Integration Not for Sale

Not for Sale 5 Integration

Chapter Preview We are now at a critical point in the calculus story. Many would 5.1 Approximating Areas argue that this chapter is the cornerstone of calculus because it explains the relationship under Curves between the two processes of calculus: differentiation and integration. We begin by ex- 5.2 Definite Integrals plaining why finding the area of regions bounded by the graphs of functions is such an important problem in calculus. Then you will see how antiderivatives lead to definite in- 5.3 Fundamental Theorem tegrals, which are used to solve the area problem. But there is more to the story. You will of Calculus also see the remarkable connection between derivatives and integrals, which is expressed 5.4 Working with Integrals in the Fundamental Theorem of Calculus. In this chapter, we develop key properties of 5.5 Substitution Rule definite integrals, investigate a few of their many applications, and present the first of several powerful techniques for evaluating definite integrals. 5.1 Approximating Areas under Curves

The derivative of a function is associated with rates of change and slopes of tangent lines. We also know that antiderivatives (or indefinite integrals) reverse the derivative operation. Figure 5.1 summarizes our current understanding and raises the question: What is the geometric meaning of the integral? The following example reveals a clue.

f(x)

f Ј(x) ͵ f (x)dx

Slopes of ??? tangent lines

Figure 5.1 Area under a Velocity Curve Consider an object moving along a line with a known position function. You learned in previous chapters that the slope of the line tangent to the graph of the position function at a certain time gives the velocity v at that time. We now turn the situation around. If we know the velocity function of a moving object, what can we learn about its position function?

333

M05_BRIG7345_02_SE_C05.1.indd 333 21/10/13 9:18 PM 334 Chapter 5 • Integration Not for Sale

➤ Recall from Section 3.5 that the Imagine a car traveling at a constant velocity of 60 mi hr along a straight high- displacement of an object moving along way over a two-hour period. The graph of the velocity function v = 60 on the interval a line is the difference between its initial 0 … t … 2 is a horizontal line (Figure 5.2). The displacement> of the car between t = 0 and final position. If the velocity of an and t = 2 is found by a familiar formula: object is positive, its displacement equals the distance traveled. displacement = rate # time = 60 mi hr # 2 hr = 120 mi. This product is the area of the rectangle formed> by the velocity curve and the t-axis between t = 0 and t = 2 (Figure 5.3). In the case of constant positive velocity, we see that the area between the velocity curve and the t-axis is the displacement of the moving object.

v v Elapsed time ➤ The side lengths of the rectangle in 70 70 ϭ 2 hr Figure 5.3 have units mi hr and hr. 60 60 Therefore, the units of the area are v ϭ 60 v ϭ 60 > 50 50 mi hr # hr = mi, which is a unit of 40 (velocity function) 40 Area ϭ displacement Velocity displacement. 30 30 ϭ (60 mi/hr)(2 hr) > ϭ 60 mi/hr 20 20 ϭ 120 mi Velocity (mi/hr) Velocity 10 (mi/hr) Velocity 10

0 0.5 1.0 1.5 2.0 t 0 0.5 1.0 1.5 2.0 t Time (hr) Time (hr) Figure 5.2 Figure 5.3

Quick C heck 1 What is the displacement of an object that travels at a constant velocity of 10 mi hr for a half hour, 20 mi hr for the next half hour, and 30 mi hr for the next hour? > > > Because objects do not necessarily move at a constant velocity, we first extend these ideas to positive velocities that change over an interval of time. One strategy is to divide the time interval into many subintervals and approximate the velocity on each subinterval with a constant velocity. Then the displacements on each subinterval are calculated and summed. This strategy produces only an approximation to the displacement; however, this approximation generally improves as the number of subintervals increases.

EXAMPLE 1 Approximating the displacement Suppose the velocity in m s of an object moving along a line is given by the function v = t2, where 0 … t … 8. Approximate the displacement of the object by dividing the time interval 0, 8 into n subintervals> of equal length. On each subinterval, approximate the velocity with a constant equal to the value of v evaluated at the midpoint of the subinterval. 3 4 a. Begin by dividing 0, 8 into n = 2 subintervals: 0, 4 and 4, 8 . b. Divide 0, 8 into n 4 subintervals: 0, 2 , 2, 4 , 4, 6 , and 6, 8 . 3 = 4 3 4 3 4 c. Divide 0, 8 into n 8 subintervals of equal length. 3 4 = 3 4 3 4 3 4 3 4 So lution 3 4 a. We divide the interval 0, 8 into n = 2 subintervals, 0, 4 and 4, 8 , each with length 4. The velocity on each subinterval is approximated by evaluating v at the midpoint of that subinterval3 (Figure4 5.4a). 3 4 3 4 • We approximate the velocity on 0, 4 by v 2 = 22 = 4 m s. Traveling at 4 m s for 4 s results in a displacement of 4 m s # 4 s = 16 m. 3 4 1 2 > > • We approximate the velocity on 4, 8 by v 6 62 36 m s. Traveling at 36 m s > = = for 4 s results in a displacement of 36 m s # 4 s = 144 m. 3 4 1 2 > > Therefore, an approximation to the displacement over the entire interval 0, 8 is > v 2 # 4 s + v 6 # 4 s = 4 m s # 4 s + 36 m s # 4 s = 1603 m.4 1 1 2 2 1 1 2 2 1 > 2 1 > 2

M05_BRIG7345_02_SE_C05.1.indd 334 21/10/13 11:09 AM Not for Sale 5.1 Approximating Areas under Curves 335 b. With n = 4 (Figure 5.4b), each subinterval has length 2. The approximate displacement over the entire interval is 1 m s # 2 s + 9 m s # 2 s + 25 m s # 2 s + 49 m s # 2 s = 168 m. (1)1* (1)1* (+)+* (+)+* 1 v 1> 2 1 v 3> 2 1 v 5 > 2 1 v 7 > 2 c. With n =1 82 subintervals (1Figure2 5.4c), the approximation1 2 to the1 2 displacement is 170 m. In each case, the approximate displacement is the sum of the areas of the rectangles under the velocity curve.

The midpoint of each subinterval is used to approximate the velocity over that subinterval.

v v v

70 70 70 v ϭ t2 v ϭ t2 v ϭ t2 60 60 60 50 50 50 40 40 40 30 30 30 20 20 20 10 10 10

0 1 2 3 4 5 6 7 8 t 0 1 2 3 4 5 6 7 8 t 0 1 2 3 4 5 6 7 8 t n ϭ 2 n ϭ 4 n ϭ 8 (a) (b) (c) Figure 5.4 Related Exercises 9–16

Quick C heck 2 In Example 1, if we used n = 32 subintervals of equal length, what would be the length of each subinterval? Find the midpoint of the first and last subinterval. v The progression in Example 1 may be continued. Larger values of n mean more rectan- 70 gles; in general, more rectangles give a better fit to the region under the curve (Figure 5.5). 60 With the help of a calculator, we can generate the approximations in Table 5.1 using 2 v ϭ t n = 1, 2, 4, 8, 16, 32, and 64 subintervals. Observe that as n increases, the approxima- 50 tions appear to approach a limit of approximately 170.7 m. The limit is the exact displace- 40 ment, which is represented by the area of the region under the velocity curve. This strategy 30 of taking limits of sums is developed fully in Section 5.2.

10 T able 5.1 Approximations to the area under the velocity curve v t2 on 0, 8 0 12345678t Number of Length of 3 4 Approximate displacement n ϭ 64 subintervals each subinterval (area under curve) Figure 5.5 1 8 s 128.0 m 2 4 s 160.0 m 4 2 s 168.0 m 8 1 s 170.0 m 16 0.5 s 170.5 m 32 0.25 s 170.625 m 64 0.125 s 170.65625 m

M05_BRIG7345_02_SE_C05.1.indd 335 21/10/13 11:09 AM 336 Chapter 5 • Integration Not for Sale ➤ The language “the area of the region Approximating Areas by Riemann Sums bounded by the graph of a function” is We wouldn’t spend much time investigating areas under curves if the idea applied only often abbreviated as “the area under the curve.” to computing displacements from velocity curves. However, the problem of finding areas under curves arises frequently and turns out to be immensely important—as you will see in the next two chapters. For this reason, we now develop a systematic method for y approximating areas under curves. Consider a function f that is continuous and nonnega- y ϭ f(x) tive on an interval a, b . The goal is to approximate the area of the region R bounded by the graph of f and the x-axis from x = a to x = b (Figure 5.6). We begin by dividing the interval a, b into 3n subintervals4 of equal length, x , x , x , x , , x , x , R 3 4 0 1 1 2 c n - 1 n where a = x0 and b = xn (3Figure4 5.73 ). The4 length3 of each4 subinterval, denoted ∆x, is found by dividing the length of the entire interval by n: O abx b - a ∆x . Figure 5.6 = n

⌬x ⌬x ⌬x ⌬x …

x x0 ϭ axx1 x2 x3 … xnϪ1 n ϭ b Figure 5.7

Definition Regular Partition Suppose a, b is a closed interval containing n subintervals

3 4 x0, x1 , x1, x2 , c, xn - 1, xn b - a of equal length ∆x 3 with4 a3 x 4 and b 3 x . The4 endpoints x , x , x , , = n = 0 = n 0 1 2 c

xn - 1, xn of the subintervals are called grid points, and they create a regular partition of the interval a, b . In general, the kth grid point is

3 4 xk = a + k∆x, for k = 0, 1, 2, c, n.

Quick C heck 3 If the interval 1, 9 is partitioned into 4 subintervals of equal length, what is ∆x? List the grid points x0, x1, x2, x3, and x4. 3 4 * In the kth subinterval xk - 1, xk , we choose any point x k and build a rectangle whose * * height is f x k , the value of f at x k (Figure 5.8). The area of the rectangle on the kth subinterval is 3 4 1 2 * height # base = f x k ∆x, where k = 1, 2, c, n. ➤ Although the idea of integration was developed in the 17th century, it was Summing the areas of the rectangles1 in2 Figure 5.8, we obtain an approximation to the area almost 200 years later that the German of R, which is called a Riemann sum: mathematician Bernhard Riemann * * * f x 1 ∆x + f x 2 ∆x + + f x n ∆x. (1826–1866) worked on the mathematical g theory underlying integration. Three notable Riemann sums1 are2 the left1 , right2 , and midpoint1 2 Riemann sums.

M05_BRIG7345_02_SE_C05.1.indd 336 21/10/13 11:09 AM Not for Sale 5.1 Approximating Areas under Curves 337 Area of kth rectangle y ϭ height ؒ base * ⌬x ϭ f(xk) ⌬x

y ϭ f(x)

* f(xk)

x ϭ axx x x x x ϭ b x ➤ For the left Riemann sum, O 0 1 2 ……kϪ1 k nϪ1 n * * x a 0 # ∆x, x a 1 # ∆x, ** ** 1 = + 2 = + x1 x2 xk xn * x 3 = a + 2 # ∆x, * Figure 5.8 and in general, x k = a + k - 1 ∆x, for k = 1, c, n. 1 2 For the right Riemann sum, Definition Riemann Sum * * x 1 = a + 1 # ∆x, x 2 = a + 2 # ∆x, Suppose f is defined on a closed interval a, b , which is divided into n subin- * x 3 = a + 3 # ∆x, * * tervals of equal length ∆x. If x k is any point in the kth subinterval xk - 1, xk , for and in general x k = a + k∆x, 3 4 k = 1, 2, c, n, then for k = 1, c, n. 3 4 * * * For the midpoint Riemann sum, f x 1 ∆x + f x 2 ∆x + g + f x n ∆x is called a Riemann sum for f on a, b . This sum is called * 1 * 3 1 2 1 2 1 2 x 1 = a + ∆x, x 2 = a + ∆x, 2 2 * • a left Riemann sum if x k is the 3left endpoint4 of xk - 1, xk (Figure 5.9); and in general, • a right Riemann sum if x * is the right endpoint of x , x (Figure 5.10); and k 3 k - 14 k * 1 xk + xk - 1 * x k = a + k - ∆x = , • a midpoint Riemann sum if x k is the midpoint of xk 1, xk (Figure 5.11), for 2 2 3 - 4 k = 1, 2, c, n. 3 4 for k = 1, c1 , n. 2

y y y Left Riemann Sum Right Riemann Sum Midpoint Riemann Sum ⌬x y ϭ f(x) y ϭ f(x) y ϭ f(x) ⌬x ⌬x

* f(x*) f(x*) f(xk) k k

O * ** **x O * * * **O * * * ……**x x1 ϭ a x2 x3 ……xk xn b a x1 x2 x3 ……xk xn ϭ b a x1 x2 x3 xk xn b Figure 5.9 Figure 5.10 Figure 5.11

We now use this definition to approximate the area under the curve y = sin x.

EXAMPLE 2 Left and right Riemann sums Let R be the region bounded by the graph

of f x = sin x and the x-axis between x = 0 and x = p 2. a. Approximate1 2 the area of R using a left Riemann sum with> n = 6 subintervals. Illustrate the sum with the appropriate rectangles. b. Approximate the area of R using a right Riemann sum with n = 6 subintervals. Illustrate the sum with the appropriate rectangles. c. Do the area approximations in parts (a) and (b) underestimate or overestimate the actual area under the curve?

M05_BRIG7345_02_SE_C05.1.indd 337 21/10/13 11:09 AM 338 Chapter 5 • Integration Not for Sale So lution Dividing the interval a, b = 0, p 2 into n = 6 subintervals means the length of each subinterval is 3 4 3 > 4 b - a p 2 - 0 p ∆x = = = . n 6 12 > * * * a. To find the left Riemann sum, we set x 1, x 2, c, x 6 equal to the left endpoints of the * six subintervals. The heights of the rectangles are f x k , for k = 1, c, 6.

y 1 2 f (x) ϭ sin x Left Riemann Sum 1

Width of each First rectangle rectangle is has height 0. ␲ f (x*) ⌬x ϭ . * 6 12 f (x5) f (x*) f (x*) ϭ 0 4 1 * f (x3) * f (x2)

* ␲ ␲ ␲ ␲ 5␲ ␲ x1 ϭ 0 * * * ** x x2 ϭ 12 x3 ϭ 6 x4 ϭ 4 x5 ϭ 3 x6 ϭ 12 2 Figure 5.12

The resulting left Riemann sum (Figure 5.12) is * * * f x 1 ∆x + f x 2 ∆x + g + f x 6 ∆x p p p p p 1 =2 sin 0 1# 2 + sin #1 2+ sin # 12 12 12 6 12 1 2p p a bp p a b5p p + sin # + sin # + sin # 4 12 3 12 12 12 ≈ 0.863.a b a b a b * * * b. In a right Riemann sum, the right endpoints are used for x 1, x 2, c, x 6, and the * heights of the rectangles are f x k , for k = 1, c, 6.

y 1 2 f (x) ϭ sin x Right Riemann Sum 1

Width of each rectangle is ␲ * f (x*) f (x ) ⌬x ϭ 12 . f (x*) * 5 6 1 f (x4) * f (x3) * f (x2)

00 * ␲ * ␲ * ␲ * ␲ * 5␲ * ␲ x x1 ϭ 12 x2 ϭ 6 x3 ϭ 4 x4 ϭ 3 x5 ϭ 12 x6 ϭ 2 Figure 5.13

The resulting right Riemann sum (Figure 5.13) is * * * f x 1 ∆x + f x 2 ∆x + g + f x 6 ∆x p p p p p p 1= 2 sin 1 # 2 + sin 1# 2 + sin # 12 12 6 12 4 12 a pb p a 5bp p a bp p + sin # + sin # + sin # 3 12 12 12 2 12 ≈ 1.125.a b a b a b

M05_BRIG7345_02_SE_C05.1.indd 338 21/10/13 11:09 AM Not for Sale 5.1 Approximating Areas under Curves 339 Quick C heck 4 If the function c. Looking at the graphs, we see that the left Riemann sum in part (a) underestimates the in Example 2 is replaced with actual area of R, whereas the right Riemann sum in part (b) overestimates the area of

f x = cos x, does the left Riemann R. Therefore, the area of R is between 0.863 and 1.125. As the number of rectangles sum or the right Riemann sum increases, these approximations improve. overestimate1 2 the area under the Related Exercises 17–26 curve? EXAMPLE 3 A midpoint Riemann sum Let R be the region bounded by the graph

of f x = sin x and the x-axis between x = 0 and x = p 2. Approximate the area of R using a midpoint Riemann sum with n = 6 subintervals. Illustrate the sum with the ap- propriate1 2 rectangles. > So lution The grid points and the length of the subintervals ∆x = p 12 are the same * * * as in Example 2. To find the midpoint Riemann sum, we set x 1, x 2, c, x 6 equal to the > midpoints of the subintervals. The midpoint of the first subinterval is the average of x0 and x1, which is

* x1 + x0 p 12 + 0 p x 1 = = = . 2 2 24 > The remaining midpoints are also computed by averaging the two nearest grid points.

y f (x) ϭ sin x Midpoint Riemann Sum 1

Width of each rectangle is ␲ ⌬x ϭ 12 . f (x*) f (x*) 6 * * 5 f (x1) f (x4) * f (x3) * f (x2)

0 ␲ ␲ ␲ ␲ 5␲ ␲ x 12 6 4 3 12 2

* ␲ * 3␲ * 5␲ * 7␲ * 9␲ * 11␲ x1 ϭ 24 x2 ϭ 24 x3 ϭ 24 x4 ϭ 24 x5 ϭ 24 x6 ϭ 24 Figure 5.14

The resulting midpoint Riemann sum (Figure 5.14) is * * * f x 1 ∆x + f x 2 ∆x + g + f x 6 ∆x p p 3p p 5p p 1= 2 sin 1 # 2 + sin 1 # 2 + sin # 24 12 24 12 24 12 a 7bp p a 9bp p a 11b p p + sin # + sin # + sin # 24 12 24 12 24 12 ≈ 1.003.a b a b a b Comparing the midpoint Riemann sum (Figure 5.14) with the left (Figure 5.12) and right (Figure 5.13) Riemann sums suggests that the midpoint sum is a more accurate estimate

T able 5.2 of the area under the curve. Related Exercises 27–34

x f x EXAMPLE 4 Riemann sums from tables Estimate the area A under the graph of f on 0 1 1 2 the interval 0, 2 using left and right Riemann sums with n = 4, where f is continuous 0.5 3 but known only at the points in Table 5.2. 1.0 4.5 3 4 So lution With n = 4 subintervals on the interval 0, 2 , ∆x = 2 4 = 0.5. Using the 1.5 5.5 left endpoint of each subinterval, the left Riemann sum is 2.0 6.0 3 4 > A ≈ f 0 + f 0.5 + f 1.0 + f 1.5 ∆x = 1 + 3 + 4.5 + 5.5 0.5 = 7.0. 1 1 2 1 2 1 2 1 22 1 2

M05_BRIG7345_02_SE_C05.1.indd 339 21/10/13 11:09 AM 340 Chapter 5 • Integration Not for Sale Using the right endpoint of each subinterval, the right Riemann sum is

A ≈ f 0.5 + f 1.0 + f 1.5 + f 2.0 ∆x = 3 + 4.5 + 5.5 + 6.0 0.5 = 9.5. With only1 1five2 function1 values,2 1 these2 estimates1 22 of the 1area are necessarily crude.2 Better estimates are obtained by using more subintervals and more function values. Related Exercises 35–38

Sigma (Summation) Notation Working with Riemann sums is cumbersome with large numbers of subintervals. There- fore, we pause for a moment to introduce some notation that simplifies our work. Sigma (or summation) notation is used to express sums in a compact way. For ex- 10 ample, the sum 1 + 2 + 3 + g + 10 is represented in sigma notation as a k. Here k = 1 is how the notation works. The symbol Σ (sigma, the Greek capital S) stands for sum. The index k takes on all integer values from the lower limit k = 1 to the upper limit k = 10 . The expression that immediately follows Σ (the summand) is evaluated for each value of k, and the resulting values are summed. Here are 1some examples.2 1 2 99 n a k = 1 + 2 + 3 + g + 99 = 4950 a k = 1 + 2 + g + n k = 1 k = 1 3 4 2 2 2 2 2 a k = 0 + 1 + 2 + 3 = 14 a 2k + 1 = 3 + 5 + 7 + 9 = 24 k = 0 k = 1 2 1 2 2 2 2 2 2 a k + k = -1 + -1 + 0 + 0 + 1 + 1 + 2 + 2 = 8. k = -1 1 2 11 2 1 22 1 2 1 2 1 2 The index in a sum is a dummy variable. It is internal to the sum, so it does not matter what symbol you choose as an index. For example,

99 99 99 a k = a n = a p. k = 1 n = 1 p = 1 Two properties of sums and sigma notation are useful in upcoming work. Suppose that a1, a2, c, an and b1, b2, c, bn are two sets of real numbers, and suppose that c is a real number. Then we can factor multiplicative constants out of a sum: 5 6 5 6 n n Constant Multiple Rule a cak = c a ak. k = 1 k = 1 We can also split a sum into two sums:

n n n Addition Rule a ak + bk = a ak + a bk. k = 1 k = 1 k = 1 1 2 In the coming examples and exercises, the following formulas for sums of powers of integers are essential.

theore m 5.1 Sums of Powers of Integers n p ➤ Formulas for a k , where p is a positive Let n be a positive integer and c a real number. k = 1 integer, have been known for centuries. n n n n + 1 The formulas for p 0, 1, 2, and 3 are c = cn k = = a a 2 relatively simple. The formulas become k = 1 k = 1 1 2 complicated as p increases. 2 2 n n n + 1 2n + 1 n n n + 1 k2 = k3 = a 6 a 4 k = 1 1 21 2 k = 1 1 2

Related Exercises 39–42

M05_BRIG7345_02_SE_C05.1.indd 340 21/10/13 11:09 AM Not for Sale 5.1 Approximating Areas under Curves 341 Riemann Sums Using Sigma Notation With sigma notation, a Riemann sum has the convenient compact form

n * * * * f x 1 ∆x + f x 2 ∆x + g + f x n ∆x = a f x k ∆x. k = 1 1 2 1 2 1 2 1 2 To express left, right, and midpoint Riemann sums in sigma notation, we must identify the * points x k. * • For left Riemann sums, the left endpoints of the subintervals are x k = a + k - 1 ∆x, for k = 1, c, n. * 1 2 • For right Riemann sums, the right endpoints of the subintervals are x k = a + k∆x, for k = 1, c, n. • For midpoint Riemann sums, the midpoints of the subintervals are * 1 x k = a + k - 2 ∆x, for k = 1, c, n. The three Riemann1 2sums are written compactly as follows.

