The Integral - Department of Mathematics | Penn Mathdeturck/m103/rogawski/chap5odd.pdf · The Integral 5.1 Approximating and Computing Area Preliminary Questions 1. The interval [2,5]

CHAPTER

5 The Integral

5.1 Approximating and Computing Area

Preliminary Questions

1. The interval [2, 5] is divided into 6 subintervals in order to calculate R6 for some function.What are the right-endpoints of those subintervals? What are the left-endpoints?

2. If f (x) = x−2 on [3, 7], which is larger: RN or L N ?

3. Which of the following pairs of sums are not equal?(a)

∑4i=1 i,

∑4�=1 �

(b)∑4

j=1 j 2,∑5

k=2 k2

(c)∑4

j=1 j,∑5

i=2(i − 1)

(d)∑4

i=1 i(i + 1),∑5

j=2( j − 1) j

4. The interval [1, 5] is divided into 16 subintervals.

(a) What are the left endpoints of the first and last subintervals?(b) What are the right endpoints of the first two subintervals?

5. True or False:(a) The right-endpoint rectangles lie below the graph of an increasing function.(b) If f is monotonic, then the area under the graph lies in between RN and L N .(c) If f is not monotonic, then L N and RN may converge to different limits as N → ∞.(d) If f (x) is constant, then the right-endpoint rectangles all have the same height.

Exercises

1. An athlete runs with velocity 4 mph for half an hour, 6 mph for the next hour, and 5 mphfor another half-hour. Compute the total distance traveled and indicate on a graph how thisquantity can be interpreted as an area.

The total distance traveled is (4) ( 12) + (6) (1) + (5) ( 1

2) = 10 1

2miles.

1

2 Chapter 5 The Integral

0

1

2

3

4

5

6

mph

0.5 1 1.5 2hours

2.Figure 1 shows the velocity of an object over a 3-minute interval. Determine the distancetraveled over the intervals [0, 3] and [1, 2.5] (remember to convert from mph to miles perminute).

Figure 1

3. Assume that the velocity of an object is 32t ft/s. Use Eq. (??) to determine the distancetraveled by the object over the time intervals (in seconds) [0, 2] and [2, 5].The total distance traveled is given by the area under the graph of v = 32t .

During the interval [0, 2], the object travels 12 (2)(64) = 64 ft.

During the interval [2, 5], the object travels 12 (3)(160 − 64) + (3)(64) = 336 ft.

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

20

40

60

80

100

120

140

160

4.

Consider f (x) = 2x + 3 on [0, 3].(a) Determine the left- and right-endpoints if [0, 3] is divided into 6 subintervals.(b) Compute R6 and L6.(c) Find the exact area using geometry and compute the error in the two calculations of (b).

5. Let f (x) = x2 + x − 2.(a) Calculate R3 and L3 for the interval [2, 5].(b) Sketch the graph of f and the rectangles that make up each of the approximations.

Let f (x) = x2 + x − 2 and set a = 2, b = 5, n = 3,h = �x = (b − a) /n = (5 − 2) /3 = 1.(a) Let xk = a + kh, k = 0, 1, 2, 3.

Selecting left endpoints of subintervals, xk , k = 0, 1, 2, or {2, 3, 4}, we have

L3 =2∑

k=0

f (xk)�x = h2∑

k=0

f (xk) = (1) (4 + 10 + 18) = 32

Selecting right endpoints of subintervals, xk , k = 1, 2, 3, or {3, 4, 5}, we have

R3 =3∑

k=1

f (xk)�x = h3∑

k=1

f (xk) = (1) (10 + 18 + 28) = 56

(b) Here are figures of the three sets of rectangles that approximate the area under thecurve f (x) over the interval [2, 5].

5.1 Approximating and Computing Area 3

2 2.5 3 3.5 4 4.5 5

5

10

15

20

25

30

x

x2+x−2

2 2.5 3 3.5 4 4.5 5

5

10

15

20

25

30

x

x2+x−2

6.

Let f (x) = cos x .(a) Calculate R4 and L4 for the interval [0, π

2 ].(b) Sketch the graph of f and the rectangles that make up each of the approximations.(c) Is the area under the graph larger or smaller than R4? Than L4?

7. Consider f (x) = 5x on [0, 3].(a) Find a formula for RN (use formula (??)).(b) Is RN bigger or smaller than the area under the graph?(c) Calculate limN→∞ RN .(d) Calculate the area under the graph using geometry and verify your answer to (c).

Let f (x) = 5x on [0, 3]. Let n be a (symbolic) positive integer and set a = 0, b = 3,h = �x = (b − a) /n = (3 − 0) /n = 3/n.(a) Let xk = a + kh = 3k/n, k = 1, 2, . . . , n be the right endpoints of the n subintervals

of [0, 3]. Then

Rn = hn∑

k=1

f (xk) = 3

n

n∑k=1

5

(3k

n

)= 45

n2

n∑k=1

k

= 45

n2

(n2

2+ n

2

)= 45

2+ 45

2n

(b) Since f (x) = 5x is increasing, the right endpoint approximation Rn is greater than thearea under the graph.

(c) The area under the graph is limn→∞

Rn = limn→∞

(45

2+ 45

2n

)= 45

2= 22.5.

(d) The area under the graph is the area of a triangle: A = 12

B H = 12 (3) (15) = 45

2, which

agrees with the answer in (c).8.

Estimate R6 and L6 over [0, 1.5] for the function shown in Figure 2.

Figure 2

9. Estimate R6 and L6 for the graph in Figure 3.

13

23

1 43

53

2

1

.5

0

Figure 3

Let f (x) on [0, 2] be given by the figure in the text. For n = 6, h = (2 − 0)/6 = 13,

{xk}6k=0 = {

0, 13, 2

3, 1, 4

3, 5

3, 2}.


Therefore

L6 = 1

3

5∑k=0

f (xk) = 1

3

(1

2+ 5

8+ 7

8+ 9

8+ 1 + 3

4

)= 1.625

R6 = 1

3

6∑k=1

f (xk) = 1

3

(5

8+ 7

8+ 9

8+ 1 + 3

4+ 1

)≈ 1.792

In Exercises 10–19, calculate the approximation for the given function and interval.10.

R8, 7 − x , [3, 5]11. R6, 2x2 − x + 2, [1, 4]

Let f (x) = 2x2 − x + 2 on [1, 4]. For n = 6, h = (4 − 1)/6 = 12,

{xk}6k=0 = {

1, 1 12 , 2, 2 1

2 , 3, 3 12 , 4

}. Therefore

R6 = 1

2

6∑k=1

(2x2k − xk + 2) = 47.5

12.L6, 2x2 − x + 2, [1, 4]

13. R4, x3 − 2x + 5, [0, 1]Let f (x) = x3 − 2x + 5 on [0, 1]. For n = 4, h = (1 − 0)/4 = 1

4,

{xk}4k=0 = {

0, 14 ,

12 ,

34 , 1

}. Therefore

R4 = 1

4

4∑k=1

(x3k − 2xk + 5) ≈ 4.140625

14.R4, x3 − 2x + 5, [−2, 2]

15. R5, sin x , [ π

4 , 3π

4 ]Let f (x) = sin x on [ π

4 , 3π

4 ]. For n = 5, h = (3π/4−π/4)

5 = π

10 ,{xk}5

k=0 = {π

4, 7π

20, 9π

20, 11π

20, 13π

20, 3π

4

}. Therefore

R5 = π

10

5∑k=1

sin xk ≈ 1.402563

16.L5, x−1, [1, 2]

17. L4, x−2, [1, 3]Let f (x) = x−2 on [1, 3]. For n = 4, h = (3 − 1)/4 = 1

2 , {xk}4k=0 = {

1, 32 , 2, 5

2 , 3}.

Therefore

L4 = 1

2

3∑k=0

x−2k ≈ 0.927222

18.L4, cos x , [ π

4 , π

2 ]19. L5, x2 + 3|x |, [−3, 2]

Let f (x) = x2 + 3 |x | on [−3, 2]. For n = 5, h = (2 − (−3))/5 = 1,{xk}5

k=0 = {−3, −2, −1, 0, 1, 2}. Therefore

L5 = 14∑

k=0

(x2k + 3 |x |) = 36


20.R & W Compute the average of R5 and L5 to approximate the area under the graph of

f (x) = x−1 over [3.5, 5]. Explain why the average is more accurate than either endpointapproximation.

21. Calculate the sums:

(a)5∑

i=1

3

(b)5∑

i=0

3

(c)4∑

k=2

k3

(d)4∑

j=3

sin(

jπ

2

)

When the number of summands is small, we may compute the sum directly. Whereapplicable (especially when the number of summands is large), we may use summationlaws. For illustration, both ways are shown in this exercise.

(a) Do this directly:5∑

i=1

3 = 3 + 3 + 3 + 3 + 3 = 15. Or use the law for summing a

constant:5∑

i=1

3 = (3)(5) = 15.

(b) Here5∑

i=0

3 = 3 + 3 + 3 + 3 + 3 + 3 = 18 or5∑

i=0

3 = (3)(6) = 18.

(c) Again,4∑

k=2

k3 = 23 + 33 + 43 = 99 or

4∑k=2

k3 =(

4∑k=1

k3

)−(

1∑k=1

k3

)

=(

44

4+ 43

2+ 42

4

)−(

14

4+ 13

2+ 12

4

)= 99.

(d) Finally,4∑

j=1

sin

(jπ

2

)= 1 + 0 + (−1) + 0 = 0.

In Exercises 22–34, use the Linearity Rules and formulas (??)–(??) to rewrite and evaluate thesums.22.

6∑j=1

j 3

23.10∑j=1

4 j 3

We have∑10

j=1 4 j 3 = 4∑10

j=1 j 3 = 4(

104

4+ 103

2+ 102

4

)= 12100.

24.10∑j=1

2 j25.

20∑k=1

(2k + 1)

We have∑20

k=1 (2k + 1) = 2∑20

k=1 k +∑20k=1 1 = 2

(202

2+ 20

2

)+ 20 = 440.

26.15∑

i=1

(6 − 3i)


27.10∑

�=1

(�3 − 2�2)

We have∑10�=1 (�3 −2�2) = ∑10

�=1 �3 −2∑10

�=1 �2 =(

104

4+ 103

2+ 102

4

)−2

(103

3+ 102

2+ 10

6

)= 2255.

28.30∑

s=1

(3s2 − 4s − 1)29.

18∑i=1

(4i − 2i 2 + 9)

We have∑18

i=1 (4i − 2i 2 + 9) = 4∑18

i=1 i − 2∑18

i=1 i 2 + 9∑18

i=1 1 =4(

182

2+ 18

2

)− 2

(183

3+ 182

2+ 18

6

)+ (9)(18) = −3372.

30.10∑j=5

j 2 Hint: write as a difference of two sums.31.

200∑j=101

j

We have∑200

j=101 j = ∑200j=1 j −∑100

j=1 j =(

2002

2+ 200

2

)−(

1002

2+ 100

2

)= 15050.

32.50∑j=4

(7 j − 9)33.

20∑j=10

6 j 2

We have∑20j=10 6 j 2 = 6

∑20j=1 j 2 − 6

∑9j=1 j 2 = 6

(203

3+ 202

2+ 20

6

)− 6

(93

3+ 92

2+ 9

6

)= 15510.

34.10∑j=3

(6 j 3 − j 2)35. Write the sum of the cubes of the whole numbers from 100 to 250 in summation notation.

The sum of the first 100 cubes may be written in summation notation as250∑

k=100

k3.

36.Write the sum √

1 + 13 +√

2 + 23 + · · · +√

n + n3

in summation notation.

In Exercises 37–41, calculate the sum, assuming that a1 = −1,∑10

i=1 ai = 10, and∑10

i=1 bi = 7.

37.10∑

i=1

2ai

We have10∑

i=1

2ai = 210∑

i=1

ai = (2)(10) = 20.

38.10∑

i=1

(ai − bi )39.

10∑�=1

(3a� + 4b�)

We have

10∑i=1

(3ai + 4bi) =(

310∑

i=1

ai

)+(

410∑

i=1

bi

)

= (3)(10) + (4)(7) = 58.

40.10∑

i=2

ai41. Can you calculate∑10

i=1 ai bi from the information given?

From the information given,10∑

i=1

ai bi cannot be computed.


42.

Evaluate limn→∞

n∑i=1

i

n2.

43. Evaluate limn→∞

n∑i=1

i 2 − i + 1

n3.

Now

sn =n∑

i=1

i 2 − i + 1

n3= 1

n3

((n∑

i=1

i 2

)−(

n∑i=1

i

)+(

n∑i=1

1

))

= 1

n3

((n3

3+ n2

2+ n

6

)−(

n2

2+ n

2

)+ (n)

)= 1

3+ 2

3n2.

Therefore, limn→∞

sn = 1

3.

In Exercises 44–59, use formulas (??)–(??) to find a formula for RN for the given function andinterval. Then compute the area under the graph as a limit.44.

x ; [0, 3]45. x ; [2, 7]

Let f (x) = x on the interval [2, 7]. Then �x = 7 − 2

N= 5

Nand a = 2. Hence,

RN = �xN∑

j=1

f (2 + j�x) = 5

N

N∑j=1

(2 + j

5

N

)= 10

N

N∑j=1

1 + 25

N 2

N∑j=1

j

= 10

NN + 25

N 2

(N 2

2+ N

2

)= 10 + 25

2+ 25

2N

and

limN→∞

RN = limN→∞

(10 + 25

2+ 25

2N

)= 45

2= 22.5.

46.3 − x ; [1, 2]

47. 2x + 7; [3, 6]Let f (x) = 2x + 7 on the interval [3, 6]. Then �x = 6 − 3

N= 3

Nand a = 3. Hence,

RN = �xN∑

j=1

f (3 + j�x) = 3

N

N∑j=1

(2

(3 + j

3

N

)+ 7

)

= 39

N

N∑j=1

1 + 18

N 2

N∑j=1

j = 39

NN + 18

N 2

(N 2

2+ N

2

)= 39 + 9 + 9

N

and

limN→∞

RN = limN→∞

(48 + 9

N

)= 48.

48.x2; [0, 1]

49. x2; [2, 4]


Let f (x) = x2 on the interval [2, 4]. Then �x = 4−2N

= 2N

and a = 2. Hence,

RN = �xN∑

j=1

f (2 + j�x) = 2

N

N∑j=1

(4 + j

8

N+ j 2 4

N 2

)

= 8

N

N∑j=1

1 + 16

N 2

N∑j=1

j + 8

N 3

N∑j=1

j 2 = 8

NN + 16

N 2

(N 2

2+ N

2

)

+ 8

N 3

(N 3

3+ N 2

2+ N

6

)

= 8 + 8 + 8

N+ 8

3+ 4

N+ 4

3N 2

and

limN→∞

RN = limN→∞

(56

3+ 12

N+ 4

3N 2

)≈ 18.6667.

