1. This document contains examples of finding antiderivatives (indefinite integrals) of various functions. 2. The examples demonstrate using basic integration rules like power rule, reverse derivative rule, trigonometric integral formulas to find antiderivatives. 3. Additional examples show using initial conditions to determine an unknown constant term when an antiderivative is given in terms of an arbitrary constant.

1. 7.5
5.0
2.5
Chapter 4 0.0
−2 −1 0 1 2
x
−2.5
Integration 4. sin x, sin x + 2, sin x − 5
2
x
−3 −2 −1 0 1 2 3
4.1 Antiderivatives 0
−2
x4 x4 x4
1. , + 3, −2
4 4 4
−4
20
−6
15
3 5 3 2
5. (3x4 − 3x)dx = x − x +c
10 5 2
1 4
5 6. (x3 − 2)dx = x − 2x + c
4
√ 1 x−3
−3 −2 −1 1 2 3 7. 3 x− 4 dx = 2x3/2 + +c
x 3
1
8. 2x−2 + √ dx
x
x4 x2 x4 x2 x4 x2 = −2x−1 + 2x1/2 + c
2. − , − − 1, − +4
4 2 4 2 4 2
x1/3 − 3
6 9. dx = (x−1/3 − 3x−2/3 )dx
x2/3
5 3
= x2/3 − 9x1/3 + c
4
2
x + 2x3/4
3
10. dx = (x−1/4 + 2x−1/2 )dx
x5/4
2
4
= x3/4 + 4x1/2 + c
1 3
0
−2 −1 0 1 2
11. (2 sin x + cos x)dx = −2 cos x + sin x + c
x −1
12. (3 cos x − sin x)dx = 3 sin x + cos x + c
3. ex , ex + 1, ex − 3 13. 2 sec x tan xdx = 2 sec x + c
240

2. 4.1. ANTIDERIVATIVES 241
4 d
14. √ dx = 4 arcsin x + c 30. ln |sin x · 2|
1 − x2 dx
1 d
= (sin x · 2)
15. 5 sec2 xdx = 5 tan x + c sin x · 2 dx
2 cos x
= = cot x
2 sin x
4 cos x
16. dx = −4 csc x + c
sin2 x 31. (a) N/A
(b) By Power Formula,
17. (3ex − 2)dx = 3ex − 2x + c
√ 2
( x3 + 4)dx = x5/2 + 4x + c.
5
18. (4x − 2ex )dx = 2x2 − 2ex + c
32. (a) By Power Formula,
3x2 − 4
19. (3 cos x − 1/x)dx = 3 sin x − ln |x| + c dx = (3 − 4x−2 )dx
x2
= 3x + 4x−1 + c
20. (2x−1 + sin x)dx = 2 ln |x| − cos x + c (b) N/A
33. (a) N/A
4x
21. dx = 2 ln |x2 + 4| + c (b) By Reversing derivative formula,
x2 + 4
sec2 xdx = tan x + c
3 3
22. dx = tan−1 x + c
4x2 + 4 4 34. (a) By Power Formula,
1 1
cos x − 1 dx = − − x + c
23. dx = ln | sin x| + c x 2 x
sin x
(b) N/A
24. (2 cos x − ex )dx = 2 sin x − ex + c 35. Finding the antiderivative,
x2
f (x) = 3ex + + c.
ex 2
25. dx = ln | ex + 3| + c
ex + 3 Since f (0) = 4,
ex + 3 we have 4 = f (0) = 3 + c.
26. dx = (1 + 3e−x )dx Therefore,
ex
= x − 3e−x + c x2
f (x) = 3ex + + 1.
2
27. x1/4 (x5/4 − 4)dx = (x3/2 − 4x1/4 )dx 36. Finding the antiderivative,
2 5/2 16 5/4 f (x) = 4 sin x + c.
= x − x +c Since f (0) = 3,
5 5
we have 3 = f (0) = c.
Therefore,
28. x2/3 (x−4/3 − 3)dx = (x−2/3 − 3x2/3 )dx
f (x) = 4 sin x + 3.
9
= 3x1/3 − x5/3 + c 37. Finding the antiderivative
5
f (x) = 4x3 + 2ex + c1 .
d Since, f (0) = 2.