Definition Left, Right, and Midpoint Riemann Sums in Sigma Notation Suppose f is defined on a closed interval a, b , which is divided into n subintervals of * equal length ∆x. If x k is a point in the kth subinterval xk - 1, xk , for k = 1, 2, c, n, 3 n 4 * then the Riemann sum for f on a, b is a f x k ∆x3. Three cases4 arise in practice. k = 1 * 3 4 1 2 • Left Riemann sum if x k = a + k - 1 ∆x • Right Riemann sum if x * a k x k = +1 ∆ 2 * 1 • Midpoint Riemann sum if x k = a + k - 2 ∆x 1 2 y EXAMPLE 5 Calculating Riemann sums Evaluate the left, right, and midpoint 10 3 3 f (x) ϭ x ϩ 1 Riemann sums for f x = x + 1 between a = 0 and b = 2 using n = 50 subintervals. Make a conjecture about the exact area of the region under the curve (Figure 5.15). 8 1 2 So lution With n = 50, the length of each subinterval is 6 b - a 2 - 0 1 ∆x = = = = 0.04. 4 n 50 25

2 * The value of x k for the left Riemann sum is * x k = a + k - 1 ∆x = 0 + 0.04 k - 1 = 0.04k - 0.04, 0 1 2 x Figure 5.15 for k = 1, 2, c, 50. Therefore,1 2 the left Riemann1 sum, evaluated2 with a calculator, is n 50 * a f x k ∆x = a f 0.04k - 0.04 0.04 = 5.8416. k = 1 k = 1 1 2 1 * 2 To evaluate the right Riemann sum, we let x k = a + k∆x = 0.04k and find that

n 50 * a f x k ∆x = a f 0.04k 0.04 = 6.1616. k = 1 k = 1 1 2 1 2

M05_BRIG7345_02_SE_C05.1.indd 341 21/10/13 11:09 AM 342 Chapter 5 • Integration Not for Sale For the midpoint Riemann sum, we let

* 1 1 x k = a + k - ∆x = 0 + 0.04 k - = 0.04k - 0.02. 2 2 a b a b The value of the sum is n 50 * a f x k ∆x = a f 0.04k - 0.02 0.04 = 5.9992. k = 1 k = 1 1 2 1 2 Because f is increasing on 0, 2 , the left Riemann sum underestimates the area of the shaded region in Figure 5.15 and the right Riemann sum overestimates the area. There- fore, the exact area lies between3 45.8416 and 6.1616. The midpoint Riemann sum usually gives the best estimate for increasing or decreasing functions. T able 5.3 Left, right, and midpoint Table 5.3 shows the left, right, and midpoint Riemann sum approximations for val- R iemann sum approximations ues of n up to 200. All three sets of approximations approach a value near 6, which is a

n Ln Rn Mn reasonable estimate of the area under the curve. In Section 5.2, we show rigorously that the limit of all three Riemann sums as n S ∞ is 6. 20 5.61 6.41 5.995 40 5.8025 6.2025 5.99875 Altern ative So lution It is worth examining another approach to Example 5. Con- sider the right Riemann sum given previously: 60 5.86778 6.13444 5.99944 n 50 80 5.90063 6.10063 5.99969 * a f x k ∆x = a f 0.04k 0.04. 100 5.9204 6.0804 5.9998 k = 1 k = 1 120 5.93361 6.06694 5.99986 1 2 1 2 3 Rather than evaluating this sum with a calculator, we note that f 0.04k = 0.04k + 1 140 5.94306 6.05735 5.9999 and use the properties of sums: 160 5.95016 6.05016 5.99992 1 2 1 2 n 50 180 5.95568 6.04457 5.99994 * 3 a f x k ∆x = a 0.04k + 1 0.04 e 200 5.9601 6.0401 5.99995 k = 1 k = 1 e * ∆ x 1 2 11 f 2x k 2 50 50 13 2 = a 0.04k 0.04 + a 1 # 0.04 a ak + bk = a ak + a bk k = 1 k = 1 1 2 1 502 50 4 3 = 0.04 a k + 0.04 a 1. a cak = c a ak k = 1 k = 1 1 2 Using the summation formulas for powers of integers in Theorem 5.1, we find that

50 50 2 2 3 50 # 51 a 1 = 50 and a k = . k = 1 k = 1 4 Substituting the values of these sums into the right Riemann sum, its value is

50 * 3851 a f x k ∆x = = 6.1616, k = 1 625 1 2 confirming the result given by a calculator. The idea of evaluating Riemann sums for arbitrary values of n is used in Section 5.2, where we evaluate the limit of the Riemann sum as n S ∞. Related Exercises 43–52

M05_BRIG7345_02_SE_C05.1.indd 342 21/10/13 11:10 AM Not for Sale 5.1 Approximating Areas under Curves 343 Section 5.1 Exerci ses

Review Questions T 10. Approximating displacement The velocity in ft s of an object moving along a line is given by v 10t on the interval 1. Suppose an object moves along a line at 15 m s, for 0 … t 6 2, = 1 t 7. > and at 25 m s, for 2 … t … 5, where t is measured in seconds. … … 1 Sketch the graph of the velocity function and find> the displace- a. Divide the time interval 1, 7 into n = 3 subintervals, 1, 3 , ment of the >object for 0 … t … 5. 3, 5 , and 5, 7 . On each subinterval, assume the object moves at a constant velocity3 equal4 to v evaluated at the mid-3 4 2. Given the graph of the positive velocity of an object moving along point3 4 of the3 subinterval4 and use these approximations to esti- a line, what is the geometrical representation of its displacement mate the displacement of the object on 1, 7 (see part (a) of over a time interval a, b ? the figure). 3 4 3. Suppose you want to3 approximate4 the area of the region bounded b. Repeat part (a) for n = 6 subintervals (see part (b) of the

by the graph of f x = cos x and the x-axis between x = 0 and figure). x = p 2. Explain a possible strategy. v v 1 2 v ϭ 10t v ϭ 10t 4. Explain> how Riemann sum approximations to the area of a region 8 8 under a curve change as the number of subintervals increases. 6 6 5. Suppose the interval 1, 3 is partitioned into n 4 subintervals. = 4 4 What is the subinterval length ∆x? List the grid points x0, x1, x2, 3 4 2 2 x3, and x4. Which points are used for the left, right, and midpoint Riemann sums? 0 1 2 3 4 5 6 7 t 0 1 2 3 4 5 6 7 t 6. Suppose the interval 2, 6 is partitioned into n = 4 subintervals (a) (b) with grid points x0 = 2, x1 = 3, x2 = 4, x3 = 5, and x4 = 6. 3 4 Write, but do not evaluate, the left, right, and midpoint Riemann The velocity of an object is 2 11–16. Approximating displacement sums for f x = x . given by the following functions on a specified interval. Approximate the displacement of the object on this interval by subdividing the in- 7. Does a right1 2Riemann sum underestimate or overestimate the area of the region under the graph of a function that is positive and terval into n subintervals. Use the left endpoint of each subinterval to decreasing on an interval a, b ? Explain. compute the height of the rectangles. 11. v 2t 1 m s , for 0 t 8; n 2 8. Does a left Riemann sum 3underestimate4 or overestimate the = + … … = area of the region under the graph of a function that is positive t T 12. v = e m s1, for> 2 0 … t … 3; n = 3 and increasing on an interval a, b ? Explain. 11 > 2 3 4 13. v = m s , for 0 … t … 8; n = 4 Basic Skills 2t + 1 9. Approximating displacement The velocity in ft s of an object 2 1 > 2 14. v = t 2 + 4 ft s , for 0 … t … 12; n = 6 moving along a line is given by v = 3t2 + 1 on the interval 0 … t … 4. > T 15. v = 4 > t + 1 1mi> 2hr , for 0 … t … 15; n = 5 1 a. Divide the interval 0, 4 into n = 4 subintervals, 0, 1 , t + 3 1 > 2 16. v = m s , for 0 … t … 4; n = 4 1, 2 , 2, 3 , and 3, 4 . On each subinterval, assume the 6 object moves at a constant3 4 velocity equal to v evaluated3 4 at the 1 > 2 midpoint3 4 3 of4 the subinterval3 4 and use these approximations to 17–18. Left and right Riemann sums Use the figures to calculate estimate the displacement of the object on 0, 4 (see part (a) the left and right Riemann sums for f on the given interval and for the of the figure). given value of n. b. Repeat part (a) for n = 8 subintervals (see3 part 4(b) of the 17. f x = x + 1 on 1, 6 ; n = 5 figure). 1y2 3 4 y v v f (x) ϭ x ϩ 1 f (x) ϭ x ϩ 1 7 7 v ϭ 3t2 ϩ 1 v ϭ 3t2 ϩ 1 50 50 6 6 40 40 5 5 4 4 30 30 3 3 20 20 2 2 10 10 1 1

0 0 0 1 2 3 4 t 0 1 2 3 4 t 1 2 3 4 5 6 x 1 2 3 4 5 6 x

(a) (b)

M05_BRIG7345_02_SE_C05.1.indd 343 21/10/13 11:10 AM 344 Chapter 5 • Integration Not for Sale 1 29–34. Midpoint Riemann sums Complete the following steps for the 18. f x = on 1, 5 ; n = 4 x given function, interval, and value of n. y 1 2 3 4 y a. Sketch the graph of the function on the given interval. b. Calculate ∆x and the grid points x0, x1, c, xn. f (x) ϭ 1/xf(x) ϭ 1/x c. Illustrate the midpoint Riemann sum by sketching the appropriate 1 1 rectangles. d. Calculate the midpoint Riemann sum.

29. f x = 2x + 1 on 0, 4 ; n = 4 -1 T 30. f 1x2 = 2 cos x on3 0, 41 ; n = 5 0 1 2 3 4 5 x 0 1 2 3 4 5 x T 31. f 1x2 = x on 1, 3 3; n =4 4 12 19–26. Left and right Riemann sums Complete the following steps 32. f 1x2 = x on 0,3 4 ;4 n = 4 for the given function, interval, and value of n. 1 2 1 3 4 33. f x = on 1, 6 ; n = 5 a. Sketch the graph of the function on the given interval. x b. Calculate ∆x and the grid points x , x , , x . 0 1 c n 34. f 1x2 = 4 - x3 on 4-1, 4 ; n = 5 c. Illustrate the left and right Riemann sums. Then determine which Riemann sum underestimates and which sum overestimates the area 35–36.1 Riemann2 sums from3 tables4 Evaluate the left and right under the curve. Riemann sums for f over the given interval for the given value of n. d. Calculate the left and right Riemann sums. 35. n = 4; 0, 2

19. f x = x 1 on 0, 4 ; n = 4 + 3 4 x 0 0.5 1 1.5 2

20. f x 9 x on 3, 8 ; n 5 1 2 = - 3 4 = f x 5 3 2 1 1

21. f 1x2 = cos x on 0,3 p 42 ; n = 4 1 2 36. n = 8; 1, 5 -1 T 22. f 1x2 = sin x 33 on> 0,4 3 ; n = 6 2 3x 4 1 1.5 2 2.5 3 3.5 4 4.5 5 23. f 1x2 = x - 11 >on2 2, 43 ; n4= 4

f x 0 2 3 2 2 1 0 2 3 2 24. f 1x2 = 2x on 1, 63 ; n4 = 5 1 2 x 2 37. Displacement from a table of velocities The velocities (in mi hr) T 25. f 1x2 = e on 31, 44; n = 6 > of an automobile moving along a straight highway over a two-hour > T 26. f 1x2 = ln 4x on3 1, 34 ; n = 5 period are given in the following table.

27. A1 midpoint2 Riemann3 4 sum Approximate the area of the region t hr 0 0.25 0.5 0.75 1 1.25 1.5 1.75 2 2 bounded by the graph of f x = 100 - x and the x-axis on v mi hr 50 50 60 60 55 65 50 60 70 0, 10 with n = 5 subintervals. Use the midpoint of each subin- 1 , 2 1 2 terval to determine the height of each rectangle (see figure). a.1 Sketch2 a smooth curve passing through the data points. 3 4 y b. Find the midpoint Riemann sum approximation to the displacement on 0, 2 with n = 2 and n = 4. 100 f (x) ϭ 100 Ϫ x2 80 38. Displacement from3 a table4 of velocities The velocities (in m s)

60 of an automobile moving along a straight freeway over a four- second period are given in the following table. > 40

20 t s 0 0.5 1 1.5 2 2.5 3 3.5 4 v m1 ,2s 20 25 30 35 30 30 35 40 40 0 2 4 6 8 10 x a. Sketch1 2a smooth curve passing through the data points. T 28. A midpoint Riemann sum Approximate the area of the region b. Find the midpoint Riemann sum approximation to the dis-

bounded by the graph of f t = cos t 2 and the t-axis placement on 0, 4 with n = 2 and n = 4 subintervals. on 0, p with n 4 subintervals. Use the midpoint of each sub- = 39. Sigma notation Express the following sums using sigma notation. interval to determine the height1 2 of each1 > rectangle2 (see figure). 3 4 (Answers are not unique.) 3 4 y a. 1 + 2 + 3 + 4 + 5 b. 4 + 5 + 6 + 7 + 8 + 9 c. 12 22 32 42 d. 1 1 1 1 1 + + + + 2 + 3 + 4 40. Sigma notation Express the following sums using sigma notation. f (t) ϭ cos (t/2) (Answers are not unique.) a. 1 + 3 + 5 + 7 + g + 99 b. 4 + 9 + 14 + g + 44 c. 3 + 8 + 13 + g + 63 1 1 1 1 0 3 t ` ␲ d. + + + g + dq4 1 # 2 2 # 3 3 # 4 49 # 50

M05_BRIG7345_02_SE_C05.1.indd 344 21/10/13 11:10 AM Not for Sale 5.1 Approximating Areas under Curves 345 41. Sigma notation Evaluate the following expressions. Further Explorations 10 6 53. Explain why or why not Determine whether the following state- a. a k b. a 2k + 1 k = 1 k = 1 ments are true and give an explanation or counterexample.

4 5 1 2 a. Consider the linear function f x = 2x + 5 and the region 2 2 c. a k d. a 1 + n bounded by its graph and the x-axis on the interval 3, 6 . k = 1 n = 1 Suppose the area of this region1 is2 approximated using midpoint 3 3 1 2 2m + 2 Riemann sums. Then the approximations give the exact3 4area of e. f. 3j 4 a a - the region for any number of subintervals. m = 1 3 j = 1 b. A left Riemann sum always overestimates the area of a region 5 4 1 2 2 np bounded by a positive increasing function and the x-axis on an g. a 2p + p h. a sin p = 1 n = 0 2 interval a, b . 1 2 c. For an increasing or decreasing nonconstant function on an T 42. Evaluating sums Evaluate the following expressions by two interval 3a, b4 and a given value of n, the value of the midpoint methods. Riemann sum always lies between the values of the left and (i) Use Theorem 5.1. (ii) Use a calculator. right Riemann3 4 sums. 45 45 75 2 2 T 54. Riemann sums for a semicircle Let f x 1 x . a. a k b. a 5k - 1 c. a 2k = - k = 1 k = 1 k = 1 a. Show that the graph of f is the upper half of2 a circle of radius 50 75 1 2 20 1 2 2m + 2 1 centered at the origin. d. 1 n2 e. f. 3j 4 a + a a - b. Estimate the area between the graph of f and the x-axis on the n = 1 m = 1 3 j = 1 35 1 2 40 1 2 interval -1, 1 using a midpoint Riemann sum with n = 25. g. 2p p2 h. n2 3n 1 c. Repeat part (b) using n = 75 rectangles. a + a + - 3 4 p = 1 n = 0 d. What happens to the midpoint Riemann sums on -1, 1 as 1 2 1 2 n S ∞? T 43–46. Riemann sums for larger values of n Complete the following 3 4 steps for the given function f and interval. T 55–58. Sigma notation for Riemann sums Use sigma notation to a. For the given value of n, use sigma notation to write the left, right, write the following Riemann sums. Then evaluate each Riemann sum and midpoint Riemann sums. Then evaluate each sum using a using Theorem 5.1 or a calculator. calculator. 55. The right Riemann sum for f x = x + 1 on 0, 4 with n = 50 b. Based on the approximations found in part (a), estimate the area x of the region bounded by the graph of f and the x-axis on the 56. The left Riemann sum for f x1 2= e on 0, ln 32 with4 n = 40 interval. 3 57. The midpoint Riemann sum1 for2 f x = x3 on 3,4 11 with n 32 43. f x x on 0, 4 ; n 40 = = = 1 2 3 4 12 58. The midpoint Riemann sum for f x 1 cos px on 0, 2 44. f 1x2 = x + 1 3on 4-1, 1 ; n = 50 = + with n 50 2 = 45. f 1x2 = x - 1 on 32, 7 ; 4n = 75 1 2 3 4 59–62. Identifying Riemann sums Fill in the blanks with right or 46. f 1x2 = cos 2x on 30, p 44 ; n = 60 midpoint, an interval, and a value of n. In some cases, more than one answer may work. T 47–52.1 Approximating2 3 areas> with4 a calculator Use a calculator and right Riemann sums to approximate the area of the given region. 4 Present your calculations in a table showing the approximations 59. a f 1 + k # 1 is a ______Riemann sum for f on the k = 1 for n = 10, 30, 60, and 80 subintervals. Comment on whether your interval1 ___,2 ___ with n = ______. approximations appear to approach a limit. 4 2 3 4 47. The region bounded by the graph of f x 4 x and the = - 60. a f 2 + k # 1 is a ______Riemann sum for f on the x-axis on the interval -2, 2 k = 1 1 2 interval ___, ___ with n ______. 2 1 2 = 48. The region bounded by3 the graph4 of f x = x + 1 and the x-axis on the interval 0, 2 4 3 4 1 2 61. a f 1.5 + k # 1 is a ______Riemann sum for f on the 49. The region bounded by3 the4 graph of f x = 2 - 2 sin x and the k = 1 x-axis on the interval -p 2, p 2 interval1 ___, 2___ with n = ______. 1 2 x 8 50. The region bounded by3 the> graph> of4 f x = 2 and the x-axis on 3 k 41 62. f 1.5 + # is a ______Riemann sum for f on the the interval 1, 2 a 2 2 1 2 k = 1 intervala ___, ___b with n = ______. 51. The region bounded3 4 by the graph of f x = ln x and the x-axis on the interval 1, e 1 2 3 4

52. The region bounded3 4 by the graph of f x = x + 1 and the x-axis on the interval 0, 3 1 1 2 3 4

M05_BRIG7345_02_SE_C05.1.indd 345 21/10/13 11:10 AM 346 Chapter 5 • Integration Not for Sale 63. Approximating areas Estimate the area of the region bounded c. Use geometry to find the displacement of the object between 2 by the graph of f x = x + 2 and the x-axis on 0, 2 in the fol- t = 3 and t = 5. lowing ways. d. Assuming that the velocity remains 30 m s, for t Ú 4, find the 1 2 3 4 a. Divide 0, 2 into n 4 subintervals and approximate the area function that gives the displacement between t = 0 and any = > of the region using a left Riemann sum. Illustrate the solution time t Ú 5. 3 4 geometrically. v b. Divide 0, 2 into n = 4 subintervals and approximate the area of the region using a midpoint Riemann sum. Illustrate the 40 solution3 geometrically.4 30 c. Divide 0, 2 into n = 4 subintervals and approximate the area 20 of the region using a right Riemann sum. Illustrate the solution 10 geometrically.3 4 (m/s) Velocity 0 123456 t 64. Approximating area from a graph Approximate the area of the Time (s) region bounded by the graph (see figure) and the x-axis by dividing the interval 0, 6 into n = 3 subintervals. Use a left and right 67. Displacement from a velocity graph Consider the velocity func- Riemann sum to obtain two different approximations. tion for an object moving along a line (see figure). 3 4 y a. Describe the motion of the object over the interval 0, 6 . b. Use geometry to find the displacement of the object between 11 t = 0 and t = 2. 3 4 10 c. Use geometry to find the displacement of the object between 9 t = 2 and t = 5. 8 d. Assuming that the velocity remains 10 m s, for t Ú 5, find the 7 function that gives the displacement between t = 0 and any time t Ú 5. > 6 5 v 4 40 3 30 2 20 1 10 Velocity (m/s) Velocity 0 t 0 1 23456 x 123456 Time (s) 65. Approximating area from a graph Approximate the area of the region bounded by the graph (see figure) and the x-axis by 68. Flow rates Suppose a gauge at the outflow of a reservoir measures the flow rate of water in units of ft3 hr. In Chapter 6, we dividing the interval 1, 7 into n = 6 subintervals. Use a left and right Riemann sum to obtain two different approximations. show that the total amount of water that flows out of the reservoir 3 4 is the area under the flow rate curve. Consider> the flow-rate y function shown in the figure. 11 a. Find the amount of water (in units of ft3) that flows out of the 10 reservoir over the interval 0, 4 . 9 b. Find the amount of water that flows out of the reservoir over the interval 8, 10 . 3 4 8 c. Does more water flow out of the reservoir over the interval 7 0, 4 or 4, 36 ? 4 6 d. Show that the units of your answer are consistent with the 5 units3 4 of the3 variables4 on the axes. 4 y 3 /hr) 5000 3 2 4000 1 3000 2000 0 x 1234567 1000 Water flow rate (ft flow Water Applications 0 246 8 10 t 66. Displacement from a velocity graph Consider the velocity func- Time (hr) tion for an object moving along a line (see figure). 69. Mass from density A thin 10-cm rod is made of an alloy whose a. Describe the motion of the object over the interval 0, 6 . density varies along its length according to the function shown b. Use geometry to find the displacement of the object between in the figure. Assume density is measured in units of g cm. t = 0 and t = 3. 3 4 >

M05_BRIG7345_02_SE_C05.1.indd 346 21/10/13 11:10 AM Not for Sale 5.1 Approximating Areas under Curves 347 In Chapter 6, we show that the mass of the rod is the area under T 72–75. Functions with absolute value Use a calculator and the the density curve. method of your choice to approximate the area of the following regions. Present your calculations in a table, showing approximations a. Find the mass of the left half of the rod 0 … x … 5 . using n 16, 32, and 64 subintervals. Comment on whether your b. Find the mass of the right half of the rod 5 … x … 10 . = approximations appear to approach a limit. c. Find the mass of the entire rod 0 … x …1 10 . 2 d. Find the point along the rod at which it will1 balance (called2 the 2 72. The region bounded by the graph of f x ͉ 25 x ͉ and the 1 2 = - center of mass). x-axis on the interval 0, 10 y 1 2 2 73. The region bounded by3 the graph4 of f x = ͉ x x - 1 ͉ and the 6 x-axis on the interval -1, 1 5 1 2 1 2

4 74. The region bounded by3 the graph4 of f x = ͉ cos 2x ͉ and the x-axis on the interval 0, p 3 1 2 2 3 75. The region bounded by3 the 4graph of f x = ͉ 1 - x ͉ and the Density (g/cm) 1 x-axis on the interval -1, 2 1 2 0 246 8 10 x Additional Exercises 3 4 Length (cm) 76. Riemann sums for constant functions Let f x = c, where 70–71. Displacement from velocity The following functions describe c 7 0, be a constant function on a, b . Prove that any Riemann 1 2 the velocity of a car (in mi hr) moving along a straight highway for sum for any value of n gives the exact area of the region between 3 4 a 3-hr interval. In each case, find the function that gives the displace- the graph of f and the x-axis on a, b . ment of the car over the interval> 0, t , where 0 t 3. … … 77. Riemann sums for linear functions3 4Assume that the linear func-

tion f x mx c is positive on the interval a, b . Prove that 40 if 0 … t …31.54 = + 70. v t = the midpoint Riemann sum with any value of n gives the exact 50 if 1.5 t 3 6 … area of1 the2 region between the graph of f and the3 x-axis4 on a, b . 1 2 e v 78. Shape of the graph for left Riemann sums Suppose a left 3 4 60 Riemann sum is used to approximate the area of the region 50 bounded by the graph of a positive function and the x-axis on the 40 interval a, b . Fill in the following table to indicate whether the 30 resulting approximation underestimates or overestimates the exact 3 4 20 area in the four cases shown. Use a sketch to explain your reasoning in each case. Velocity (mi/hr) Velocity 10 Increasing on a, b Decreasing on a, b 0 0.5 1.0 1.5 2.0 2.5 3.0 t Concave up on a, b Time (hr) 3 4 3 4 Concave down on3 a4, b 30 if 0 … t … 2 3 4 71. v t = 50 if 2 6 t … 2.5 79. Shape of the graph for right Riemann sums Suppose a right 44 if 2.5 6 t … 3 Riemann sum is used to approximate the area of the region 1 2 • bounded by the graph of a positive function and the x-axis on the v interval a, b . Fill in the following table to indicate whether the 60 resulting approximation underestimates or overestimates the exact 50 area in the3 four4 cases shown. Use a sketch to explain your reason- 40 ing in each case. 30 Increasing on a, b Decreasing on a, b 20 Concave up on a, b 3 4 3 4 Velocity (mi/hr) Velocity 10 Concave down on3 a4, b 0 0.5 1.0 1.5 2.0 2.5 3.0 t 3 4 Time (hr) Quick C heck Answer s 1. 45 mi 2. 0.25, 0.125, 7.875 3. ∆x = 2; 1, 3, 5, 7, 9 4. The left sum overestimates the area. 5 6

M05_BRIG7345_02_SE_C05.1.indd 347 21/10/13 11:10 AM 348 Chapter 5 • Integration Not for Sale 5.2 Definite Integrals

We introduced Riemann sums in Section 5.1 as a way to approximate the area of a region

bounded by a curve y = f x and the x-axis on an interval a, b . In that discussion, we assumed f to be nonnegative on the interval. Our next task is to discover the geometric meaning of Riemann sums 1when2 f is negative on some or all3 of a4, b . Once this matter is settled, we proceed to the main event of this section, which is to define the definite integral. With definite integrals, the approximations given by Riemann sums3 become4 exact.

Net Area How do we interpret Riemann sums when f is negative at some or all points of a, b ? The answer follows directly from the Riemann sum definition. 3 4 EXAMPLE 1 Interpreting Riemann sums Evaluate and interpret the following Riemann 2 sums for f x = 1 - x on the interval a, b with n equally spaced subintervals. a. A midpoint1 2 Riemann sum with a, b 3 = 41, 3 and n = 4 b. A left Riemann sum with a, b 0, 3 and n 6 3= 4 3 4 = So lution 3 4 3 4 b - a 3 - 1 a. The length of each subinterval is ∆x = = = 0.5. So the grid points are n 4

x0 = 1, x1 = 1.5, x2 = 2, and x3 = 2.5, and x4 = 3. To compute the midpoint Riemann sum, we evaluate f at the midpoints of the subintervals, which are * * * * x1 = 1.25, x2 = 1.75, x3 = 2.25, and x4 = 2.75. The resulting midpoint Riemann sum is y n 4 2 1.25 2.251.752.75 * * f x x f x 0.5 a k ∆ = a k k 1 k 1 1 1.5 2 2.5 3 = = 1 2 1 21 2 f 1.25 0.5 f 1.75 0.5 f 2.25 0.5 f 2.75 0.5 0 0.5 x = + + + = -0.5625 - 2.0625 - 4.0625 - 6.5625 0.5 Ϫ2 1 21 2 1 21 2 1 21 2 1 21 2 6.625. = 1- 2 Ϫ4 * All values of f x k are negative, so the Riemann sum is also negative. Because area The midpoint Riemann is always a nonnegative quantity, this Riemann sum does not approximate the area of Ϫ6 sum for f (x) ϭ 1 Ϫ x2 the region between1 2 the curve and the x-axis on 1, 3 . Notice, however, that the values on [1, 3] is Ϫ6.625. * of f x k are the negative of the heights of the corresponding rectangles (Figure 5.16). f (x) ϭ 1 Ϫ x2 Therefore, the Riemann sum approximates the negative3 4 of the area of the region 1 2 Figure 5.16 bounded by the curve. b - a 3 - 0 b. The length of each subinterval is ∆x = = = 0.5, and the grid points are n 6

x0 = 0, x1 = 0.5, x2 = 1, x3 = 1.5, x4 = 2, x5 = 2.5, and x6 = 3. * * * To calculate the left Riemann sum, we set x 1, x 2, c , x 6 equal to the left endpoints of the subintervals: * * * * * * x 1 = 0, x 2 = 0.5, x 3 = 1, x 4 = 1.5, x 5 = 2, and x 6 = 2.5.

M05_BRIG7345_02_SE_C05.2.indd 348 21/10/13 11:10 AM Not for Sale 5.2 Definite Integrals 349 The left Riemann The left Riemann The resulting left Riemann sum is y sum on [0, 1.5] sum on [1.5, 3] is 0.875. is Ϫ4.75. n 6 2 * * f x x f x 0.5 a k ∆ = a k k = 1 k = 1 1 2 1 21 2 0 x = f 0 + f 0.5 + f 1 + f 1.5 + f 2 + f 2.5 0.5 y 0.5 1 1.5 2 2.5 3 y

Ϫ2 1 nonnegative1 2 1 contribution2 1 2 1 negative2 contribution1 2 1 22 = 1 + 0.75 + 0 - 1.25 - 3 - 5.25 0.5 Ϫ4 3.875. The resulting left = 1- 2 Riemann sum on * In this case, the values of f x are nonnegative for k 1, 2, and 3, and negative for Ϫ6 [0, 3] is Ϫ3.875. k = k = 4, 5, and 6 (Figure 5.17). Where f is positive, we get positive contributions to the f (x) ϭ 1 Ϫ x2 Riemann sum, and where f1 is negative,2 we get negative contributions to the sum. Figure 5.17 Related Exercises 11–20

Let’s recap what we learned in Example 1. On intervals where f x 6 0, Riemann sums approximate the negative of the area of the region bounded by the curve (Figure 5.18). 1 2 y * * * * * x1 x2 x3 xnϪ1 xn

a ϭ x0 x1 x2 x3 … xnϪ2 xnϪ1 xn ϭ b O x

y ϭ f(x)

y n * ab The Riemann sum ⌺ f (xk)⌬x kϭ1 O x approximates the negative of the area of the region bounded y by the x-axis and the curve. y ϭ f(x) Regions above the x-axis Figure 5.18

In the more general case that f is positive on only part of a, b , we get positive O abx contributions to the sum where f is positive and negative contributions to the sum where f is negative. In this case, Riemann sums approximate the area of the3 regions4 that lie above Region below the x-axis minus the area of the regions that lie below the x-axis (Figure 5.19). This dif- the x-axis ference between the positive and negative contributions is called the net area; it can be Figure 5.19 positive, negative, or zero.