50.4 − x2; [0, 2]

51. 3x2 − x + 4; [0, 1]Let f (x) = 3x2 − x + 4 on the interval [0, 1]. Then �x = 1 − 0

N= 1

Nand a = 0. Hence,

RN = �xN∑

j=1

f (0 + j�x) = 1

N

N∑j=1

(3 j 2 1

N 2− j

1

N+ 4

)

= 3

N 3

N∑j=1

j 2 − 1

N 2

N∑j=1

j + 4

N

N∑j=1

1

= 3

N 3

(N 3

3+ N 2

2+ N

6

)− 1

N 2

(N 2

2+ N

2

)+ 4

NN

= 1 + 3

2N+ 1

2N 2− 1

2− 1

2N+ 4

and

limN→∞

RN = limN→∞

(4.5 + 1

N+ 1

2N 2

)= 4.5.

52.3x2 − x + 4; [1, 5]

53. 4x3 − 3x ; [0, 2]Let f (x) = 4x3 − 3x on the interval [0, 2]. Then �x = 2 − 0

N= 2

Nand a = 0. Hence,

RN = �xN∑

j=1

f (0 + j�x) = 2

N

N∑j=1

(4 j 3 8

N 3− 3 j

2

N

)

= 64

N 4

N∑j=1

j 3 − 12

N 2

N∑j=1

j = 64

N 4

(N 4

4+ N 3

2+ N 2

4

)− 12

N 2

(N 2

2+ N

2

)

= 16 + 32

N+ 16

N 2− 6 − 6

N

and

limN→∞

RN = limN→∞

(10 + 26

N+ 16

N 2

)= 10.


54.x3; [0, 1]

55. x3 + x ; [0, 4]Let f (x) = x3 + x on the interval [0, 4]. Then �x = 4 − 0

N= 4

Nand a = 0. Hence,

RN = �xN∑

j=1

f (0 + j�x) = 4

N

N∑j=1

(j 3 64

N 3+ j

4

N

)

= 256

N 4

N∑j=1

j 3 + 16

N 2

N∑j=1

j

= 256

N 4

(N 4

4+ N 3

2+ N 2

4

)+ 16

N 2

(N 2

2+ N

2

)

= 64 + 128

N+ 64

N 2+ 8 + 8

N

and

limN→∞

RN = limN→∞

(72 + 136

N+ 64

N 2

)= 72.

56.x3 + 2x2; [0, 3]

57. 1 − x3; [0, 1]Let f (x) = 1 − x3 on the interval [0, 1]. Then �x = 1 − 0

N= 1

Nand a = 0. Hence,

RN = �xN∑

j=1

f (0 + j�x) = 1

N

N∑j=1

(1 − j 3 1

N 2

)

= 1

N

N∑j=1

1 − 1

N 4

N∑j=1

j 3 = 1

NN − 1

N 4

(N 4

4+ N 3

2+ N 2

4

)

= 1 − 1

4− 1

2N− 1

4N 2

and

limN→∞

RN = limN→∞

(3

4− 1

2N− 1

4N 2

)= 3

4.

58.2x + 1, [a, b] (a, b constants with a < b)

59. x2, [a, b] (a, b constants with a < b)

Let f (x) = x2 on the interval [a, b]. Then �x = b − a

N. Hence,

RN = �xN∑

j=1

f (a + j�x) = (b − a)

N

N∑j=1

(a2 + 2aj

(b − a)

N+ j 2 (b − a)2

N 2

)

= a2(b − a)

N

N∑j=1

1 + 2a(b − a)2

N 2

N∑j=1

j + (b − a)3

N 3

N∑j=1

j 2

= a2(b − a)

NN + 2a(b − a)2

N 2

(N 2

2+ N

2

)+ (b − a)3

N 3

(N 3

3+ N 2

2+ N

6

)

= a2(b − a) + a(b − a)2 + a(b − a)2

N+ (b − a)3

3+ (b − a)3

2N+ (b − a)3

6N 2


and

limN→∞

RN = limN→∞

(a2(b − a) + a(b − a)2 + a(b − a)2

N

+ (b − a)3

3+ (b − a)3

2N+ (b − a)3

6N 2

)

= a2(b − a) + a(b − a)2 + (b − a)3

3.

60.Let f (x) = √

x2 + 1 and �x = 13. Explain what the following sum represents (in terms of

areas of rectangles) but do not evaluate it.

6∑i=1

�x · f (1 + i�x)

In Exercises 61–66, use the approximation indicated (in summation notation) to express the areaunder the graph as a limit but do not evaluate the limit.

61. RN , sin x over [0, π ]Let f (x) = sin x over [0, π ] and set a = 0, b = π , h = �x = (b − a) /n = π/n. Then

Rn = hn∑

k=1

f (xk) = π

n

n∑k=1

sin

(kπ

n

)

Hence

limn→∞

Rn = limn→∞

(π

n

n∑k=1

sin

(kπ

n

))

is the area between the graph of f (x) = sin x and the x-axis over [0, π ].62.

RN ; x−1 over [1, 7]63. RN ; tan x over [1, 4]

Let f (x) = tan x over the interval [1, 4]. Then �x = 4 − 1

N= 3

Nand a = 1. Hence,

RN = �xN∑

j=1

f (1 + j�x) = 3

N

N∑j=1

tan

(1 + j

3

N

)

and

limN→∞

RN = limN→∞

3

N

N∑j=1

tan

(1 + j

3

N

)

64.L N ; x−2 over [3, 5]

65. L N ; cos x over [ π

8, π ]

Let f (x) = cos x over the interval[

π

8 , π]. Then �x = π − π

8

N= 7π

8Nand a = π

8 . Hence,

L N = �xN−1∑j=0

f(π

8+ j�x

)= 7π

8N

N−1∑j=0

cos

(π

8+ j

7π

8N

)

and

limN→∞

L N = limN→∞

7π

8N

N−1∑j=0

cos

(π

8+ j

7π

8N

)

66.L N ; cos x over [ π

8, π

4]


67. Show that the area A under the graph of f (x) = x−1 over [1, 8] satisfies

1

2+ 1

3+ 1

4+ 1

5+ 1

6+ 1

7+ 1

8≤ A ≤ 1 + 1

2+ 1

3+ 1

4+ 1

5+ 1

6+ 1

7

Hint: use R7 and L7.

Let f (x) = x−1, 1 ≤ x ≤ 8. Since f is decreasing, the left endpoint approximation L7

overestimates the true area between the graph of f and the x-axis, whereas the rightendpoint approximation R7 underestimates it. Accordingly,

12 + 1

3 + 14 + 1

5 + 16 + 1

7 + 18 = R7 < A < L7 = 1 + 1

2 + 13 + 1

4 + 15 + 1

6 + 17

Left endpoint approximation, n = 7

0

0.2

0.4

0.6

0.8

1

1 2 3 4 5 6 7 8

Right endpoint approximation, n = 7

0

0.2

0.4

0.6

0.8

1

1 2 3 4 5 6 7 868.

Which two of the following five expressions represent endpoint approximations to the areaunder the graph of f (x) = x−2 over [3, 7]?

(a)4

100

100∑k=1

(k

100

)−2

(b)4

100

100∑j=0

(3 + 4

k

100

)−2

(c)4

100

100∑k=1

(3 + 4 · k

100

)−2

(d)4

100

99∑�=0

(3 + �

100

)−2

(e)4

100

99∑k=0

(3 + 4 · k

100

)−2

69. Which area is represented by the following limits?

(a) limN→∞

1

N

N∑j=1

(j

N

)4

(b) limN→∞

3

N

N∑j=1

(2 + 3 j

N

)4

(c) limN→∞

5

N

N∑j=1

(−2 + 5

j

N

)4

Be apprised that the upper limits of summation below are N (not ∞).

(a) The limit limN→∞

RN = limN→∞

1

N

N∑j=1

(j

N

)4

represents the area between the graph of

f (x) = x4 and the x-axis over the interval [0, 1].(b) The limit lim

N→∞RN = lim

N→∞3

N

N∑j=1

(2 + 3 · j

N

)4

represents the area between the graph

of f (x) = x4 and the x-axis over the interval [2, 5].(c) The limit lim

N→∞RN = lim

N→∞5

N

N∑j=1

(−2 + 5 · j

N

)4

represents the area between the

graph of f (x) = x4 and the x-axis over the interval [−2, 1].70.

Evaluate the limit

limN→∞

1

N

N∑j=1

√1 −

(j

N

)2

by interpreting it as the area of part of a familiar geometric figure.

71. Use R5 and L5 to show that the area A under the graph of y = x1/3 over the interval [1, 1.5]satisfies .530 ≤ A ≤ .545.

Let f (x) = x1/3 on [1, 1.5]. For n = 5, h = (1.5 − 5)/5 = .1,{xk}5

k=0 = {1, 1.1, 1.2, 1.3, 1.4, 1.5}. Therefore

R5 = (.1)

5∑k=1

(xk)1/3 ≈ .545


and

L5 = (.1)

4∑k=0

(xk)1/3 ≈ .530

Hence, the area A must be .530 ≤ A ≤ .545 as R5 is an overestimate and L5 is anunderestimate of the exact value for A.

72.Use R6 and L6 to show that the area A under y = x−2 over the interval [10, 12] satisfies.0161 ≤ A ≤ .0195.

Midpoint Approximation The midpoint approximation MN , which is similar to RN and MN ,approximates the area under the graph using rectangles whose heights are the function valuesf (c j ) where c j is the midpoint of the j th interval.

73. Calculate M2 and M4 for x2 for the interval [0, 1].Let f (x) = x2 on [0, 1].

For n = 2, h = (1 − 0)/2 = 12 , {xk}2

k=0 = {0, 1

2 , 1}

and{x∗

k

}1

k=0= {

14 ,

34

}. Therefore

M2 = 1

2

1∑k=0

(x∗k )

2 = .3125

For n = 4, h = (1 − 0)/4 = 14, {xk}4

k=0 = {0, 1

4, 1

2, 3

4, 1}

and{x∗

k

}3

k=0= {.125, .375, .625, .875}. Therefore

M4 =3∑

k=0

(x∗k )

2 = .328125

74.Calculate M3 and M6 for

√x for the interval [2, 5].

75. Calculate M2 and M4 for x for the interval [0, 1]. What do you notice about the accuracy ofthese approximations?

Let f (x) = x on [0, 1].

For n = 2, h = (1 − 0)/2 = 12, {xk}2

k=0 = {0, 1

2, 1}

and{x∗

k

}1

k=0= {

14, 3

4

}. Therefore

M2 = 1

2

1∑k=0

x∗k = 1

2

For n = 4, h = (1 − 0)/4 = 14 , {xk}4

k=0 = {0, 1

4 ,12 ,

34 , 1

}and{

x∗k

}3

k=0= {.125, .375, .625, .875}. Therefore

M4 =3∑

k=0

x∗k = 1

2

Note that both M2 and M4 are exact values for the area under f (x) = x on the interval[0, 1].

76.Calculate M6 to estimate the area under the graph of sin(x2) over the interval [0, π

2].

Figure 4 Graph of sin(x2).


Further Insights and Challenges

77. Although the accuracy of RN generally improves as N increases, this need not be true forsmall values of N . Draw the graph of a positive continuous function f (x) on an intervalsuch that R1 is closer to the exact area under the curve than R2.

Let δ be a small positive number less than 14. (In the figures below, δ = 1

10. But imagine δ

being very tiny.) Define f (x) on [0, 1] by

f (x) =

1 if 0 ≤ x < 12

− δ12δ

− xδ

if 12− δ ≤ x < 1

2xδ

− 12δ

if 12 ≤ x < 1

2 + δ

1 if 12+ δ ≤ x ≤ 1

.

Then f is continous on [0, 1]. (Again, just look at the figures.)

The exact area between f and the x-axis is A = 1 − 12bh = 1 − 1

2(2δ)(1) = 1 − δ. (For

δ = 110

, we have A = 910

.)

With R1 = 1, the absolute error is |E1| = |R1 − A| = |1 − (1 − δ)| = δ. (For δ = 110

,this absolute error is |E1| = 1

10.)

With R2 = 12 , the absolute error is |E2| = |R2 − A| = ∣∣ 1

2 − (1 − δ)∣∣ = ∣∣δ − 1

2

∣∣ = 12 − δ.

(For δ = 110 , we have |E2| = 2

5 .)Accordingly, R1 is closer to the exact area A that is R2. Indeed, the tinier δ is, the moredramatic the effect.

Graph of f(x)

0

0.2

0.4

0.6

0.8

1

0.2 0.4 0.6 0.8 1x

Right endpt approx, n = 1

0

0.5

1

0.5 1

Right endpt approx, n = 2

0

0.5

1

0.5 178.

Draw the graph of a positive continuous function on an interval such that R2 and L2 areboth smaller than the exact area under the graph.79. R & W Explain the following statement graphically: the endpoint approximations areless accurate when f ′(x) is large.

When f ′ is large, the graph of f is steeper and hence there is more gap between f and Ln

or Rn . Recall that the top line segments of the subrectangles involved in an endpointapproximation constitute a piecewise constant function. If f ′ is large, then f is increasingmore rapidly and hence is less like a constant function.

Smaller f’

0

1

1 2 3 4

Larger f’

0

2

1 2 3 480.

Prove that if f (x) is monotonic, then MN always lies in between RN and L N . Explain whyMN must then be closer to the actual area.81. R & W In Exercise 48, one shows that RN for the function f (x) = x2 over [0, 1] is givenby the formula RN = 1

3 + 12N

+ 6N 2 and taking the limit as N → ∞, one concludes that the


area under the graph is 13. Can you interpret the quantity 1

2N+ 6

N 2 as the area of a region?Which region?

Let f (x) = x2 on [0, 1]. The quantity

1

2N+ 6

N 2in RN = 1

3+ 1

2N+ 6

N 2

represents the collective area of the parts of the subrectangles that lie above the graph off (x). It is the error between RN and the true area A = 1

3 .

0

0.2

0.4

0.6

0.8

1

0.2 0.4 0.6 0.8 1x82.

Prove the formula

RN − L N = (b − a)

N· ( f (b) − f (a)) (1)

for any function f (x) on [a, b].Hint: observe that the sums defining RN and L N only differ in their first and last terms.