29. ln |sec x + tan x|
dx We have 2 = f (0) = 2 + c1
1 d
= (sec x + tan x) and therefore
sec x + tan x dx f (x) = 4x3 + 2ex .
2
sec x tan x + sec x
= Finding the antiderivative,
sec x + tan x f (x) = x4 + 2ex + c2 .
sec x (tan x + sec x)
= Since f (0) = 3,
sec x + tan x
= sec x We have 3 = f (0) = 2 + c2
Therefore,

3. 242 CHAPTER 4. INTEGRATION
f (x) = x4 + 2ex + 1. 1
f (x) = −3 sin x + x4 + c1 x + c2 .
3
42. Taking antiderivatives,
38. Finding the antiderivative, f (x) = x1/2 − 2 cos x
f (x) = 5x4 + e2x + c1 . 2
Since f (0) = −3, f (x) = x3/2 − 2 sin x + c1
3
we have −3 = f (0) = 1 + c1 4 5/2
f (x) = x + 2 cos x + c1 x + c2 .
Therefore, 15
f (x) = 5x4 + e2x − 4. 43. Taking antiderivatives,
Finding the antiderivative, f (x) = 4 − 2/x3
e2x
f (x) = x5 + − 4x + c2 . f (x) = 4x + x−2 + c1
2 f (x) = 2x2 − x−1 + c1 x + c2
Since f (0) = 2,
1 2 c1
We have 2 = f (0) = + c2 f (x) = x3 − ln |x| + x2 + c2 x + c3
2 3 2
Therefore, 44. Taking antiderivatives,
e2x 3 f (x) = sin x − ex
f (x) = x5 + − 4x + .
2 2 f (x) = − cos x − ex + c1
f (x) = − sin x − ex + c1 x + c2
c1
39. Taking antiderivatives, f (x) = cos x − ex + x2 + c2 x + c3
f (t) = 2t + t2 + c1 2
t3 45. Position is the antiderivative of velocity,
f (t) = t2 + + c1 t + c2
3 s(t) = 3t − 6t2 + c.
Since f (0) = 2, Since s(0) = 3, we have c = 3. Thus,
we have 2 = f (0) = c2 s(t) = 3t − 6t2 + 3.
Therefore,
t3 46. Position is the antiderivative of velocity,
f (t) = t2 + + c1 t + 2.
3 s(t) = −3e−t − 2t + c.
Since f (3) = 2, Since s(0) = 0, we have −3 + c = 0 and there-
we have fore c = 3. Thus,
2 = f (3) = 9 + 9 + 3c1 + 2 s(t) = −3e−t − 2t + 3.
− 6 = c1
Therefore, 47. First we find velocity, which is the antideriva-
t3 tive of acceleration,
f (t) = + t2 − 6t + 2.
3 v(t) = −3 cos t + c1 .
Since v(0) = 0 we have
−3 + c1 = 0, c1 = 3 and
40. Taking antiderivatives,
v(t) = −3 cos t + 3.
f (t) = 4t + 3t2 + c1
Position is the antiderivative of velocity,
f (t) = 2t2 + t3 + c1 t + c2
s(t) = −3 sin t + 3t + c2 .
Since f (1) = 3,
Since s(0) = 4, we have c2 = 4. Thus,
we have 3 = f (1) = 2 + 1 + c1 + c2
s(t) = −3 sin t + 3t + 4.
Therefore,
c1 + c2 = 0 48. First we find velocity, which is the antideriva-
Since f (−1) = −2, tive of acceleration,
we have −2 = f (−1) = 2 − 1 − c1 + c2 1
v(t) = t3 + t + c1 .
Therefore, −c1 + c2 = −3. 3
So, c1 = 2 and c2 = − 3
3
2 Since v(0) = 4 we have c1 = 4 and
Hence, 1
3 3 v(t) = t3 + t + 4.
f (t) = t3 + 2t2 + t − . 3
2 2
Position is the antiderivative of velocity,
1 4 1 2
41. Taking antiderivatives, s(t) = t + t + 4t + c2 .
12 2
f (x) = 3 sin x + 4x2 Since s(0) = 0, we have c2 = 0. Thus,
4 1 4 1 2
f (x) = −3 cos x + x3 + c1 s(t) = t + t + 4t.