Definition Net Area ➤ Net area suggests the difference between Consider the region R bounded by the graph of a continuous function f and the x-axis positive and negative contributions much between x = a and x = b. The net area of R is the sum of the areas of the parts of R like net change or net profit. Some texts that lie above the x-axis minus the sum of the areas of the parts of R that lie below the use the term signed area for net area. x-axis on a, b .

3 4

Quick C heck 1 Suppose f x = -5. What is the net area of the region bounded by the graph of f and the x-axis on the interval 1, 5 ? Make a sketch of the function and the region. 1 2 3 4

M05_BRIG7345_02_SE_C05.2.indd 349 21/10/13 11:10 AM 350 Chapter 5 • Integration Not for Sale Quick C heck 2 Sketch a continuous The Definite Integral function f that is positive over the Riemann sums for f on a, b give approximations to the net area of the region bounded interval 0, 1 and negative over the by the graph of f and the x-axis between x = a and x = b, where a 6 b. How can we interval 1, 2 , such that the net area 3 4 3 2 make these approximations exact? If f is continuous on a, b , it is reasonable to expect of the region bounded by the graph of the Riemann sum approximations to approach the exact value of the net area as the num- f and the1 x-axis4 on 0, 2 is zero. ber of subintervals n S ∞ and as the length of the subintervals3 4 ∆x S 0 (Figure 5.20). In 3 4 terms of limits, we write n * net area = lim a f x k ∆x. nS ∞ k = 1 1 2 y y y ϭ f(x) y ϭ f(x) 6 40 Net area f (x*)⌬x Net area f (x*)⌬x Ϸ ⌺ k Ϸ ⌺ k kϭ1 kϭ1

* * * x2 x3 x4 *** O abx1 x5 x6 x O abx

n ϭ 6 subintervals n ϭ 40 subintervals

As the number of subintervals n increases, the Riemann sums approach the net area of the region between the curve y ϭ f(x) and the x-axis on [a, b].

y y y ϭ f(x) y ϭ f(x) 75 n Net area Ϸ f (x*)⌬x Net area ϭ lim f (x*)⌬x ⌺ k n ϱ ⌺ k kϭ1 ៬ kϭ1

OOabx a b x

n ϭ 75 subintervals n ៬ ϱ and ⌬x ៬ 0 Figure 5.20

The Riemann sums we have used so far involve regular partitions in which the subintervals have the same length ∆x. We now introduce partitions of a, b in which the lengths of the subintervals are not necessarily equal. A general partition of a, b consists of the n subintervals 3 4 3 4 x0, x1 , x1, x2 , c, xn - 1, xn ,

where x0 = a and xn = b. 3The length4 3 of 4the kth3 subinterval4 is ∆xk = xk - xk - 1, for * k = 1, c, n. We let x k be any point in the subinterval xk - 1, xk . This general partition is used to define the general Riemann sum. 3 4

M05_BRIG7345_02_SE_C05.2.indd 350 21/10/13 11:10 AM Not for Sale 5.2 Definite Integrals 351

Definition General Riemann Sum

Suppose x0, x1 , x1, x2 , c, xn - 1, xn are subintervals of a, b with

3 4 3a = x40 6 x13 6 x2 64 g 6 xn - 1 6 xn 3= b.4 * Let ∆xk be the length of the subinterval xk - 1, xk and let x k be any point in xk - 1, xk , for k = 1, 2, c, n. 3 4

3 4 ⌬x1 ⌬x2 ⌬xk ⌬xn

……x x0 x1 x2 xkϪ1 xk xnϪ1 xn * * * * x1 x2 xk xn

If f is defined on a, b , the sum

n * 3 4 * * * a f x k ∆xk = f x 1 ∆x1 + f x 2 ∆x2 + g + f x n ∆xn k = 1 1 2 1 2 1 2 1 2 is called a general Riemann sum for f on a, b .

3 4 * As was the case for regular Riemann sums, if we choose x k to be the left endpoint of xk - 1, xk , for k = 1, 2, c, n, then the general Riemann sum is a left Riemann sum. * Similarly, if we choose x k to be the right endpoint xk - 1, xk , for k = 1, 2, c, n, then 3 4 * the general Riemann sum is a right Riemann sum, and if we choose x k to be the midpoint 3 4 of the interval xk - 1, xk , for k = 1, 2, c, n, then the general Riemann sum is a midpoint Riemann sum. 3 4 n * Now consider the limit of a f x k ∆xk as n S ∞ and as all the ∆xk S 0. We let ∆ k = 1 denote the largest value of ∆xk; that1 is,2 ∆ = max ∆x1, ∆x2, c, ∆xn . Observe that if n ➤ * Note that ∆ S 0 forces all ∆xk S 0, 0, then x 0, for k 1, 2, , n. For the 5limit lim f x 6x to exist, it must ∆ S ∆ k S = c a k ∆ k which forces n S ∞. Therefore, it ∆S0 k = 1 * suffices to write ∆ S 0 in the limit. have the same value over all general partitions of a, b and for all choices1 2 of x k on a partition. 3 4 Definition Definite Integral

n * A function f defined on a, b is integrable on a, b if lim a f x k ∆xk exists and ∆S0 k = 1 * is unique over all partitions3 of4 a, b and all choices3 of4 x k on a partition.1 2 This limit is the definite integral of f from a to b, which we write 3 4 b n * f x dx = lim a f x k ∆xk. Upper limit Upper limit La ∆S0 k = 1 of integration of summation 1 2 1 2

b n When the limit defining the definite integral of f exists, it equals the net area of the region ϭ * ⌬ ͵ f (x) dx lim f (xk) xk bounded by the graph of f and the x-axis on a, b . It is imperative to remember that the a ⌬ 0 ⌺ ៬ kϭ1 indefinite integral 1f x dx is a family of functions of x (the antiderivatives of f ) and that b 3 4 Integrand the definite integral 1a f x dx is a real number (the net area of a region). 1 2 Lower limit Lower limit Notation The notation1 2for the definite integral requires some explanation. There is a of integration of summation direct match between the notation on either side of the equation in the definition (Figure 5.21). In the limit as ∆ S 0, the finite sum, denoted g, becomes a sum with an infinite x is the variable of integration. number of terms, denoted 1. The integral sign 1 is an elongated S for sum. The limits of Figure 5.21 integration, a and b, and the limits of summation also match: The lower limit in the sum,

M05_BRIG7345_02_SE_C05.2.indd 351 21/10/13 11:10 AM 352 Chapter 5 • Integration Not for Sale k = 1, corresponds to the left endpoint of the interval, x = a, and the upper limit in the sum, k = n, corresponds to the right endpoint of the interval, x = b. The function under the integral sign is called the integrand. Finally, the differential dx in the integral (which corresponds to ∆xk in the sum) is an essential part of the notation; it tells us that the variable of integration is x. The variable of integration is a dummy variable that is completely internal to the integral. It does not matter what the variable of integration is called, as long as it does not conflict with other variables that are in use. Therefore, the integrals in Figure 5.22 all have the same meaning.

➤ For Leibniz, who introduced this notation b b b in 1675, dx represented the width of an ͵ f(x) dx ϭ ͵ f(␻) d␻ ϭ ͵ f(s) ds infinitesimally thin rectangle and f x dx a a a represented the area of such a rectangle. b 1 2 y y y He used 1a f x dx to denote the sum of these areas from a to b. y ϭ f (x) y ϭ f (␻) y ϭ f (s) 1 2

O abx O ab␻ O abs Figure 5.22 The strategy of slicing a region into smaller parts, summing the results from the parts, and taking a limit is used repeatedly in calculus and its applications. We call this strategy the slice-and-sum method. It often results in a Riemann sum whose limit is a definite integral.

Evaluating Definite Integrals ➤ A function f is bounded on an interval Most of the functions encountered in this text are integrable (see Exercise 83 for an excep- I if there is a number M such that tion). In fact, if f is continuous on a, b or if f is bounded on a, b with a finite number

͉ f x ͉ 6 M for all x in I. of discontinuities, then f is integrable on a, b . The proof of this result goes beyond the 3 4 3 4 1 2 scope of this text. 3 4 th eor em 5.2 Integrable Functions If f is continuous on a, b or bounded on a, b with a finite number of discontinuities, then f is integrable on a, b . 3 4 3 4 3 4 b b When f is continuous on a, b , we have seen that the definite integral 1a f x dx is Net area ϭ ͵ f (x) dx a the net area bounded by the graph of f and the x-axis on a, b . Figure 5.23 illustrates how ϭ area above x-axis (Regions 1 and 3) the idea of net area carries over3 to piecewise4 continuous functions (Exercises 76–80).1 2 Ϫ area below x-axis (Region 2) 3 4 1 y Quick C heck 3 Graph f x x and use geometry to evaluate x dx. = 1-1

EXAMPLE 2 Identifying1 2 the limit of a sum Assume that n lim 3x*2 2x * 1 x 1 a k + k + ∆ k ∆S0 k = 1 3 1 2 is the limit of a Riemann sum for a function f on 1, 3 . Identify the function f and express O a 2 b x the limit as a definite integral. What does the definite integral represent geometrically? 3 4 A bounded piecewise continuous n *2 * function is integrable. So lution By comparing the sum a 3x k + 2x k + 1 ∆xk to the general Riemann k = 1 Figure 5.23 n * 1 2 2 sum a f x k ∆xk, we see that f x = 3x + 2x + 1. Because f is a polynomial, it is k = 1 continuous1 on2 1, 3 and is, therefore,1 2 integrable on 1, 3 . It follows that

M05_BRIG7345_02_SE_C05.2.indd 352 21/10/13 11:10 AM Not for Sale 5.2 Definite Integrals 353 3 2 Because f is positive on 1, 3 , the definite integral 11 3x + 2x + 1 dx is the area of the region bounded by the curve y = 3x 2 + 2x + 1 and the x-axis on 1, 3 (Figure 5.24). 3 4 1 2 y 3 4

2 40 y ϭ 3x ϩ 2x ϩ 1

0 1 2 3 x

n 3 lim (3x*2 ϩ 2x* ϩ 1)⌬x ϭ (3x2 ϩ 2x ϩ 1)dx ⌬ 0 ⌺ k k k ͵ ៬ kϭ1 1 Figure 5.24 Related Exercises 21–24 y y ϭ 2x ϩ 3 EXAMPLE 3 Evaluating definite integrals using geometry Use familiar area formu- 12 Area ϭ 18 10 las to evaluate the following definite integrals. 8 4 6 4 6 11 a. 2x + 3 dx b. 2x - 6 dx c. 1 - x - 3 2 dx 4 L2 L1 L3 7 2 2 1 2 1 2 1 2 So lution To evaluate these definite integrals geometrically, a sketch of the correspond- 0 1 2345x ing region is essential. 4 2 a. The definite integral 12 2x + 3 dx is the area of the trapezoid bounded by the x-axis Figure 5.25 and the line y = 2x + 3 from x = 2 to x = 4 (Figure 5.25). The width of its base is 2 1 2 and the lengths of its two parallel sides are f 2 = 7 and f 4 = 11. Using the area formula for a trapezoid, we have 1 2 1 2 ➤ A trapezoid and its area When a = 0, 4 1 we get the area of a triangle. When 2x + 3 dx = # 2 11 + 7 = 18. 2 a = b, we get the area of a rectangle. L2 1 2 1 2 b. A sketch shows that the regions bounded by the line y = 2x - 6 and the x-axis are 1 triangles (Figure 5.26). The area of the triangle on the interval 1, 3 is 2 # 2 # 4 = 4. 1 b Similarly, the area of the triangle on 3, 6 is # 3 # 6 = 9. The definite integral is the A ϭ h(a ϩ b) 2 a q net area of the entire region, which is the area of the triangle above3 4the x-axis minus the area of the triangle below the x-axis:3 4 h 6 2x - 6 dx = net area = 9 - 4 = 5. L1 1 2 y y ϭ 2x Ϫ 6 6 Area ϭ 9 4 6 2 2

0 1 2 3 4 5 6 x Ϫ2 4 3 Ϫ4

Ϫ6 Area ϭ 4

Figure 5.26

M05_BRIG7345_02_SE_C05.2.indd 353 21/10/13 11:10 AM 354 Chapter 5 • Integration Not for Sale y 2 c. We first let y = 1 - x - 3 and observe that y Ú 0 when 2 … x … 4. Squaring ϭ Ϫ Ϫ 2 2 2 y ͙1 (x 3) both sides leads to2 the equation x - 3 + y = 1, whose graph is a circle of radius 1 1 2 1 centered at (3, 0). Because y 0, the graph of y 1 x 3 2 is the upper 1 Ú1 2 = - - 4 2 2 half of the circle. It follows that the integral 13 1 - x - 3 dx is the area of a 0 1 2 3 4 x 1 2 quarter circle of radius 1 (Figure 5.27). Therefore,2 2 1 2 Area of shaded region ϭ ~␲(1) ϭ ~␲ 4 2 1 2 p Figure 5.27 1 - x - 3 dx = p 1 = . L3 4 4 2 1 2 1 2 Related Exercises 25–32

3 Quick C heck 4 Let f x = 5 and use geometry to evaluate f x dx. What is the value b 11

of 1a c dx, where c is a real number? 1 2 1 2 EXAMPLE 4 Definite integrals from graphs Figure 5.28 shows the graph of a func- y tion f with the areas of the regions bounded by its graph and the x-axis given. Find the y ϭ f(x) values of the following definite integrals.

12 8 b c c d

a. f x dx b. f x dx c. f x dx d. f x dx a b c d x La Lb La Lb 10 1 2 1 2 1 2 1 2 So lution a. Because f is positive on a, b , the value of the definite integral is the area of the b Figure 5.28 region between the graph and the x-axis on a, b ; that is, f x dx 12. 3 4 1a = b. Because f is negative on b, c , the value of the definite integral is the negative of the c3 4 1 2 area of the corresponding region; that is, 1b f x dx = -10. 3 4 c. The value of the definite integral is the area of the region on a, b (where f is 1 2 positive) minus the area of the region on b, c (where f is negative). Therefore, c 3 4 1a f x dx = 12 - 10 = 2. d 3 4 d. Reasoning as in part (c), we have f x dx 10 8 2. 1 2 1b = - + = - Related Exercises 33–40 1 2 Properties of Definite Integrals b Recall that the definite integral 1a f x dx was defined assuming that a 6 b. There are, however, occasions when it is necessary to reverse the limits of integration. If f is integrable on a, b , we define 1 2 a b 3 4 f x dx = - f x dx. Lb La 1 2 1 2 In other words, reversing the limits of integration changes the sign of the integral. Another fundamental property of integrals is that if we integrate from a point to itself, then the length of the interval of integration is zero, which means the definite integral is also zero.

Definition Reversing Limits and Identical Limits of Integration Suppose f is integrable on a, b . a b a 1. 1b f x dx = - 1a f x3 dx 4 2. 1a f x dx = 0 1 2 1 2 1 2

b a

Quick C heck 5 Evaluate 1a f x dx + 1b f x dx assuming f is integrable on a, b . 1 2 1 2 3 4

M05_BRIG7345_02_SE_C05.2.indd 354 21/10/13 11:10 AM Not for Sale 5.2 Definite Integrals 355 Integral of a Sum Definite integrals possess other properties that often simplify their evaluation. Assume f and g are integrable on a, b . The first property states that their sum f + g is integrable on a, b and the integral of their sum is the sum of their integrals: 3 4 b3 4 b b

f x + g x dx = f x dx + g x dx. La La La 1 1 2 1 22 1 2 1 2 We prove this property assuming that f and g are continuous. In this case, f + g is continuous and, therefore, integrable. We then have

b n * * Definition of f x + g x dx = lim a f x k + g x k ∆xk La ∆S0 k = 1 definite integral

1 1 2 1 22 n1 1 2 1 22n * * Split into two finite = lim a f x k ∆xk + a g x k ∆xk ∆S0 k = 1 k = 1 sums.

a n 1 2 1n 2 b * * Split into two = lim a f x k ∆xk + lim a g x k ∆xk ∆S0 k = 1 ∆S0 k = 1 limits. b 1 2 b 1 2 Definition of = f x dx + g x dx. La La definite integral 1 2 1 2 C onstants in Integrals Another property of definite integrals is that constants can be factored out of the integral. If f is integrable on a, b and c is a constant, then cf is integrable on a, b and 3 4 3 4 b b

c f x dx = c f x dx. La La 1 2 1 2 The justification for this property (Exercise 81) is based on the fact that for finite sums, n n * * a c f x k ∆xk = c a f x k ∆xk. k = 1 k = 1 1 2 1 2 Integrals over Subintervals If the point p lies between a and b, then the integral on a, b may be split into two integrals. As shown in Figure 5.29, we have the property

3 4 b p b ͵ f (x) dx ϭϩ͵ f (x) dx ͵ f (x) dx. a a p y y y ϭ f (x) y ϭ f (x)

O abx O apb x Figure 5.29

M05_BRIG7345_02_SE_C05.2.indd 355 21/10/13 11:10 AM 356 Chapter 5 • Integration Not for Sale It is surprising that this property also holds when p lies outside the interval a, b . For example, if a 6 b 6 p and f is integrable on a, p , then it follows (Figure 5.30) that 3 4 b p 3 p 4 ͵ f (x) dx ϭϪ͵ f (x) dx ͵ f (x) dx. a a b

y y y y ϭ f (x) y ϭ f (x) y ϭ f (x)

O abp x O apb x O abp x Figure 5.30

b p Because 1p f x dx = - 1b f x dx, we have the original property: 1 2 b 1 2 p b

f x dx = f x dx + f x dx. La La Lp 1 2 1 2 1 2 b y Integrals of Absolute Values Finally, how do we interpret 1a ͉ f x ͉ dx, the integral y ϭ f (x) of the absolute value of an integrable function? The graphs f and ͉ f ͉ are shown in b * 1 2 * Figure 5.31. The integral 1a ͉ f x ͉ dx gives the area of regions R1 and R2. But R1 and R1 b have the same area; therefore, 1a ͉ f x ͉ dx also gives the area of R1 and R2. The conclusion b 1 2 is that ͉ f x ͉ dx is the area of the entire region (above and below the x-axis) that lies R 1a 1 2 2 between the graph of f and the x-axis on a, b . x All these1 properties2 will be used frequently in upcoming work. It’s worth collecting a R1 b them in one table (Table 5.4). 3 4

y T able 5.4 Properties of definite integrals

y ϭ ͉f (x)͉ Let f and g be integrable functions on an interval that contains a, b, and p. a

1. f x dx = 0 Definition La a 1 2 b

2. f x dx = - f x dx Definition * R2 R1 Lb La b b b a b x 1 2 1 2 3. f x + g x dx = f x dx + g x dx La La La b1 1 2 1 22 b 1 2 1 2

4. cf x dx = c f x dx For any constant c b La La b b f (x) dx ϭ area of R * ϩ area of R 1 2 p 1 2 ͵ ͉ ͉ 1 2 a 5. f x dx = f x dx + f x dx ϭ area of R1 ϩ area of R2 La La Lp b 6. The function1 2 ͉ f ͉ is1 integrable2 on 1a,2 b , and ͉ f x ͉ dx is the sum of the areas of the regions Figure 5.31 1a bounded by the graph of f and the x-axis on a, b . 3 4 1 2 3 4 5 EXAMPLE 5 Properties of integrals Assume that 10 f x dx = 3 and 7 f x dx 10. Evaluate the following integrals, if possible. 10 = - 1 2

71 2 7 0 0 7 a. 10 2 f x dx b. 15 f x dx c. 15 f x dx d. 17 6 f x dx e. 10 ͉ f x ͉ dx 1 2 1 2 1 2 1 2 1 2

M05_BRIG7345_02_SE_C05.2.indd 356 21/10/13 11:11 AM Not for Sale 5.2 Definite Integrals 357 So lution a. By Property 4 of Table 5.4, 7 7

2 f x dx = 2 f x dx = 2 # -10 = -20. L0 L0 1 2 7 1 2 5 1 2 7 b. By Property 5 of Table 5.4, 10 f x dx = 10 f x dx + 15 f x dx. Therefore, 7 7 1 2 5 1 2 1 2 f x dx = f x dx - f x dx = -10 - 3 = -13. L5 L0 L0 1 2 1 2 1 2 c. By Property 2 of Table 5.4,

0 5

f x dx = - f x dx = -3. L5 L0 1 2 1 2 d. Using Properties 2 and 4 of Table 5.4, we have

0 7 7

6 f x dx = - 6 f x dx = -6 f x dx = -6 -10 = 60. L7 L0 L0 1 2 1 2 1 2 1 21 2 e. This integral cannot be evaluated without knowing the intervals on which f is positive and negative. It could have any value greater than or equal to 10. Related Exercises 41–46

2 2 Quick C heck 6 Evaluate x dx and x dx using geometry. 1-1 1-1 ͉ ͉ Evaluating Definite Integrals Using Limits In Example 3, we used area formulas for trapezoids, triangles, and circles to evaluate definite integrals. Regions bounded by more general functions have curved boundaries for which conventional geometrical methods do not work. At the moment, the only way to handle such integrals is to appeal to the definition of the definite integral and the summation formulas given in Theorem 5.1. n b * We know that if f is integrable on a, b , then 1a f x dx = lim a f x k ∆xk, ∆S0 k = 1 * for any partition of a, b and any points3 x k. 4To simplify1 these2 calculations,1 we2 use equally spaced grid points and right Riemann sums. That is, for each value of n, we let b - a 3 4 ∆x ∆x and x * a k ∆x, for k 1, 2, , n. Then as n S ∞ and k = = n k = + = c ∆ S 0,

b n n * f x dx = lim a f x k ∆xk = lim a f a + k∆x ∆x. La ∆S0 k = 1 nS ∞ k = 1 1 2 1 2 1 2 2 3 EXAMPLE 6 Evaluating definite integrals Find the value of 10 x + 1 dx by evaluating a right Riemann sum and letting n ∞. 4 S * 1 2 x2 ϭ n 2 2k So lution Based on approximations found in Example 5, Section 5.1, we x* ϭ x* ϭ 1 n k n conjectured that the value of this integral is 6. To verify this conjecture, we now a ϭ x ϭ 0 b ϭ x ϭ 2 evaluate the integral exactly. The interval a, b = 0, 2 is divided into n subinter- 0 2 n ⌬x ϭ b - a 2 n vals of length ∆x , which produces the grid points = n = n 3 4 3 4 2k x* ϭ a ϩ k⌬x ϭ k n 2 2k x * a k∆x 0 k , for k 1, 2, , n. k ϭ 1, ..., n k = + = + # n = n = c

M05_BRIG7345_02_SE_C05.2.indd 357 21/10/13 11:11 AM 358 Chapter 5 • Integration Not for Sale 3 Letting f x = x + 1, the right Riemann sum is n n 3 * 1 2 2k 2 a f x k ∆x = a + 1 k = 1 k = 1 n n 1 2 a a b b 2 n 8k3 n n = a 3 + 1 a cak = c a ak n k = 1 n k = 1 k = 1 ➤ An analogous calculation could be done a n bn n n n using left Riemann sums or midpoint 2 8 3 Riemann sums. = 3 a k + a 1 a ak + bk = a ak + a bk n n k = 1 k = 1 k = 1 k = 1 k = 1 a b 1 2 n2 n 1 2 n n2 n 1 2 n 2 8 + 3 + = 3 + n a k = and a 1 = n; n n 4 k 1 4 k 1 1 2 = 1 2 = a a b b Theorem 5.1 4 n2 + 2n + 1 = + 2. Simplify. n2 1 2 2 3 Now we evaluate 10 x + 1 dx by letting n S ∞ in the Riemann sum:

21 2 n 3 * x + 1 dx = lim a f x k ∆x L0 nS∞ k = 1 1 2 1 2 4 n2 + 2n + 1 = lim 2 + 2 nS∞ n 1 2 a b n2 + 2n + 1 = 4 lim 2 + lim 2 nS∞ n nS∞ x a b 1 = 4 1 + 2 = 6. 2 3 1 2 Therefore, 10 x + 1 dx = 6, confirming our conjecture in Example 5, Section 5.1. Related Exercises 47–52 1 2 The Riemann sum calculations in Example 6 are tedious even if f is a simple function. For polynomials of degree 4 and higher, the calculations are more challenging, and for rational and transcendental functions, advanced mathematical results are needed. The next section introduces more efficient methods for evaluating definite integrals.