83. Assume that f (x) is positive, continuous, and monotonic increasing or decreasing and letA be the area under graph over an interval [a, b].(a) Use Eq. (1) to prove that

|RN − A| ≤ (b − a)

N| f (b) − f (a)| (2)

Hint: A lies between RN and L N if f (x) is monotonic.(b) Use Eq. (2) to find a value of N such that |RN − A| < 10−2 for f (x) = √

x on [1, 4].Convergence for Monotonic Functions. Let f (x) be continuous, positive, and monotonicon [a, b]. Let A be the area between the graph of f and the x-axis over [a, b]. Forspecificity, say f is increasing. (The case for f decreasing on [a, b] is similar.)(a) As noted in the text, we have Ln ≤ A ≤ Rn . By Exercise ?? and the fact that A lies

between Ln and Rn , we therefore have

0 ≤ Rn − A ≤ Rn − Ln = b − a

n( f (b) − f (a)) .

Hence

|Rn − A| ≤ b − a

n( f (b) − f (a)) = b − a

n| f (b) − f (a)| ,

where f (b) − f (a) = | f (b) − f (a)| because f is increasing on [a, b]. Therefore,

|Rn − A| ≤ b − a

n| f (b) − f (a)| .

(b) Let f (x) = √x on [1, 4]. Then b = 4, a = 1, and

|RN − A| < 4−1N

( f (4) − f (1)) = 3N(2 − 1) = 3

N. We need 3

N< 10−2, which gives

3 < N · 10−2 and consequently N > 300. Thus |R301 − A| < 10−2 for f (x) = √x on

[1, 4].84.

Use Eq. (2) to find a value of N such that |RN − A| < 10−2 for f (x) = √9 − x2 on [0, 3].

85. The area of the unit circle. By definition, π is equal to one-half of the circumference ofthe unit circle. We know that π is also equal to the area of the unit circle, but this requiresproof. In this exercise, triangles are used (instead of rectangles) to compute the area of aunit circle. Consider a regular polygon with n sides inscribed in the unit circle. Figure 5shows the case n = 8. In general, θ = 2π/N .

5.2 The Definite Integral 15

(a) Show that the triangles O Q P and O P R have areas 12

sin(2π/N ) and 12

sin(2π/N ),respectively.

(b) Show that

N

2sin

(2π

N

)≤ area of unit circle ≤ N

2tan

(2π

N

)

(c) Let N → ∞ and use the Squeeze Theorem to show that the area of the unit circle is π .Hint: take h = 2π/N in the formula lim

h→0sin h

h= 1.

P

Q

R

θ

Figure 5 Area of the unit circle.

The area of the unit circle.

(a) The area of �O P Q is S = 12bh = 1

2 (1) (sin θ) = 12

sin( 2π

n); that of �O P R is

T = 12 B H = 1

2 (1) tan θ = 12 tan( 2π

n).

(b) From the figure in the text, the area A of the circular sector O P Q is greater than that of�O P Q, yet less than that of �O P R. Accordingly, we have S ≤ A ≤ T . Thus

nS ≤ n A ≤ nT , or n2

sin( 2π

n) ≤ A ≤ n

2tan( 2π

n). Hence

sin(

2π

n

)2π

n

≤ A

π≤ tan

(2π

n

)2π

n

.

(c) Let h = 2π

n. Then from (b) we have

sin h

h≤ A

π≤ tan h

h= (sin h) /h

cos h.

As h → 0, this gives 1 ≤ Aπ

≤ 1. Thus, Aπ

= 1 or A = π . The area of the unit circle istherefore π .

5.2 The Definite Integral


1. What is∫ b

a dx? (here the function is f (x) = 1)

2. True or false:(a) If f (x) is continuous and positive on [a, b], then

∫ b

a f (x) dx is the area under the graphof f (x) over [a, b].


(b) For all continuous functions f (x) on [a, b], ∫ b

a f (x) dx is the area between the graphand the x-axis over [a, b].

(c) If f (x) ≤ 0 on [a, b], then − ∫ b

a f (x) dx is the area between the graph of f (x) and thex-axis over [a, b].

3. Explain the equality∫ π

0 cos x dx = 0 graphically.

4. True or false:∫ −1

−5 5 dx is negative.

5. True or false:∫ 0

8 x2 dx is positive because x2 is a positive function.

6. Which statement is correct:(a) If f (x) ≤ 1

2then

∫ 6

0 f (x) dx ≤ 12.

(b) If f (x) ≤ 12 then

∫ 6

0 f (x) dx ≤ 3.

7. What is the largest possible value of∫ 2

0 f (x) dx if f (x) ≤ 13?

8. Theorem ?? says that if f (x) ≤ g(x), then the integral of f over a given interval is smallerthan the integral of g over the same interval. Is the same true for derivatives? In other words,does f (x) ≤ g(x) imply that f ′(x) ≤ g′(x)?

Exercises

In Exercises 1–10, draw a graph of the signed area represented by the integral and compute itusing geometry.

1.∫ 3

−3 2x dx

Using geometry, the value of the integral∫ 3

−3 2x dx = 12 (3)(6) + 1

2 (−3)(6) = 0.

−3 −2 −1 0 1 2 3−6

−4

−2

0

2

4

6

2. ∫ 3

−2(2x + 4) dx3.∫ 1

−2 (3x + 4) dx


−2 3x + 4 dx is 12

(73

)(7)− 1

2

(23

)(2) = 15

2= 7.5.

–2

7

–2 –1 1

4. ∫ 1

−2 4 dx


5.∫ 8

6 (7 − x) dx


6 (7 − x) dx = 12(1)(1) + 1

2(1)(−1) = 0.

6 6.2 6.4 6.6 6.8 7 7.2 7.4 7.6 7.8 8−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

6. ∫ 3π/2π2

sin x dx7.∫ 5

0

√25 − x2 dx


0

√25 − x2 dx is 1

4π (5)2 = 25

4π .

0

12345

1 2 3 4 58. ∫ 3

−2 |x | dx9.∫ 2

−2 (2 − |x |) dx


−2 (2 − |x |) dx = 12 (2)(2) + 1

2 (2)(2) = 4.

−2 −1.5 −1 −0.5 0 0.5 1 1.5 20

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

10. ∫ 1

−1 (2x − |x |) dx

11. Calculate∫ 6

0 (4 − x) dx in two ways:(a) limN→∞ RN

(b) Sketch the relevant signed area and use geometry.

Let f (x) = 4 − x over [0, 6]. Consider the integral∫ 6

0 f (x) dx = ∫ 6

0 4 − x dx .(a) Let n be a (symbolic) positive integer and set a = 0, b = 6,

h = �x = (b − a) /n = 6/n.


Let xk = a + kh = 6k/n, k = 1, 2, . . . , n be the right endpoints of the nsubintervals of [0, 6]. Then

Rn = hn∑

k=1

f (xk) = 6

n

n∑k=1

(4 − 6k

n

)= 6

n

(4

(n∑

k=1

1

)− 6

n

(n∑

k=1

k

))

= 6

n

(4n − 6

n

(n2

2+ n

2

))= 6 − 18

n

Hence limn→∞

Rn = limn→∞

(6 − 18

n

)= 6.

(b) Using geometry, the value of the integral∫ 6

0 4 − x dx is 12 (4) (4) − 1

2 (2) (2) = 6.

–2

0

2

4

2 4 6

12.Calculate

∫ 5

2 (2x + 1) dx in two ways: as a limit limN→∞ RN , and using geometry.13. Evaluate the following integrals, where f (x) is the function shown in Figure 1.

2 4 6

Figure 1 The two parts of the graph are semi-circles.

(a)∫ 2

0 f (x) dx

(b)∫ 6

0 f (x) dx

(c)∫ 4

1 f (x) dx

(d)∫ 6

1 | f (x)| dx

Let f (x) be given by the figure in the text.

(a) We have∫ 2

0 f (x) dx = − 12π (1)2 = − π

2.

(b) We have∫ 4

1 f (x) dx = 14π (2)2 − 1

4π (1)2 = 3

4π .

(c) We have∫ 6

0 f (x) dx = 12 π (2)2 − 1

2 π (1)2 = 3π

2 .

(d) We have∫ 6

1 | f (x)| dx = 12π (2)2 + 1

4π (1)2 = 9π

4.

In Exercises 14–17, sketch the signed area represented by the integral. Indicate the regions thatare treated as positive and negative.14. ∫ 2

0 (x − x2) dx


15.∫ 3

0 (2x − x2) dx

Here is a sketch of the signed area represented by the integral∫ 3

0 2x − x2 dx .

- - -

+ + +

–3

–2

–10

12 3

16. ∫ 2π

πsin x dx

17.∫ 3π

0 sin x dx

Here is a sketch of the signed area represented by the integral∫ 3π

0 sin x dx .

- - -

+ + +

–2

–1

01 2

In Exercises 18–21, calculate the Riemann sum R( f, P, C) for the given function, partition, andchoice of intermediate points. Also, sketch the graph of f and the rectangles corresponding toR( f, P, C).

18.f (x) = x , P = {1, 1.2, 1.5, 2}, C = {1.1, 1.4, 1.9}

19. f (x) = x2 + x , P = {2, 3, 4.5, 5}, C = {2, 3.5, 5}Let f (x) = x2x and P = {x0 = 2, x1 = 3, x3 = 4.5, x4 = 5} and

C = {c1 = 2, c2 = 3.5, c3 = 5}. Then R( f, P, C) = �x1 f (c1) + �x2 f (c2) + �x3 f (c3) =(3 − 2)(6) + (4.5 − 3)(15.75) + (5 − 4.5)(30) = 44.625. Here is a sketch of the graph of fand the rectangles.

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.50

5

10

15

20

25

30

35

40

x

x2+x

20.f (x) = x + 1, P = {−2, −1.6, −1.2, −.8, −.4, 0}, C = {−1.7, −1.3, −.9, −.5, 0}

21. f (x) = sin x , P = {0, π

6, π

3, π

2}, C = {.4, .7, 1.2}

Let f (x) = sin x and P = {x0 = 0, x1 = π

6 , x3 = π

3 , x4 = π

2 } andC = {c1 = .4, c2 = .7, c3 = 1.2}. ThenR( f, P, C) = �x1 f (c1) + �x2 f (c2) + �x3 f (c3) =( π

6 − 0)(sin .4) + ( π

3 − π

6 )(sin .7) + ( π

2 − π

3 )(sin 1.2) = 1.029225. Here is a sketch of thegraph of f and the rectangles.


0 0.5 1 1.50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

sin(x)

In Exercises 22–29, use the linearity properties of the integral and the formulas in Example ??to calculate the integrals.22. ∫ 4

1 x2 dx23.

∫ 3

0 (3x + 4) dx

We have∫ 3

0 3x + 4 dx = 3∫ 3

0 x dx + 4∫ 3

0 1 dx = 3 · 12 (3)2 + 4(3 − 0) = 51

2 = 25.5.24. ∫ 3

−2 (3x + 4) dx25.

∫ 3

−2 (3x2 − 4x) dx

We have∫ 3

−2 3x2 − 4x dx = ∫ 0

−2 3x2 − 4x dx + ∫ 3

0 3x2 − 4x dx = ∫ 3

0 3x2 − 4x dx − ∫ −2

0 3x2 − 4x dx

= 3∫ 3

0 x2 dx − 4∫ 3

0 x dx −(

3∫ −2

0 x2 dx − 4∫ −2

0 x dx)

=3 · 1

3 (3)3 − 4 · 12 (3)2 − (

3 · 13 (−2)3 − 4 · 1

2 (−2)2) = 25.

26. ∫ 1

0 (x2 − 2x) dx27.

∫ 3

0 (6x2 + 7x + 1) dx

We have∫ 3

0 6x2 + 7x + 1 dx = 6∫ 3

0 x2 dx + 7∫ 3

0 x + ∫ 3

0 1 dx =6 · 1

3 (3)3 + 7 · 12 (3)2 + (3 − 0) = 177

2= 88.5.

28. ∫ 1

−a (x2 + x) dx29.

∫ a2

a x2 dx

We have∫ a2

a x2 dx = ∫ a2

0 x2 dx − ∫ a

0 x2 dx = 13

(a2)3 − 1

3 (a)3 = 13a6 − 1

3a3.

30.Using the formula

∑Nj=1 j 3 = N 4

4+ N 3

2+ N 2

4, compute the limit of right-endpoint

approximations to show that for all b (positive or negative):∫ b

0

x3 = b4

4(1)

In Exercises 31–36, use Eq. (1) to evaluate the integral.

31.∫ 3

0 x3 dx

We have∫ 3

0 x3 dx = 34

4 = 20.25 using Eq. (1).32. ∫ 0

−3 x3 dx33.

∫ 0

2 (2x3 − 3x) dx

We have∫ 0

2 (2x3 − 3x) dx = − ∫ 2

0 (2x3 − 3x) dx = −2∫ 2

0 x3 dx + 3∫ 2

0 x dx =−2

(24

4

)+ 3

(22

2

)= −2 using Eq. (1).

34. ∫ 2

0 (x − x3) dx35.

∫ 3

1 x3 dx

We have∫ 3

1 x3 dx = ∫ 3

0 x3 dx − ∫ 1

0 x3 dx = 34

4 − 14

4 = 20 using Eq. (1).36. ∫ 2

1 (x − x3) dx


37. What is∫ 5

2 ( f (x) + g(x)) dx , assuming that∫ 5

2 f (x) dx = 7 and∫ 5

2 g(x) dx = 3?

We have∫ 5

2 ( f (x) + g(x)) dx = ∫ 5

2 f (x) dx + ∫ 5

2 g(x) dx = 7 + 3 = 10.

In Exercises 38–44, calculate the integral, assuming that∫ 5

0f (x) dx = 5,

∫ 5

0g(x) dx = 12

38. ∫ 5

0 ( f (x) + g(x)) dx39.

∫ 5

0 ( f (x) + 4g(x)) dx

We have∫ 5

0 f (x) + 4g(x) dx = ∫ 5

0 f (x) dx + 4∫ 5

0 g(x) dx = 5 + 4(12) = 53.40. ∫ 5

0 ( f (x) − g(x)) dx41.

∫ 5

0 4 f (x) dx

We have∫ 5

0 4 f (x) dx = 4∫ 5

0 f (x) dx = 4(5) = 20.42. ∫ 0

5 g(x) dx43.

∫ 5

0 (3 f (x) − 5g(x)) dx

We have∫ 5

0 3 f (x) − 5g(x) dx = 3∫ 5

0 f (x) dx − 5∫ 5

0 g(x) dx = 3(5) − 5(12) = −45.44.

Is it possible to calculate∫ 5

0 g(x) f (x) dx from the information given?45. Calculate

∫ 6

0 f (x) dx , assuming that∫ 2

0 f (x) dx = 7 and∫ 6

2 f (x) dx = 5.

Let∫ 2

0 f (x) dx = 7 and∫ 6

2 f (x) dx = 5. Then∫ 6

0 f (x) dx cannot be computed from theinformation given. However, we do have∫ 6

0 f (x) dx = ∫ 2

0 f (x) dx + ∫ 6

2 f (x) dx = 7 + 5 = 12.