3 12 2

4. 4.1. ANTIDERIVATIVES 243
49. (a) There are many correct answers, but any 15
correct answer will be a vertical shift of
these answers. 10
10.0
5
7.5
0
5.0 y −3 −2 −1 0 1 2 3
x
−5
2.5
0.0 −10
−4.0 −3.2 −2.4 −1.6 −0.8 0.0 0.8 1.6 2.4 3.2
x
−2.5
51. We start by taking antiderivatives:
−5.0
f (x) = x2 /2 − x + c1
f (x) = x3 /6 − x2 /2 + c1 x + c2 .
Now, we use the data that we are given. We
(b) There are many correct answers, but any know that f (1) = 2 and f (1) = 3, which gives
correct answer will be a vertical shift of us
these answers. 3 = f (1) = 1/2 − 1 + c1 ,
and
8.8
1 = f (1) = 1/6 − 1/2 + c1 + c2 .
8.0
Therefore c1 = 7/2 and c2 = −13/6 and the
7.2
function is
6.4 x3 x2 7x 13
f (x) = − + − .
5.6
6 2 2 6
4.8
4.0
52. We start by taking antiderivatives:
3.2
f (x) = 3x2 + 4x + c1
2.4
f (x) = x3 + 2x2 + c1 x + c2 .
Now, we use the data that we are given. We
−3 −2 −1 0 1 2 3
x
know that f (−1) = 1 and f (−1) = 2, which
gives us
2 = f (−1) = −1 + c1 ,
and
1 = f (−1) = 1 − c1 + c2 .
50. (a) There are many correct answers, but any
Therefore c1 = 3 and c2 = 3 and the function
correct answer will be a vertical shift of
is
these answers.
f (x) = x3 + 2x2 + 3x − 3.
14
d
12 53. sin x2 = 2x cos x2
dx
10
Therefore,
8 y
6 2x cos x2 dx = sin x2 + c
4
2
d 9
54. (x3 + 2)3/2 = x2 (x3 + 2)1/2
−4 −2 0
0
2
dx 2
x
−2
Therefore,
−4 2
x2 x3 + 2dx = (x3 + 2)3/2 + c
9
(b) There are many correct answers, but any d
55. x2 sin 2x = 2(x sin 2x + x2 cos 2x)
correct answer will be a vertical shift of dx
these answers. Therefore,

5. 244 CHAPTER 4. INTEGRATION
1 36 x2
x sin 2x + x2 cos 2x dx = 3− = 2
9 12 − 9x 2 3x − 4
1 2 For 33(a): Almost the same as in Exercise 59,
= x sin 2x + c
2 example 1.11 (b).
1 x−1
d x2 2xe3x − 3x2 e3x For 34(b): ln +c
56. = 2 x+1
dx e 3x e6x Verify:
Therefore, d 1 x−1
ln
2xe3x − 3x2 e3x x2 dx 2 x + 1
6x
dx = 3x + c 1 x + 1 (x + 1) − (x − 1)
e e = · ·
2 x−1 (x + 1)2
x cos(x2 ) 1
57. dx = sin(x2 ) + c = 2
sin(x2 ) x −1
61. Use a CAS to find antiderivatives and verify by
d √ √ 1 computing the derivatives:
58. 2 x sin x = 2 x cos x + √ sin x
dx x
√ 1 3 1 3
2 x cos x + √ sin x dx (a) x2 e−x dx = − e−x + c
x 3
√ Verify:
= 2 x sin x + c d 1 3
− e−x
dx 3
59. Use a CAS to find antiderivatives and verify by 1 3
= − e−x · (−3x2 )
computing the derivatives: 3 3
For 11.1(b): = x2 e−x
1
sec xdx = ln | sec x + tan x| + c (b) dx = ln |x − 1| − ln |x| + c Verify:
x2 − x
Verify: d
(ln |x − 1| − ln |x|)
d dx
ln | sec x + tan x| 1 1 x − (x − 1)
dx = − =
sec x tan x + sec2 x x−1 x x(x − 1)
= = sec x 1 1
sec x + tan x = = 2
For 11.1(f): x(x − 1) x −x
sin 2x x cos 2x
x sin 2xdx = − +c (c) sec xdx = ln | sec x + tan x| + c
4 2
Verify: Verify:
d sin 2x x cos 2x d
− [ln | sec x + tan x|]
dx 4 2 dx
2 cos 2x cos 2x − 2x sin 2x sec x tan x + sec2 x
= − =
4 2 sec x + tan x
= x sin 2x sec x(sec x + tan x)
= = sec x
sec x + tan x
60. Use a CAS to find antiderivatives and verify by
computing the derivatives: 62. Use a CAS to find antiderivatives and verify
For 31(a): The answer is too complicated to be by computing the derivatives:
presented here.