Section 5.2 Exerci ses Review Questions a 7. Give a geometrical explanation of why 1a f x dx = 0. 1. What does net area measure? 6 3 8. Use Table 5.4 to rewrite 11 2x - 4x dx as1 the2 difference of 2. What is the geometric meaning of a definite integral if the two integrals. 1 2 integrand changes sign on the interval of integration? a 9. Use geometry to find a formula for 10 x dx, in terms of a. 3. Under what conditions does the net area of a region equal the area b 10. If f is continuous on a, b and ͉ f x ͉ dx 0, what can you of a region? When does the net area of a region differ from the 1a = conclude about f ? area of a region? 3 4 1 2

4. Suppose that f x 6 0 on the interval a, b . Using Riemann Basic Skills b sums, explain why the definite integral 1a f x dx is negative. 11–14. Approximating net area The following functions are negative 1 2 3 4 2p 2p on the given interval. 5. Use graphs to evaluate 10 sin x dx and 10 1 cos2 x dx. a. Sketch the function on the given interval. n * b. Approximate the net area bounded by the graph of f and the x-axis 6. Explain how the notation for Riemann sums, a f x k ∆x, k = 1 on the interval using a left, right, and midpoint Riemann sum with 1 2b n = 4. corresponds to the notation for the definite integral, 1a f x dx. 3 11. f x = -2x - 1 on 0, 4 T 12. f x = -4 - x on 3, 7 1 2 Copyright Pearson. All rights1 2 reserved. 3 4 1 2 3 4

M05_BRIG7345_02_SE_C05.2.indd 358 21/10/13 11:11 AM Not for Sale 5.2 Definite Integrals 359

T 13. f x = sin 2x on p 2, p 33–36. Net area from graphs The figure shows the areas of regions 3 bounded by the graph of f and the x-axis. Evaluate the following T 14. f 1x2 = x - 1 on3 -> 2, 04 integrals. T 15–20. Approximating net area The following functions are positive 1 2 3 4 y and negative on the given interval. a. Sketch the function on the given interval. y ϭ f(x) b. Approximate the net area bounded by the graph of f and the x-axis on the interval using a left, right, and midpoint Riemann sum with 16 n = 4. 11 c. Use the sketch in part (a) to show which intervals of a, b make O a 5 b c x positive and negative contributions to the net area. 3 4

15. f x = 4 - 2x on 0, 4 2 16. f x 8 2x on 0, 4 1 2 = - 3 4 a b

17. f 1x2 = sin 2x on 0,3 3p 44 33. f x dx 34. f x dx L0 L0 3 18. f 1x2 = x on -1,3 2 > 4 1 2 1 2 c c - 1 19. f 1x2 = tan 33x - 41 on 0, 1 35. f x dx 36. f x dx - x La L0 20. f x = xe on 1, 1 1 2 1 - 2 3 4 1 2 1 2 37–40. Net area from graphs The accompanying figure shows four 21–24.1 Identifying2 definite3 4integrals as limits of sums Consider the following limits of Riemann sums for a function f on a, b . Identify f regions bounded by the graph of y = x sin x: R1, R2, R3, and R4, whose and express the limit as a definite integral. areas are 1, p - 1, p + 1, and 2p - 1, respectively. (We verify these 3 4 results later in the text.) Use this information to evaluate the following n *2 integrals. 21. lim a x k + 1 ∆xk on 0, 2 ∆S0 k = 1 y 1 2 3 4 n Area ϭ 1 Area ϭ ␲ ϩ 1 *2 22. lim a 4 - x k ∆xk on -2, 2 2 y ϭ x sin x ∆S0 k = 1 R 1 2 3 4 R1 2 n x * * ␲ 2␲ 23. lim a x k ln x k∆xk on 1, 2 q w ∆S0 R3 k = 1 Ϫ2 R Area ϭ ␲ Ϫ 1 4 3 4 Area ϭ 2␲ Ϫ 1 n Ϫ4 *2 24. lim a ͉ x k - 1 ͉ ∆xk on -2, 2 ∆S0 k = 1 3 4 25–32. Net area and definite integrals Use geometry (not Riemann p 3p 2 sums) to evaluate the following definite integrals. Sketch a graph of the 37. x sin x dx 38. > x sin x dx L0 L0 integrand, show the region in question, and interpret your result. 2p 2p 4 2 39. x sin x dx 40. x sin x dx 25. 8 2x dx 26. 2x 4 dx - + L0 Lp 2 L0 L-4 1 2 1 2 41. Properties of integrals Use only the fact> that 2 2 4 3x 4 - x dx = 32 and the definitions and properties of 27. - ͉ x ͉ dx 28. 1 - x dx 10 L L integrals to evaluate the following integrals, if possible. -1 0 1 2 1 2 1 2 0 4 4 3 2 2 a. 3x 4 x dx b. x x 4 dx 29. 16 - x dx 30. 4 - x - 1 dx - - L4 L0 L0 L 1 2 - 2 1 2 1 2 1 2 0 8 4 5 if x … 2 c. 6x 4 - x dx d. 3x 4 - x dx 31. f x dx, where f x = L L L 3x - 1 if x 7 2 4 0 0 1 2 1 2 1 2 1 2 e 4 10 4x if 0 … x … 2 42. Properties of integrals Suppose f x dx = 8 and 6 11 32. g x dx, where g x = -8x + 16 if 2 6 x … 3 11 f x dx = 5. Evaluate the following integrals. L 1 2 1 -8 if x 7 3 1 2 1 2 • 1 42 4

a. -3 f x dx b. 3 f x dx L1 L1 1 1 22 1 2 4 6

M05_BRIG7345_02_SE_C05.2.indd 359 21/10/13 11:11 AM 360 Chapter 5 • Integration Not for Sale 3 6 2p a 2p a 43. Properties of integrals Suppose 10 f x dx = 2, 13 f x dx = -5, c. 10 sin ax dx = 10 cos ax dx = 0. (Hint: Graph the 6 functions> and use properties> of trigonometric functions.) and 13 g x dx = 1. Evaluate the following1 2 integrals. 1 2 b a d. If 1a f x dx = 1b f x dx, then f is a constant function. 3 1 2 6 b b e. Property 4 of Table 5.4 implies that 1a x f x dx = x 1a f x dx. a. 5f x dx b. -3g x dx 1 2 1 2 L0 L3 T 54–57. Approximating definite integrals Complete1 2 the following1 2 1 2 1 1 22 6 3 steps for the given integral and the given value of n.

c. 3f x - g x dx d. f x + 2g x dx a. Sketch the graph of the integrand on the interval of integration. L3 L6 b. Calculate ∆x and the grid points x0, x1, c, xn, assuming a regular 1 1 2 1 22 1 1 2 1 22 partition. 44. Properties of integrals Suppose f x Ú 0 on 0, 2 , f x … 0 2 5 c. Calculate the left and right Riemann sums for the given value of n. on 2, 5 , f x dx 6, and f x dx 8. Evaluate the 10 = 12 1 2 = -3 4 1 2 d. Determine which Riemann sum (left or right) underestimates the following integrals. 3 4 1 2 1 2 value of the definite integral and which overestimates the value of 5 5 the definite integral.

a. f x dx b. ͉ f x ͉ dx L L 2 6 0 0 2 1 2 1 2 54. x - 2 dx; n = 4 55. 1 - 2x dx; n = 6 5 5 L0 L3 1 2 1 2 c. 4 f x dx d. f x f x dx ͉ ͉ ͉ ͉ + p 2 7 L2 L0 1 56. cos x dx; n = 4 57. dx; n = 6 1 2 1 1 2 1 2 2 > x 45–46. Using properties of integrals Use the value of the first integral L0 L1 I to evaluate the two given integrals. T 58–62. Approximating definite integrals with a calculator Consider 1 the following definite integrals. 3 3 45. I = x - 2x dx = - 4 a. Write the left and right Riemann sums in sigma notation, for L 0 n = 20, 50, and 100. Then evaluate the sums using a calculator. 1 2 1 0 b. Based on your answers to part (a), make a conjecture about the value of the definite integral. a. 4x - 2x 3 dx b. 2x - x 3 dx L L 0 1 9 1 1 2 1 2 2 p 2 58. 3 x dx 59. x + 1 dx L4 L0 46. > 2 sin u - cos u du 1 L0 1 2 e 1 1 2 - 1 p 2 0 60. ln x dx 61. cos x dx L1 L0 a. > 2 sin u - cos u du b. 4 cos u - 8 sin u du L0 Lp 2 1 1 2 1 2 p x > 62. p cos dx 47–52. Limits of sums Use the definition of the definite integral to 2 L- 1 evaluate the following definite integrals. Use right Riemann sums and a b Theorem 5.1. T 63–66. Midpoint Riemann sums with a calculator Consider the following definite integrals. 2 5 a. Write the midpoint Riemann sum in sigma notation for an arbitrary 47. 2x 1 dx 48. 1 x dx + - value of n. L0 L1 1 2 1 2 b. Evaluate each sum using a calculator with n = 20, 50, and 100. 7 2 Use these values to estimate the value of the integral. 49. 4x + 6 dx 50. x 2 - 1 dx 4 2 L3 L0 px 1 2 1 2 63. 2 x dx 64. sin dx L L 4 4 2 1 1 -1 51. x 2 - 1 dx 52. 4x 3 dx a b 4 1 2 L1 L0 2 -1 1 2 65. 4x - x dx 66. > sin x dx Further Explorations L0 L0 1 2 53. Explain why or why not Determine whether the following state- 67. More properties of integrals Consider two functions f and g on ments are true and give an explanation or counterexample. 6 6 1, 6 such that 11 f x dx = 10, 11 g x dx = 5, a. If f is a constant function on the interval a, b , then the right 6 4

b 14 f x dx = 5, and 11 g x dx = 2. Evaluate the following and left Riemann sums give the exact value of 1a f x dx, for 3 4 1 2 1 2 3 4 integrals. any positive integer n. 1 2 1 2 b. If f is a linear function on the interval a, b , then a1 midpoint2 4 6 b a. 3 f x dx b. f x g x dx Riemann sum gives the exact value of f x dx, for any - 31a 4 L L positive integer n. 1 1 1 2 1 2 1 1 2 1 22

M05_BRIG7345_02_SE_C05.2.indd 360 21/10/13 11:11 AM Not for Sale 5.2 Definite Integrals 361 4 6 4 x y c. f x - g x dx d. g x - f x dx 80. dx L1 L4 L0 x 1.0 1 1 2 1 22 1 1 2 1 22 6 1 0.8

e. 8g x dx f. 2 f x dx 0.6 L4 L4 x y ϭ Ϫx 1 2 1 2 0.4 68–71. Area versus net area Graph the following functions. Then use geometry (not Riemann sums) to find the area and the net area of the 0.2 region described. 041 2 3 x 68. The region between the graph of y = 4x - 8 and the x-axis, for 4 x 8 - … … 81. Constants in integrals Use the definition of the definite integral b b 69. The region between the graph of y = -3x and the x-axis, to justify the property 1a c f x dx = c 1a f x dx, where f is for -2 … x … 2 continuous and c is a real number. 1 2 1 2 70. The region between the graph of y 3x 6 and the x-axis, 82. Zero net area If 0 6 c 6 d, then find the value of b (in terms of = - d for 0 … x … 6 c and d) for which 1c x + b dx = 0. 71. The region between the graph of y = 1 - ͉ x ͉ and the x-axis, 83. A nonintegrable function1 Consider2 the function defined on for -2 … x … 2 0, 1 such that f x = 1 if x is a rational number and f x = 0 if x is irrational. This function has an infinite number of disconti- 72–75. Area by geometry Use geometry to evaluate the following integrals. 3 4 1 2 1 1 2 nuities, and the integral 10 f x dx does not exist. Show that the 3 6 right, left, and midpoint Riemann sums on regular partitions with 1 2 72. ͉ x + 1 ͉ dx 73. ͉ 2x - 4 ͉ dx n subintervals equal 1 for all n. (Hint: Between any two real num- L-2 L1 bers lie a rational and an irrational number.)

6 4 84. Powers of x by Riemann sums Consider the integral 1 2 p 74. 3x - 6 dx 75. 24 - 2x - x dx I p = 10 x dx, where p is a positive integer. L1 L-6 a. Write the left Riemann sum for the integral with n 2 1 2 1 2 subintervals. Additional Exercises b. It is a fact (proved by the 17th-century mathematicians Fermat 76. Integrating piecewise continuous functions Suppose f is con- 1 n - 1 k p 1 and Pascal) that lim = . Use this fact to tinuous on the intervals a, p and p, b , where a 6 p 6 b, n n a n p 1 evaluate I p . S ∞ k = 0 + with a finite jump at p. Form a uniform partition on the inter- a b val a, p with n grid points3 and4 another1 4 uniform partition on b 1 2 dx the interval p, b with m grid points, where p is a grid point 85. An exact integration formula Evaluate 2, where 3 4 b La x of both partitions. Write a Riemann sum for 1a f x dx and separate it into3 two4 pieces for a, p and p, b . Explain why 0 6 a 6 b, using the definition of the definite integral and the b p b 1 2 following steps. f x dx = f x dx + f x dx. 1a 1a 13p 4 3 4 a. Assume x , x , , x is a partition of a, b with 77–78. Integrating1 2 piecewise1 2 continuous1 2 functions Use geometry 0 1 c n and the result of Exercise 76 to evaluate the following integrals. ∆xk = xk - xk - 1, for k = 1, 2, c, n. Show that 5 6 3 4 xk - 1 … xk - 1xk … xk, for k = 1, 2, c, n. 10 1 2 if 0 … x … 5 1 1 ∆xk 77. f x dx, where f x = b. Show that - = , for k = 1, 2, c, n. L0 3 if 5 6 x … 10 xk - 1 xk xk - 1xk 1 2 1 2 e bdx 6 c. Simplify the general Riemann sum for using 2x if 1 … x 6 4 2 78. f x dx, where f x = La x 10 2x if 4 x 6 * L1 - … … x k = xk - 1xk. 1 2 1 2 e 1 79–80. Integrating piecewise continuous functions Recall that the floor bdx 1 1 function x is the greatest integer less than or equal to x and that the d. Conclude that 2 = - . La x a b ceiling function x is the least integer greater than or equal to x. Use the result of Exercise 76 and the graphs to evaluate the following integrals. (Source: The College Mathematics Journal, 32, 4, Sep 2001) 5 y 79. x x dx Quick C heck Answers L1 15 1. -20 2. f x = 1 - x is one possibility. 3. 0 3 5 4. 10; c b - a 5. 0 6. 2; 2 10 y ϭ x x 1 2 1 2

041 2 35x

M05_BRIG7345_02_SE_C05.2.indd 361 31/10/13 9:51 AM 362 Chapter 5 • Integration Not for Sale 5.3 Fundamental Theorem of Calculus

Evaluating definite integrals using limits of Riemann sums, as described in Section 5.2, is usually not possible or practical. Fortunately, there is a powerful and practical method for evaluating definite integrals, which is developed in this section. Along the way, we discover the inverse relationship between differentiation and integration, expressed in the most important result of calculus, the Fundamental Theorem of Calculus. The first step in this process is to introduce area functions (first seen in Section 1.2).

x Area Functions A(x) ϭ ͵ f (t) dt The concept of an area function is crucial to the discussion about the connection between a derivatives and integrals. We start with a continuous function y = f t defined for t a, y Ú y ϭ f (t) where a is a fixed number. The area function for f with left endpoint a is denoted A x ; it gives the net area of the region bounded by the graph of f and the 1t-axis2 between t = a and t = x (Figure 5.32). The net area of this region is also given by the definite integral1 2 A(x)

Independent variable O axt of the area function

constant variable x Figure 5.32 A(x) ϭ ͵ f (t) dt. a

Variable of integration (dummy variable) ➤ Suppose you want to devise a function A whose value is the sum of the first n Notice that x is the upper limit of the integral and the independent variable of the positive integers, 1 + 2 + g + n. The most compact way to write this function area function: As x changes, so does the net area under the curve. Because the symbol x is n already in use as the independent variable for A, we must choose another symbol for the is A n = a k, for n Ú 1. In this k = 1 variable of integration. Any symbol—except x—can be used because it is a dummy vari- function,1 2 n is the independent variable of able; we have chosen t as the integration variable. the function A, and k, which is internal Figure 5.33 gives a general view of how an area function is generated. Suppose that f to the sum, is a dummy variable. This is a continuous function and a is a fixed number. Now choose a point b 7 a. The net area function involving a sum is analogous to an area function involving an integral. c d b c A(d) ϭ ͵ f (t) dt ϩ ͵ f (t) dt y A(b) ϭ ͵ f (t) dt y A(c) ϭ ͵ f (t) dt y a c a a

d ab t a c t a c t

y ϭ f (t) y ϭ f (t) y ϭ f (t) ➤ Notice that t is the independent variable when we plot f and x is the independent Net area increases Net area decreases Values of the net variable when we plot A. area appear on from x ϭ a to x ϭ c for x Ͼ c the graph of the y area function. (c, A(c)) (d, A(d)) (b, A(b)) y ϭ A(x)

(a, A(a)) a b c d x

a A(a) ϭ ͵ f (t) dt ϭ 0 Figure 5.33 a

M05_BRIG7345_02_SE_C05.3.indd 362 21/10/13 11:11 AM Not for Sale 5.3 Fundamental Theorem of Calculus 363 of the region between the graph of f and the t-axis on the interval a, b is A b . Moving the right endpoint to c, 0 or d, 0 produces different regions with net areas A c and 3 4 x 1 2 A d , respectively. In general, if x 7 a is a variable point, then A x = 1a f t dt is the net area of the region 1between2 the1 graph2 of f and the t-axis on the interval a, x . 1 2 1 2Figure 5.33 shows how A x varies with respect to x.1 Notice2 that1 2 A a = a 3 4 1a f t dt = 0. Then for x 7 a, the net area increases for x 6 c, at which point 1 2 1 2 f c = 0. For x 7 c, the function f is negative, which produces a negative contribution to the1 area2 function. As a result, the area function decreases for x 7 c. 1 2

Definition Area Function Let f be a continuous function, for t Ú a. The area function for f with left endpoint a is

A x = f t dt, La 1 2 1 2 where x Ú a. The area function gives the net area of the region bounded by the graph of f and the t-axis on the interval a, x .

3 4 The following two examples illustrate the idea of area functions.

y EXAMPLE 1 Comparing area functions The graph of f is shown in Figure 5.34 y ϭ f (t) x x with areas of various regions marked. Let A x = f t dt and F x = f t dt be 10 1-1 13 two area functions for f (note the different left endpoints). Evaluate the following area Area ϭ 3 functions. 1 2 1 2 1 2 1 2

Ϫ2 2 4 6 8 10 t a. A 3 and F 3 b. A 5 and F 5 c. A 9 and F 9 So lution Area ϭ 35 1 2 1 2 1 2 1 2 1 2 1 2 3 a. The value of A 3 = 1 1 f t dt is the net area of the region bounded by the Ϫ15 - Area ϭ 27 graph of f and the t-axis on the interval -1, 3 . Using the graph of f , we see that A 3 = -27 (because1 2 this 1region2 has an area of 27 and lies below the t-axis). On Figure 5.34 3 3 4 the other hand, F 3 = 13 f t dt = 0 by Property 1 of Table 5.4. Notice that A132 - F 3 = -27. 1 2 5 1 2 b. The value of A 5 = f t dt is found by subtracting the area of the region that 1 2 1 2 1-1 lies below the t-axis on -1, 3 from the area of the region that lies above the t-axis 1 2 1 2 on 3, 5 . Therefore, A 5 = 3 - 27 = -24. Similarly, F 5 is the net area of the region bounded by the3 graph4 of f and the t-axis on the interval 3, 5 ; therefore, 3 4 1 2 1 2 F 5 = 3. Notice that A 5 - F 5 = -27. 3 4 c. Reasoning as in parts (a) and (b), we see that A 9 27 3 35 59 and 1 2 1 2 1 2 = - + - = - F 9 = 3 - 35 = -32. As before, observe that A 9 - F 9 = -27. 1 2 Related Exercises 11–12 1 2 1 2 1 2 Example 1 illustrates the important fact (to be explained shortly) that two area functions of the same function differ by a constant; in Example 1, the constant is -27. y x A(x) ϭ ͵ (2t ϩ 3) dt Quick C heck 1 In Example 1, let B x be the area function for f with left endpoint 5. 2 f (t) ϭ 2t ϩ 3 Evaluate B 5 and B 9 . 1 2 EXAMPLE1 2 2 Area 1of2 a trapezoid Consider the trapezoid bounded by the line

f t = 2t + 3 and the t-axis from t = 2 to t = x (Figure 5.35). The area function A(x) x A x = 12 f t dt gives the area of the trapezoid, for x Ú 2. 1 2 a.1 Evaluate2 A1 2 . b. Evaluate A 5 . 0 2 x t c. Find and graph the area function y A x , for x 2. 1 2 = Ú 1 2 Figure 5.35 d. Compare the derivative of A to f . 1 2

M05_BRIG7345_02_SE_C05.3.indd 363 21/10/13 11:11 AM 364 Chapter 5 • Integration Not for Sale So lution 2 a. By Property 1 of Table 5.4, A 2 = 2t 3 dt = 0. b 12 + A ϭ qh(a ϩ b) b. Notice that A 5 is the area of the trapezoid (Figure 5.35) bounded by the line a 1 2 1 2 y = 2t + 3 and the t-axis on the interval 2, 5 . Using the area formula for a trap- h ezoid (Figure1 5.362 ), we find that Figure 5.36 3 4 5 1 1 A 5 = 2t + 3 dt = 5 - 2 # f 2 + f 5 = # 3 7 + 13 = 30. 2 2 c L2 e 1 2 1 2 distance1 between2 1 sum1 2 of parallel1 22 1 2 parallel sides side lengths

c. Now the right endpoint of the base is a variable x Ú 2 (Figure 5.37). The distance between the parallel sides of the trapezoid is x - 2. By the area formula for a trapezoid, the area of this trapezoid for any x Ú 2 is 1 A x = x - 2 # f 2 + f x

2 c e 1 2 distance1 between2 1 sum1 2 of parallel1 22 parallel sides side lengths 1 = x - 2 7 + 2x + 3 2 = x1- 2 21x + 5 2 x 2 3x 10. = 1 + 21- 2

x y A(x) ϭ ͵(2t ϩ 3)dt ϭ x2 ϩ 3x Ϫ 10 2 f(t) ϭ 2t ϩ 3

2x ϩ 3 A(x) 7

0 2 x t x Ϫ 2 Figure 5.37

Area function: Expressing the area function in terms of an integral with a variable upper limit we have y A(x) ϭ x2 ϩ 3x Ϫ 10 80 x A x = 2t + 3 dt = x 2 + 3x - 10. 60 L2 1 2 1 2 40 Because the line f t = 2t + 3 is above the t-axis, for t Ú 2, the area function A x = x 2 + 3x - 10 is an increasing function of x with A 2 = 0 (Figure 5.38). 20 1 2 d. Differentiating the area function, we find that 1 2 1 2

0 2 4 6 8 x d 2 A′ x = x + 3x - 10 = 2x + 3 = f x . dx Figure 5.38 1 2 1 2 1 2 Therefore, A′ x = f x , or equivalently, the area function A is an antiderivative of f. ➤ Recall that if A′ x = f x , then f is We soon show that this relationship is not an accident; it is the first part of the Funda- the derivative of A; equivalently, A is an mental Theorem1 2 of Calculus.1 2 1 2 1 2 antiderivative of f . Related Exercises 13–22

Quick C heck 2 Verify that the area function in Example 2c gives the correct area when x = 6 and x = 10.

M05_BRIG7345_02_SE_C05.3.indd 364 21/10/13 11:11 AM Not for Sale 5.3 Fundamental Theorem of Calculus 365 Fundamental Theorem of Calculus Example 2 suggests that the area function A for a linear function f is an antiderivative of

f; that is, A′ x = f x . Our goal is to show that this conjecture holds for more general functions. Let’s start with an intuitive argument; a formal proof is given at the end of the section. 1 2 1 2 Assume that f is a continuous function defined on an interval a, b . As before, x A x = 1a f t dt is the area function for f with a left endpoint a: It gives the net area of the region bounded by the graph of f and the t-axis on the interval 3 a, x4 , for x Ú a. Figure1 2 5.39 is1 the2 key to the argument. 3 4

Approximately A(x ϩ h) Ϫ A(x) ഠ hf (x) rectangular when h is small Area ഠ hf(x)

y y y y ϭ f (t) y ϭ f (t) y ϭ f (t)

f (x) A(x ϩ h) A(x)

O axx ϩ h t O axx ϩ h t O axx ϩ h t h Figure 5.39

Note that with h 7 0, A x + h is the net area of the region whose base is the interval a, x + h and A x is the net area of the region whose base is the interval a, x . So the difference A x + h - A1 x is 2the net area of the region whose base is the interval x, x3 + h . If4 h is small,1 2 the region in question is nearly rectangular with a base 3of length4 1 2 1 2 h and a height f x . Therefore, the net area of this region is 3 4 1 2 A x + h - A x ≈ h f x . Dividing by h, we have 1 2 1 2 1 2 A x + h - A x ≈ f x . h 1 2 1 2 1 2 ➤ Recall that An analogous argument can be made with h 6 0. Now observe that as h tends to zero, this approximation improves. In the limit as h S 0, we have f x + h - f x f ′ x = lim . hS0 h A x h A x 1 2 1 2 + - 1 2 lim = lim f x . If the function f is replaced with A, hS0 h hS0 h 1 2 1 2 d then 1 2 A′ x f x A x + h - A x x 1 2 1 2 A′ x = lim . We see that indeed A′ x = f x . Because A x = 1a f t dt, the result can also be h 0 h S 1 2 1 2 written 1 2 1 2 1 2 1 2 1 2 d x A′ x = f t dt = f x , dx La e 1 2 A1x2 1 2 which says that the derivative of the integral of1 2f is f. This conclusion is the first part of the Fundamental Theorem of Calculus.

M05_BRIG7345_02_SE_C05.3.indd 365 21/10/13 11:11 AM 366 Chapter 5 • Integration Not for Sale

theore m 5.3 (Part 1) Fundamental Theorem of Calculus If f is continuous on a, b , then the area function

3 4 x

A x = f t dt, for a … x … b, La 1 2 1 2 is continuous on a, b and differentiable on a, b . The area function satisfies

A′ x = f x . Equivalently, 3 4 x 1 2 1 2 1 2 d A′ x = f t dt = f x , dx La 1 2 1 2 1 2 which means that the area function of f is an antiderivative of f on a, b .

3 4 Given that A is an antiderivative of f on a, b , it is one short step to a powerful method for evaluating definite integrals. Remember (Section 4.9) that any two antiderivatives of f differ by a constant. Assuming that F is any other3 4antiderivative of f on a, b , we have

F x = A x + C, for a … x … b. 3 4 Noting that A a = 0, it follows1 2 that1 2

1 2F b - F a = A b + C - A a + C = A b . Writing A b in terms1 2 of a definite1 2 1 integral1 2 leads2 to1 the1 remarkable2 2 result1 2 1 2 b

A b = f x dx = F b - F a . La 1 2 1 2 1 2 1 2 We have shown that to evaluate a definite integral of f, we • find any antiderivative of f, which we call F; and

• compute F b - F a , the difference in the values of F between the upper and lower limits of integration. 1 2 1 2 This process is the essence of the second part of the Fundamental Theorem of Calculus.

theore m 5.3 (Part 2) Fundamental Theorem of Calculus If f is continuous on a, b and F is any antiderivative of f on a, b , then

3 4 b 3 4

f x dx = F b - F a . La 1 2 1 2 1 2

b It is customary and convenient to denote the difference F b - F a by F x ͉ a. Using this shorthand, the Fundamental Theorem is summarized in Figure 5.40. 1 2 1 2 1 2

x is variable Antiderivative of f of integration. evaluated at a and b.

b b Limits of f (x) dx ϭ F(b) Ϫ F(a) ϭ F(x) integration ͵ ͯa a

Integrand Shorthand notation Figure 5.40

M05_BRIG7345_02_SE_C05.3.indd 366 21/10/13 11:11 AM Not for Sale 5.3 Fundamental Theorem of Calculus 367 T he Inverse R elationship between Differentiation and Integration It is worth pausing to observe that the two parts of the Fundamental Theorem express the inverse relationship between differentiation and integration. Part 1 of the Fundamental Theorem says

d x f t dt = f x , dx La 1 2 1 2 or the derivative of the integral of f is f itself.