In Exercises 46–49, calculate the following integrals, assuming that∫ 1

0

f (x) dx = 1,

∫ 2

0

f (x) dx = 4,

∫ 4

1

f (x) dx = 7

46. ∫ 4

0 f (x) dx47.

∫ 2

1 f (x) dx

Given∫ 1

0 f (x) dx = 1,∫ 2

0 f (x) dx = 4, and∫ 4

1 f (x) dx = 7, we have∫ 2

1 f (x) dx = ∫ 2

0 f (x) dx − ∫ 1

0 f (x) dx = 4 − 1 = 3.48. ∫ 1

4 f (x) dx49.

∫ 4

2 f (x) dx

Given∫ 1

0 f (x) dx = 1,∫ 2

0 f (x) dx = 4, and∫ 4

1 f (x) dx = 7, recall from Exercise ?? that∫ 4

0 f (x) dx = 8. Accordingly, we have∫ 4

2 f (x) dx = ∫ 4

0 f (x) dx − ∫ 2

0 f (x) dx = 8 − 4 = 4.

In Exercises 50–55, express each of the following as a single integral.50. ∫ 3

0 f (x) dx + ∫ 7

3 f (x) dx51.

∫ 9

2 f (x) dx − ∫ 9

4 f (x) dx

We have∫ 9

2 f (x) dx − ∫ 9

4 f (x) dx = ∫ 4

2 f (x) dx .52. ∫ 9

2 f (x) dx − ∫ 5

2 f (x) dx


53.∫ 3

7 f (x) dx + ∫ 9

3 f (x) dx

We have ∫ 3

7

f (x) dx +∫ 9

3

f (x) dx = −∫ 7

3

f (x) dx +∫ 9

3

f (x) dx

=∫ 9

7

f (x) dx .

54. ∫ 3

1 f (x) dx + ∫ 1

−4 f (x) dx55.

∫ 6

−4 f (x) dx + ∫ −4

0 f (x) dx

We have∫ 6

−4 f (x) dx + ∫ −4

0 f (x) dx = ∫ 0

−4 f (x) dx + ∫ 6

0 f (x) dx − ∫ 0

−4 f (x) dx = ∫ 6

0 f (x) dx .

In Exercises 56–60, calculate the integral assuming that f is a function such that∫ b

1 f (x) dx = 1 − b−1 for all b > 0.56. ∫ 3

1 f (x) dx57.

∫ 112

4 f (x) dx

We have∫ 1

12

4 f (x) dx = −4∫ 1/2

1 f (x) dx = −4(

1 − (12

)−1)

= 4.58. ∫ 4

1 (4 f (x) − 2) dx59.

∫ 1

3 f (x) dx

We have∫ 1

3 f (x) dx = − ∫ 3

1 f (x) dx = −(1 − 3−1) = − 23 .

60. ∫ 2c

c f (x) dx (c any positive number)61. Which of the following statements is false? Explain.

(a) If f (x) > 0 for x ∈ [a, b], then the integral∫ b

a f (x) dx is also positive.

(b) If∫ b

a f (x) dx ≥ 0, then f (x) ≥ 0 for x ∈ [a, b].Sketch the graph of a function providing a counterexample for the statement which is notvalid.

(a) It is true that if f (x) ≥ 0 for x ∈ [a, b], then∫ b

a f (x) dx ≥ 0 by the ComparisonTheorem with g(x) = 0.

(b) It is false that if∫ b

a f (x) dx ≥ 0, then f (x) ≥ 0 for x ∈ [a, b]. Indeed, in Exercise 3,

we saw that∫ 1

−2 3x + 4 dx = 7.5 ≥ 0, yet f (−2) = −2 < 0. Here is the graph fromthat exercise.

–2

7

–2 –1 1

62.Explain the difference in graphical interpretation between

∫ b

a f (x) dx and∫ b

a | f (x)| dxwhen f (x) takes on both positive and negative values.63. Verify the inequalities:∫ 1

0x3 dx ≤

∫ 1

0x2 dx,

∫ 2

1x2 dx ≤

∫ 2

1x3 dx

We have∫ 1

0 x3 dx = 14 (1)4 = 1

4≤ 1

3= 1

3 (1)3 = ∫ 1

0 x2 dx ; that is,∫ 1

0 x3 dx ≤ ∫ 1

0 x2 dx .


We have∫ 2

1 x2 dx = ∫ 2

0 x2 dx − ∫ 1

0 x2 dx = 13(2)3 − 1

3(1)3 = 7

3= 2 1

3≤ 3 3

4= 15

4=

14 (2)4 − 1

4 (1)4 = ∫ 2

0 x3 dx − ∫ 1

0 x3 dx = ∫ 2

1 x3 dx ; that is,∫ 2

1 x2 dx ≤ ∫ 2

1 x3 dx .

64.Prove that

.0198 ≤∫ .3

.2

sin x dx ≤ .0296

Hint: show that .198 ≤ sin x ≤ .296 for x in [.2, .3].

65. Prove that

.27 ≤∫ .3

0

cos x dx ≤ .3

Hint: show that .9 ≤ cos x ≤ 1 for x in [0, .3].For 0 ≤ x ≤ π

6≈ 0.52, we have d/dx (cos x) = − sin x ≤ 0. Hence cos x is nonincreasing

on [0.0, 0.3]. Accordingly, for 0.0 ≤ x ≤ 0.3, we have

m = 0.9 ≤ 0.955 ≈ cos 0.3 ≤ cos x ≤ cos 0.0 = 1 ≤ 1 = M

Therefore, by the Comparison Theorem, we have0.27 = m(0.3 − 0.0) = ∫ 0.3

0.0 m dx ≤ ∫ 0.3

0.0 cos x dx ≤ ∫ 0.3

0.0 M dx = M(0.3 − 0.0) = 0.30. In

other words, 0.27 ≤ ∫ 0.3

0.0 cos x dx ≤ 0.30.

Further Insights and Challenges66.

Evaluate∫ a

−a f (x) dx assuming that f is an odd function, that is, f (−x) = − f (x) forall x . Explain.67. Let k > 0 be a positive exponent. Show that∫ b

0xk dx = bk+1

∫ 1

0xk dx

for all b > 0. Hint: Compare the right-endpoint approximations for the integrals of xk over[0, a] and [0, 1].Let k > 0 be a positive exponent and b be any positive number. Let f (x) = xk on [0, b].Since f is continuous, both

∫ b

0 f (x) dx and∫ 1

0 f (x) dx exist.

Let n be a (symbolic) positive integer and set A = 0, B = b,h = �x = (B − A) /n = b/n. Let x j = A + jh = bj/n, j = 1, 2, . . . , n be the rightendpoints of the n subintervals of [0, b]. Then the right endpoint approximation to∫ b

0 f (x) dx = ∫ b

0 xk dx is

Rn = hn∑

j=1

f (x j ) = b

n

n∑j=1

(bj

n

)k

= bk+1

(1

nk+1

n∑j=1

j k

)

In particular, if b = 1 above, then the right endpoint approximation to∫ 1

0 f (x) dx = ∫ 1

0 xk dx is

Sn = hn∑

j=1

f (x j ) = 1

n

n∑j=1

(j

n

)k

= 1

nk+1

n∑j=1

j k = 1

bk+1Rn

In other words, Rn = bk+1 Sn. Therefore,∫ b

0 xk dx = limn→∞ Rn = limn→∞ bk+1 Sn = bk+1 limn→∞ Sn = bk+1∫ 1

0 xk dx . In other

words,∫ b

0 xk dx = bk+1∫ 1

0 xk dx .

68.Verify the formula by interpreting the integral as an area:∫ b

0

√1 − x2 dx = 1

2b√

1 − b2 + 1

2θ

Here 0 ≤ b ≤ 1 and θ is the angle between 0 and π

2 such that sin θ = b.


69. Prove the inequality ∫ π/4

0sin(x2) dx ≤ .162

Hint: The inequality sin x/x ≤ 1 implies that sin(x2) ≤ x2 for x in [0, π

4].

Note that 0 < π

4< 1 impies ( π

4)2 < π

4. Accordingly, for x ∈ [0, π

4], the inequality

sin u/u ≤ 1 implies sin(x2) ≤ x2 for 0 ≤ x ≤ π

4 . By the Comparison Theorem, we have∫ π/4

0 sin(x2) dx ≤ ∫ π/4

0 x2 dx = 13( π

4)3 ≈ 0.16149 < 0.162. We therefore conclude that∫ π/4

0 sin(x2) dx ≤ 0.162.

1

2

4π

2π

70.Theorem ?? states the additivity property of definite integrals for adjacent intervalsassuming that a ≤ b ≤ c. Show that the conclusion remains true for arbitrary a, b and c.Hint: use the relation

∫ b

a f (x) dx = − ∫ a

b f (x) dx .

71. The midpoint approximation MN is generally more accurate than either RN or L N . In thisexercise, we verify this for f (x) = x2 on [0, 1].(a) Show that for all N ≥ 1,

RN ( f ) = 1

3+ 1

2N+ 1

6N 2

L N ( f ) = 1

3− 1

2N+ 1

6N 2

MN ( f ) = 1

3− 1

12N 2

(b) What is the size of the error (in terms of N ) in the three approximations? Rank thethree approximations in order of increasing accuracy.

(c) How large must N be in order to make the error in MN smaller than .001?(d) Find an N such that the error in RN and L N is smaller than .001.

Let f (x) = x2 on [0, 1].(a) Let n be a (symbolic) positive integer and set a = 0, b = 1,

h = �x = (b − a) /n = 1/n. Let xk = a + kh = k/n, k = 0, 1, . . . , n and letx∗

k = a + (k + 12)h = (k + 1

2)/n, k = 0, 1, . . . , n − 1. Then for n ≥ 1 we have

Rn = hn∑

k=1

f (xk) = 1

n

n∑k=1

(k

n

)2

= 1

n3

n∑k=1

k2

= 1

n3

(n3

3+ n2

2+ n

6

)= 1

3+ 1

2n+ 1

6n2

Ln = hn−1∑k=0

f (xk) = 1

n

n−1∑k=0

(k

n

)2

= 1

n3

n−1∑k=1

k2

= 1

n3

((n − 1)3

3+ (n − 1)2

2+ n − 1

6

)= 1

3− 1

2n+ 1

6n2

5.3 The Fundamental Theorem of Calculus, Part I 25

Mn = hn−1∑k=0

f (x∗k ) = 1

n

n−1∑k=0

(k + 1

2

n

)2

= 1

n3

n−1∑k=0

(k2 + k + 1

4

)

= 1

n3

((n−1∑k=1

k2

)+(

n−1∑k=1

k

)+ 1

4

(n−1∑k=0

1

))

= 1

n3

(((n − 1)3

3+ (n − 1)2

2+ n − 1

6

)

+(

(n − 1)2

2+ n − 1

2

)+ 1

4n

)= 1

3− 1

12n2

(b) The error of Rn is given by1

2n+ 1

6n2, the error of Ln is given by − 1

2n+ 1

6n2and the

error of Mn is given by − 1

12n2. Of the three approximations, Rn is the least accurate,

then Ln and finally Mn is the most accurate.

(c) The absolute error in Mn is1

12n2. In order to have

1

12n2<

1

1000, we require

n2 >1000

12or n >

√1000/12 ≈ 22.36; i.e., n ≥ 23 (since n is a positive integer).

(d) The absolute error in Rn is1

2n+ 1

6n2. In order to have

1

2n+ 1

6n2<

1

1000, take

n > 500; i.e., n ≥ 501 (since n is a positive integer).

5.3 The Fundamental Theorem of Calculus, Part I


1. What is the area under the graph of a positive function f over [0, 2], assuming that f has anantiderivative F(x) such that F(0) = 3 and F(2) = 7?

2. Find∫ 5

2 f (x) dx assuming that F(x) is an antiderivative of f (x) and F(5) = F(2) + 12.

3. Suppose that f is a negative function with the antiderivative F such that F(1) = 7 andF(3) = 4. Does the quantity F(3) − F(1) have an interpretation in terms of area?

4. Evaluate∫ 7

0 f (x) dx and∫ 7

2 f (x) dx assuming that f (x) has an antiderivative F(x) withvalues in the table.

x 0 2 7F(x) 3 7 9

5. Answer true or false and explain.(a) The FTC I is only valid for positive functions.(b) To use the FTC I, you have to choose the right antiderivative.(c) If you cannot find an antiderivative of f (x), then the definite integral does not exist.


Exercises

In Exercises 1–35, evaluate the definite integral using the FTC.

1.∫ 6

3 x dx

We have∫ 6

3 x dx = 12 x2∣∣6

3= 1

2 (6)2 − 12 (3)2 = 27

2 = 13.5.2. ∫ 9

0 2 dx3.∫ 3

1 (x3 − x2) dx

We have∫ 3

1 (x3 − x2) dx = 14x4 − 1

3x3∣∣3

1= (

1434 − 1

333)+ (

14− 1

3

) ≈ 11.333.4. ∫ π

20 cos x dx

5.∫ 3π

20 cos x dx

We have∫ 3π/2

0 cos x dx = sin x |3π/20 = sin( 3π

2 ) − sin(0) = −1.6. ∫ 1

0 (4 − 5x4) dx7.∫ 4

0

√x dx

We have∫ 4

0

√x dx = ∫ 4

0 x1/2 dx = 23 x3/2

∣∣40= 2

3 (4)3/2 − 23 (0)3/2 = 16

3 = 5 13 .

8. ∫ 4

−3(x2 + 2) dx9.∫ 4

0 (3x5 + x2 − 2x) dx

We have∫ 4

0 3x5 + x2 − 2x dx = (12x6 + 1

3x3 − x2

)∣∣40=(

12(4)6 + 1

3(4)3 − (4)2

)− (12(0)6 + 1

3(0)3 − (0)2

) = 61603

= 2053 13.

10. ∫ 4

1

1

t2dt11.

∫ 2

1 (10x9 − 3x4 + 9) dx

We have∫ 2

1 10x9 − 3x4 + 9 dx = (x10 − 3

5x5 + 9x

)∣∣21=(

(2)10 − 35(2)5 + 9(2)

)− ((1)10 − 3

5(1)5 + 9(1)

) = 50675

= 1013 25.

12. ∫ 2

−2(10x9 − 3x4 + 9) dx13.

∫ 2

−3 u2 du

We have∫ 2

−3 u2 du = 13u3∣∣2−3

= 13(2)3 − 1

3(−3)3 = 35

3= 11 2

3.

14. ∫ −1

−3 (4t3/2 + t7/2) dt15.