√
1 √ 2 3 − 3x x 1
For 32(b): 3x + 3 ln √ +c (a) dx = arctan x2 + c
9 2 3 + 3x x4 + 1 2
Verify: Verify:
√ d 1
d 1 √ 2 3 − 3x arctan x2
3x + 3 ln √ dx 2
dx 9 2 3 + 3x
√ 1 1 x
1 2 3 + 3x = · 4 · 2x = 4
= 3+ √ · 2 x +1 x +1
9 2 3 − 3x
√ √ (b) 3x sin 2xdx
−3(2 3 + 3x) − 3(2 3 − 3x)
√ 3 3x
(2 3 + 3x)2 = sin 2x − cos 2x + c
4 2

6. 4.1. ANTIDERIVATIVES 245
Verify: 67. The key is to find the velocity and position
d 3 3x functions. We start with constant acceleration
sin 2x − cos 2x
dx 4 2 a, a constant. Then, v(t) = at + v0 where v0
3 3 is the initial velocity. The initial velocity is 30
= cos 2x − cos 2x + 3x sin 2x
2 2 miles per hour, but since our time is in seconds,
= 3x sin 2x it is probably best to work in feet per second
(30mph = 44ft/s). v(t) = at + 44.
(c) ln xdx = x ln x − x + c We know that the car accelerates to 50 mph
Verify: (50mph = 73ft/s) in 4 seconds, so v(4) = 73.
d 29
(x ln x − x) = ln x + 1 − 1 Therefore, a · 4 + 44 = 73 and a = ft/s
dx 4
= ln x So,
29
−1 v(t) = t + 44 and
63. √ dx = cos−1 (x) + c1 4
1 − x2 29 2
s(t) = t + 44t + s0
−1 8
√ dx = − sin−1 (x) + c2 where s0 is the initial position. We can assume
1 − x2 the the starting position is s0 = 0.
Therefore, 29 2
cos−1 x + c1 = − sin−1 x + c2 Then, s(t) = t + 44t and the distance
8
Therefore, traveled by the car during the 4 seconds is
sin−1 x + cos−1 x = constant s(4) = 234 feet.
To find the value of the constant, let x be any
convenient value. 68. The key is to find the velocity and position
Suppose x = 0; then sin−1 0 = 0 and cos−1 0 = functions. We start with constant acceleration
π/2, so a, a constant. Then, v(t) = at + v0 where v0
π is the initial velocity. The initial velocity is 60
sin−1 x + cos−1 x =
2 miles per hour, but since our time is in seconds,
it is probably best to work in feet per second
64. To derive these formulas, all that needs to be
(60mph = 88ft/s). v(t) = at + 88.
done is to take the derivatives to see that the
We know that the car comes to rest in 3 sec-
integrals are correct:
d onds, so v(3) = 0.
(tan x) = sec2 x Therefore,
dx
d a(3) + 88 = 0 and a = −88/3ft/s (the accelera-
(sec x) = sec x tan x tion should be negative since the car is actually
dx
decelerating.
65. To derive these formulas, all that needs to be So,
done is to take the derivatives to see that the 88
integrals are correct: v(t) = − t + 88 and
3
d x 44
(e ) = ex s(t) = − t2 + 88t + s0 where s0 is the initial
dx 3
d position. We can assume the the starting po-
−e−x = e−x
dx sition is s0 = 0.