Noting that f is an antiderivative of f ′, Part 2 of the Fundamental Theorem says

f ′ x dx = f b - f a , La 1 2 1 2 1 2 Quick C heck 4 Explain why f is an or the definite integral of the derivative of f is given in terms of f evaluated at two points.

antiderivative of f ′. In other words, the integral “undoes” the derivative. This last relationship is important because it expresses the integral as an accumulation

operation. Suppose we know the rate of change of f (which is f ′) on an interval a, b . The Fundamental Theorem says that we can integrate (that is, sum or accumulate) the rate of change over that interval and the result is simply the difference in f evaluated 3at the4 endpoints. You will see this accumulation property used many times in the next chapter. Now let’s use the Fundamental Theorem to evaluate definite integrals.

EXAMPLE 3 Evaluating definite integrals Evaluate the following definite integrals using the Fundamental Theorem of Calculus, Part 2. Interpret each result geometrically. y 10 2p 1 4 t 1 y ϭ x Ϫ x2 2 - 150 60 6 a. 60x - 6x dx b. 3 sin x dx c. > dt L0 L0 L1 16 1 t 120 1 2 Area ϭ So lution > 90 10 2 ͵ (60x Ϫ 6x ) dx a. Using the antiderivative rules of Section 4.9, an antiderivative of 60x 6x 2 is 60 0 - 30x 2 2x 3. By the Fundamental Theorem, the value of the definite integral is 30 -

10 10 0 x 2 4 6 8 10 60x - 6x 2 dx = 30x 2 - 2x 3 Fundamental Theorem L0 0 Figure 5.41 1 2 1 2 2 ` 3 2 3 Evaluate at x = 10 = 30 # 10 - 2 # 10 - 30 # 0 - 2 # 0 y and x = 0. y ϭ 3 sin x = 3000 - 2000 - 0 3 1 2 1 2 1000. Simplify. = 1 2 10 2 Because f is positive on 0, 10 , the definite integral 10 60x - 6x dx is ␲ 2␲ x the area of the region between the graph of f and the x-axis on the interval 0, 10 (Figure 5.41). 3 4 1 2 Ϫ3 b. As shown in Figure 5.42, the region bounded by the graph of f x = 3 sin x 2␲ 3 4 and the x-axis on 0, 2p consists of two parts, one above the x-axis and Net area ϭ ͵ 3 sin x dx ϭ 0. 0 one below the x-axis. By the symmetry of f, these two regions1 have2 the 3 4 Figure 5.42 same area, so the definite integral over 0, 2p is zero. Let’s confirm this

fact. An antiderivative of f x = 3 sin x is -3 cos x. Therefore, the value of the definite integral is 3 4 1 2 2p 2p 3 sin x dx = -3 cos x Fundamental Theorem L0 0 ` = -3 cos 2p - -3 cos 0 Substitute. 3 3 0. Simplify. = 1- - -1 22= 1 1 22 c. Although the variable of integration1 2is t, rather than x, we proceed as in parts (a) and (b) after simplifying the integrand: t - 1 1 1 = - . 1 t t t Copyright Pearson. All rights reserved. 1

M05_BRIG7345_02_SE_C05.3.indd 367 21/10/13 11:12 AM 368 Chapter 5 • Integration Not for Sale ➤ We know that Finding antiderivatives with respect to t and applying the Fundamental Theorem, we have d 1 t1 2 = t -1 2. dt 2 > > 1 4 t 1 1 4 1 1 2 - -1 2 Simplify the 1 > dt = > t - dt -1 2 1 2 integrand. Therefore, t dt = t + C L1 16 1 t L1 16 > t L 2 > > dt > > a 1b 4 and = t -1 2 dt = 2t1 2 + C. 1 2 Fundamental t = 2t ln ͉ t ͉ L L > > - > Theorem 1 > 1 16 1 2 1 2 1 1 2 ` 1> 1 1 y = 2 - ln - 2 - ln Evaluate. 1 3 4 > 4 16 > 16 16 Ω 16 ~ a a b1 1 b 1 a a b b t = 1 - ln - + ln Simplify. 4 2 16 Ϫ4 1 Ϫ8 ͙t Ϫ 1 = - ln 4 ≈ -0.8863. y ϭ 2 t Ϫ12 The definite integral is negative because the graph of f lies below the t-axis (Figure 5.43). Related Exercises 23–50 Ϫ16

EXAMPLE 4 Net areas and definite integrals The graph of f x = 6x x + 1 x - 2 Figure 5.43 is shown in Figure 5.44. The region R1 is bounded by the curve and the x-axis on the 1 2 1 21 2 interval -1, 0 , and R2 is bounded by the curve and the x-axis on the interval 0, 2 . y y ϭ 6x(x ϩ 1)(x Ϫ 2) a.3 Find the4 net area of the region between the curve and the x-axis on -31, 24. 10 b. Find the area of the region between the curve and the x-axis on 1, 2 . - 3 4 5 So lution R1 3 4 Ϫ1 1 2 x a. The net area of the region is given by a definite integral. The integrand f is Ϫ5 first expanded to find an antiderivative: R2 Ϫ10 2 2 3 2 f x dx = 6x - 6x - 12x dx. Expand f . Figure 5.44 L-1 L-1 1 2 3 1 2 = x 4 - 2x 3 - 6x 2 Fundamental Theorem 2 -1 a 27 b ` = - . Simplify. 2 27 The net area of the region between the curve and the x-axis on -1, 2 is - 2 , which is the area of R1 minus the area of R2 (Figure 5.44). Because R2 has a larger area than 3 4 R1, the net area is negative.

b. The region R1 lies above the x-axis, so its area is

0 3 0 5 6x 3 - 6x 2 - 12x dx = x 4 - 2x 3 - 6x 2 = . L-1 2 -1 2 1 2 a b ` The region R2 lies below the x-axis, so its net area is negative:

2 3 2 6x 3 - 6x 2 - 12x dx = x 4 - 2x 3 - 6x 2 = -16. L0 2 0 1 2 a b ` Therefore, the area of R2 is - -16 = 16. The combined area of R1 and R2 is 5 37 2 + 16 = 2 . We could also find the area of this region directly by evaluating 2 1 2 f x dx. Related Exercises 51–60 1-1 ͉ ͉ Examples1 2 3 and 4 make use of Part 2 of the Fundamental Theorem, which is the most potent tool for evaluating definite integrals. The remaining examples illustrate the use of the equally important Part 1 of the Fundamental Theorem.

M05_BRIG7345_02_SE_C05.3.indd 368 21/10/13 11:12 AM Not for Sale 5.3 Fundamental Theorem of Calculus 369 EXAMPLE 5 Derivatives of integrals Use Part 1 of the Fundamental Theorem to simplify the following expressions. 2 d x d 5 d x a. sin2 t dt b. t2 + 1 dt c. cos t2 dt dx L1 dx Lx dx L0 2 So lution a. Using Part 1 of the Fundamental Theorem, we see that

d x sin2 t dt = sin2 x. dx L1 b. To apply Part 1 of the Fundamental Theorem, the variable must appear in the upper b a limit. Therefore, we use the fact that 1a f t dt = - 1b f t dt and then apply the Fundamental Theorem: 1 2 1 2 d 5 d x t2 + 1 dt = - t2 + 1 dt = - x 2 + 1. dx Lx dx L5 2 2 2 c. The upper limit of the integral is not x, but a function of x. Therefore, the function to be differentiated is a composite function, which requires the Chain Rule. We let u = x 2 to produce

u y = g u = cos t2 dt. L0 1 2 By the Chain Rule,

2 d x dy dy du cos t2 dt = = Chain Rule dx L0 dx du dx

➤ Example 5c illustrates one case of u d 2 Leibniz’s Rule: = cos t dt 2x Substitute for g; note that u′ x = 2x. du L0 g x d 2 1 2 a b1 2 1 2f t dt = f g x g′ x . = cos u 2x Fundamental Theorem dx La 2x cos x 4. Substitute u x 2. 1 2 1 1 22 1 2 = 1 21 2 = Related Exercises 61–68

y EXAMPLE 6 Working with area functions Consider the function f shown in x y ϭ f (t) Figure 5.45 and its area function A x = 10 f t dt, for 0 … x … 17. Assume that the four regions R1, R2, R3, and R4 have the same area. Based on the graph of f, do the following. R1 R3 1 2 1 2 a. Find the zeros of A on 0, 17 . 0 4 8 12 16 t b. Find the points on 0, 17 at which A has local maxima or local minima. R R 3 4 2 4 c. Sketch a graph of A, for 0 x 17. 3 4 … … So lution Figure 5.45 x a. The area function A x = f t dt gives the net area bounded by the graph of f and 10 0 the t-axis on the interval 0, x (Figure 5.46a). Therefore, A 0 = 10 f t dt = 0. 1 2 1 2 Because R1 and R2 have the same area but lie on opposite sides of the t-axis, it follows 8 3 4 16 1 2 1 2 that A 8 = 10 f t dt = 0. Similarly, A 16 = 10 f t dt = 0. Therefore, the zeros of A are x = 0, 8, and 16. 1 2 1 2 1 2 1 2 b. Observe that the function f is positive, for 0 6 t 6 4, which implies that A x increases as x increases from 0 to 4 (Figure 5.46b). Then as x increases from 4 to 8, ➤ Recall that local extrema occur only at A x decreases because f is negative, for 4 6 t 6 8 (Figure 5.46c). Similarly,1 2 A x

interior points of the domain. increases as x increases from x = 8 to x = 12 (Figure 5.46d) and decreases from x 1=212 to x = 16. By the First Derivative Test, A has local minima at x = 8 and1 2 x = 16 and local maxima at x = 4 and x = 12 (Figure 5.46e).

M05_BRIG7345_02_SE_C05.3.indd 369 21/10/13 11:12 AM 370 Chapter 5 • Integration Not for Sale c. Combining the observations in parts (a) and (b) leads to a qualitative sketch of A (Figure 5.46e). Note that A x Ú 0, for all x Ú 0. It is not possible to determine function values (y-coordinates) on the graph of A. 1 2

x A(x) ϭ ͵ f (t) dt ϭ net area over [0, x]. The net area A(x) increases The net area decreases 0 as x increases from 0 to 4. as x increases from 4 to 8. y y y y ϭ f (t) y ϭ f (t) y ϭ f (t)

x x 0 t 0 4 8 12 16 t 0 x 4 8 12 16 t 4 8 12 16

(a) (b) (c)

The net area A(x) increases Graph of area function on [0, 17] as x increases from 8 to 12. y y y ϭ f (t) y ϭ A(x)

0 4 8 x 12 16 t

0 4 8 12 16 x

(d) (e) Figure 5.46 Related Exercises 69–80

EXAMPLE 7 The sine integral function Let sin t if t 7 0 g t = t 1 if t = 0. 1 2 • x Graph the sine integral function S x = 10 g t dt, for x Ú 0. So lution Notice that S is an area1 function2 for1 2 g. The independent variable of S is x, and t has been chosen as the (dummy) variable of integration. A good way to start is by graphing the integrand g (Figure 5.47a). The function oscillates with a decreasing amplitude with g 0 = 1. Beginning with S 0 = 0, the area function S increases until x = p because g is positive on 0, p . However, on p, 2p , g is negative and the net area decreases. On1 2 2p, 3p , g is positive again,1 2 so S again increases. Therefore, the graph of S has alternating local maxima1 2 and minima. Because1 2 the amplitude of g decreases, each maximum of S1 is less than2 the previous maximum and each minimum of S is greater than the previous minimum (Figure 5.47b). Determining the exact value of S at these maxima and minima is difficult.

y y

2 2 y ϭ S(x)

1 y ϭ g(t) 1

0 x ␲ 2␲ 3␲ 4␲ t 0 ␲ 2␲ 3␲ 4␲ 5␲ 6␲ x

(a) (b)

Figure 5.47

M05_BRIG7345_02_SE_C05.3.indd 370 21/10/13 11:12 AM Not for Sale 5.3 Fundamental Theorem of Calculus 371 Appealing to Part 1 of the Fundamental Theorem, we find that

d x sin x S′ x = g t dt = , for x 7 0. dx L0 x 1 2 1 2 ➤ Note that As anticipated, the derivative of S changes sign at integer multiples of p. Specifically, S′ is positive and S increases on the intervals 0, p , 2p, 3p , c, lim S′ x = lim g x = 0. xS∞ xS∞ 2np, 2n + 1 p , c, while S′ is negative and S decreases on the remaining inter- 1 2 1 2 1 2 1 2 vals. Clearly, S has local maxima at x = p, 3p, 5p, c, and it has local minima at 1 1 2 2 x = 2p, 4p, 6p, c. One more observation is helpful. It can be shown that although S oscillates for increasing x, its graph gradually flattens out and approaches a horizontal asymptote. (Finding the exact value of this horizontal asymptote is challenging; see Exercise 111.) Assembling all these observations, the graph of the sine integral function emerges (Figure 5.47b). Related Exercises 81–84

We conclude this section with a formal proof of the Fundamental Theorem of Calculus. Proof of the Fundamental Theorem: Let f be continuous on a, b and let A be the area function for f with left endpoint a. The first step is to prove that A is differentiable on 3 4 a, b and A′ x = f x , which is Part 1 of the Fundamental Theorem. The proof of Part 2 then follows. 1 2 1 2 1 2 Step 1. We assume that a 6 x 6 b and use the definition of the derivative, A x + h - A x A′ x = lim . hS 0 h 1 2 1 2 1 2 First assume that h 7 0. Using Figure 5.48 and Property 5 of Table 5.4, we have

x + h x x + h

A x + h - A x = f t dt - f t dt = f t dt. La La Lx 1 2 1 2 1 2 1 2 1 2 That is, A x + h - A x is the net area of the region bounded by the curve on the interval x, x + h . 1 2 1 2 3 4 xϩh A(x ϩ h) Ϫ A(x) ϭ ͵ f (t) dt x

y y y y ϭ f (t) y ϭ f (t) y ϭ f (t)

xϩh A(x ϩ h) A(x) ͵ f (t) dt x

0 axx ϩ h t 0 axx ϩ h t 0 axx ϩ h t Figure 5.48

➤ The quantities m and M exist for any Let m and M be the minimum and maximum values of f on x, x + h , respectively, h 0; however, their values depend on h. 7 which exist by the continuity of f . Suppose x = t0 6 t1 6 t2 6 c 6 tn = x + h is a n 3 4 * general partition of x, x + h and let a f tk ∆tk be a corresponding general Riemann k = 1

sum, where ∆tk = tk3 - tk - 1. 4Because m …1f 2t … M on x, x + h , it follows that n n n *1 2 3 4 a m∆tk … a f tk ∆tk … a M∆tk k = 1 k = 1 k = 1 d 1 2 d mh Mh

M05_BRIG7345_02_SE_C05.3.indd 371 21/10/13 11:12 AM 372 Chapter 5 • Integration Not for Sale or

n * mh … a f tk ∆tk … Mh. k = 1 1 2 n n n We have used the facts that a m∆tk = m a ∆tk = mh and similarly, a M∆tk = Mh. k = 1 k = 1 k = 1 Notice that these inequalities hold for every Riemann sum for f on x, x + h ; that is, for all partitions and for all n. Therefore, we are justified in taking the limit as n S ∞ across these inequalities to obtain 3 4

n * lim mh … lim a f tk ∆tk … lim Mh. nS∞ nS∞k = 1 nS∞ h x + h 1 2 1x f t dt Evaluating each of these three limits results in1 2

x + h

mh … f t dt … Mh. Lx 1d 2 A x + h - A x Substituting for the integral, we find that1 2 1 2 mh … A x + h - A x … Mh. Dividing these inequalities by h 7 01, we have2 1 2 A x + h - A x m … … M. h 1 2 1 2 The case h 6 0 is handled similarly and leads to the same conclusion. We now take the limit as h S 0 across these inequalities. As h S 0, m and M

approach f x , because f is continuous at x. At the same time, as h S 0, the quotient that is sandwiched between m and M approaches A′ x : 1 2 A x + h1 2- A x lim m = lim = lim M. hS0 hS0 h hS0 c c (++++++)++++++*1 2 1 2

f x A′ x f x By the Squeeze Theorem (Theorem1 2 2.5), we conclude1 2 that A′ x1 exists2 and A is differen-

tiable for a 6 x 6 b. Furthermore, A′ x = f x . Finally, because A is differentiable on a, b , A is continuous on a, b by Theorem 3.1. Exercise 1161 shows2 that A is also right- and left-continuous at the endpoints a and1 2 b, respectively.1 2 1 2 1 2 Step 2. Having established that the area function A is an antiderivative of f, we know that

➤ Once again we use an important fact: F x = A x + C, where F is any antiderivative of f and C is a constant. Noting that Two antiderivatives of the same function A a = 0, it follows that differ by a constant. 1 2 1 2 1 2 F b - F a = A b + C - A a + C = A b . Writing A b in terms1 2of a definite1 2 integral,1 1 2 we 2have1 1 2 2 1 2 1 2 b

A b = f x dx = F b - F a , La 1 2 1 2 1 2 1 2 which is Part 2 of the Fundamental Theorem.

M05_BRIG7345_02_SE_C05.3.indd 372 21/10/13 11:12 AM Not for Sale 5.3 Fundamental Theorem of Calculus 373 Section 5.3 Exerci ses Review Questions 13–16. Area functions for constant functions Consider the following functions f and real numbers a (see figure). 1. Suppose A is an area function of f . What is the relationship x between f and A? a. Find and graph the area function A x = 1a f t dt for f. b. Verify that A′ x = f x . 2. Suppose F is an antiderivative of f and A is an area function of f. 1 2 1 2 y What is the relationship between F and A? 1 2 1 2 y ϭ f (t) 3. Explain in words and write mathematically how the Fundamental Theorem of Calculus is used to evaluate definite integrals.

4. Let f x = c, where c is a positive constant. Explain why an area A(x) ϭ area function of f is an increasing function. 1 2

5. The linear function f x = 3 - x is decreasing on the interval 0, 3 . Is the area function for f (with left endpoint 0) increasing 0 a x t or decreasing on the interval1 2 0, 3 ? Draw a picture and explain. 3 4 2 2 6. Evaluate 3x 2 dx and 3x 2 dx. 13. f t = 5, a = 0 14. f t = 10, a = 4 10 1-2 3 4 7. Explain in words and express mathematically the inverse relation- 15. f 1t2 = 5, a = -5 16. f 1t2 = 2, a = -3 ship between differentiation and integration as given by Part 1 of 17. Area1 2 functions for the same linear function1 2 Let f t = t the Fundamental Theorem of Calculus. x and consider the two area functions A x = 10 f t dt and x 1 2 F x = 12 f t dt. 8. Why can the constant of integration be omitted from the antide- 1 2 1 2 a. Evaluate A 2 and A 4 . Then use geometry to find an expres- rivative when evaluating a definite integral? 1 2 1 2 x b sion for A x , for x Ú 0. d d 1 2 1 2 b. Evaluate F 4 and F 6 . Then use geometry to find an expres- 9. Evaluate f t dt and f t dt, where a and b are dx dx 1 2 La La sion for F x , for x Ú 2. 1 2 1 2 constants. 1 2 1 2 c. Show that A x - F x is a constant and that 1 2 b A′ x = F ′ x = f x . 10. Explain why 1a f ′ x dx = f b - f a . 1 2 1 2

18. Area functions1 2 1for2 the 1same2 linear function Let f t = 2t - 2 1 2 1 2 1 2 x Basic Skills and consider the two area functions A x = 11 f t dt and x 1 2 11. Area functions The graph of f is shown in the figure. Let F x = 14 f t dt. x x 1 2 1 2 A x = f t dt and F x = f t dt be two area functions a. Evaluate A 2 and A 3 . Then use geometry to find an expres- 1-2 14 1 2 1 2 for f . Evaluate the following area functions. sion for A x , for x Ú 1. 1 2 1 2 1 2 1 2 1 2 1 2

a. A -2 b. F 8 c. A 4 d. F 4 e. A 8 b. Evaluate F 5 and F 6 . Then use geometry to find an expres- 1 2 sion for F x , for x Ú 4. y y ϭ f (t) 1 2 1 2 1 2 1 2 1 2 1 2 1 2 c. Show that A x - F x is a constant and that 1 2 8 A′ x = F ′ x = f x . 1 2 1 2 Area ϭ 17 19–22. Area1 2functions1 2 for linear1 2 functions Consider the following Area ϭ 8 4 functions f and real numbers a (see figure). x a. Find and graph the area function A x = 1a f t dt.

Ϫ4 4 8 t b. Verify that A′ x = f x . 1 2 1 2 Ϫ4 Area ϭ 9 y1 2 1 2 y ϭ f (t)

12. Area functions The graph of f is shown in the figure. Let x x A x = 10 f t dt and F x = 12 f t dt be two area functions for f. Evaluate the following area functions. A(x) ϭ area 1 2 1 2 1 2 1 2 a. A 2 b. F 5 c. A 0 d. F 8

e. A 8 f. A 5 g. F 2 1 2 1 2 1 2 1 2 0 a x t 1 2 y 1 2 y ϭ1 f2(t)

19. f t t 5, a 5 20. f t 2t 5, a 0 Area ϭ 8 = + = - = + =

21. f t 3t 1, a 2 22. f t 4t 2, a 0 2 1 2 = + = 1 2 = + = 1 2 1 2 1 3 5 7 t Ϫ2

Area ϭ 5

Area ϭ 11

M05_BRIG7345_02_SE_C05.3.indd 373 21/10/13 11:12 AM 374 Chapter 5 • Integration Not for Sale 23–24. Definite integrals Evaluate the following integrals using the 52. The region above the x-axis bounded by y = 4 - x 2 Fundamental Theorem of Calculus. Explain why your result is consis- 4 tent with the figure. 53. The region below the x-axis bounded by y = x - 16 1 7p 4 54. The region bounded by y = 6 cos x and the x-axis between 2 23. x - 2x + 3 dx 24. > sin x + cos x dx x = -p 2 and x = p L0 L-p 4 1 2 > 1 2 55–60. Areas >of regions Find the area of the region bounded by the y y graph of f and the x-axis on the given interval.