∫ 3

−3(4u3 − 7u) du

We have∫ 3

−3 4u3 − 7u du = (u4 − 7

2 u2)∣∣3

−3= (

(3)4 − 72 (3)2

)− ((−3)4 − 7

2 (−3)2) = 0.

16. ∫ 9

1 3x1/2 dx17.

∫ 4

2 π 2 dx

We have∫ 4

2 π 2 dx = π 2x∣∣4

2= π 2(4) − π 2(2) = 2π 2.

18. ∫ 2

1 (x2 − x−2) dx19.

∫ 4

0 (3x5 + x2 − 2x) dx

We have∫ 4

0 3x5 + x2 − 2x dx = (12x6 + 1

3x3 − x2

)∣∣40=(

12(4)6 + 1

3(4)3 − (4)2

)− (12(0)6 + 1

3(0)3 − (0)2

) = 61603

= 2053 13.

20. ∫ 4

1 x−4 dx21.

∫ 9

1 3x−1/2 dx

We have∫ 9

1 3x−1/2 dx = 6x1/2∣∣9

1= 6(9)1/2 − 6(1)1/2 = 12.

22. ∫ 9

4

8

x3dx


23.∫ −1

−2

1

x3dx

We have∫ −1

−2

1

x3dx = −1

2x−2

∣∣∣∣−1

−2

= −1

2(−2)−2 + 1

2(−1)−2 = −1

4.

24. ∫ 3π4

π4

cos x dx25.

∫ π2

π4

(cos θ − sin θ) dθ

We have∫ π/2

π/4 cos θ − sin θ dθ = (sin θ + cos θ)|π/2π/4 = (

sin π

2 + cos π

2

)− (sin π

4 + cos π

4

) = 1 − √2.

26. ∫ 4π

2πcos x dx

27.∫ π

40 sec2 t dt

We have∫ π/4

0 sec2 t dt = tan t |π/40 = tan π

4− tan 0 = 1.

28. ∫ π4

0 sec x tan x dx29.

∫ π3

π6

csc x cot x dx

We have∫ π/3

π/6 csc x cot x dx = (− csc x)|π/3π/6 = (− csc π

3) − (− csc π

6) = 2 − 2

3

√3.

30. ∫ π2

π6

csc2 x dx

31.∫ 5

0 |3 − x | dx Hint: write as a sum of two integrals.

We have∫ 5

0 |3 − x | dx = ∫ 3

0 (3 − x) dx + ∫ 5

3 (x − 3) dx =(3x − 1

2x2)∣∣3

0+ ( 1

2x2 − 3x)

∣∣53= (9 − 9

2) − 0 + ( 25

2− 15) − ( 9

2− 9) = 6.5.

32. ∫ 3

0 |x2 − 1| dx33.

∫ 3

−2 |x3| dx

We have∫ 3

−2 |x3| dx = ∫ 0

−2 −x3 dx +∫ 3

0 x3 dx = − 14x4∣∣0−2

+ 14x4∣∣3

0= 0+ 1

4(−2)4 + 1

434 −0 = 24.25.

34. ∫ π

0 | cos x | dx35.

∫ 5

0 |x2 − 4x + 3| dx

We have∫ 5

0 |x2−4x +3| dx = ∫ 5

0 |(x −3)(x −1)| dx = ∫ 1

0 (x2−4x +3) dx +∫ 3

1 −(x2−4x +3) dx +∫ 5

3 (x2 − 4x + 3) dx = ( 13 x3 − 2x2 + 3x)

∣∣10− ( 1

3 x3 − 2x2 + 3x)∣∣31+ ( 1

3 x3 − 2x2 + 3x)∣∣53=

( 13− 2 + 3) − 0 − (9 − 18 + 9) + ( 1

3− 2 + 3) + ( 125

3− 50 + 15) − (9 − 18 + 9) ≈ 9.333.

36.Use the FTC to show that

∫ 1

−1 xn dx = 0 if n is odd. Explain graphically.

In Exercises 37–42, evaluate the integral in terms of the constant.

37.∫ 4

2 c dx

We have∫ 4

2 c dx = cx |42 = c · 4 − c · 2 = 2c for any constant c.

38. ∫ a

b x4 dx39.

∫ b

1 x5 dx

We have∫ b

1 x5 dx = 16x6∣∣b

1= 1

6b6 − 1

6(1)6 = 1

6(b6 − 1) for any number b.

40. ∫ 4

a x dx41.

∫ a

b c dx

We have∫ b

a c dx = cx |ba = c · b − c · a = (b − a)c for any constant c.

42. ∫ 4

a (3x2 − x) dx


43. Suppose that f is a negative function with an antiderivative F such that F(1) = 7 andF(3) = 4. Can you compute the area (a positive number) between the x-axis and the graphon [1, 3]?The area is A = ∫ 1

3 f (x) dx = − ∫ 3

1 f (x) dx = −(F(3) − F(1)) = −(4 − 7) = 3.44.

Does the integral∫ 1

0 xn dx get larger or smaller as n increases? Make a guess based ongraphs and then check by calculation.45. Calculate

∫ 3

−2 f (x) dx , where

f (x) ={

12 − x2 for x ≤ 2x3 for x ≥ 2

Hint: break up the integral into two integrals corresponding to the two parts of the graph.

–2 –1 1 2 3

10

20

Figure 1 Graph of f (x) in Exercise 45.

Let f (x) ={

12 − x2 for x ≤ 2

x3 for x ≥ 2. Then

∫ 3

−2 f (x) dx = ∫ 2

−2 f (x) dx + ∫ 3

2 f (x) dx = ∫ 2

−2 12 − x2 dx + ∫ 3

2 x3 dx .

Now∫ 2

−2 12 − x2 dx = (12x − 13 x3)

∣∣2−2

= (12(2) − 13 (2)3) − (12(−2) − 1

3 (−2)3) = 1283 .

Moreover,∫ 3

2 x3 dx = 14 x4∣∣3

2= 1

4 (3)4 − 14 (2)4 = 65

4 .

Therefore,∫ 3

−2 f (x) dx = ∫ 2

−2 f (x) dx + ∫ 3

2 f (x) dx = 1283

+ 654

= 70712

= 58 1112

.


In Exercises 46–49, express the following integrals in terms of the constant (a, c, etc).46. ∫ π

0 cos(ax) dx Hint: −a−1 sin ax is an antiderivative.47.

∫ 1

0 (x + k)3 dx Hint: 14(x + k)4 is an antiderivative.

We have∫ 1

0 (x + k)3 dx = 14(x + k)4

∣∣10= 1

4(k + 1)4 − 1

4k4 = k3 + 3

2k2 + k + 1

4.

48. ∫ c

0 cos(ct) dt49.

∫ 2c

−c x4 dx

We have∫ 2c

−c x4 dx = 15x5∣∣2c

−c= 1

5(2c)5 − 1

5(−c)5 = 33

5c5.

50.Show that

limn→∞

1

nk+1

n∑j=0

j k = 1

k + 1

Hint: interpret the left-hand side as an integral and evaluate using the FTC.


51. Let c > 0 and let A be the area under the graph of the inverted parabola y = c − ax2. Showthat A = 2

3R where R is the area of the circumscribed rectangle.

c

Figure 2 Graph of y = c − ax2.

Let c > 0 and a > 0. The inverted parabola y = f (x) = c − ax2 has x-intercepts ±√c/a.

Accordingly, the area between the inverted parabola and the x-axis is

A =∫ √

c/a

−√c/a

c − ax2 dx =(

cx − 1

3ax3

)∣∣∣∣√

c/a

−√c/a

= x

(c − 1

3ax2

)∣∣∣∣√

c/a

−√c/a

= 2

√c

a

(c − 1

3a( c

a

))=

43c3/2

a1/2

52.

Applying Theorem ?? to the inequality cos t ≤ 1 gives∫ x

0 cos t dt ≤ ∫ x

0 1 dt for all x .(a) Use this to conclude that sin x ≤ x .(b) Integrate the inequality in (a) to prove that − cos x + 1 ≤ x2/2.(c) Continuing to integrate, show that − sin x + x ≤ x3/6 and cos x − 1 + x2/2 ≤ x4/24.(d) Deduce the inequalities

1 − x2

2≤ cos x ≤ 1 − x2

2+ x4

24, 1 − x3

3≤ sin x ≤ x

(e) Use (d) to show that .9687 < cos 14

< .969.

53. (a) Use the method of the previous exercise to prove that

x − x3

6+ x5

120− x7

5040≤ sin x ≤ x − x3

6+ x5

120

1 − x2

2+ x4

24− x6

720≤ cos x ≤ 1 − x2

2+ x4

24

(b) Use this to show that .9689124 < cos 14

< .9689128.(c) Can you guess the general of inequalities for sin x and cos x?

(a) Integrating the inequality cos x − 1 + x2

2≤ x4

24gives sin x + x3

6− x ≤ x5

120and

equivalently, sin x ≤ x − x3

6+ x5

120. Integrating again, we have

− cos x + x4

24 − x2

2 + 1 ≤ x6

720 and hence 1 − x2

2 + x4

24 − x6

720 ≤ cos x . Finally, integrating

a third time gives − sin x + x5

120 − x3

6 + x ≤ x7

5040 and equivalently

x − x3

6+ x5

120− x7

5040≤ sin x . Thus by combining these inequalities we have

x − x3

6+ x5

120− x7

5040≤ sin x ≤ x − x3

6+ x5

120and

1 − x2

2+ x4

24− x6

720≤ cos x ≤ 1 − x2

2+ x4

24.

(b) For x = 14 , 1 − (1/4)2

2 + (1/4)4

24 − (1/4)6

120 ≈ .9689124 and 1 − (1/4)2

2 + (1/4)4

24 ≈ .9689128.Thus .9689124 < cos 1

4 < .9689128.(c) The general statement of the inequalities is

1 − x2

2+ x4

24− x6

720+ · · · − 1

(2n+2)! x2n+2 ≤ cos x ≤ 1 − x2

2+ x4

24+ · · · + 1

(2n)! x2n and

x − x3

6+ x5

120− x7

5040+ · · · − 1

(2n+1)! x2n+1 ≤ sin x ≤ x − x3

6+ x5

120+ · · · + 1

(2n−1)! x2n−1.


5.4 The Fundamental Theorem of Calculus, Part II


1. What is A(−2) where A(x) = ∫ x

−2 f (t) dt?

2. Let G(x) = ∫ x

4

√t3 + 1dt . Answer true or false and explain:

(a) The FTC is needed to calculate G(4).(b) The FTC is needed to calculate G ′(4).(c) G(2) = − ∫ 4

2

√t3 + 1 dt .

3. Which of the following is an antiderivative of√

x satisfying F(2) = 0?(a)

∫ x

212 x−1/2 dx

(b)∫ 2

0 x1/2 dx

(c)∫ x

2 x1/2 dx

4. Find F(1) where F(x) = ∫ x

1 sin(sin t) dt . Is it possible to compute F ′(π)?

5. Let G(x) = ∫ x3

4 sin(t) dt . Which of the following statements are correct?(a) G(x) is the composite function sin(x3).(b) G(x) is the composite function A(x3) where A(x) = ∫ x

4 sin(t) dt .(c) G(x) is too complicated to differentiate.(d) The Product Rule is used to differentiate G(x).(e) The Chain Rule is used to differentiate G(x).(f) G ′(x) = sin(x3) · (3x2)

6. (Adapted from: Calculus Problems for a New Century, p. 103.) Trick question: find thederivative of

∫ 3

1 t3 dt at x = 2.

7. What is the value of∫ 9

2 f ′(x) dx if f is a differentiable function such that f (2) = f (9) = 4?

Exercises

1. Let f (x) = 2x + 4.(a) Write the area function with lower limit a = −2 as an integral.(b) Find a formula for this area function.

–2 4

4

12

Figure 1

Let f (x) = 2x + 4.(a) The area function with lower limit a = −2 is A(x) = ∫ x

a f (t) dt = ∫ x

−2 2t + 4 dt .

(b) We have∫ x

−2 2t + 4 dt = (t2 + 4t)∣∣x−2

= (x2 + 4x) − ((−2)2 + 4(−2)) = x2 + 4x + 4or (x + 2)2. Therefore, A(x) = (x + 2)2.

5.4 The Fundamental Theorem of Calculus, Part II 31

2.Find a formula for the area function of f (x) = 2x + 4 with lower limit a = 0.

3. Let G(x) = ∫ x

1 (t2 − 2) dt .(a) What is G(1)?(b) Use the FTC Part II to find G ′(1) and G ′(2).(c) Find a formula for G(x) and use it to verify your answers to (a) and (b).

Let G(x) = ∫ x

1 t2 − 2 dt .

(a) Then G(1) = ∫ 1

1 t2 − 2 dt = 0.(b) Now G ′(x) = x2 − 2, whence G ′(1) = −1 and G ′(2) = 2.(c) We have

∫ x

1 t2 − 2 dt = ( 13t3 − 2t)

∣∣x1

= ( 13x3 − 2x) − ( 1

3(1)3 − 2(1)) = 1

3x3 − 2x + 5

3.

Thus G(x) = 13x3 − 2x + 5

3and G ′(x) = x2 − 2. Moreover,

G(1) = 13(1)3 − 2(1) + 5

3= 0, as in (a), and G ′(1) = −1 and G ′(2) = 2, as in (b).

4.True or false: some continuous functions don’t have antiderivatives. Explain.

5. Find F(0), F ′(0) and F ′(3) for the function F(x) = ∫ x

0

√t3 + 1 dt .

Let F(x) = ∫ x

0

√t3 + 1 dt . Then F ′(x) = √

x3 + 1. Thus F(0) = 0, F ′(0) = 1, andF ′(3) = √

28.6.

Let G(x) be the antiderivative of x−2 defined by G(x) = ∫ x

3 t−2 dt . Which initial conditiondoes G(x) satisfy?7. Let F(x) be the antiderivative of f (x) = √

x3 + 1 such that F(3) = 0.(a) Represent F as a definite integral.(b) What is F ′(3)?

Let F(x) be the antiderivative of f (x) = √x3 + 1 satisfying F(3) = 0.

(a) Then F(x) = ∫ x

3

√t3 + 1 dt .

(b) Since F ′(x) = √x3 + 1, we have F ′(3) = √

28.

In Exercises 8–15, find formulas for the functions represented by the integrals.

8. ∫ x

2 u3 du9.∫ x

π4

cos u du

We have F(x) = ∫ x

π/4 cos u du = sin u|xπ/4 = sin x −

√2

2 .10. ∫ x2

1 t dt11.

∫ x

2 (t2 − t) dt


2 t2 − t dt = ( 13t3 − 1

2t2)∣∣ = 1

3x3 − 1

2x2 − 2

3.

12. ∫ 5

x (4t − 1) dt13.

