y = 0,and the vertical lines x = 1 and x = 2. Compute the volume of the solidobtained by revolving this region around the x-axis (that is, the horizontalline y = 0).

Solution We will compute the volume of this solid using both the discmethod (that is, volume by slices) and the shell method. Using the discmethod, we see that for each x in the interval [1, 2], the corresponding dischas radius x2. It follows that the volume V of the solid is given by

V =

∫ 2

1π(x2)2 dx

= π

∫ 2

1x4 dx

= πx5

5

∣∣∣∣21

= π

(25

5− 15

5

)= π

(32

5− 1

5

)=

31π

5. (3.43)

Using the shell method, we integrate with respect to y because we arerevolving the region around a horizontal line. For each y in the interval[1, 4], we have a cylindrical shell centered at the line y = 0 with thicknessdy, average radius y, and height 2−√y, since that is the horizontal distancebetween the line x = 2 and the curve y = x2, or, equivalently, x =

√y.

Furthermore, for each y in the interval [0, 1], we have a cylindrical shellcentered at the line y = 0 with thickness dy, average radius y, and height 1.It follows that the volume V is given by

V =

∫ 1

02π(y)(1) dy +

∫ 4

12π(y)(2−√y) dy

= 2π

[∫ 1

0y dy +

∫ 4

12y − y3/2 dy

]


= 2π

(y2

2

)∣∣∣∣10

+

(y2 − y5/2

5/2

)∣∣∣∣∣4

1

= 2π

[(12

2

)+

(42 − 2 · 45/2

5

)−

(12 − 2 · 15/2

5

)]

= 2π

[(1

2

)+

(16− 2 · 32

5

)−(

1− 2

5

)]= 2π

[1

2+

(16− 64

5

)−(

1− 2

5

)]= 2π

[1

2+ 16− 64

5− 1 +

2

5

]= 2π

[5

10+

160

10− 128

10− 10

10+

4

10

]= 2π

[31

10

]=

31π

5. (3.44)

In this case, it is clear that the disc method is easier to use, because it is notnecessary to evaluate two separate integrals. In general, the disc or washermethod (together known as the approach of computing volume by slices) ispreferable if it is easier to find a function that describes the dimension ofthe solid that is perpendicular to the axis of revolution. On the other hand,the shell method is preferable if it is easier to find a function that describesthe dimension of the solid that is parallel to the axis of revolution.

In this example, the heights of the shells, which are measured in thedirection parallel the x-axis, can not be described by a single formula, butthe radii of the discs, which are perpendicular to the x-axis, are easy todescribe in this way. 2

3.4 Arc Length

In this section, we will learn how to use calculus to compute the length of acurve that is described by an equation of the form y = f(x), for some givenfunction f(x). Just as we learned how to compute the area under such acurve as the limit of a sum of areas of simpler regions (namely, rectangles),we can compute the length of the curve by interpreting the length as a limitof a sum of lengths of the simplest curves known, which are line segments.

3.4. ARC LENGTH 145

Suppose that we wish to compute the length of the curve y = f(x)between x = a and x = b. We can approximate this length by dividingthe interval [a, b] into subintervals of length ∆x = (b − a)/n, just as wedid when we were trying to compute the approximate area under y = f(x).Consider any subinterval [xi−1, xi]. Then, if ∆x is chosen to be sufficientlysmall, the length of the curve y = f(x) between x = xi−1 and x = xi canbe well approximated by the length of the line segment between the points(xi−1, f(xi−1)) and (xi, f(xi)). This line segment is the hypotenuse of a righttriangle with legs of length ∆x and |f(xi)−f(xi−1)|, and therefore the lengthLi of the curve y = f(x) between x = xi−1 and x = xi is approximately

Li ≈√

∆x2 + (f(xi)− f(xi−1))2 = ∆x

√1 +

(f(xi)− f(xi−1)

∆x

)2

. (3.45)

It follows that the length L of the curve between x = a and x = b isapproximated by

L ≈n∑i=1

Li =

n∑i=1

√1 +

(f(xi)− f(xi−1)

∆x

)2

∆x =

n∑i=1

√1 +

(f(xi)− f(xi−1)

xi − xi−1

)2

∆x.