3 2 2 y ϭ sin x ϩ cos x 2 y ϭ x Ϫ 2x ϩ 3 55. f x = x - 25 on 2, 4 3 56. f 1x2 = x - 1 on -3 1, 24 x Ϫd fj 1 2 1 3 4 57. f x = on -2, -1 Ϫ2 x 0 x 1 1 2 3 4 58. f x = x x + 1 x - 2 on -1, 2 25–28. Definite integrals Evaluate the following integrals using the 59. f 1x2 = sin1 x on 21-p 4, 23p 43 4 Fundamental Theorem of Calculus. Sketch the graph of the integrand

and shade the region whose net area you have found. 60. f 1x2 = cos x on 3 p 2,> p > 4 3 1 61–68.1 Derivatives2 of3 integrals> 4 Simplify the following expressions. 25. x 2 x 6 dx 26. x x dx - - - x x L-2 L0 d d 1 61. t2 + t + 1 dt 62. et dt 51 2 21 2 1 dx L3 dx L0 27. x 2 9 dx 28. 1 dx - - 2 x13 2 10 L0 L1 2 x d dp d dz 63. 64. 1 2 a b 2 2 29–50. Definite integrals Evaluate the following> integrals using the dx L2 p dx Lx2 z + 1 Fundamental Theorem of Calculus. d 1 d 0 dp 2 2 4 65. t + 1 dt 66. 2 3 2 dx Lx dx Lx p 1 29. 4x dx 30. 3x + 2x dx 2 + L0 L0 2x d x d e 1 p14 2 67. 1 + t2 dt 68. ln t2 dt dx L x dx Lex 31. x + x dx 32. > 2 cos x dx - 2 L0 L0 1 69. Matching functions with area functions Match the func- 91 2 9 2 2 + t tions f , whose graphs are given in a–d, with the area functions 33. dx 34. dt x A x f t dt, whose graphs are given in A–D. L1 x L4 t 1 = 10 2 1 ln 8 y 1 2 1 2 y 35. x 2 - 4 dx 36. ex dx L-2 L0 y ϭ f (t) 11 2 4 37. x -3 - 8 dx 38. x x - 2 x - 4 dx y ϭ f (t) L1 2 L0 p14 2 1 21 21 2 O b t O b t > dx 39. sec2 u du 40. > > 2 L0 L0 1 - x -1 p 2 -3 41. x dx 42. 1 - sin x dx (a) (b) L-2 L0 4 p12 2 y y 43. 1 - x x - 4 dx 44. > cos x - 1 dx L1 L-p 2 y ϭ f (t) 21 21 2 9 1 2 y ϭ f (t) 3 x> - x 45. dt 46. 3 dx L1 t L4 x 1 p 8 1 O b t 47. cos 2x dx 48. 10e2x dx > O b t L0 L0 3 dx p 8 49. 2 50. 8 csc2 4x dx 2 > (c) (d) L1 1 + x Lp 16 51–54. Areas Find (i) the net area and (ii) the> area of the following y y regions. Graph the function and indicate the region in question. y ϭ A(x) 51. The region bounded by y x 1 2 and the x-axis between x 1 = = O b x and x = 4 >

(A) (B) M05_BRIG7345_02_SE_C05.3.indd 374 21/10/13 11:12 AM

y y

y ϭ A(x) y ϭ A(x)

O b x O b x

y ϭ f (t) y ϭ f (t)

O b t O b t

(a) (b)

y y

y ϭ f (t) y ϭ f (t)

O b t O b t

5.3 Fundamental Theorem of Calculus 375 (c) Not(d) for Sale y y 75. Area functions from graphs The graph of f is given in the fig- x ure. Let A x = f t dt and evaluate A 2 , A 5 , A 8 , and y ϭ A(x) 10 x A 12 . O b 1 2 1 2 1 2 1 2 1 2 1 2 y 4 y ϭ f (t) O x b 2 y ϭ A(x) 4 8 12 t Ϫ2

(A) (B) ~ of circle of radius 2

y y 76–80. Working with area functions Consider the function f and the points a, b, and c. x a. Find the area function A x f t dt using the Fundamental y ϭ A(x) y ϭ A(x) = 1a Theorem. b. Graph f and A. 1 2 1 2 c. Evaluate A b and A c . Interpret the results using the graphs of part (b). O b x O b x 1 2 1 2 76. f x = sin x; a = 0, b = p 2, c = p (C) (D) x 77. f 1x2 = e ; a = 0, b = ln 2, >c = ln 4

70–73. Working with area functions Consider the function f and its 78. f 1x2 = -12x x - 1 x - 2 ; a = 0, b = 1, c = 2 graph. 1 79. f x cos px; a 0, b , c 1 x 1 2 = 1 =21 = 22 = a. Estimate the zeros of the area function A x = 10 f t dt, for 0 x 10. 1 2 1 … … 80. f x = ; a = 1, b = 4, c = 6 b. Estimate the points (if any) at which A has1 2a local maximum1 2 or x 1 2 minimum. T 81–84. Functions defined by integrals Consider the function g, which c. Sketch a graph of A, for 0 … x … 10, without a scale on the y-axis. is given in terms of a definite integral with a variable upper limit. 70. 71. a. Graph the integrand. y y b. Calculate g′ x . y ϭ f (t) c. Graph g, showing all your work and reasoning. 1 2 x y ϭ f (t) 81. g x = sin2 t dt L0 0 2 6 8 10 t 0 2 4 6 8 10 t 1 2 x 82. g x = t2 + 1 dt L0 1 2 x1 2 83. g x = sin pt2 dt (a Fresnel integral) 72. 73. L0 y y 1 2 x 1 2 y ϭ f (t) 84. g x = cos p t dt L0 y ϭ f (t) 1 1 2 1 2 Further Explorations 0 2 4 6 8 10 t 0 2 4 6 10 t 85. Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. Suppose that f is a positive decreasing function, for x 7 0. x Then the area function A x = 10 f t dt is an increasing function of x. 74. Area functions from graphs The graph of f is given in the b. Suppose that f is a negative1 2 increasing1 2 function, for x 7 0. x x figure. Let A x f t dt and evaluate A 1 , A 2 , A 4 , = 10 Then the area function A x = 10 f t dt is a decreasing and A 6 . function of x. 1 2 1 2 1 2 1 2 1 2 c. The functions p x sin1 32x and q x1 2 4 sin 3x are 1 2 y ~ of circle = = of radius 2 antiderivatives of the same function. y ϭ f (t) d. If A x = 3x 2 -1 2x - 3 is an area 1function2 for f, then 2 B x = 3x 2 - x is also an area function for f. 1 2 d b

e. 1 2 f t dt 0. t = 2 4 6 dx La 1 2

M05_BRIG7345_02_SE_C05.3.indd 375 21/10/13 11:12 AM 376 Chapter 5 • Integration Not for Sale 86–94. Definite integrals Evaluate the following definite integrals us- 110. Max , min of area functions Suppose f is continuous on 0, ∞ ing the Fundamental Theorem of Calculus. and A x is the net area of the region bounded by the graph of f ln 2 4 2 and the t-axis on 0, x . Show that the local maxima and minima3 2 1 x 2 2 4 x - of A occur1 2 at the zeros of f. Verify this fact with the function 86. e dx 87. dx 88. - 3 ds 2 L0 L1 x L1 s s 2 3 4 f x = x - 10x. p 3 p 21 8 a b sec x tan x dx 2 3 T 111. Asymptote1 2 of sine integral Use a calculator to approximate 89. > 90. > csc u du 91. y dy L0 Lp 4 L1 x 2 sin t 2 2> 2 3 lim S x = lim dt, dx z + 4 3 dx x x t 92. 93. dz 94. 1 S∞ S∞ L0 2 z 2 L 2 x x - 1 L1 L0 9 + x 1 2 where S is the sine integral function (see Example 7). Explain 1 2 T 95–98. Areas of regions Find the area of the region R bounded by the your reasoning. x graph of f and the x-axis on the given interval. Graph f and show the sin t region R. 112. Sine integral Show that the sine integral S x = dt sat- L0 t

95. f x = 2 - ͉ x ͉ on -2, 4 isfies the (differential) equation xS′ x + 21S2″ x + xS‴ x = 0. 2 -1 2 96. f 1x2 = 1 - x 3 on -41 2, 3 2 113. Fresnel integral Show that the Fresnel1 2 integral1 S2 x = 1 2 x > 2 4 1 2 10 sin t dt satisfies the (differential) equation 97. f 1x2 = 1x - 4 on2 1, 4 3 >98. f >x 4= x x - 2 on -1, 3 1 2 S″ x 2 99–102.1 2Derivatives and3 integrals4 Simplify1 2 the given1 expressions.2 3 4 S′ x 2 + = 1. 2x 8 1 2 x 1 1 22 a b 99. f ′ t dt, where f ′ is continuous on 3, 8 d L3 114. Variable integration limits Evaluate t2 + t dt. dx x12 2 3 cos x4 L-x d dt d 4 (Hint: Separate the integral into two pieces.) 1 2 100. 2 101. t + 6 dt dx L0 t 4 dx L0 + 115. Discrete version of the Fundamental Theorem In this exercise, 1 t 1 2 1 d 2 d 3 3 we work with a discrete problem and show why the relationship 102. et dt 103. dx dx b - a f ′ x dx = f b - f a makes sense. Suppose we have a set dx Lx dt L1 x Lt2 x 1 of equally spaced grid points a b d t dx 1 t dx 1 2 1 2 1 2 a x x x x x b , 104. 2 + > 2 = 0 6 1 6 2 6 c 6 n - 1 6 n = dt L0 1 + x L0 1 + x a b where the5 distance between any two grid points is ∆x. Suppose6

Additional Exercises also that at each grid point xk, a function value f xk is defined, 2 for k 0, , n. 105. Zero net area Consider the function f x = x - 4x. = c 1 2 a. Graph f on the interval x Ú 0. a. We now replace the integral with a sum and replace the deriva- b 1 2 b tive with a difference quotient. Explain why f x dx is b. For what value of b 7 0 is f x dx = 0? 1a ′ 10 n 2 f xk f xk 1 c. In general, for the function f x = x - ax, where a 7 0, for - - 1 2 b analogous to a ∆x. 1 2 what value of b 7 0 (as a function of a) is 10 f x dx = 0? k = 1 ∆x 1 2 1 2 1 2 106. Cubic zero net area Consider the graph of the cubic f 1 2 ≈f ′ xk y = x x - a x - b , where 0 6 a 6 b. Verify that the graph b. Simplify the sum in part (a) and show that it is equal to bounds a region above the x-axis, for 0 x a, and bounds a 1 2 6 6 f b f a . 1 21 2 - region below the x-axis, for a 6 x 6 b. What is the relationship c. Explain the correspondence between the integral relationship between a and b if the areas of these two regions are equal? and1 2the summation1 2 relationship. 107. Maximum net area What value of b 7 -1 maximizes the integral 116. Continuity at the endpoints Assume that f is continuous on a, b and let A be the area function for f with left endpoint a. Let b 2 m* and M* be the absolute minimum and maximum values of f on x 3 - x dx? 3 4 L-1 a, b , respectively.

1 2 2 a. Prove that m* x - a … A x … M* x - a for all x in 108. Maximum net area Graph the function f x 8 2x x and 3 4 = + - a, b . Use this result and the Squeeze Theorem to show that A determine the values of a and b that maximize the value of the integral 1 2 1 2 1 2 1 2 is continuous from the right at x = a. 3 4 b b. Prove that m* b - x … A b - A x … M* b - x for all 8 + 2x - x 2 dx. x in a, b . Use this result to show that A is continuous from La the left at x =1b. 2 1 2 1 2 1 2 1 2 3 4 109. An integral equation Use the Fundamental Theorem of Calculus, Quick C heck Answer s Part 1, to find the function f that satisfies the equation 2 1 1 1. 0, -35 2. A 6 = 44; A 10 = 120 3. 3 - 2 = 6

x 4. If f is differentiated, we get f ′. Therefore, f is an antide-

1 2 1 2 f t dt = 2 cos x + 3x - 2. rivative of f ′. L0 1 2 Verify the result by substitution into the equation.

M05_BRIG7345_02_SE_C05.3.indd 376 21/10/13 11:12 AM Not for Sale 5.4 Working with Integrals 377 5.4 Working with Integrals

With the Fundamental Theorem of Calculus in hand, we may begin an investigation of integration and its applications. In this section, we discuss the role of symmetry in integrals, we use the slice-and-sum strategy to define the average value of a function, and we explore a theoretical result called the Mean Value Theorem for Integrals.

y Integrating Even and Odd Functions y ϭ f (x) Ϫ ϭ f ( x) f (x) (even function) Symmetry appears throughout mathematics in many different forms, and its use often leads to insights and efficiencies. Here we use the symmetry of a function to simplify integral calculations. Section 1.1 introduced the symmetry of even and odd functions. An even function RR satisfies the property f -x = f x , which means that its graph is symmetric about the n Ϫa O a x y-axis (Figure 5.49a). Examples of even functions are f x = cos x and f x = x , where 1 2 1 2 n is an even integer. An odd function satisfies the property f -x = -f x , which means that its graph is symmetric about the origin (Figure 5.49b1 ).2 Examples of odd1 2 functions are a a n 1 2 1 2 f x sin x and f x x , where n is an odd integer. ͵ f (x) dx ϭ 2͵ f (x) dx = = Ϫa 0 Special things happen when we integrate even and odd functions on intervals cen- 1 2 1 2 a tered at the origin. First suppose f is an even function and consider 1-a f x dx. From (a) Figure 5.49a, we see that the integral of f on -a, 0 equals the integral of f on 0, a . Therefore, the integral on -a, a is twice the integral on 0, a , or 1 2 y 3 4 3 4 y ϭ f (x) a a (odd function) a 3 4 3 4 ͵ f (x) dx ϭ 0 f x dx = 2 f x dx. Ϫa L-a L0 1 2 1 2 a On the other hand, suppose f is an odd function and consider f x dx. As shown in 1-a R a Figure 5.49b, the integral on the interval -a, 0 is the negative of the integral on 0, a . Therefore, the integral on a, a is zero, or 1 2 Ϫa O x - R 3 4 3 4 3 4 a

f x dx = 0. L-a 1 2 f (Ϫx) ϭ Ϫf (x) We summarize these results in the following theorem. (b) Figure 5.49 theorem 5.4 Integrals of Even and Odd Functions Let a be a positive real number and let f be an integrable function on the interval -a, a . a a • If f is even, f x dx 2 f x dx. 3 4 1-a = 10 a • If f is odd, f x dx 0. 1-a 1 2 = 1 2 1 2

Quick C heck 1 If f and g are both even functions, is the product fg even or odd? Use the

facts that f -x = f x and g -x = g x .

The following1 example2 1 2 shows 1how2 symmetry1 2 can simplify integration.

EXAMPLE 1 Integrating symmetric functions Evaluate the following integrals using symmetry arguments.

2 p 2 4 3 3 a. x - 3x dx b. > cos x - 4 sin x dx L-2 L-p 2 1 2 > 1 2

M05_BRIG7345_02_SE_C05.4.indd 377 21/10/13 11:12 AM 378 Chapter 5 • Integration Not for Sale So lution a. Note that x4 - 3x3 is neither odd nor even so Theorem 5.4 cannot be applied directly. However, we can split the integral and then use symmetry:

2 2 2 x 4 - 3x 3 dx = x 4 dx - 3 x 3 dx Properties 3 and 4 of Table 5.4 L-2 L-2 L-2 s 1 2 0

2 = 2 x 4 dx - 0 x 4 is even; x 3 is odd. L0 x 5 2 = 2 Fundamental Theorem 5 0 a 32b ` 64 = 2 = . Simplify. 5 5 a b Notice how the odd-powered term of the integrand is eliminated by symmetry. Integration of the even-powered term is simplified because the lower limit is zero. ➤ There are a couple of ways to see that b. The cos x term is an even function, so it can be integrated on the interval 0, p 2 . sin3 x is an odd function. Its graph is What about sin3 x? It is an odd function raised to an odd power, which results in an symmetric about the origin, indicating odd function; its integral on -p 2, p 2 is zero. Therefore, 3 > 4 that sin3 -x = -sin3 x. Or by analogy, take an odd power of x and raise it to an p 2 3 > > 4p 2 1 2 5 3 15 3 odd power. For example, x = x , > cos x - 4 sin x dx = 2 > cos x dx - 0 Symmetry which is odd. See Exercises 53–56 for L-p 2 L0 1 2 direct proofs of symmetry in composite > 1 2 p 2 functions. = 2 sin x > Fundamental Theorem 0 = 2 1 - `0 = 2. Simplify. Related Exercises 7–20 1 2 Average Value of a Function If five people weigh 155, 143, 180, 105, and 123 lb, their average (mean) weight is 155 + 143 + 180 + 105 + 123 = 141.2 lb. 5 This idea generalizes quite naturally to functions. Consider a function f that is continu- b - a ous on a, b . Using a regular partition x a, x , x , , x b with ∆x , we 0 = 1 2 c n = = n * * select a3 point4 x k in each subinterval and compute f x k , for k = 1, c, n. The values of * f x k may be viewed as a sampling of f on a, b . The average of these function values is 1 2 * * * f x 1 + f x 2 + + f x n 1 2 3 g4 . n 1 2 1 2 1 2 b - a Noting that n = , we write the average of the n sample values as the Riemann sum ∆x * * * n f x 1 + f x 2 + g + f x n 1 * = f x k ∆x. b a ∆x b a a 1 2 1 -2 1 2 - k = 1 1 2 Now suppose we increase 1n, taking2> more and more samples of f, while ∆x decreases to zero. The limit of this sum is a definite integral that gives the average value f on a, b : n 1 * 3 4 f = lim a f x k ∆x b - a nS∞ k = 1 1 2 1 b = f x dx. b - a La Copyright Pearson. All rights reserved. 1 2

M05_BRIG7345_02_SE_C05.4.indd 378 21/10/13 11:13 AM Not for Sale 5.4 Working with Integrals 379 This definition of the average value of a function is analogous to the definition of the average of a finite set of numbers.

Definition Average Value of a Function The average value of an integrable function f on the interval a, b is

1 b 3 4 f = f x dx. b - a La 1 2

y The average value of a function f on an interval a, b has a clear geometrical interpretation. Multiplying both sides of the definition of y ϭ f (x) average value by b - a , we have 3 4 Height of b rectangle ϭ f Net area of rectangle 1 2 Խ Խ b b a f f x dx. = - ϭ f ؒ (b Ϫ a) ϭ f (x) dx ͵ La t a s 1 2 1 2 net area of net area of region O a b x rectangle bounded by curve

Figure 5.50 We see that ͉ f ͉ is the height of a rectangle with base a, b , and that rectangle has the same net area as the region bounded by the graph of f on the interval a, b (Figure 5.50). Note that f may be zero or negative. 3 4 3 4 Quick C heck 2 What is the average value of a constant function on an interval? What is the average value of an odd function on an interval -a, a ?

EXAMPLE 2 Average elevation A hiking trail has3 an elevation4 given by 3 2 f x = 60x - 650x + 1200x + 4500, where f is measured in feet1 2 above sea level and x represents horizontal distance along the trail in miles, with 0 … x … 5. What is the average elevation of the trail?

y Elevation of a hiking trail So lution The trail ranges between elevations of about 2000 and over 5 horizontal miles 5000 ft (Figure 5.51). If we let the endpoints of the trail correspond to 6000 the horizontal distances a = 0 and b = 5, the average elevation of the 5000 trail in feet is Average elevation 4000 5 1 3 2 3000 f = 60x - 650x + 1200x + 4500 dx 5 L0 2000 El ev ation (ft) 1 4 3 2 2 5 1000 1 x x x = 60 - 650 + 1200 + 4500x Fundamental Theorem 5 4 3 2 0 0 x 1 2 3 4 5 a b ` Horizontal distance (mi) 1 = 3958 3. Simplify. Figure 5.51 The average elevation of the trail is slightly less than 3960 ft. Related Exercises 21–34

Mean Value Theorem for Integrals ➤ Compare this statement to that of the Mean Value Theorem for Derivatives: The average value of a function brings us close to an important theoretical result. The There is at least one point c in a, b such Mean Value Theorem for Integrals says that if f is continuous on a, b , then there is

that f ′ c equals the average slope of f. at least one point c in the interval a, b such that f c equals the average value of f on 1 2 3 4 1 2 1 2 1 2

M05_BRIG7345_02_SE_C05.4.indd 379 21/10/13 11:13 AM 380 Chapter 5 • Integration Not for Sale a, b . In other words, the horizontal line y = f intersects the graph of f for some point c in a, b (Figure 5.52). If f were not continuous, such a point might not exist. 1 2 1 2 y y ϭ f (x)

f (c) ϭ f Height of rectangle ϭ f

O a c b x

Figure 5.52

➤ Theorem 5.5 guarantees a point c in the open interval a, b at which f equals theorem 5.5 Mean Value Theorem for Integrals its average value. However, f may also Let f be continuous on the interval a, b . There exists a point c in a, b such that 1 2 equal its average value at an endpoint of that interval. 3 4 1 b 1 2 f c = f = f t dt. b - a La 1 2 1 2

x Proof: We begin by letting F x = 1a f t dt and noting that F is continuous on a, b and differentiable on a, b (by Theorem 5.3, Part 1). We now apply the Mean Value Theorem for derivatives (Theorem1 2 4.9) to1 2F and conclude that there exists at least3 one4 point c in a, b such that1 2

1 2 F b - F a F ′ c = . s b a 1 2 - 1 2 f 1c 2 1 2 By Theorem 5.3, Part 1, we know that F ′ c = f c , and by Theorem 5.3, Part 2, we know that 1 2 1 2 b

F b - F a = f t dt. La 1 2 1 2 1 2 Combining these observations, we have

➤ A more general form of the Mean Value 1 b Theorem states that if f and g are continuous f c = f t dt, b a on a, b with g x Ú 0 on a, b , then - La there exists a number c in a, b such that 1 2 1 2 3 4 1 2 3 4 where c is a point in a, b . b 1 2b

f x g x dx = f c g x dx. 1 2 Quick C heck 3 Explain why f x 0 for at least one point of a, b if f is continuous La La = 1 2 1 2 1 2 1 2 b and f x dx 0. 1a = 1 2 1 2 EXAMPL1E 23 Average value equals function value Find the point(s) on the interval

0, 1 at which f x = 2x 1 - x equals its average value on 0, 1 . So1 lution2 The average1 2 value1 of 2f on 0, 1 is 3 4 1 1 3 4 2 1 1 f = 2x 1 - x dx = x 2 - x 3 = . 1 - 0 L0 3 0 3 1 2 a b `

M05_BRIG7345_02_SE_C05.4.indd 380 21/10/13 11:13 AM Not for Sale 5.4 Working with Integrals 381 y 1 We must find the points on 0, 1 at which f x = 3 (Figure 5.53). Using the quadratic 1 formula, the two solutions of f x = 2x 1 - x = 3 are 0.6 y ϭ 2x(1 Ϫ x) 1 2 1 2 0.4 y ϭ a 1 - 1 3 1 2 1 21 + 1 3 ≈ 0.211 and ≈ 0.789. 21 21 0.2 > > 1 These two points are located symmetrically on either side of x = 2. The two solutions, 0.4 0.8 1.2 x 0.211 and 0.789, are the same for f x = ax 1 - x for any nonzero value of a (Exercise 57). 1 2 1 2 Related Exercises 35–40

Figure 5.53

Section 5.4 Exerci ses Review Questions p 2p a 19. cos x dx 20. sin x dx 1. If f is an odd function, why is f x dx 0? 1-a = L0 L0 a a 2. If f is an even function, why is f x dx 2 f x dx? 1-a 1 2 = 10 21–30. Average values Find the average value of the following func- 12 2 tions on the given interval. Draw a graph of the function and indicate 3. Is x an even or odd function? Is sin 1x 2 an even or odd1 function?2 the average value.

4. Explain how to find the average value of a function on an interval 3 2 a, b and why this definition is analogous to the definition of the 21. f x = x on -1, 1 22. f x = x + 1 on -2, 2 average of a set of numbers. 1 2 1 3 4 1 2 3 p p 4 3 4 23. f x = 2 on -1, 1 24. f x = cos 2x on - 4 , 4 5. Explain the statement that a continuous function on an interval x + 1 a, b equals its average value at some point on a, b . 1 2 3 4 1 2 2x 3 4 25. f x = 1 x on 1, e 26. f x = e on 0, ln 2 6. Sketch3 4 the function y = x on the interval 0, 2 1and let2 R be the p p 27. f 1x2 = cos> x on3 - 24, 2 28. f 1x2 = x 1 - 3x on 40, 1 region bounded by y = x and the x-axis on 0, 2 . Now sketch a 3 4 n rectangle in the first quadrant whose base is 0, 2 and whose area 29. f 1x2 = x on 0,3 1 , for4 any positive1 integer2 1n 2 3 4 3 4 equals the area of R. 1 n 3 4 30. f 1x2 = x on3 0, 41 , for any positive integer n > Basic Skills 31. Average1 2 distance3 on4 a parabola What is the average distance 7–16. Symmetry in integrals Use symmetry to evaluate the following between the parabola y = 30x 20 - x and the x-axis on the integrals. interval 0, 20 ? 1 2 2 200 T 32. Average3 elevation4 The elevation of a path is given by 9 5 3 2 7. x dx 8. 2x dx f x = x - 5x + 30, where x measures horizontal distances. L-2 L-200 Draw a graph of the elevation function and find its average value, for1 20 … x … 4. 2 p 4 8 9. 3x - 2 dx 10. > cos x dx 33. Average height of an arch The height of an arch above the L 2 L p 4 - - ground is given by the function y = 10 sin x, for 0 … x … p. 1 2 > What is the average height of the arch above the ground? 2 11. x 9 - 3x 5 + 2x 2 - 10 dx 34. Average height of a wave The surface of a water wave is L 2 - described by y = 5 1 + cos x , for -p … x … p, where y = 0 1 2 p 2 10 corresponds to a trough of the wave (see figure). Find the average x 1 2 12. 5 sin x dx 13. dx height of the wave above the trough on -p, p . > 2 L-p 2 L-10 200 - x y 3 4 > 2 p 2 5 10 14. > cos 2x + cos x sin x - 3 sin x dx y ϭ 5(1 ϩ cos x) L-p 2 8 > 1 2 p 4 1 6 5 15. > sin x dx 16. 1 - ͉ x ͉ dx 4 L-p 4 L-1 2 > 1 2 17–20. Symmetry and definite integrals Use symmetry to evaluate 0 x Ϫ␲ Ϫq q ␲ the following integrals. Draw a figure to interpret your result.

p 2p 17. sin x dx 18. cos x dx L-p L0

M05_BRIG7345_02_SE_C05.4.indd 381 21/10/13 11:13 AM 382 Chapter 5 • Integration Not for Sale T 35–40. Mean Value Theorem for Integrals Find or approximate all x 2 y ϭ 630 1 Ϫ points at which the given function equals its average value on the given y (( 315 (( interval.

x 600 35. f x = 8 - 2x on 0, 4 36. f x = e on 0, 2 2 2 500 37. f 1x2 = 1 - x a 3on 0,4 a , where 1a is2 a positive3 real4 number 1 2 p > 3 4 400 38. f x = sin x on 0, p 39. f x = 1 - ͉ x ͉ on -1, 1 4 300 1 2 3 4 1 2 3 4 40. f x = 1 x on 1, 4 630 ft 200 Further1 2 Explorations> 3 4 630 ft 41. Explain why or why not Determine whether the following 100 statements are true and give an explanation or counterexample. 4 Ϫ300 Ϫ200 Ϫ100 0 100 200 300 x a. If f is symmetric about the line x = 2, then 10 f x dx = 2 2 f x dx. 10 1 2 b. If f has the property f a + x = -f a - x , for all x, where 48. Another Gateway Arch Another description of the Gateway 1 2 a + 2 a is a constant, then f x dx = 0. Arch is 1a1- 2 2 1 2 c. The average value of a linear function on an interval a, b is 0.00418x -0.00418x the function value at the midpoint1 2 of a, b . y = 1260 - 315 e + e , 3 4 d. Consider the function f x = x a - x on the interval 0, a , where the base of the arch is1 -315, 315 and x and2 y are 3 1 4 for a 7 0. Its average value on 0, a is 2 of its maximum measured in feet. Find the average height of the arch above the value. 1 2 1 2 3 4 ground. 3 4 3 4 42–45. Symmetry in integrals Use symmetry to evaluate the following 49. Planetary orbits The planets orbit the Sun in elliptical orbits with integrals. the Sun at one focus (see Section 10.4 for more on ellipses). The equation of an ellipse whose dimensions are 2a in the x-direction p 4 p 4 x 2 y2 42. tan x dx 43. sec2 x dx > > and 2b in the y-direction is 2 + 2 = 1. L-p 4 L-p 4 a b > > 2 2 2 a. Let d denote the square of the distance from a planet to x 3 - 4x 44. 1 ͉ x ͉ 3 dx 45. dx the center of the ellipse at 0, 0 . Integrate over the interval - 2 2 2 2 L-2 L-2 x + 1 -a, a to show that the average value of d is a + 2b 3. 1 2 b. Show that in the case of a circle1 2 a = b = R , the average Applications value3 in4 part (a) is R2. 1 2> 46. Root mean square The root mean square (or RMS) is another c. Assuming 0 6 b 6 a, the coordinates1 of the 2Sun are measure of average value, often used with oscillating functions a2 - b2, 0 . Let D2 denote the square of the distance from (for example, sine and cosine functions that describe the current, the2 planet to the Sun. Integrate over the interval -a, a to voltage, or power in an alternating circuit). The RMS of a function show1 that the average2 value of D2 is 4a2 - b2 3. f on the interval 0, T is 3 4 1 2> 3 4 T 1 2 f f t dt . RMS = y T L0 a C 1 2 Earth Compute the RMS of f t = A sin vt , where A and v are positive constants and T is any integer multiple of the period of f, b 1 2 1 2 which is 2p v. Sun x 47. Gateway Arch> The Gateway Arch in St. Louis is 630 ft high and has a 630-ft base. Its shape can be modeled by the parabola

x 2 y = 630 1 - . 315 a a b b Find the average height of the arch above the ground.