∫ x

0 sin u du


0 sin u du = (− cos u)|x0 = 1 − cos x .

14. ∫ xπ2

sin u du15.

∫ √x

2 t3 dt

We have F(x) = ∫ √x

2 t3 dt = t4

4

∣∣∣∣√

x

2

= x2

4− 4.

16.Show that

∫ x

0 |t | dt is equal to 12x |x |.

In Exercises 17–20, express the antiderivative of f (x) satisfying the given initial condition as anintegral.


17. f (x) = x−3, F(1) = 0

The antiderivative F(x) of f (x) = x−3 satisfying F(1) = 0 is F(x) = ∫ x

1 t−3 dt .18.

f (x) = x + 1

x2 + 9, F(7) = 019. f (x) = sec x , F(0) = 0

The antiderivative F(x) of f (x) = sec x satisfying F(0) = 0 is F(x) = ∫ x

0 sec t dt .20.

f (x) = sin(x3), F(−π) = 021. Let A(x) = ∫ x

0 f (t) dt where f (x) is shown in Figure 2.(a) Calculate A(2), A(3).(b) Calculate A′(2), A′(3).(c) Find a formula for A(t) (actually two formulas, one for 0 ≤ t ≤ 2 and one for t ≥ 2).

Then sketch the graph of A(t).

1 2 3 4

1

2

3

4

Figure 2

(a) A(2) = 2 · 2 = 4, the area under f (x) from x = 0 to x = 2. A(3) = 2 · 3 + 12

= 6.5,the area under f (x) from x = 0 to x = 3.

(b) A′(x) = f (x) so A′(2) = f (2) = 2 and A′(3) = f (3) = 3.(c) Here is a graph of the area function A(x) of the function f (x) depicted in Figure 2.

0 0.5 1 1.5 2 2.5 3 3.5 40

1

2

3

4

5

6

7

8

Area function F(x)

0

2468

10

1 2 3 422.

Make a rough sketch of the graph of the area function of the function shown in Figure 3.

Figure 3

23. Let G(x) = ∫ g(x)

a f (t) dt .

(a) Show that G(x) is a composite function: G(x) = A(g(x)) where A(x) = ∫ x

a f (t) dt .(b) Show that G ′(x) = A′(g(x))g′(x) = f (g(x)) g′(x)


Let G(x) = ∫ g(x)

a f (t) dt .

(a) Let A(x) = ∫ x

a f (t) dt and g(x) be specified. Then G = A ◦ g is defined by the

formula G(x) = A(g(x)) = ∫ g(x)

a f (t) dt .(b) Since A′(x) = f (x), we have by the Chain Rule that

G ′(x) = A′(g(x))g′(x) = f (g(x))g′(x).

In Exercises 24–31, use the Chain Rule to calculate the derivative.24.

G ′(x) where G(x) =∫ x3

3tan t dt

25. G ′(1) where G(x) =∫ x2

0

√x3 + 3 dt

Let G(x) = ∫ x2

0

√x3 + 3 dt . Then by Exercise 23, G ′(x) = √

x3 + 3 · 2x andG ′(1) = √

1 + 3 · 2 = 4.26.d

dx

∫ x2

0

sin2 t dt27.

d

dx

∫ 0

x

sin2 t dt

Let G(x) = ∫ 0

x sin2 t dt = − ∫ x

0 sin2 t dt . Then by Exercise 23, G ′(x) = − sin2 x .28.d

dx

∫ cos x

−6

(t4 − 3t) dt29. F ′(x) where F(x) =

∫ x4

x2

√t dt

Hint: F(x) = A(x4) − A(x2) where A(x) = ∫ x

0

√t dt (any lower limit in the integral will

work).

Let F(x) = ∫ x4

x2

√t dt = ∫ x4

0

√t dt − ∫ x2

0

√t dt . Applying Exercise 23, we have

F ′(x) = √x4 · 4x3 − √

x2 · 2x or 4x5 − 2x |x |.30.d

dx

∫ x2

√x

tan t dt .31.

d

dx

∫ 4

x

sin(t2) dt

Let G(x) = ∫ 4

x sin(t2) dt = − ∫ x

4 sin(t2) dt . Then G ′(x) = − sin(x2).32.

Let F(x) = ∫ x

0 (t2 − 5t − 6) dt .(a) Find the critical points of F(x) and determine if they are local minima or maxima.(b) Find the points of inflection of F(x).

33. Let F(x) = ∫ x

0 sin3 t dt . Find the critical points of F and determine if they are localminima or maxima.

Let F(x) = ∫ x

0 sin3 t dt . Solve F ′(x) = sin3 x = 0 to obtain x = nπ , where n is an integer.

For x = 2kπ , where k is an integer, there is a local minimum value of F(x) since F ′

changes sign from − to + as x increases through x = 2kπ .For x = (2k + 1)π , where k is an integer, there is a local maximum value of F(x) sinceF ′ changes sign from + to − as x increases through x = (2k + 1)π .

34.

Let f (x) be a differentiable function on [a, b] and let F(x) be an antiderivative of f (x).Which of the following statements is incorrect? Explain.

(a) If f (x) > 0, then F(x) has no local minimum on (a, b).(b) If c is an inflection point of F(x), then f ′(c) = 0.(c) If f ′(x) > 0, then F(x) must be concave up.(d) If f ′(c) = 0, then F(x) must have an inflection point at x = c.(e) If F(x) is decreasing on (a, b), then f (x) ≤ 0 for x ∈ (a, b).

35. Match the property of the area function A(x) with the corresponding property of thefunction f (x).

area function A(x):(a) A(x) decreasing(b) A(x) has a local maximum(c) A(x) is concave up(d) A(x) goes from concave up to concave down


graph of f (x):(i) lies below x-axis(ii) crosses x-axis from positive to negative(iii) local maximum(iv) f (x) is increasing

Let A(x) = ∫ x

a f (t) dt be an area function of f (x). Then A′(x) = f (x) andA′′(x) = f ′(x).(a) Hence A(x) is decreasing when A′(x) = f (x) < 0, i.e., when f (x) lies below the

x-axis. This is choice (i).(b) Now A(x) has a local maximum (at x0) when A′(x) = f (x) changes sign from + to 0

to − as x increases through x0, i.e., when f (x) crosses the x-axis from positive tonegative. This is choice (ii).

(c) Moreover, A(x) is concave up when A′′(x) = f ′(x) > 0, i.e., when f (x) is increasing.This corresponds to choice (iv).

(d) Finally, A(x) goes from concave up to concave down (at x0) when A′′(x) = f ′(x)

changes sign from + to 0 to − as x increases through x0, i.e., when f (x) has a localmaximum at x0. This is choice (iii).

36.

The following questions refer to A(x) = ∫ x

0 f (t)dt where f (x) is shown in Figure 4.(a) Does A(x) have a local maximum at A?(b) Where does A(x) have a local minimum?(c) Where does A(x) have a local maximum?(d) True or false: A(x) < 0 for all x .(e) Describe the change in concavity in A(x) at B.

Figure 4 Graph of f (x).

37. Let A(x) = ∫ x

0 f (t) dt where f (x) is shown in Figure 5.(a) Calculate the minimum and maximum value of A(x) on [0, 6].(b) Find a formula for A(x) valid on the interval [2, 4].(c) Find a formula for the function B(x) = ∫ x

2 f (t) dt valid on the interval [2, 4].

1 2 3 4 5 6

–3–2–101234

Figure 5

(a) At x = 1.5, f (x) changes from − to +, which corresponds to a minimum value ofA(x). HenceA(1.5) = ∫ 1.5

0 f (x) = ∫ 1

0 −1 dt + ∫ 1.5

1 (2t − 3) dt = −t |10 + t2 − 3t

∣∣1.5

1= −1.25.

At x = 4.5, f (x) changes from + to −, which corresponds to a maximum value ofA(x). Hence A(4.5) = ∫ 4.5

0 f (x) = 1.25 using geometric methods and the minimumvalue of A(x).

(b) A formula for A(x) valid on the interval [2, 4] is given byA(x) = ∫ x

0 f (t) dt = ∫ 1

0 −1 dt + ∫ 2

1 (2t − 3) dt + ∫ x

2 1 dt = −1 + 0 + (x − 2) = x − 3.

(c) A formula for B(x) = ∫ x

2 f (t) dt valid on the interval [2, 4] is given byB(x) = ∫ x

2 f (t) dt = ∫ x

2 1 dt = x − 2.38.

Let F(x) = ∫ x

0 t sin t dt .(a) Locate the local maxima and the absolute maximum of F(x) on [0, 3π ].(b) Justify using the graph of f (x): F(x) has precisely one zero in the interval [π, 2π ].

Hint: argue that F(π) > 0 and F(2π) < 0.(c) How many zeroes does F(x) have in [0, 3π ]?(d) Justify using the graph of f (x): there is an inflection point in [0, π ] where F(x)

changes from concave up to concave down.

Figure 6 Graph of y = x sin x .

39. Let F(x) = ∫ x

0 f (x) dx where f (x) is the function shown in Figure 7.


(a) Where does F(x) have local minima and maxima?(b) On which intervals is F(x) increasing and decreasing?(c) Where does F(x) have points of inflection?(d) On which intervals is F(x) concave up?

1 2 63 4 5 7 8

Figure 7

Let F(x) = ∫ x

0 f (t) dt , where f (x) is the function shown in the Figure 7. Note thatF ′(x) = f (x) and F ′′(x) = f ′(x).(a) Accordingly, F(x) will have a local minimum at x0 if F ′(x0) = f (x0) = 0 and F ′ = f

changes sign from − to 0 to + as x increases through x0. Similarly, F(x) will have alocal maximum at x0 if F ′(x0) = f (x0) = 0 and F ′ = f changes sign from + to 0 to −as x increases through x0. In light of these facts, we conclude that F has a localminimum at x = 4.5 and a local maximum at x = 2.

(b) Moreover, F(x) is increasing where F ′(x) = f (x) > 0 and decreasing whereF ′(x) = f (x) < 0. Therefore, F is increasing on (0, 2) and (4.5, 8). It is decreasing on(2, 4.5).

(c) Assuming that f is differentiable (and it looks rather smooth from the figure), we havethat A is twice differentiable on (0, 8). Thus A(x) has an inflection point at x0 providedA′′(x0) = f ′(x0) = 0 and A′′(x) = f ′(x) changes sign at x0. Thus f has a localextremum at x0. This occurs at x = 1, 3, 5, 6, 7.3.

(d) Finally, F(x) is concave up where F ′′(x) = f ′(x) > 0, i.e., where f is increasing.This occurs on (0, 1), (3, 5), and (6, 7.3).

40.

Let A(x) = ∫ x

0 f (t) dt where f (x) is shown in Figure 8. Determine:(a) the intervals on which A(x) is increasing and decreasing(b) the values x where A(x) has a local minimum or maximum(c) the intervals where A(x) is concave up or concave down(d) the inflection points of A(x)

Figure 8

41. (Exercises 41 and 42 adapted from Calculus Problems for a New Century, p. 102.)Determine f (x) assuming that

∫ x

0 f (t) dt is equal to x2 + x .

Let F(x) = ∫ x

0 f (t) dt = x2 + x . Then F ′(x) = f (x) = 2x + 1.42.

Determine g(x) and c assuming that∫ x

c g(t) dt is equal to x2 + x − 6.


43. Prove the formula

d

dx

∫ g2(x)

g1(x)

f (t) dt = f (g2(x))g′2(x) − f (g1(x))g′

1(x)

Let H(x) = ∫ g2(x)

g1(x)f (t) dt = ∫ g2(x)

0 f (t) dt − ∫ g1(x)

0 f (t) dt . Apply Exercise 23 twice toobtain

H ′(x) = f (g2(x))g′2(x) − f (g1(x))g′

1(x)


44.

Proof of the FTC, Part II The proof in the text assumes that f (x) is increasing. Toprove it in the general case, for h > 0 let m(h) and M(h) denote the minimum andmaximum values of f on [x, x + h]. The continuity of f (x) implies that

limh→0

m(h) = limh→0

M(h) = f (x)

Figure 9 Graphical interpretation of A(x + h) − A(x).

(a) Show that if h > 0, then

h m(h) ≤ A(x + h) − A(x) ≤ hM(h)

(b) Use the Squeeze Theorem to show that

limh→0+

A(x + h) − A(x)

h= f (x)

(c) Treat the case h < 0 similarly.

45. Proof of the FTC, Part I The FTC, Part I asserts that∫ b

a f (t) dt = F(b) − F(a) ifF ′(x) = f (x). Prove this using the FTC, Part II as follows. Set A(x) = ∫ x

a f (t) dt .(a) Show that F(x) = A(x) + C for some constant (recall that two functions with the

same derivative differ by a constant).

(b) Show that F(b) − F(a) = A(b) − A(a) = ∫ b

a f (t) dt .

Let F ′(x) = f (x) and A(x) = ∫ x

a f (t) dt .(a) Then by the FTC, Part II, A′(x) = f (x) and thus A′(x) and F ′(x) are both derivatives

of f (x). Hence F ′(x) = A′(x) + C .(b) F(b) − F(a) = (A(b) + C) − (A(a) + C) = A(b) − A(a) =∫ b

a f (t) dt − ∫ a

a f (t) dt = ∫ b

a f (t) dt − 0 = ∫ b

a f (t) dt which proves the FTC, Part I.46.

Can Every Antiderivative Be Expressed As an Integral? The area function∫ x

a f (t) dtis an antiderivative of f (x) for every value of a. However, not all antiderivatives areobtained in this way. The general antiderivative of f (x) = x is F(x) = 1

2x2 + C . Show that

F(x) can be expressed as an area function if C ≤ 0 but not if C > 0.

5.5 Total Change as the Integral of a Rate


1. An airplane makes the 350-mile trip from Los Angeles to San Francisco in one hour. Assum-ing the plane’s velocity at time t (in mph) is v(t), what is the value of the integral

∫ 1

0 v(t) dt?

2. A hot metal object is submerged in cold water. The rate at which it cools (in degrees perminute) is a function f (t) of time. Which quantity is represented by the integral

∫ T

0 f (t) dt?

3. Which of the following quantities would be naturally represented as derivatives and whichas integrals?(a) velocity of a train(b) total rainfall during a 6-month period(c) mileage per gallon of an automobile(d) total increase in the population of Los Angeles from 1970 to 1990

4. Two airplanes take off simultaneously from the same place and in the same direction. Theirvelocities are v1(t) and v2(t). What is the physical interpretation of the area between thegraphs of v1(t) and v2(t)?

Exercises

1. Water flows into an empty reservoir for five hours at a rate of 3000 + 5t gallons per hour.What is the quantity of water in the reservoir after five hours?