44
1 1 1 Then, s(t) = − t2 + 88t and the stopping
66. (a) dx = dx 3
kx k x distance is s(3) = 132 feet.
1
= ln |x| + c1 69. To estimate the acceleration over each inter-
k
val, we estimate v (t) by computing the slope
1 1 k
(b) dx = dx of the tangent lines. For example, for the in-
kx k kx terval [0, 0.5]:
1
= ln |kx| + c2 v(0.5) − v(0)
k a≈ = −31.6 m/s2 .
0.5 − 0
Because Notice, acceleration should be negative since
1 1
ln |kx| = (ln |k| + ln |x|) the object is falling.
k k
1 1 1 To estimate the distance traveled over the in-
= ln |x| + ln |k| = ln |x| + c terval, we estimate the velocity and multiply
k k k
The two antiderivatives are both correct. by the time (distance is rate times time). For

7. 246 CHAPTER 4. INTEGRATION
an estimate for the velocity, we will use the Time Speed Dist
average of the velocities at the endpoints. For 0 70 0
example, for the interval [0, 0.5], the time inter- 0.5 69.55 34.89
val is 0.5 and the velocity is −11.9. Therefore 1.0 70.3 69.85
the position changed is (−11.9)(0.5) = −5.95 1.5 70.35 105.01
meters. The distance traveled will be 5.95 me- 2.0 70.65 104.26
ters (distance should be positive).
Interval Accel Dist 72. To estimate the speed over the interval, we first
[0.0, 0.5] −31.6 5.95 approximate the acceleration over the interval
[0.5, 1.0] −2 12.925 by averaging the acceleration at the endpoint
[1.0, 1.5] −11.6 17.4 of the interval. Then, the velocity will be the
[1.5, 2.0] −3.6 19.3 acceleration times the length of time. the slope
of the tangent lines. For example, for the in-
terval [0.0, 0.5] the average acceleration is −0.8
and v(0.5) = 20+(−0.8)(.5) = 19.6. Of course,
speed is the absolute value of the velocity.
70. To estimate the acceleration over each inter-
And, the distance traveled is the average speed
val, we estimate v (t) by computing the slope
times the length of time. For the time t = 0.5,
of the tangent lines. For example, for the in- 20 + 19.6
terval [0, 1.0]: the distance would be × 0.5 = 9.9
2
v(1.0) − v(0) meters.
a≈ = −9.8 m/s2 . Time Speed Dist
1.0 − 0
Notice, acceleration should be negative since 0 20 0
the object is falling. 0.5 19.6 9.9
To estimate the distance traveled over the in- 1.0 17.925 19.281
terval, we estimate the velocity and multiply 1.5 16.5 27.888
by the time (distance is rate times time). For 2.0 16.125 34.044
an estimate for the velocity, we will use the av-
erage of the velocities at the endpoints. For
example, for the interval [0, 1.0], the time in- 4.2 Sums And Sigma Notation
terval is 1.0 and the velocity is −4.9. Therefore
the position changed is (−4.9)(1.0) = −4.9 me- 1. The given sum is the sum of twice the
ters. The distance traveled will be 4.9 meters squares of the integers from 1 to 14.
(distance should be positive). 14
2 2 2 2
Interval Accel Dist 2(1) + 2(2) + 2(3) + . . . + 2(14) = 2i2
i=1
[0.0, 1.0] −9.8 4.9
[1.0, 2.0] −8.8 14.2 2. The given sum is the sum of squares
[2.0, 3.0] −6.3 21.75 roots of √the integers from 1 to 14.