M05_BRIG7345_02_SE_C05.4.indd 382 21/10/13 11:13 AM Not for Sale 5.4 Working with Integrals 383 Additional Exercises 60. Unit area sine curve Find the value of c such that the region bounded 50. Comparing a sine and a quadratic function Consider the func- by y = c sin x and the x-axis on the interval 0, p has area 1. 4 61. Unit area cubic Find the value of c 7 0 such3 that4 the region tions f x = sin x and g x = 2 x p - x . p bounded by the cubic y = x x - c 2 and the x-axis on the a. Carefully1 2 graph f and1 g2 on the same1 set of2 axes. Verify that interval 0, c has area 1. 1 2 both functions have a single local maximum on the interval 62. Unit area3 4 0, p and that they have the same maximum value on 0, p . a. Consider the curve y = 1 x, for x Ú 1. For what value of b. On the interval 0, p , which is true: f x Ú g x , 3 4 3 4 b 7 0 does the region bounded by this curve and the x-axis on g x Ú f x , or neither? > 3 4 1 2 1 2 the interval 1, b have an area of 1? c. Compute and compare the average values of f and g on 0, p . p 1 2 1 2 b. Consider the curve y = 1 x , where x Ú 1, and p 6 2 with 3 4 51. Using symmetry Suppose f is an even function and 3 4 p 1. For what value of b (as a function of p) does the 8 region bounded by this curve> and the x-axis on the interval

f x dx = 18. 1, b have unit area? L-8 c. Is b p in part (b) an increasing or decreasing function of p? 1 2 Explain.3 4 8 8 1 2 a. Evaluate f x dx b. Evaluate x f x dx 63. A sine integral by Riemann sums Consider the integral L0 L-8 p 2 I sin x dx. 1 2 14 2 = 10 > a. Write the left Riemann sum for I with n subintervals. 52. Using symmetry Suppose f is an odd function, f x dx = 3, L0 8 cos u + sin u - 1 1 2 b. Show that lim u = 1. and f x dx = 9. uS0 2 1 - cos u L0 a b 1 2 1 2 p p 8 4 cos + sin - 1 n - 1 pk 2n 2n a. Evaluate f x dx b. Evaluate f x dx c. It is a fact that a sin = . L-4 L-8 k = 0 2n a b ap b 2 1 - cos 1 2 1 2 a b 2n 53–56. Symmetry of composite functions Prove that the integrand is either even or odd. Then give the value of the integral or show how it Use this fact and part (b) to evaluate I bya taking thea limitb b of the can be simplified. Assume that f and g are even functions and p and q Riemann sum as n S ∞. are odd functions. 64. Alternate definitions of means Consider the function a a b x t + 1 dx 53. f g x dx 54. f p x dx 1a L a L a f t = b . - - x t dx 1 1 22 1 1 22 1a a a 1 2 55. p g x dx 56. p q x dx Show that the following means can be defined in terms of f. L a L a - - a + b 1 1 22 1 1 22 a. Arithmetic mean: f 0 = 57. Average value with a parameter Consider the function 2

f x = ax 1 - x on the interval 0, 1 , where a is a positive 1 2 3 b. Geometric mean: f - = ab real number. 2 1 2 1 2 3 4 1 a. Find the average value of f as a function of a. a b 2ab c. Harmonic mean: f -3 = b. Find the points at which the value of f equals its average value a + b and prove that they are independent of a. 1 2 b - a d. Logarithmic mean: f -1 = ln b ln a 58. Square of the average For what polynomials f is it true that the - square of the average value of f equals the average value of the (Source: Mathematics1 Magazine2 78, 5, Dec 2005) square of f over all intervals a, b ? 65. Symmetry of powers Fill in the following table with either even 59. Problems of antiquity Several calculus problems were solved by or odd, and prove each result. Assume n is a nonnegative integer 3 4 n Greek mathematicians long before the discovery of calculus. The and f means the nth power of f. following problems were solved by Archimedes using methods f is even f is odd that predated calculus by 2000 years. n n 4 n is even f is _____ f is _____ a. Show that the area of a segment of a parabola is 3 that of its inscribed triangle of greatest area. In other words, the area n is odd f n is _____ f n is _____ 2 2 4 bounded by the parabola y = a - x and the x-axis is 3 the 2 area of the triangle with vertices {a, 0 and 0, a . Assume that a 7 0 but is unspecified. b. Show that the area bounded by the1 parabola2 y 1= a2 2- x 2 and 2 the x-axis is 3 the area of the rectangle with vertices {a, 0 2 and {a, a . Assume that a 7 0 but is unspecified. 1 2 1 2

M05_BRIG7345_02_SE_C05.4.indd 383 21/10/13 11:13 AM 384 Chapter 5 • Integration Not for Sale

66. Average value of the derivative Suppose that f ′ is a continuous 69. Generalizing the Mean Value Theorem for Integrals Suppose function for all real numbers. Show that the average value of the f and g are continuous on a, b and let

f b - f a derivative on an interval a, b is f ′ = . Interpret 3 x 4 b b - a 1 2 1 2 h x = x - b f t dt + x - a g t dt. this result in terms of secant3 lines.4 La Lx 1 2 1 2 1 2 1 2 1 2 67. Symmetry about a point A function f is symmetric about a a. Use Rolle’s theorem to show that there is a number c in (a, b) point c, d if whenever c - x, d - y is on the graph, then so is such that c + x, d + y . Functions that are symmetric about a point c, d c b are easily1 integrated2 on an1 interval with2 midpoint c. 1 2 1 2 f t dt + g t dt = f c b - c + g c c - a , a. Show that if f is symmetric about c, d and a 7 0, then La Lc c + a 1 2 1 2 1 21 2 1 21 2 1c a f x dx = 2a f c = 2ad. which is a generalization of the Mean Value Theorem for - 2 1 2 b. Graph the function f x = sin x on the interval 0, p 2 and Integrals. 1 2 1 2 p 1 show that the function is symmetric about the point 4 , 2 . b. Show that there a number c in (a, b) such that c. Using only the graph1 of2 f (and no integration), show3 that> 4 x

f t dt f c b c . p 2 p 1 2 = - 2 La 10 sin x dx = . (See the Guided Project Symmetry in > 4 1 2 1 21 2 Integrals.) c. Use a sketch to interpret part (b) geometrically. d. Use the result of part (a) to give an alternate proof of the Mean 68. Bounds on an integral Suppose f is continuous on a, b with Value Theorem for Integrals. f ″ x 7 0 on the interval. It can be shown that (Source: The College Mathematics Journal, 33, 5, Nov 2002) 3 4 b 1 2 a + b f a + f b b - a f … f x dx … b - a . 2 2 Quick C heck Answer s La 1 2 1 2 1 2 a b 1 2 1 2 1. f -x g -x = f x g x ; therefore, fg is even. a. Assuming f is nonnegative on a, b , draw a figure to illus- 2. The average value is the constant; the average value is 0. trate the geometric meaning of these inequalities. Discuss your 3. The1 average2 1 2 value1 is2 zero1 2 on the interval; by the Mean conclusions. 3 4 Value Theorem for Integrals, f x = 0 at some point on the b. Divide these inequalities by b - a and interpret the resulting inequalities in terms of the average value of f on a, b . interval. 1 2 1 2 3 4

5.5 Substitution Rule

Given just about any differentiable function, with enough know-how and persistence, you can compute its derivative. But the same cannot be said of antiderivatives. Many functions, even relatively simple ones, do not have antiderivatives that can be expressed in terms of familiar functions. Examples are sin x2, sin x x, and xx. The immediate goal of this section is to enlarge the family of functions for which we can find antiderivatives. This campaign resumes in Chapter 7, where additional1 integration2> methods are developed.

Indefinite Integrals One way to find new antiderivative rules is to start with familiar derivative rules and work backward. When applied to the Chain Rule, this strategy leads to the Substitution Rule. A few examples illustrate the technique.

EXAMPLE 1 Antiderivatives by trial and error Find 1cos 2x dx. So lution The closest familiar indefinite integral related to this problem is ➤ We assume C is an arbitrary constant without stating so each time it appears. cos x dx = sin x + C, L which is true because d sin x + C = cos x. dx 1 2

M05_BRIG7345_02_SE_C05.4.indd 384 21/10/13 11:13 AM Not for Sale 5.5 Substitution Rule 385 Therefore, we might incorrectly conclude that the indefinite integral of cos 2x is sin 2x + C. However, by the Chain Rule, d sin 2x + C = 2 cos 2x cos 2x. dx Note that sin 2x fails to be an1 antiderivative2 of cos 2x by a multiplicative factor of 2. 1 A small adjustment corrects this problem. Let’s try 2 sin 2x: d 1 1 sin 2x = # 2 cos 2x = cos 2x. dx 2 2 It works! So we have a b 1 cos 2x dx = sin 2x + C. L 2 Related Exercises 9–12

The trial-and-error approach of Example 1 is impractical for complicated integrals.

To develop a systematic method, consider a composite function F g x , where F is an

antiderivative of f ; that is, F ′ = f. Using the Chain Rule to differentiate the composite 1 1 22 function F g x , we find that

1 1 22 d F g x = F ′ g x g′ x = f g x g′ x . dx t

f g x 1 1 1 222 1 1 22 1 2 1 1 22 1 2 1 1 22 This equation says that F g x is an antiderivative of f g x g′ x , which is written

1 1 22 1 1 22 1 2 f g x g′ x dx = F g x + C, (1) L 1 1 22 1 2 1 1 22 where F is any antiderivative of f. ➤ You can call the new variable anything Why is this approach called the Substitution Rule (or Change of Variables Rule)?

you want because it is just another In the composite function f g x in equation (1), we identify the “inner function” as variable of integration. Typically, u is the u = g x , which implies that du = g′ x dx. Making this identification, the integral in standard choice for the new variable. equation (1) is written 1 1 22 1 2 1 2

f g x g′ x dx = f u du = F u + C. L s s L 1f 1u 22 1du2 1 2 1 2 1 2 We see that the integral 1f g x g′ x dx with respect to x is replaced with a new integral 1f u du with respect to the new variable u. In other words, we have substituted the new variable u for the old variable1 1 22x. Of1 2course, if the new integral with respect to u is no easier1 2 to find than the original integral, then the change of variables has not helped. The Substitution Rule requires plenty of practice until certain patterns become familiar.

theorem 5.6 Substitution Rule for Indefinite Integrals Let u = g x , where g′ is continuous on an interval, and let f be continuous on the corresponding range of g. On that interval, 1 2

f g x g′ x dx = f u du. L L 1 1 22 1 2 1 2

In practice, Theorem 5.6 is applied using the following procedure.

M05_BRIG7345_02_SE_C05.5.indd 385 21/10/13 11:13 AM 386 Chapter 5 • Integration Not for Sale

Procedure Substitution Rule (Change of Variables)

1. Given an indefinite integral involving a composite function f g x , identify an inner function u = g x such that a constant multiple of g′ x appears in the integrand. 1 1 22 1 2 1 2 2. Substitute u = g x and du = g′ x dx in the integral. 3. Evaluate the new indefinite integral with respect to u. 1 2 1 2 4. Write the result in terms of x using u = g x . Disclaimer: Not all integrals yield to the Substitution1 2 Rule.

EXAMPLE 2 Perfect substitutions Use the Substitution Rule to find the following indefinite integrals. Check your work by differentiating.

a. 2 2x + 1 3 dx b. 10e10x dx L L 1 2 So lution a. We identify u = 2x + 1 as the inner function of the composite function 2x + 1 3. du Therefore, we choose the new variable u = 2x + 1, which implies that = 2, or dx1 2 du = 2 dx. Notice that du = 2 dx appears as a factor in the integrand. The change of variables looks like this:

3 3 2x + 1 # 2 dx = u du Substitute u = 2x + 1, du = 2 dx. r L t L 1 u3 2 du ➤ Use the Chain Rule to check that u4 2x 1 4 d + 3 = + C Antiderivative + C = 2 2x + 1 . dx 4 4 1 2 a b 1 2 4 2x + 1 = + C. Replace u with 2x + 1. 4 1 2 Notice that the final step uses u = 2x + 1 to return to the original variable. b. The composite function e10x has the inner function u = 10x, which implies that du = 10 dx. The change of variables appears as

e10x 10 dx eu du Substitute u 10x, du 10 dx. r r = = = L L eu du

= eu + C Antiderivative = e10x + C. Replace u with 10x. d In checking, we see that e10x + C = e10x # 10 = 10e10x. dx 1 2 Related Exercises 13–16

3 4 10 10 Quick C heck 1 Find a new variable u so that 14x x + 5 dx = 1u du.

Most substitutions are not perfect. The remaining1 examples2 show more typical situa- tions that require introducing a constant factor.

EXAMPLE 3 Introducing a constant Find the following indefinite integrals. 4 5 9 3 a. 1x x + 6 dx b. 1cos x sin x dx 1 2

M05_BRIG7345_02_SE_C05.5.indd 386 21/10/13 11:13 AM Not for Sale 5.5 Substitution Rule 387 So lution a. The inner function of the composite function x 5 + 6 9 is x 5 + 6 and its derivative 5x 4 also appears in the integrand (up to a multiplicative factor). Therefore, we use the 1 2 1 substitution u = x 5 + 6, which implies that du = 5x 4 dx, or x 4 dx = du. By the Substitution Rule, 5

1 Substitute u = x 5 + 6, x 5 + 6 9 x 4 dx = u9 # du r t 5 4 4 1 L L du = 5x dx 1 x dx = du. 1 5 1 u9 2 du 5

1 9 = u du c f x dx = c f x dx 5 L L L 1 2 1 2 1 u10 = # + C Antiderivative 5 10

1 5 10 = x + 6 + C. Replace u with x 5 + 6. 50 b. The integrand can be written as 1cos x 3 2sin x. The inner function in the composition cos x 3 is cos x, which suggests the substitution u = cos x. Note that du = -sin x dx or sin x dx = -du. The change of1 variables2 appears as 1 2 cos3 x sin x dx u3 du Substitute u cos x, du sin x dx. s s = - = = - L L u3 -du u4 = - + C Antiderivative 4

cos4 x = - + C. Replace u with cos x. 4 Related Exercises 17–32

Quick C heck 2 In Example 3a, explain why the same substitution would not work as well 3 5 9 for the integral 1x x + 6 dx.

Sometimes the1 choice2 for a u-substitution is not so obvious or more than one u-substitution works. The following example illustrates both of these points.

x EXAMPLE 4 Variations on the substitution method Find dx. L x + 1 1 So lution Substitution 1 The composite function x + 1 suggests the new variable u = x + 1. You might doubt whether this choice will 1work because du = dx, which leaves the x in the numerator of the integrand unaccounted for. But let’s proceed. Letting u = x + 1, we have x = u - 1, du = dx, and x u - 1 dx = du Substitute u = x + 1, du = dx. L x + 1 L u 1 1 1 = u - du Rewrite integrand. L u 1 a 1 b = u1 2 - u-1 2 du. Fractional powers L > > 1 2

M05_BRIG7345_02_SE_C05.5.indd 387 21/10/13 11:13 AM 388 Chapter 5 • Integration Not for Sale We integrate each term individually and then return to the original variable x: 2 u1 2 - u-1 2 du = u3 2 - 2u1 2 + C Antiderivatives L > > 3 > > 1 2 2 3 2 1 2 = x + 1 - 2 x + 1 + C Replace u with x + 1. 3 > > 1 2 2 1 21 2 1 2 Factor out x + 1 = x + 1 x - 2 + C. 3 and simplify. > > 1 2 1 2 1 2 2 ➤ In Substitution 2, you could also use the Substitution 2 Another possible substitution is u = x + 1. Now u = x + 1, 2 fact that x = u - 1, and dx = 2u du. Making these substitutions1 leads to 1 x u2 - 1 u′ x = , dx = 2u du Substitute u = x + 1, x = u2 - 1. 2 x + 1 L x + 1 L u 1 2 1 1 which implies 1 = 2 u2 - 1 du Simplify the integrand. 1 L du = dx. 2 x + 1 u13 2 1 = 2 - u + C Antiderivatives 3 a b 2 3 2 1 2 = x + 1 - 2 x + 1 + C Replace u with x + 1. 3 > > 1 2 1 21 2 1 2 = x + 1 x - 2 + C. Factor out x + 1 1 2 and simplify. 3 > > 1 2 Observe that the same1 indefinite2 1 integral2 is found using either substitution. Related Exercises 33–38 Definite Integrals The Substitution Rule is also used for definite integrals; in fact, there are two ways to proceed. • You may use the Substitution Rule to find an antiderivative F and then use the

Fundamental Theorem to evaluate F b - F a . • Alternatively, once you have changed variables from x to u, you also may change the 1 2 1 2 limits of integration and complete the integration with respect to u. Specifically, if u = g x , the lower limit x = a is replaced with u = g a and the upper limit x = b is replaced with u = g b . 1 2 1 2 The second option tends1 2to be more efficient, and we use it whenever possible. This approach is summarized in the following theorem, which is then applied to several definite integrals.

theorem 5.7 Substitution Rule for Definite Integrals Let u = g x , where g′ is continuous on a, b , and let f be continuous on the range of g. Then 1 2 3 4 b g b

f g x g′ x dx = 1 2f u du. La Lg a

1 1 22 1 2 1 2 1 2

EXAMPLE 5 Definite integrals Evaluate the following integrals.

2 4 p 2 dx x 4 a. 3 b. 2 dx c. > sin x cos x dx L0 x + 3 L0 x + 1 L0 1 2

M05_BRIG7345_02_SE_C05.5.indd 388 21/10/13 11:13 AM Not for Sale 5.5 Substitution Rule 389 So lution ➤ When the integrand has the form a. Let the new variable be u = x + 3 and then du = dx. Because we have changed the

f ax + b , the substitution u = ax + b variable of integration from x to u, the limits of integration must also be expressed in is often effective. 1 2 terms of u. In this case, x = 0 implies u = 0 + 3 = 3, Lower limit x = 2 implies u = 2 + 3 = 5. Upper limit The entire integration is carried out as follows:

2 dx 5 -3 3 = u du Substitute u = x + 3, du = dx. L0 x + 3 L3 1 2 u-2 5 = - Fundamental Theorem 2 3 1 ` 8 = - 5-2 - 3-2 = . Simplify. 2 225 b. Notice that a multiple of the1 derivative 2of the denominator appears in the numerator; 2 1 therefore, we let u = x + 1, which implies that du = 2x dx, or x dx = 2 du. Changing limits of integration, x = 0 implies u = 0 + 1 = 1, Lower limit x = 4 implies u = 42 + 1 = 17. Upper limit Changing variables, we have

4 x 1 17 -1 2 2 dx = u du Substitute u = x + 1, du = 2x dx. L0 x + 1 2 L1 1 17 = ln ͉ u ͉ Fundamental Theorem 2 1 1 ` = ln 17 - ln 1 Simplify. 2 1 1 2 = ln 17 ≈ 1.417. ln 1 = 0 2 c. Let u = sin x, which implies that du = cos x dx. The lower limit of integration becomes u = 0 and the upper limit becomes u = 1. Changing variables, we have

p 2 1 4 4 > sin x cos x dx = u du u = sin x, du = cos x dx L0 L0 u5 1 1 = = . Fundamental Theorem 5 0 5 a b ` Related Exercises 39–52

The Substitution Rule enables us to find two standard integrals that appear frequently in 2 2 practice, 1sin x dx and 1cos x dx. These integrals are handled using the identities 1 - cos 2x 1 + cos 2x sin2 x = and cos2 x = . 2 2

M05_BRIG7345_02_SE_C05.5.indd 389 21/10/13 11:13 AM 390 Chapter 5 • Integration Not for Sale p 2 2 2 EXAMPLE 6 Integral of cos U Evaluate > cos u du. L0 So lution Working with the indefinite integral first, we use the identity for cos2 u: 1 + cos 2u 1 1 cos2 u du = du = du + cos 2u du. L L 2 2 L 2 L ➤ See Exercise 102 for a generalization The change of variables u = 2u (or Table 4.9) is now used for the second integral, and of Example 6. Trigonometric integrals we have involving powers of sin x and cos x are explored in greater detail in Section 7.3. 2 1 1 cos u du = du + cos 2u du L 2 L 2 L 1 1 1 = du + # cos u du u = 2u, du = 2 du 2 L 2 2 L u 1 = + sin 2u + C. Evaluate integrals; u = 2u. 2 4 Using the Fundamental Theorem of Calculus, the value of the definite integral is

p 2 p 2 2 u 1 > cos u du = + sin 2u > L0 2 4 0 a p 1 b ` 1 p = + sin p - 0 + sin 0 = . 4 4 4 4 a b a Relatedb Exercises 53–60

y y ϭ 2(2x ϩ 1) Geometry of Substitution 10 The Substitution Rule has a geometric interpretation. To keep 2 matters simple, consider the integral 0 2 2x + 1 dx. The graph 8 1 of the integrand y = 2 2x + 1 on the interval 0, 2 is shown Area of R in Figure 5.54a, along with the region R 1whose area2 is given by 6 2 y 1 2 3 4 ϭ 2(2x ϩ 1)dx the integral. The change of variables u = 2x + 1, du = 2 dx, ͵ 5 y ϭ u 0 u 0 = 1, and u 2 = 5 leads to the new integral 4 Area of RЈ 2 5 R 5 1 2 1 2 2 ϭ ͵ u du 2 2x + 1 dx = u du. RЈ 1 1 L0 L1 1 2 Figure 5.54b also shows the graph of the new integrand y u on 0 2 x 0 15u = the interval 1, 5 and the region R′ whose area is given by the (a) (b) new integral. You can check that the areas of R and R′ are equal. Area of R ϭ Area of RЈ An analogous3 interpretation4 may be given to more complicated Figure 5.54 integrands and substitutions.

Quick C heck 3 Changes of variables occur frequently in mathematics. For example, suppose you want to solve the equation x 4 - 13x 2 + 36 = 0. If you use the substitution u = x 2, what is the new equation that must be solved for u? What are the roots of the original equation?

Section 5.5 Exerci ses

Review Questions 5. When using a change of variables u = g x to evaluate the b 1. On which derivative rule is the Substitution Rule based? definite integral 1a f g x g′ x dx, how are the limits of integration transformed? 1 2 2. Why is the Substitution Rule referred to as a change of variables? 1 1 22 1 2 6. If the change of variables u = x 2 - 4 is used to evaluate the 4 3. The composite function f g x consists of an inner function g definite integral 12 f x dx, what are the new limits of integration?

and an outer function f . If an integrand includes f g x , which 2 function is often a likely choice1 1 22 for a new variable u? 7. Find 1cos x dx. 1 2 1 1 22 2 8. What identity is needed to find sin2 x dx? 4. Find a suitable substitution for evaluating 1tan x sec x dx and 1 explain your choice. Copyright Pearson. All rights reserved.