The quantity of water in the resevoir after five hours is∫ 5

0 3000 + 5t dt = (3000t + 52t2)∣∣5

0= 30125

2= 15,062.5 gallons.

2.Find the total displacement over the time interval [2, 5] of a particle moving in a straightline whose velocity is v(t) = 4t − 3 ft/s.3. A population of insects increases at a rate of 200 + 10t + .25t2 insects per day. Find theinsect population after three days assuming there are 35 insects at t = 0.

The increase in the insect population over three days is∫ 3

0 200 + 10t + 14t2 dt = (200t + 5t2 + 1

12t3)∣∣3

0= 2589

4= 647.25. Accordingly, the

population after 3 days is 35 + 647.25 = 682.25 or 682 insects.

5.5 Total Change as the Integral of a Rate 37

4.A survey shows that a candidate in a mayoral election is gaining votes at a rate of2000t + 1000 votes per day, where t is the number of days after the candidacy wasannounced. How many votes will the candidate receive after 60 days?

5. In a 4-week period, a factory produces bicycles at a rate of 95 + .1t2 − t bicycles per week(t in weeks). How many bicycles were produced during the 4-week period?

The rate of production is r(t) = 95 + 110

t2 − t bicycles per week. Accordingly, during theinterval [0, 4], the number of bikes produced is∫ 4

0 r(t) dt = ∫ 4

0 95 + 110

t2 − t dt = (95t + 130

t3 − 12t2)∣∣4

0= 5612

15≈ 374.13 or 374 bicycles.

6.Find the change in height over the time interval [1, 6] of a helicopter whose (vertical)velocity at time t is v(t) = .02t2 + t ft/s.7. A cat falls from a tree (with zero initial velocity) at time t = 0. How many feet does the catfall between t = .5 and t = 1 s? Use Galileo’s formula v(t) = −32t ft/s.

Given v(t) = −32 ft/s, the total distance the cat falls during the interval [ 12 , 1] is∫ 1

1/2 |v(t)| dt = ∫ 1

1/2 32t dt = 16t2∣∣1

1/2= 16 − 4 = 12 ft.

8.A projectile is released with initial (vertical) velocity 100 m/s. Use the formulav = 100 − 9.8t to determine the distance traveled during the first 15 seconds.

In Exercises 9–12, a particle moves in a straight line with the given velocity. Find the total dis-placement and total distance traveled over the time interval, and draw a displacement diagramlike Figure ?? (with distance and time labels).

9. v(t) = 12 − 4t ft/s, [0, 5]

Total displacement is given by∫ 5

0 (12 − 4t) dt = 12t − 2t2∣∣5

0= 10 ft.

Total distance is given by∫ 5

0 |12 − 4t | dt = ∫ 3

0 (12 − 4t) dt + ∫ 5

3 (4t − 12) dt = 12t − 2t2∣∣3

0+ 2t2 − 12t

∣∣53= 26 ft.

The displacement diagram is given here.

10.v(t) = 32 − 2t2 ft/s, [0, 6]

11. v(t) = t−2 − 1 m/s, [.5, 2]

Total displacement is given by∫ 2

.5(t−2 − 1) dt = −t−1 − t

∣∣2.5

= 0 m.Total distance is given by∫ 2

.5

∣∣t−2 − 1∣∣ dt = ∫ 1

.5(t−2 − 1) dt + ∫ 2

1 (1 − t−2) dt = −t−1 − t∣∣1.5

+ t + t−1∣∣2

1= 1 m.

The displacement diagram is given here.


12.v(t) = cos t m/s, [0, 4π ]

13. Carbon Tax To encourage manufacturers to reduce pollution, a carbon tax on each tonof CO2 released into the atmosphere has been proposed. To model the effects of such a tax,policymakers study the marginal cost of abatement B(x), defined as the cost of increasingCO2 reduction from x to x + 1 tons (Figure 1). What quantity is represented by

∫ x

0 B(t) dt?

25

50

75

100 dollars/ton

tons reduced(in ten thousands)

1 2 3

Figure 1 Marginal cost of abatement B(x).

The quantity∫ x

0 B(t) dt represents the amount of tax (in dollars) imposed on themanufacturer for releasing x tons of CO2 into the atmosphere.

14.Figure 2 shows the power (defined as the rate of energy consumption per unit time)supplied by the California power grid over a typical one-day period. Which quantity isrepresented by the area under the graph? A megawatt of power is 106 watts or 3.6 × 109

joules/h. Use the graph to make a very rough estimate of the number of joules consumedduring the day in California.

Figure 2 Power consumption over one-day period in California.

15. Figure 3 shows the migration rate (call it M(t)) in and out of Ireland during the period1988–1998. This is the number of people (in thousands per year) who move in or out of thecountry.

(a) What does∫ 1991

1988 M(t) dt represent (in words)?(b) Did migration over the eleven-year period 1988–1998 result in a net influx or outflow

of people from Ireland? Base your answer on a rough estimate of the positive andnegative areas involved.

(c) During which year could the Irish Prime Minister announce: “We are still losingpopulation but we’ve hit an inflection point—the trend is now improving.”

Figure 3 Irish migration (in thousands).

We note that M(t) is piecewise linear.

(a) The amount∫ 1991

1988 M(t) dt represents the net migration in thousands of people duringthe period from 1988–1991.

5.5 Total Change as the Integral of a Rate 39

(b) Via linear interpolation and using the midpoint approximation with n = 10, themigration (in thousands of people) over the period 1988–1998 is estimated to be

1 · (−43 − 33.5 − 12 + 0.5 − 2.5 − 6 − 3.5 + 3 + 11.5 + 19) = −66.5

That is, there was a net outflow of 66,500 people from Ireland during this period.(c) “The trend is now improving” implies that the population is decreasing, but that the

rate of decrease is approaching zero. The population is decreasing with an improvingtrend in part of the years 1989, 1990, 1991, 1993, and 1994.

“We’ve hit an inflection point” implies that the rate of population has changed fromdecreasing to increasing. There are two years in which the trend improves after it wasgetting worse: 1989 and 1993. During only one of these, 1989, was the populationdeclining for the entire previous year.16.

Heat Capacity The heat capacity C(T ) of a substance is the amount of energy (in joules)required to raise the temperature of one gram of the substance by one degree (◦C) when itstemperature is T .

(a) Explain in words why the energy required to raise the temperature from T1 to T2 is thearea under the graph of C(T ).

(b) How much energy is required to raise the temperature from 50 to 100◦C ifC(T ) = 6 + .2

√T .

17. The traffic flow rate between 8 AM and 10 AM at a certain point on a highway isq(t) = −10000 + 6800t − 400t2. How many cars pass this point between t = 8 AM andt = 10 AM?

The number of cars is given by∫ 10

8 q(t) dt = ∫ 10

8 −10000 + 6800t − 400t2 dt = −10000t + 3400t2 − 4003

t3∣∣10

8≈ 37333.3

cars.18.Suppose the marginal cost of producing x video recorders is .001x2 − .6x + 350. What isthe total cost of producing 300 units? If production is set at 300 units, what would be thecost of producing 20 additional units?

19. Figure 4 shows the graph of Q(t), the rate of retail truck sales in the U.S. (in thousandssold per year).(a) What does the area under the graph over the interval [1995, 1997] represent?(b) Express the total number of trucks sold in the period 1994–1997 as an integral (but do

not compute).(c) Use the data and the average of the right- and left-endpoint approximations to estimate

the total number of trucks sold during 2-year period 1995–1996.

year(qtr) sales year(qtr) sales

1995(1) 6484 1996(1) 7216

1995(2) 6255 1996(2) 6850

1995(3) 6424 1996(3) 7322

1995(4) 6818 1996(4) 7537

Figure 4 Quarterly retail sales of trucks in the U.S. in thousands (data supplied by Econo-magic.com).

The graph of Q(t) depicted in the exercise gives the quarterly rate of retail truck sales inthousands sold per quarter.


(a) The area under the graph over the interval [1995, 1997] represents number (inthousands) of trucks sold during the 1995–1997 period.

(b) The number of trucks sold in the period 1994–1997 is given by the integral∫ 1997

1994 Q(t) dt .(c) We note that Q(t) is piecewise linear. Recall that the unit of time is one quarter year;

hence �t = 1. Using n = 7, the number of trucks sold during the period 1995–1997 isestimated to be the average of the right- and left-endpoint approximations.

RN = 1 · (6255 + 6424 + 6818 + 7216 + 6850 + 7322 + 7537) = 48422 trucks

L N = 1 · (6484 + 6255 + 6424 + 6818 + 7216 + 6850 + 7322) = 47369 trucks

The average of RN and L N is 47895.5 trucks.20.

The velocity of a car is recorded at half-second intervals (in ft/s). Use the average of theleft- and right-endpoint approximations to estimate the total distance traveled during thefirst four seconds.

t 0 .5 1 1.5 2 2.5 3 3.5 4

v(t) 0 12 20 29 38 44 32 35 30Further Insights and Challenges

21. (From Calculus, H. Flanders, R. Korfhage, J. Price, Academic Press (1970).) A particlemoving along the x-axis with velocity v(t) = (t + 1)−2 is located at the origin at t = 0.Show that the particle will never pass the point x = 1.

The particle’s velocity is v(t) = s ′(t) = (t + 1)−2, an antiderivative for which isF(t) = −(t + 1)−1. Hence its position at time t is

s(t) =∫ t

0

s ′(u) du = F(u)∣∣t

0= F(t) − F(0) = 1 − 1

t + 1< 1

for all t ≥ 0. Thus the particle will never pass the point x = 1.22.

A particle moving along the x-axis with velocity v(t) = (t + 1)−1/2 is at the origin at t = 0.Will the particle be at the point x = 1 at any time t? If so, find t .

5.6 Substitution Method


1. Which of these functions does not appear to be of the form g(u(x))u′(x)?(a) 5x4 sin(x5)

(b) sin5 x cos x(c) x5 sin x

2. Write each of the following functions in the form cg(u(x))u′(x) where c is a constant.

(a) x(x2 + 9)4

(b) x2 sin(x3)

(c) sin x cos2 x

3. Which of the following is equal to∫ 2

0 x2(x3 + 1) dx for a suitable substitution?

(a)∫ 2

0 u du

(b)∫ 9

0 u du

(c) 13

∫ 9

1 u du

5.6 Substitution Method 41

Exercises

In Exercises 1–6, calculate du for the given function.

1. u = 1 − x2

Let u = 1 − x2. Then du = −2x dx .2.

u = sin x3. u = x3 − 2

Let u = x3 − 2. Then du = 3x2 dx .4.

u = 2x4 + 8x5. u = cos(x2)

Let u = cos(x2). Then du = − sin(x2) · 2x dx = −2x sin(x2) dx .6.

u = tan x

In Exercises 7–20, write the integral in terms of u and du. Then evaluate.

7.∫

(x − 7)3 dx ; u = x − 7

Let u = x − 7. Then du = dx . Hence∫(x − 7)3 dx = ∫

u3 du = 14 u4 + C = 1

4 (x − 7)4 + C .8. ∫

2x√

x2 + 1 dx ; u = x2 + 19.∫(x + 1)−2 dx ; u = x + 1

Let u = x + 1. Then du = dx . Hence∫(x + 1)−2 dx = ∫

u−2 du = −u−1 + C = −(x + 1)−1 + C = − 1

x + 1+ C .

10. ∫sin(2x − 4) dx ; u = 2x − 4

11.∫

x3

(x4 + 1)4dx; u = x4 + 1

Let u = x4 + 1. Then du = 4x3 dx or 14du = x3 dx . Hence

∫x3

x4 + 1dx = 1

4

∫1

udu = 1

4ln u + C = 1

4ln(x4 + 1) + C.

12. ∫x(x + 1)9 dx ; u = x + 1

13.∫

x + 1

(x2 + 2x)3dx ; u = x2 + 2x

Let u = x2 + 2x . Then du = 2x dx or 12 du = x dx . Hence∫

x + 1

(x2 + 2x)3dx = 1

2

∫1

u3du = 1

2· −1

2u−2 + C

= −1

4(x2 + 2x)−2 + C = −1

4(x2 + 2x)2+ C.

14. ∫x

(8x + 5)3dx; u = 8x + 515.

∫x cos(x2) dx; u = x2

Let u = x2. Then du = 2x dx or 12 du = x dx . Hence,∫

x cos(x2) dx = 12

∫cos u du = 1

2 sin u + C = 12 sin(x2) + C .

16. ∫ √4x − 1 dx; u = 4x − 1


17.∫

x√

4x − 1 dx; u = 4x − 1

Let u = 4x − 1 and x = 14(u + 1). Then du = 4 dx or 1

4du = dx . Hence,∫

x√

4x − 1 dx = 116

∫(u + 1)u1/2 du = 1

16

∫(u3/2 + u1/2) du =

116 · 2

5 u5/2 + 116 · 2

3 u3/2 + C = 140 (4x − 1)5/2 + 1

24 (4x − 1)3/2 + C .18. ∫

x2√

4x − 1 dx; u = 4x − 119.

∫sin2 x cos x dx ; u = sin x

Let u = sin x . Then du = cos x dx . Hence∫sin2 x cos x dx = ∫

u2 du = 13 u3 + C = 1

3 sin3 x + C .20. ∫

sec2 x tan x dx; u = tan x

In Exercises 21–24, show that each of the following integrals is equal to a multiple of sin(u(x))

for the appropriate choice of u(x).

21.∫

x cos(x2) dx

Let u = x2. Then du = 2x dx or 12

du = dx . Hence∫x cos(x2) dx = 1

2

∫cos u du = 1

2sin u + C , which is a multiple of sin(u(x)).

22. ∫x2 cos(x3 + 1) dx

23.∫

x1/2 cos(x3/2) dx

Let u = x3/2. Then du = 32 x1/2 dx or 2

3 du = x1/2 dx . Hence∫x1/2 cos(x3/2) dx = 2

3

∫cos u du = 2

3 sin u + C , which is a multiple of sin(u(x)).24. ∫

cos x cos(sin x) dx

In Exercises 25–54, evaluate the indefinite integral.

25.∫

(4x + 3)4 dx

Let u = 4x + 3. Then du = 4 dx or 14 du = dx . Hence∫

(4x + 3)4 dx = 14

∫u4 du = 1

4· 1

5u5 + C = 1

20(4x + 3)5.

26. ∫x2(x3 + 1)3 dx

27.∫

x√

x2 − 4 dx

Let u = x2 − 4. Then du = 2x dx or 12

du = x dx . Hence∫x√

x2 − 4 dx = 12

∫ √u du = 1

2· 2

3u3/2 + C = 1

3(x2 − 4)3/2.