√ √ √
[3.0, 4.0] −3.6 26.7 2 − 1 + 3 − 1 + 4 − 1 + . . . + 15 − 1
√ √ √ √ √
= 1 + 2 + 3 + ... + 13 + 14
14
√
= i
71. To estimate the speed over the interval, we i=1
first approximate the acceleration over the in-
terval by averaging the acceleration at the end- 50
(50)(51)(101)
point of the interval. Then, the velocity will be 3. (a) i2 = = 42, 925
6
the acceleration times the length of time. The i=1
slope of the tangent lines. For example, for the 50 2
2
50(51)
interval [0, 0.5] the average acceleration is −0.9 (b) i = = 1, 625, 625
and v(0.5) = 70 + (−0.9)(0.5) = 69.55. i=1
2
And, the distance traveled is the speed times
10
the length of time. For the time t = 0.5, the √
70 + 69.55 4. (a) i
distance would be ×0.5 ≈ 34.89 me- i=1
2 √ √ √ √ √
ters. =1+ 2+ 3+ 4+ 5+ 6

8. 4.2. SUMS AND SIGMA NOTATION 247
√ √ √ √
+ 7 + 8 + 9 + 10 = 338, 350 − 15, 150 + 200 = 323, 400
≈ 22.47
10 140
10(11) √
(b) i= = 55 14. n2 + 2n − 4
i=1
2
n=1
140 140 140
6
= n2 + 2 n− 4
5. 3i2 = 3 + 12 + 27 + 48 + 75 + 108 n=1 n=1 n=1
i=1 (140)(141)(281) 140(141)
= 273 = +2 − 4 (140)
6 2
7 = 943, 670
6. i2 + i = 12 + 20 + 30 + 42 + 56
i=3
30
= 160 2
15. (i − 3) + i − 3
10
i=3
7. (4i + 2) 30 30
2
i=6 = (i − 3) + (i − 3)
= (4(6) + 2) + (4(7) + 2) + (4(8) + 2) i=3 i=3
+ (4(9) + 2) + (4(10) + 2) 27 27
= 26 + 30 + 34 + 38 + 42 = n2 + n (substitute i − 3 = n)
= 170 n=0 n=0
27 27
8
2 =0+ n2 + 0 + n
8. (i + 2) n=1 n=1
i=6 27 (28) (55) 27 (28)
= (62 + 2) + (72 + 2) + (82 + 2) = + = 7308
6 2
= 38 + 51 + 66 = 155
70 70 20 20
9. (3i − 1) = 3 · i − 70 16. (i − 3) (i + 3) = i2 − 9
i=1 i=1 i=4 i=4
70(71) 20 20
=3· − 70 = 7, 385
2 = i2 − 9 1
45 45 45 i=4 i=4
10. (3i − 4) = 3 i−4 1 20 3 20
i=1 i=1 i=1 = i2 − i2 −9 1
45(46) i=1 i=1 i=4
=3 − 4(45) = 2925
2 20 (21) (41)
= − 1 − 4 − 9 − 9 (17)
40 40 6
11. (4 − i2 ) = 160 − i2 = 2703
i=1 i=1
(40)(41)(81) n
= 160 −
6 17. k2 − 3
= 160 − 22, 140 = −21, 980
k=3
n n
50 50 50
12. (8 − i) = 8 1− i = k2 + (−3)
k=3 k=3
i=1 i=1 i=1
n 2
50(51)
= 8(50) − = −875 = k2 − k2
2 k=1 k=1
100 n 2
13. n2 − 3n + 2 + (−3) − (−3)
n=1 k=1 k=1
100 100 100 n (n + 1) (2n + 1)
= −1−4
= n2 − 3 n+ 2 6
n=1 n=1 n=1 + (−3) n − (−3) (2)
(100)(101)(201) 100(101) n (n + 1) (2n + 1)
= −3 + 200 = − 5 − 3n + 6
6 2 6

9. 248 CHAPTER 4. INTEGRATION
n (n + 1) (2n + 1) = ((2.05)3 + 4)(0.1) + . . .
= − 3n + 1
6 + ((2.95)3 + 4)(0.1)
n = (202.4375)(0.1)
18. k2 + 5 = 20.24375
k=0 n 2
n n 1 i i
23. +2
= k2 + 5 n n n
i=1
k=0 k=0
n n n n
1 i2 i
=0+ k2 + 5 + 5 = +2
k=1 k=1
n i=1
n2 i=1
n
n (n + 1) (2n + 1) n n
= + 5 + 5n 1 1 2
6 = i2 + i
n n2 i=1
n i=1
n
19. f (xi )∆x 1 1 n(n + 1)(2n + 1)
=
i=1 n n2 6
5
= (x2 + 4xi ) · 0.2
i
2 n(n + 1)
+
i=1 n 2
= (0.22 + 4(0.2))(0.2) + . . .