M05_BRIG7345_02_SE_C05.5.indd 390 21/10/13 11:13 AM Not for Sale 5.5 Substitution Rule 391 Basic Skills 39–52. Definite integrals Use a change of variables to evaluate the 9–12. Trial and error Find an antiderivative of the following functions following definite integrals. by trial and error. Check your answer by differentiating. 1 2 2x 39. 2x 4 x 2 dx 40. dx 12 3x + 1 - 2 2 9. f x = x + 1 10. f x = e L0 L0 x + 1 p 2 1 2 p 4 11. f 1x2 = 1 2x +21 12. f 1x2 = cos 2x + 5 2 1 sin x 2 41. > sin u cos u du 42. > 2 dx 1 L0 L0 cos x 13–16. 1Substitution2 given Use the given substitution1 2 1 to find the2 fol- 2 4 lowing indefinite integrals. Check your answer by differentiating. 3 p 43. x 2ex + 1 dx 44. dp 2 L 1 L0 9 p 13. 2x x 2 + 1 4 dx, u = x 2 + 1 - + L p 2 cos x p24 sin u 1 2 45. > 2 dx 46. > 3 du Lp 4 sin x L0 cos u 14. 8x cos 4x 2 + 3 dx, u = 4x 2 + 3 L 2 5 3 2 > dx v + 1 47. 48. dv 1 2 > 2 3 L2 5 3 L0 15. sin3 x cos x dx, u sin x x 25x - 1 v + 3v + 4 = 1 L >14 x2 2 1 24 x 49. dx 50. dx 2 > 2 2 2 L0 x + 1 L0 1 16x 16. 6x + 1 3x + x dx, u = 3x + x - L 1 3 4 ln 4 2ex 2 2 1 2 51. > 2 dx 52. x dx 17–32. Indefinite integrals Use a change of variables to find the fol- L1 3 9x + 1 L0 3 + 2e lowing indefinite integrals. Check your work by differentiating. 53–60. Integrals> with sin2 x and cos2 x Evaluate the following

2 integrals. 17. 2x x 2 - 1 99 dx 18. xex dx L L p 2 2 1 2 4 53. cos x dx 54. sin x dx 2x 2 x + 1 19. dx 20. dx L-p L L 1 4x 3 L 12 x p 4 - 1 2 2 p 2 2 1 55. sin u + du 56. > cos 8u du 1 L 6 L0 21. x 2 + x 10 2x + 1 dx 22. dx L L 10x - 3 p 4 a b 2 2 2 1 2 1 2 57. > sin 2u du 58. x cos x dx 3 4 6 10 L p 4 L 23. x x + 16 dx 24. sin u cos u du - p 6 1 2 L L > sin 2 y 1 2 59. dy (Hint: sin 2y = 2 sin y cos y.) dx > sin2 y 2 25. 26. x 9 sin x 10 dx L0 + L 1 - 9x 2 L p 2 4 2 60. > sin u du 6 2 4 5 L0 27. x - 3x x - x dx L 1 2 1 2 Further Explorations x 28. dx (Hint: Let u = x - 2.) 61. Explain why or why not Determine whether the following x 2 L - statements are true and give an explanation or counterexample.

dx 3 Assume that f, f ′, and f ″ are continuous functions for all real 29. 30. dy numbers. L 1 + 4x 2 L 1 + 25y2

2 1 8x 6 1 2 + 31. dx, x 32. dx a. f x f ′ x dx = f x + C. 2 7 2 L 2 L x 4x - 1 2 L 2x + 3x 1 2 1 2 1 1 122 n n + 1 33–38. Variations2 on the substitution method Find the following b. f x f ′ x dx = f x + C, n -1. n 1 integrals. L + 1 1 22 1 2 1 1 22 y2 c. sin 2x dx = 2 sin x dx. x L L 33. dx 34. 4 dy L x - 4 L y + 1 x 2 + 1 10 d. x 2 + 1 9dx = + C. 1 x -x 10 x 1e - e 2 L 1 2 35. dx 36. x x dx b L 3 x 4 L e + e- 1 2 + e. f ′ x f″ x dx = f ′ b - f ′ a . 2 La 3 37. x 2x + 1 dx 38. z + 1 3z + 2 dz 1 2 1 2 1 2 1 2 L L 2 1 1 2

M05_BRIG7345_02_SE_C05.5.indd 391 21/10/13 11:14 AM 392 Chapter 5 • Integration Not for Sale 62–78. Additional integrals Use a change of variables to evaluate the Applications following integrals. T 86. Periodic motion An object moves along a line with a velocity in m s given by v t = 8 cos pt 6 . Its initial position is s 0 = 0. 62. sec 4w tan 4w dw 63. sec2 10x dx > a. Graph the velocity function. L L 1 2 1 > 2 1 2 b. As discussed in Chapter 6, the position of the object is given t 5 3 by s t v y dy, for t 0. Find the position function, 64. sin x + 3 sin x - sin x cos x dx = 10 Ú L for t Ú 0. 1 2 c. What1 2is the period1 2 of the motion—that is, starting at any csc2 x 65. dx 66. x 3 2 + 8 5 x dx point, how long does it take the object to return to that L cot3 x L > 1 position? 1 2x 2 8 e 87. Population models The population of a culture of bacteria has 67. sin x sec x dx 68. 2x dx L L e + 1 200 2 a growth rate given by p′ t = bacteria per hour, for 1 e t 1 r 2 ln p + 69. x 1 - x dx 70. dp t 0, where r 1 is a real number. In Chapter 6 it is shown L0 L1 p Ú 7 1 2 2 that the increase in the population1 over 2the time interval 0, t 3 6 5 t x dx is given by p s ds. (Note that the growth rate decreases in 71. dx 72. 10 ′ 3 2 > 2 3 4 L2 x - 1 L0 25x + 36 time, reflecting competition for space and food.) 1 2 2 2 1 a. Using the population model with r = 2, what is the increase 3 4 2 7 73. x 16 - x dx 74. x - 1 x - 2x dx in the population over the time interval 0 … t … 4? L0 L 1 2 - b. Using the population model with r = 3, what is the increase 0 1 21 2 sin x in the population over the time interval 0 … t … 6? 75. dx c. Let ∆P be the increase in the population over a fixed time L-p 2 + cos x 1 interval 0, T . For fixed T, does ∆P increase or decrease with v + 1 v + 2 76. dv the parameter r? Explain. 2v3 9v2 12v 36 d. A lab technician3 4 measures an increase in the population of L0 1+ 21+ +2 2 p 4 350 bacteria over the 10-hr period 0, 10 . Estimate the value 4 2 sin x of r that best fits this data point. 77. 2 dx 78. > e sin 2x dx L1 9x + 6x + 1 L0 e. Looking ahead: Use the population3 model4 in part (b) to find 79–82. Areas of regions Find the area of the following regions. the increase in population over the time interval 0, T , for 2 any T 0. If the culture is allowed to grow indefinitely 79. The region bounded by the graph of f x = x sin x and the 7 T , does the bacteria population increase without3 4 x-axis between x = 0 and x = p S ∞ bound? Or does it approach a finite limit? 1 1 2 80. The region bounded by the graph of f u = cos u sin u and the 1 2 88. Average distance on a triangle Consider the right triangle with u-axis between u = 0 and u = p 2 1 2 vertices 0, 0 , 0, b , and a, 0 ,where a 7 0 and b 7 0. Show 4 81. The region bounded by the graph of> f x = x - 4 and the that the average vertical distance from points on the x-axis to the x-axis between x = 2 and x = 6 hypotenuse1 is2 b12, for2 all a1 7 0.2 1 2 1 2 x T 89. Average value of> sine functions Use a graphing utility to verify 82. The region bounded by the graph of f x = and the 2 x - 9 that the functions f x = sin kx have a period of 2p k, where x-axis between x = 4 and x = 5 1 2 2 k = 1, 2, 3, c. Equivalently, the first “hump” of f x = sin kx occurs on the interval1 2 0, p k . Verify that the average> value of 83. Morphing parabolas The family of parabolas y 1 a x 2 a3, 1 2 = - the first hump of f x = sin kx is independent of k. What is the where a 7 0, has the property that for x Ú 0, the x-intercept is average value? 3 > 4 a, 0 and the y-intercept is 0, 1 a . Let A a be the1 area> 2 of the> 1 2 region in the first quadrant bounded by the parabola and the x-axis. Additional Exercises Find1 2A a and determine whether1 > it 2is an increasing,1 2 decreasing, or 90. Looking ahead: Integrals of tan x and cot x Use a change of constant function of a. 1 2 variables to verify each integral. 84. Substitutions Suppose that f is an even function with 8 a. tan x dx ln ͉ cos x ͉ C ln ͉ sec x ͉ C f x dx 9. Evaluate each integral. = - + = + 10 = L 1 2 1 2 2 2 3 b. cot x dx ln sin x C a. x f x dx b. x f x dx = ͉ ͉ + L-1 L-2 L 1 2 1 2 85. Substitutions Suppose that p is a nonzero real number and f is 91. Looking ahead: Integrals of sec x and csc x 1 an odd integrable function with 10 f x dx = p. Evaluate each a. Multiply the numerator and denominator of sec x by integral. sec x + tan x; then use a change of variables to show that 1 2 p 2 p

a. >1 2 cos px f sin px dx sec x dx = ln ͉ sec x + tan x ͉ + C. L0 L p 2 1 2 b. Use a change of variables to show that b. > cos x f sin x dx L-p 2 csc x dx = -ln ͉ csc x + cot x ͉ + C. > 1 2 L

M05_BRIG7345_02_SE_C05.5.indd 392 21/10/13 11:14 AM Not for Sale 5.5 Substitution Rule 393 92. Equal areas The area of the shaded region under the curve 102. sin2 ax and cos2 ax integrals Use the Substitution Rule to prove y = 2 sin 2x in (a) equals the area of the shaded region under the that curve y = sin x in (b). Explain why this is true without computing areas. x sin 2ax sin2 ax dx = - + C and 2 4a y y L 1 2 sin 2ax y ϭ 2 sin 2x 2 x cos ax dx = + + C. 2 2 2 4a L 1 2 103. Integral of sin2 x cos2 x Consider the integral y ϭ sin x 2 2 1 1 I = 1sin x cos x dx. a. Find I using the identity sin 2x = 2 sin x cos x. b. Find I using the identity cos2 x = 1 - sin2 x. 0 ␲ x 0 ␲ ␲ x c. Confirm that the results in parts (a) and (b) are consistent and 2 2 compare the work involved in each method. (a) (b) 104. Substitution: shift Perhaps the simplest change of variables is 93. Equal areas The area of the shaded region under the curve the shift or translation given by u = x + c, where c is a real x - 1 2 number. y = on the interval 4, 9 in (a) equals the area of 12 x a. Prove that shifting a function does not change the net area 1 2 the shaded1 region under the curve3 y =4 x 2 on the interval 1, 2 in under the curve, in the sense that (b). Without computing areas, explain why. b b c 3 4 + y y f x + c dx = f u du. La La + c 2 4 4 y ϭ x 1 2 1 2 b. Draw a picture to illustrate this change of variables in the case 3 3 that f x = sin x, a = 0, b = p, and c = p 2. 2 (͙x Ϫ 1)2 2 y ϭ 105. Substitution:1 2 scaling Another change of variables> that can be 1 2͙x 1 interpreted geometrically is the scaling u = cx, where c is a real number. Prove and interpret the fact that 0 x 0 x 4 9 21 b 1 bc (a) (b) f cx dx = f u du. La c Lac 94–98. General results Evaluate the following integrals in which the Draw a picture to illustrate1 2 this change of1 variables2 in the case that p 1 function f is unspecified. Note that f is the pth derivative of f and f x = sin x, a = 0, b = p, and c = 2. f p is the pth power of f . Assume f and1 2 its derivatives are continuous for all real numbers. 106–109.1 2 Multiple substitutions If necessary, use two or more substitutions to find the following integrals.

3 2 94. 5f x + 7f x + f x f ′ x dx L 106. x sin4 x 2 cos x 2 dx Hint: Begin with u = x 2, then use L 12 1 2 1 2 1 22 1 2 v = sin u. 3 2 1 95. 5f x + 7f x + f x f ′ x dx, where f 1 = 4, L1 dx 2 107. Hint: Begin with u = 1 + x. 1 1 2 1 2 1 22 1 2 1 2 L f 2 = 5 1 + 1 + x 2 12 1 1 2 1 12 108. x 1 - x dx

L0 96. f ′ x f ″ x dx, where f ′ 0 = 3 and f ′ 1 = 2 2 1 L0 1 1 2 1 2 1 2 1 2 109. x - x x dx p n p + 1 L0 97. f x f x dx, where p is a positive integer, n -1 2 1 L 1 2 1 2 1 1 22 1 2 110. tan10 4x sec2 4x dx Hint: Begin with u 4x. 2 = 98. 2 f x + 2 f x f x f ′ x dx L L p 2 1 2 cos u sin u 1 1 2 1 22 1 2 1 2 111. du Hint: Begin with u = cos u. 99–101. More than one way Occasionally, two different substitutions > 2 L0 do the job. Use each substitution to evaluate the following integrals. cos u + 16 2 1 2 1 99. x x + a dx; a 7 0 u = x + a and u = x + a Quick C heck Answer s L0 4 5 4 1 1 1. u = x + 5 2. With u = x + 6, we have du = 5x , 1 1 2 4 p p and x does not appear in the integrand. 3. New equation: 100. x x + a dx; a 7 0 u = x + a and u = x + a 2 L0 u - 13u + 36 = 0; roots: x = 2, 3 2 2 { { 1 2 101. sec3 u tan u du u = cos u and u = sec u L 1 2

M05_BRIG7345_02_SE_C05.5.indd 393 21/10/13 11:14 AM 394 Chapter 5 • Integration Not for Sale CHAPTER 5 Review Exercises

1. Explain why or why not Determine whether the following state- 4 2 5. Area by geometry Use geometry to evaluate 10 8x - x dx. ments are true and give an explanation or counterexample. As- (Hint: Complete the square.) 2 sume f and f′ are continuous functions for all real numbers. x 6. Bagel output The manager of a bagel bakery collects the follow- a. If A x = 1a f t dt and f t = 2t - 3, then A is a quadratic function. ing production rate data (in bagels per minute) at seven different 1 2 1 2 1 2 x times during the morning. Estimate the total number of bagels b. Given an area function A x f t dt and an antideriva- = 1a produced between 6:00 and 7:30 a.m., using a left and right tive F of f , it follows that A′ x = F x . b 1 2 1 2 Riemann sum. c. 1a f ′ x dx = f b - f a . 1 2 b 1 2 d. If f is continuous on a, b and 1a ͉ f x ͉ dx = 0, then Production rate 1 2 1 2 1 2 a m f x = 0 on a, b . Time of day ( . .) (bagels/min) 3 4 1 2 e. If the average value of f on a, b is zero, then f x = 0 6:00 45 1 2 3 4 on a, b . 6:15 60 b 3 b 4 a 1 2 f. 2f x 3g x dx 2 f x dx 3 g x dx. 6:30 75 1a 3 4 - = 1a + 1b 6:45 60 g. f ′ g x g′ x dx f g x C. 1 1 1 2 1 22= +1 2 1 2 7:00 50 2. Velocity1 to1 displacement22 1 2 An1 1object22 travels on the x-axis with a 7:15 40 velocity given by v t = 2t + 5, for 0 … t … 4. 7:30 30 a. How far does the object travel, for 0 t 4? 1 2 … … b. What is the average value of v on the interval 0, 4 ? 7. Integration by Riemann sums Consider the integral 4 c. True or false: The object would travel as far as in part (a) if it 11 3x - 2 dx. 3 4 traveled at its average velocity (a constant), for 0 … t … 4. a. Evaluate the right Riemann sum for the integral with n 3. 1 2 = 3. Area by geometry Use geometry to evaluate the following defi- b. Use summation notation to express the right Riemann sum in nite integrals, where the graph of f is given in the figure. terms of a positive integer n. 4 4 c. Evaluate the definite integral by taking the limit as n S ∞ of

a. f x dx b. f x dx the Riemann sum of part (b). L0 L6 d. Confirm the result of part (c) by graphing y = 3x - 2 7 1 2 7 1 2 and using geometry to evaluate the integral. Then evaluate 4 c. f x dx d. f x dx 11 3x - 2 dx with the Fundamental Theorem of Calculus. L5 L0 1 2 y 1 2 8–11. Limit1 definition2 of the definite integral Use the limit definition of the definite integral with right Riemann sums and a regular partition 5 to evaluate the following definite integrals. Use the Fundamental Theo- 4 rem of Calculus to check your answer.

3 1 2 y ϭ f (x) 2 2 8. 4x - 2 dx 9. x - 4 dx L0 L0 1 21 2 41 2 10. 3x 2 x dx 11. x 3 x dx 1 23456 7 8 x + - Ϫ1 L1 L0 1 2 1 2 Ϫ2 12. Evaluating Riemann sums Consider the function f x = 3x + 4 Ϫ3 on the interval 3, 7 . Show that the midpoint Riemann sum with n = 4 gives the exact area of the region bounded by1 the2 graph. 3 4 4. Displacement by geometry Use geometry to find the displace- 13. Sum to integral Evaluate the following limit by identifying the ment of an object moving along a line for the time intervals integral that it represents: (i) 0 … t … 5, (ii) 3 … t … 7, and (iii) 0 … t … 8, where the n 5 graph of its velocity v = g t is given in the figure. 4k 4 lim a + 1 . nS∞ k 1 n n v 1 2 = aa b b a b 5 14. Area function by geometry Use geometry to find the area A x

4 that is bounded by the graph of f t = 2t - 4 and the t-axis be- v ϭ g(t) tween the point 2, 0 and the variable point x, 0 , where x Ú1 22. 3 1 2 Verify that A′ x = f x . 2 1 2 1 2 1 1 2 1 2

0 123456 7 8 t

M05_BRIG7345_02_SE_C05EOC.indd 394 21/10/13 11:14 AM Not for Sale Review Exercises 395 15–30. Evaluating integrals Evaluate the following integrals. 38. Properties of integrals The figure shows the areas of regions 2 bounded by the graph of f and the x-axis. Evaluate the following 15. 3x 4 - 2x + 1 dx 16. cos 3x dx integrals. L-2 L c d b

21 2 1 a. f x dx b. f x dx c. 2 f x dx La Lb Lc 17. x + 1 3 dx 18. 4x 21 - 2x 16 + 1 dx L0 L0 d 1 2 b 1 2 d 1 2

1 2 21 2 d. 4 f x dx e. 3 f x dx f. 2 f x dx 8 6 4x 8 La La Lb 19. 9x - 7x dx 20. e + dx L L 2 1 2 1 2 1 2 - y 1 2 Area ϭ 15 1 y2 21. x x + 1 dx 22. 3 dy L0 L y 27 1 1 + Area ϭ 20 y ϭ f(x) 1 1 2 dx Area ϭ 12 23. 24. y2 3y3 1 4 dy 2 + L0 4 x L - a b c d x 1 2 3 2 x 25. dx 26. x sin x 2 cos8 x 2 dx 4 2 39–44. Properties of integrals Suppose that f x dx = 6, L0 25 - x L 4 4 11 11 g x dx = 4, and 13 f x dx = 2. Evaluate the following integrals p2 p 1 2 2 2 or state that there is not enough information. 27. sin 5u du 28. 1 - cos 3u du 1 2 1 2 L0 L0 4 1

3 2 ln1 2 x 2 39. 3 f x dx 40. - 2 f x dx x + 2x - 2 e L1 L4 29. 3 2 dx 30. 2x dx L2 x + 3x - 6x L0 1 + e 4 1 2 4 1 2

31–34. Area of regions Compute the area of the region bounded by 41. 3 f x - 2g x dx 42. f x g x dx L1 L1 the graph of f and the x-axis on the given interval. You may find it 31 1 2 1 22 1 1 2 1 2 useful to sketch the region. f x 43. dx 44. f x - g x dx 2 L1 g x L4 31. f x = 16 - x on -4, 4 1 2 1 1 2 1 22 3 45. Displacement1 2 from velocity A particle moves along a line with a 32. f 1x2 = x - x on -3 1, 0 4 velocity given by v t = 5 sin pt starting with an initial position

33. f 1x2 = 2 sin x 4 3on 0, 42p s 0 = 0. Find the displacement of the particle between t = 0 1 2 2 2 and t = 2, which is given by s t = v t dt. Find the distance 34. f 1x2 = 1 x1 +> 21 on3 -1, 4 3 1 2 10 2 traveled by the particle during this interval, which is 10 ͉ v t ͉ dt. 35–36. Area versus net area Find2 (i) the net area and (ii) the area of 1 2 1 2 1 2 >1 2 3 4 46. Average height A baseball is launched into the outfield on a para- the region bounded by the graph of f and the x-axis on the given 1 2 bolic trajectory given by y 0.01x 200 x . Find the average interval. You may find it useful to sketch the region. = - height of the baseball over the horizontal extent of its flight. 4 2 1 2 35. f x x x on 1, 1 = - - 47. Average values Integration is not needed. 2 36. f 1x2 = x - x on 0,3 3 4 a. Find the average value of f shown in the figure on the interval 4 1, 6 and then find the point(s) c in 1, 6 guaranteed to exist 37. Symmetry1 2 properties3 Suppose4 that 10 f x dx = 10 and 4 by the Mean Value Theorem for Integrals. g x dx = 20. Furthermore, suppose that f is an even func- 10 1 2 3 4 1 2 tion and g is an odd function. Evaluate the following integrals. y 1 2 4 4 5

a. f x dx b. 3g x dx 4 L-4 L-4 4 1 2 1 1 2 3 2 y ϭ f (x) c. 4f x - 3g x dx d. 8x f 4x dx 2 L-4 L0 1 21 1 2 1 22 1 2

e. 3x f x dx 0 123456 7 8 x L-2 1 2 (a)

M05_BRIG7345_02_SE_C05EOC.indd 395 21/10/13 11:14 AM 396 Chapter 5 • Integration Not for Sale b. Find the average value of f shown in the figure on the interval c. Interpret the physical meaning of the area under the velocity 2, 6 and then find the point(s) c in 2, 6 guaranteed to exist curve. by the Mean Value Theorem for Integrals. 3 4 1 2 54. Area functions Consider the graph of the continuous function f y x x

in the figure and let F x f t dt and G x f t dt. 5 = = L0 L1 4 Assume the graph consists1 2 of a line1 2segment from1 2 0, -21 2 to 3 2, 2 and two quarter circles of radius 2. y ϭ f (x) 1 2 2 1 2 y 1

0 123456 7 8 x 2 (b) 1 y ϭ f (t) 48. An unknown function The function f satisfies the equation 4 x 3x 48 f t dt. Find f and check your answer by - = 12 t substitution. Ϫ2 Ϫ1 1 32 4 1 2 Ϫ1

49. An unknown function Assume f ′ is continuous on 2, 4 , 2 11 f ′ 2x dx = 10, and f 2 = 4. Evaluate f 4 . Ϫ2 3 4 x 2 50. Function1 2 defined by an integral1 2 Let H x = 1102 4 - t dt, for -2 … x … 2. 2 1 2 a. Evaluate H 0 . a. Evaluate F 2 , F -2 , and F 4 . b. Evaluate G 2 , G 0 , and G 4 . b. Evaluate H′ 1 . - 1 2 c. Explain why1 there2 1 is a2 constant1 C2 such that F x = G x + C, c. Evaluate H′ 2 . 1 2 1 2 1 2 d. Use geometry1 2to evaluate H 2 . for -2 … x … 4. Fill in the blank with a number: F x = G x _____, for 2 x 4. 1 2 1 2 e. Find the value1 2of s such that H x = sH -x . + - … … 1 2 1 2 55–56. Area functions and the Fundamental Theorem Consider the 51. Function defined by an integral 1Make2 a graph1 2of the function 1 2 xdt function f x = , for x Ú 1. Be sure to include all of the evidence L1 t t if -2 … t 6 0 2 you1 2 used to arrive at the graph. f t = t if 0 … t … 2 2 52. Identifying functions Match the graphs A, B, and C in the figure 1 2 • x

with the functions f x , f ′ x , and 10 f t dt. and its graph shown below. y 1 2 1 2 1 2 y 2 1.5

1.0 A 1 y ϭ f (t)

0.5 B Ϫ2 Ϫ1 21 t 1 2 3 4 5 6 7 8 x Ϫ1 Ϫ0.5 C

Ϫ1.0 Ϫ2

x x 53. Ascent rate of a scuba diver Divers who ascend too quickly in Let F x = f t dt and G x = f t dt. the water risk decompression illness. A common recommendation L-1 L-2 for a maximum rate of ascent is 30 feet/minute with a 5-minute 1 2 1 2 1 2 1 2 55. a. Evaluate F 2 and F 2 . safety stop 15 feet below the surface of the water. Suppose that a - b. Use the Fundamental Theorem to find an expression for F x , diver ascends to the surface in 8 minutes according to the velocity ′ for 2 x1 20. 1 2 function - … 6 c. Use the Fundamental Theorem to find an expression for F′1x2, 30 if 0 … t … 2 for 0 … x … 2. 1 2 v t = 0 if 2 6 t … 7 d. Evaluate F′ -1 and F′ 1 . Interpret these values. 15 if 7 6 t … 8. e. Evaluate F″ -1 and F″ 1 . 1 2 • f. Find a constant1 C2 such that1 2 F x = G x + C. 1 2 1 2 a. Graph the velocity function v. 1 2 1 2 b. Compute the area under the velocity curve.

M05_BRIG7345_02_SE_C05EOC.indd 396 21/10/13 11:14 AM Not for Sale Guided Projects 397 56. a. Evaluate G 1 and G 1 . 69. Equivalent equations Explain why if a function u satisfies the - x b. Use the Fundamental Theorem to find an expression for G′ x , equation u x + 2 10 u t dt = 10, then it also satisfies the equa- for -2 … x1 6 20. 1 2 tion u′ x + 2u x = 0. Is it true that if u satisfies the second c. Use the Fundamental Theorem to find an expression for G′1x2, equation, then1 2 it satisfies1 2the first equation? 1 2 1 2 for 0 … x … 2. 1 2 T 70. Area function properties Consider the function f t = d. Evaluate G′ 0 and G′ 1 . Interpret these values. 2 x t - 5t + 4 and the area function A x = 10 f t dt. e. Find a constant C such that F x = G x + C. 1 2 1 2 1 2 a. Graph f on the interval 0, 6 . 1 2 1 2 57–58. Limits with integrals Evaluate1 the2 following1 2 limits. b. Compute and graph A on the interval 0, 6 . c. Show that the local extrema3 of4 A occur at the zeros of f . 2 x t 2 x t 3 d. Give a geometrical and analytical explanation3 4 for the observa- 12 e dt 11 e dt 57. lim 58. lim tion in part (c). xS2 x - 2 xS1 x - 1 e. Find the approximate zeros of A, other than 0, and call them x1 59. Geometry of integrals Without evaluating the integrals, explain and x2, where x1 6 x2. why the following statement is true for positive integers n: f. Find b such that the area bounded by the graph of f and the t-axis on the interval 0, t1 equals the area bounded by the 1 1 n graph of f and the t-axis on the interval t1, b . x n dx x dx 1. 3 4 x + = g. If f is an integrable function and A x = f t dt, is it L0 L0 1a 1 always true that the local extrema of A occur3 4at the zeros of f ? 1 2 1 2 60. Change of variables Use the change of variables u3 = x 2 - 1 to Explain. 3 3 2 evaluate the integral 11 x x - 1 dx. 71. Function defined by an integral x 15 9 Let f x t 1 t 2 dt. 61. Inverse tangent integral Prove2 that for nonzero constants a and b, = 10 - - a. Find the intervals on which f is increasing and the intervals on dx 1 1 ax 1 2 1 2 1 2 = tan- + C. which f is decreasing. L a2x 2 + b2 ab b b. Find the intervals on which f is concave up and the intervals a b 62–67. Additional integrals Evaluate the following integrals. on which f is concave down. c. For what values of x does f have local minima? Local sin 2x 62. dx Hint: sin 2x = 2 sin x cos x. maxima? L 1 + cos2 x d. Where are the inflection points of f ? 1 -1 25 t 1 1 tan x 72. Exponential inequalities Sketch a graph of f t = e on an arbi- 63. sin dx 64. dx L x 2 x L 1 + x 2 trary interval a, b . Use the graph and compare areas of regions 1 2 to prove that 1 2 dx sin-1 x 3 4 65. 66. dx eb ea ea eb -1 2 a + b 2 - + L tan x 1 + x L 1 - x 2 e 6 6 . 1 2> b - a 2 x x 2 1e - e-21 2 67. x x dx L e + e- (Source: Mathematics Magazine 81, 5, Dec 2008) 68. Area with a parameter Let a 7 0 be a real number and consider

the family of functions f x = sin ax on the interval 0, p a . a. Graph f , for a 1, 2, 3. = 1 2 3 > 4 b. Let g a be the area of the region bounded by the graph of f and the x-axis on the interval 0, p a . Graph g for 0 6 a1 62 ∞. Is g an increasing function, a decreasing function, or neither? 3 > 4

Chapter 5 Guided Projects Applications of the material in this chapter and related topics can be found in the following Guided Projects. For additional information, see the Preface. • Limits of sums • Symmetry in integrals • Distribution of wealth

M05_BRIG7345_02_SE_C05EOC.indd 397 21/10/13 11:14 AM