28. ∫(2x + 1)(x2 + x)3 dx

29.∫

x√x2 + 9

dx

Let u = x2 + 9. Then du = 2x dx or 12du = x dx . Hence∫

x√x2 + 9

dx = 1

2

∫1√u

du = 1

2

(1

2

)√u + C = 1

4

√x2 + 9 + C

or ln√

x2 + 9 + C .30. ∫

dx

(x + 9)231.

∫dx

(4x + 9)3

Let u = 4x + 9, then du = 4dx or 14

du = dx . Hence∫dx

(4x + 9)3= 1

4

∫du

u3= 1

4· −1

2

1

u2+ C = −1

8(4x + 9)2+ C.

32. ∫x2 sin(x3) dx


33.∫

1√x − 7

dx

Let u = x − 7. Then du = dx . Hence∫(x − 7)−1/2 dx = ∫

u−1/2 du = 2u1/2 + C = 2√

x − 7 + C .34. ∫

2x2 + x

(4x3 + 3x2)2dx35.

∫(3x2 + 1)(x3 + x)2 dx

Let u = x3 + x . Then du = (3x2 + 1) dx . Hence∫(3x2 + 1)(x3 + x)2 dx = ∫

u2 du = 13u3 + C = 1

3(x3 + x)3 + C .

36. ∫5x4 + 2x

(x5 + x2)3dx37.

∫x2(x3 + 1)4 dx

Let u = x3 + 1. Then du = 3x2 dx or 13 du = 3x2 dx . Hence∫

x2(x3 + 1)4 dx = 13

∫u4 du = 1

3 · 15 u5 + C = 1

15 (x3 + 1)5 + C .38. ∫

(x − 9)−2/3 dx39.

∫(x + 1)(x2 + 2x)3 dx

Let u = x2 + 2x . Then du = (2x + 2) dx or 12 du = (x + 1) dx . Hence∫

(x + 1)(x2 + 2x)3 dx = 12

∫u3 du = 1

2· 1

4u4 + C = 1

8(x2 + 2x)4 + C .

40. ∫(x + 1)7 dx

41.∫

x2(x + 1)7 dx

Let u = x + 1 and u − 1 = x . Then du = dx . Hence∫

x2(x + 1)7 dx = ∫(u − 1)2u7 du =∫

(u9 − 2u8 + u7) du = 110 u10 − 2

9 u9 + 18 u8 + C = 1

10 (x + 1)10 − 29 (x + 1)9 + 1

8 (x + 1)8 + C .42. ∫

(3x + 9)10 dx43.

∫x(3x + 9)10 dx

Let u = 3x + 9 and 13 (u − 9) = x . Then du = 3 dx or 1

3 du = dx . Hence∫x(3x + 9)10 dx = 1

9

∫(u − 9)u10 du = 1

9

∫(u11 − 9u10) du = 1

9 · 112 u12 − 1

11 u11 + C =1

108 (3x + 9)12 − 111 (3x + 9)11 + C .

44. ∫x(x + 1)1/4 dx

45.∫

x3(x2 − 1)3/2 dx

Let u = x2 − 1 and u + 1 = x2. Then du = 2x dx or 12 du = x dx . Hence∫

x3(x2 − 1)3/2 dx = ∫x2 · x(x2 − 1)3/2 dx = 1

2

∫(u + 1)u3/2 du = 1

2

∫(u5/2 + u3/2) du =

12

· 27u7/2 + 1

2· 2

5u5/2 + C = 1

7(x2 − 1)7/2 + 1

5(x2 − 1)5/2 + C .

46. ∫sin5 x cos x dx

47.∫

x2 sin(x3 + 1) dx

Let u = x3 + 1. Then du = 3x2 dx or 13du = x2 dx . Hence∫

x2 sin(x3 + 1) dx = 13

∫sin u du = − 1

3cos u + C = − 1

3cos(x3 + 1) + C .

48. ∫sec2(4x + 9) dx

49.∫

sec2 x tan4 x dx

Let u = tan x . Then du = sec2 x dx . Hence∫sec2 x tan4 x dx = ∫

u4 du = 15u5 + C = 1

5tan5 x + C .

50. ∫cos 2x

(1 + sin 2x)2dx51.

∫sin 4x

√cos 4x + 1 dx

Let u = cos 4x + 1. Then du = −4 sin 4x or − 14du = sin 4x . Hence∫

sin 4x√

cos 4x + 1 dx = − 14

∫u1/2 du = − 1

4 · 23 u3/2 + C = − 1

6 (cos 4x + 1)3/2 + C .52. ∫

cos x(3 sin x − 1) dx

53.∫

cos√

x√x

dx

Let u = x1/2. Then du = 12x−1/2 dx or 2du = x−1/2 dx . Hence∫

cos√

x√x

dx = 2∫

cos u du = 2 sin u + C = 2 sin√

x + C .


54. ∫sec2 x(4 tan3 x − 3 tan2 x) dx

55. Evaluate∫ √

x3 + 1 x5 dx using u = x3 + 1Hint: write x5 dx = x3 · x2 dx and observe that x3 = u − 1.

Let u = x3 + 1. Then x3 = u − 1 and du = 3x2 dx or 13 du = x2 dx . Hence∫

x5√

x3 + 1 dx = 13

∫u1/2(u − 1) du = 1

3

∫u3/2 − u1/2 du = 1

3( 2

5u5/2 − 2

3u3/2) + C =

215

(x3 + 1)5/2 − 29(x3 + 1)3/2 + C .

56.Evaluate

∫(x3 + 1)1/4 x5 dx using u = x3 + 1.

57. What are the new limits of integration if the substitution u = 3x + π is applied to theintegral

∫ π

0 sin(3x + π) dx?

The new limits of integration are u(π) = 3π + π = 4π and u(0) = 3 · 0 + π = π .58.

Which of the following is the result of applying the substitution u = 4x − 9 to the integral∫ 8

2 (4x − 9)20 dx?

(a)∫ 32

8 u20 du

(b) 14

∫ 8

2 u20 du

(c) 14

∫ 32

8 u20 du

(d) 4∫ 32

8 u20 du

In Exercises 59–68, use the change of variables formula to evaluate the definite integral.

59.∫ π/2

0 cos 3x dx

Let u = 3x . Then du = 3 dx or 13du = dx . Hence∫ π/2

0 cos 3x dx = 13

∫ 3π/2

0 cos u du = 13 sin u

∣∣3π/2

0= − 1

3 − 0 = − 13 .

60. ∫ 3

1 (x + 2)3 dx61.

∫ π/2

0 cos(3x + π

2) dx

Let u = 3x + π

2 . Then du = 3 dx or 13 du = dx . Hence∫ π/2

0 cos(3x + π

2) dx = 1

3

∫ 2π

π/2 cos u du = 13

sin u∣∣2π

π/2= 0 − 1

3= − 1

3.

62. ∫ 6

1

√x + 3 dx

63.∫ 1

0

x

(x2 + 1)3dx

Let u = x2 + 1. Then du = 2x dx or 12

du = x dx . Hence

∫ 1

0

x

(x2 + 1)3dx =

∫ 2

1

1

u3du = 1

2· −1

2u−2

∣∣∣∣2

1

= − 1

16+ 1

4= 1

16= 0.1875.

64. ∫ 2

−1

√5x + 6 dx

65.∫ 4

0 x√

x2 + 9 dx

Let u = x2 + 9. Then du = 2x dx or 12

du = x dx . Hence∫ 4

0

√x2 + 9 dx = 1

2

∫ 25

9

√u du = 1

2· 2

3u3/2

∣∣25

9= 1

3(125 − 27) = 32.6667.

66. ∫ 2

0

x + 3

(x2 + 6x + 1)3dx67.

∫ π/2

0 cos3 x sin x dx

Let u = cos x . Then du = − sin x dx . Hence∫ π/2

0 cos3 x sin x dx = − ∫ 0

1 u3 du = ∫ 1

0 u3 du = 14 u4∣∣1

0= 1

4 − 0 = 14 .

68. ∫ π/4

0 tan2 x sec2 x dx69. Some Choices Are Better Than Others One way of evaluating

∫sin x cos2 x dx is using

the substitution u = sin x . Show that this gives∫sin x cos2 x dx =

∫u√

1 − u2 du

Complete the evaluation by using another substitution. Then show that u = cos x is a betterchoice.

Consider the integral∫

sin x cos2 x dx .


Let u = sin x . Then cos x = √1 − u2 and du = cos x dx . Hence∫

sin x cos2 x dx = ∫u√

1 − u2 du. Now let w = 1 − u2. Then dw = −2u du or− 1

2dw = u du. Therefore,

∫u√

1 − u2 du = − 12

∫w1/2 dw = − 1

2· 2

3w3/2 + C =

− 13 w

3/2 + C = − 13 (1 − u2)3/2 + C = − 1

3 (1 − sin2 x)3/2 + C .A better substitution choice is u = cos x . Then du = − sin u du or −du = sin u du.Hence

∫sin x cos2 x dx = − ∫

u2 du = − 13u3 + C = − 1

3cos3 x + C .

70.Can They Both Be Right? (Adapted from Calculus Problems for a New Century,p. 121.) Joan uses the substitution u = tan x to conclude∫

tan x sec2 x dx = 1

2tan2 x + C

while Frank, noting that u = sec x gives du = sec x tan x , concludes that∫tan x sec2 x dx =

∫sec x(tan x sec x) dx

= 1

2sec2 x + C

Explain the apparent contradiction.

71. Evaluate∫

sin x cos x dx using substitution in two different ways: first using u = sin x andthen using u = cos x . Reconcile the two different answers.

Let u = sin x . Then du = cos x dx . Hence∫sin x cos x dx = ∫

u du = 12 u2 + C1 = 1

2 sin2 x + C1.Let u = cos x . Then du = − sin x dx or −du = sin x dx . So∫

sin x cos x dx = − ∫u du = − 1

2u2 + C2 = − 1

2cos2 x + C2.

To reconcile these two seemingly different answers, recall that any two antiderivatives ofa specified function differ by a constant. To show that this is true here, note that( 1

2sin2 x + C1) − (− 1

2cos2 x + C2) = 1

2+ C1 − C2, a constant. Here we used the

trigonometric identity sin2 x + cos2 x = 1.

72.Use the substitution u = f (x) to evaluate

∫f (x)3 f ′(x) dx where f (x) is differentiable.

The answer is an expression in f (x).73. Evaluate∫

f (x) f ′(x) dx .

Let u = f (x). Then du = f ′(x) dx . Hence∫f (x) f ′(x) dx = ∫

u du = 12 u2 + C = 1

2 f (x)2 + C .74.

Evaluate∫

f ′(x)

f (x)2dx .

75. Evaluate∫ π/2

0 sinn x cos x dx , where n is an integer, n �= −1.

Let u = sin x . Then du = cos x dx . Hence∫ π/2

0sinn x cos x dx =

∫ 1

0un du = un+1

n + 1

∣∣∣∣1

0

= 1

n + 1.

Further Insights and Challenges76.

(a) Use the substitution u = 1 + x1/n to show that∫ √1 + x1/n dx = n

∫u1/2(u − 1)n−1 du

(b) Use this to evaluate the left-hand side for n = 2, 3.

77. Show that if f is an odd function ( f (−x) = − f (x)) then∫ a

−a

f (x) dx = 0

Hint: show that∫ a

0 f (x) dx = − ∫ 0

−a f (x) dx using substitution.

We assume that f is continuous. If f (x) is an odd function, then f (−x) = − f (x). Letu = −x . Then x = −u and du = −dx or −du = dx . Accordingly,∫ a

−a

f (x) dx =∫ 0

−a

f (x) dx +∫ a

0f (x) dx

= −∫ 0

a

f (−u) du +∫ a

0

f (x) dx

=∫ a

0f (x) dx −

∫ a

0f (u) du = 0


78.Show that if f is an even function ( f (−x) = f (x)), then∫ a

0f (x) dx =

∫ 0

−a

f (x) dx

79. (a) Use the substitution u = x/a to prove that the hyperbola y = x−1 has a specialproperty: if a, b > 0, then

∫ b

a1x

dx = ∫ b/a

11x

dx .(b) Show that the area under the hyperbola over the intervals [1, 2], [2, 4], [4, 8], . . . are

all equal.

(a) Let u = xa

and au = x . Then du = 1a

dx or a du = dx . Hence

∫ b

a

1

xdx =

∫ b/a

1

a

audu =

∫ b/a

1

1

udu.

Note that∫ b/a

1

1

udu =

∫ b/a

1

1

xdx after the substitution x = u.

(b)∫ 2

11x

dx = ln x |21 = ln 2 − ln 1 = ln 2.∫ 4

21x

dx = ln x |42 = ln 4 − ln 2 = ln 4

2= ln 2.∫ 8

41x

dx = ln x |84 = ln 8 − ln 4 = ln 8

4= ln 2.

80.Show the two regions in Figure 1 have the same area. They correspond to the areas underthe graphs of

√1 − u2 and cos2 x .

Figure 1 (A) Graph of√

1 − u2, (B) Graph of cos2 x .

81. Area of a Circle Although we usually take it for granted that the area of a circle is πr2,this formula requires proof. The formula is a consequence of two distinct facts: first, thatthe area is proportional to the square of the radius and second, that the area of the unitcircle is π (one-half the circumference of the unit circle). Prove the first part.(a) Use the Change of Variables formula to show that∫ r

0

√r2 − x2 dx = r2

∫ 1

0

√1 − x2 dx .

(b) Show the area of a circle is proportional to the square of its radius.

The integral∫ r

0

√r2 − x2 dx is equal to 1

4the area of a circle of radius r > 0. (It is the area

of the quarter circular disk x2 + y2 ≤ r2 in the first quadrant.)(a) Let u = 1

rx . Then x = ru and du = 1

rdx or r du = dx . So∫ r

0

√r2 − x2 dx = r

∫ 1

0

√r2 − r2u2 du = r2

∫ 1

0

√1 − u2 du = r2( π

4 ) = 14 πr2 from

Exercise ??.(b) In (a) it was shown that 1

4of the area bounded by a circle of radius r is 1

4πr2.

Therefore, the area bounded by the full circle is πr2. Hence the area of a circle isproportional to the square of its radius.

82.Area of an Ellipse Use the change of variables formula and the fact that the unit circlehas area π to show that the area of the ellipse with equation

x2

a2+ y2

b2= 1

is equal to πab.

Hint: Show that the area of the ellipse is equal to twice

b∫ a

−a

√1 − (x/a)2 dx

and use the Change of Variables formula to relate the integral on the right to the area of theunit circle.

Figure 2 Graph ofx2

a2+ y2

b2= 1.

Documents

The Integral - Department of Mathematics | Penn Mathdeturck/m103/rogawski/chap5odd.pdf · The Integral 5.1 Approximating and Computing Area Preliminary Questions 1. The interval [2,5]