n(n + 1)(2n + 1) n(n + 1)
+ (12 + 4)(0.2) = +
6n3 n2
= (0.84)(0.2) + (1.76)(0.2)
n 2
+ (2.76)(0.2) + (3.84)(0.2) 1 i i
lim +2
+ (5)(0.2) n→∞ n n n
i=1
= 2.84
n n(n + 1)(2n + 1) n(n + 1)
= lim +
20. f (xi )∆x n→∞ 6n3 n2
i=1 2 4
5 = +1=
= (3xi + 5) · 0.4 6 3
i=1 n 2
= (3(0.4) + 5)(0.4) + . . . 1 i i
24. −5
+ (3(2) + 5)(0.4) i=1
n n n
= (6.2)(0.4) + (7.4)(0.4) n n
+ (8.6)(0.4) + (9.8)(0.4) 1 i2 i
= −5
+ (11)(0.4) n i=1
n2 i=1
n
= 17.2 n n
n 1 1 5
= i2 − i
21. f (xi )∆x n n2 i=1
n i=1
i=1
10 1 1 n(n + 1)(2n + 1)
=
= (4x2 − 2) · 0.1
i n n2 6
i=1
= (4(2.1)2 − 2)(0.1) + . . . 5 n(n + 1)
−
+ (4(3)2 − 2)(0.1) n 2
= (15.64)(0.1) + (17.36)(0.1) n(n + 1)(2n + 1) 5n(n + 1)
+ (19.16)(0.1) + (21.04)(0.1) = −
6n3 2n2
+ (23)(0.1) + (25.04)(0.1) 2
−13n − 12n + 1
+ (27.16)(0.1) + (29.36)(0.1) =
6n2
+ (31.64)(0.1) + (34)(0.1) n 2
1 i i
= 24.34 lim −5
n
n→∞
i=1
n n n
22. f (xi )∆x −13n2 − 12n + 1
i=1 = lim
10 n→∞ 6n2
= (x3 + 4) · 0.1 13 12 1
= lim − − + 2
i=1 n→∞ 6 6n 6n

10. 4.2. SUMS AND SIGMA NOTATION 249
13 n
=− n2 (n + 1)2
6 i3 =
i=1
4
n
1 2i
2
2i is true for all integers n ≥ 1.
25. 4 − For n = 1, we have
n n n 1
i=1 12 (1 + 1)2
n n i3 = 1 = ,
1 i2 i i=1
4
= 16 −2
n i=1
n2 i=1
n as desired.
So the proposition is true for n = 1.
n n
1 16 2 Next, assume that
= i2 − i k
n n2 n k 2 (k + 1)2
i=1 i=1 i3 = ,
1 16 n(n + 1)(2n + 1) i=1
4
= for some integer k ≥ 1.
n n2 6
In this case, we have by the induction assump-
2 n(n + 1) tion that for n = k + 1,
−
n 2 n k+1 k
i3 = i3 = i3 + (k + 1)3
16n(n + 1)(2n + 1) n(n + 1)
= − i=1 i=1 i=1
6n3 n2 k 2 (k + 1)2
n 2 = + (k + 1)3
1 2i 2i 4
lim 4 − k 2 (k + 1)2 + 4(k + 1)3
n→∞
i=1
n n n =
4
16n(n + 1)(2n + 1) n(n + 1) (k + 1)2 (k 2 + 4k + 4)
= lim − =
n→∞ 6n3 n2 4
(k + 1)2 (k + 2)2
16 13 =
= −1= 4
3 3 n2 (n + 1)2
=
n 2 4
1 2i i as desired.
26. +4
i=1
n n n
28. Want to prove that
n n n
1 4i2 i n2 (n + 1)2 (2n2 + 2n − 1)
= +4 i5 =
n i=1
n2 i=1
n 12
i=1
n n is true for all integers n ≥ 1.
1 4 4 For n = 1, we have
= i2 + i
n n2 i=1
n i=1
1
12 (1 + 1)2 (2 + 2 − 1)
i3 = 1 = ,
1 4 n(n + 1)(2n + 1) i=1
12
=
n n2 6 as desired.