(3.46)As n, the number of line segments, approaches ∞, ∆x approaches zero,

so the length of each subinterval [xi−1, xi] tends to zero. It follows from thedefinition of the derivative that

lim∆x→0

f(xi)− f(xi−1)

xi − xi−1= lim

∆x→0

f(xi−1 + ∆x)− f(xi−1)

∆x= f ′(xi−1) (3.47)

and therefore the sum converges to the definite integral

L = limn→∞

n∑i=1

Li =

∫ b

a

√1 + [f ′(x)]2 dx. (3.48)

The value of this integral is called the arc length of the curve y = f(x) fromx = a to x = b. Similarly, if a curve is defined by the equation x = f(y)from y = c to y = d, the arc length of the curve is given by the definiteintegral ∫ d

c

√1 + [f ′(y)]2 dy. (3.49)

Example 96 Compute the arc length of the curve y = 2x + 3, where 0 ≤x ≤ 2.


Solution Since the curve is just a line segment, we can simply use thedistance formula to compute the arc length, since the arc length is thedistance between the endpoints of the segment. The endpoints are (0, 3)and (2, 7), and therefore the arc length is√

(2− 0)2 + (7− 3)2 =√

22 + 42 =√

20 = 2√

5. (3.50)

Using the arc length formula, we have y′ = 2, and therefore the arc lengthis given by the integral∫ 2

0

√1 + 22 dx =

∫ 2

0

√5 dx =

√5x∣∣∣20

= 2√

5. (3.51)

2

Example 97 Compute the arc length of the curve y = sinx from x = 0 tox = π.

Solution Since y′ = cosx, the arc length is given by the integral∫ π

0

√1 + cos2 x dx. (3.52)

Unfortunately, this integral cannot be evaluated using the Fundamental The-orem of Calculus. Using an approximation method such as Simpson’s Rule,the value of the integral is seen to be approximately 3.8202. 2

Example 98 Compute the arc length of the astroid described by the equa-tion x2/3 + y2/3 = 1.

Solution We consider only the portion of the astroid in the upper quadrantx ≥ 0, y ≥ 0, which has endpoints (0, 1) and (1, 0). In this quadrant, theastroid can be described by the equation

y = (1− x2/3)3/2. (3.53)

It follows that the arc length L of this segment of the astroid is given by theintegral

L =

∫ 1

0

√1 + (y′)2 dx

=

∫ 1

0

√1 + ((3/2)(1− x2/3)1/2(−(2/3)x−1/3)2 dx

3.4. ARC LENGTH 147

=

∫ 1

0

√1 + (−(1− x2/3)1/2x−1/3)2 dx

=

∫ 1

0

√1 + (1− x2/3)x−2/3 dx

=

∫ 1

0

√1 + (x−2/3 − 1) dx

=

∫ 1

0

√x−2/3 dx

=

∫ 1

0x−1/3 dx

=3

2x2/3

∣∣∣∣10

=3

2. (3.54)

2

Example 99 Prove that the shortest distance between two given points isa straight line.

Solution For simplicity, we assume that the two points lie on the samehorizontal line; specifically, the points are (a, k) and (b, k). Let y = f(x)describe a curve connecting the two points. Then, the arc length of thecurve is given by ∫ b

a

√1 + [f ′(x)]2 dx. (3.55)

Since the integrand√

1 + [f ′(x)]2 is always positive, we can minimize thearc length by choosing f(x) so that the integrand itself is minimized. Thisis the case when f ′(x) = 0; i.e., f(x) is constant. Therefore the arc lengthis minimized when f(x) = k and the corresponding curve is a straight lineconnecting the two points. 2

In some cases, it is desirable to compute the arc length of a curve y =f(x) as a function of its endpoints. For example, if the left endpoint of thecurve is fixed at the point (a, f(a)) and we wish to know the arc length alongthis curve from the left endpoint to any other point (x, f(x)), then we canobtain this length as a function of x from the integral

s(x) =

∫ x

a

√1 + [f ′(t)]2 dt. (3.56)

The function s(x) is known as the arc length function.

MAT 168: Calculus II with Analytic Geometry · MAT 168: Calculus II with Analytic Geometry James V....

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Transcript of MAT 168: Calculus II with Analytic Geometry · MAT 168: Calculus II with Analytic Geometry James V....