So the proposition is true for n = 1.
4 n(n + 1)
+ Next, assume that
n 2 k
k 2 (k + 1)2 (2k 2 + 2k − 1)
4n(n + 1)(2n + 1) 4n(n + 1) i5 = ,
= + i=1
12
6n3 2n2 for some integer k ≥ 1.
2
10n + 12n + 2 In this case, we have by the induction assump-
=
3n2 tion that for n = k + 1,
n 2 n k+1 k
1 2i i
lim +4 i5 = i5 = i5 + (k + 1)5
n→∞
i=1
n n n i=1 i=1 i=1
k 2 (k + 1)2 (2k 2 + 2k − 1)
10n + 12n + 2 2
= + (k + 1)5
= lim 12
n→∞ 3n2
10 12 2 10 k 2 (k + 1)2 (2k 2 + 2k − 1) + 12(k + 1)5
= lim + + = =
n→∞ 3 3n 3n2 3 12
(k + 1)2 [k 2 (2k 2 + 2k − 1) + 12(k + 1)3 ]
27. Want to prove that =
12

11. 250 CHAPTER 4. INTEGRATION
(k + 1)2 [2k 4 + 14k 3 + 35k 2 + 36k + 12] a − ar
= 34. When n = 0, a = .
12 1−r
Assume the formula holds for n = k − 1, which
(k + 1) (k + 4k + 4)(2k 2 + 6k + 3)
2 2
gives
=
12 a − ark
a + ar + · · · ark−1 = .
n2 (n + 1)2 (2n2 + 2n − 1) 1−r
= Then for n = k,
12
as desired. we have a + ar + · · · ark
10
= a + ar + · · · ark−1 + ark
a − ark
29. (i3 − 3i + 1) = + ark
1−r
i=1
10 10 a − ark + ark (1 − r)
=
= i3 − 3 i + 10 1−r
i=1 i=1 a − ark + ark − ark+1
=
100(11)2 10(11) 1−r
= −3 + 10 a − ark+1
4 2 =
= 2, 870 1−r
a − arn+1
=
20 1−r
30. (i3 + 2i) as desired.
i=1
20 20
n
= i3 + 2 i 6
35. e6i/n
i=1 i=1
i=1
n
400(21)2 20(21) n
= +2 = 44, 520 6
4 2 = e6i/n
n i=1
100
31. (i5 − 2i2 ) 6 e6/n − e6
=
i=1 n 1 − e6/n
100 100
= 5
i −2 i2 6 1 − e6
= −1
i=1 i=1 n 1 − e6/n
(100 )(1012 )[2(1002 ) + 2(100) − 1]
2 6 1 − e6 6
= = −
12 n 1 − e6/n n
6
100(101)(201) Now lim = 0, and
−2 x→∞ n
6
= 171, 707, 655, 800 6 1 − e6
lim
x→∞ n 1 − e6/n
100
1/n
32. (2i5 + 2i + 1) = 6(1 − e6 ) lim
x→∞ 1 − e6/n
i=1
100 100 1
= 6(1 − e6 ) lim
=2 i5 + 2 i + 100 x→∞ −6e6/n
i=1 i=1
= e6 − 1.
2 2 2 n
(100 )(101 )[2(100 ) + 2(100) − 1] 6
=2 Thus lim e6i/n = e6 − 1.
12 x→∞ n
i=1
100(101)
+2· + 100
2 n
= 343, 416, 675, 200 2
36. e(2i)/n
i=1
n
n n n
33. (cai + dbi ) = cai + dbi 2 e2/n − e2
=
i=1 i=1 i=1 n 1 − e2/n
n n
=c ai + d bi 2 1 − e2
= −1
i=1 i=1 n 1 − e2/n

12. 4.3. AREA 251
2 1 − e2 2 Notice that ∆x = 0.25.
= −
n 1 − e2/n n A4 = [f (0.125) + f (0.375) + f (0.625)
2 + f (0.875)](0.25)
Now lim = 0, and
x→∞ n = [(0.125)2 + 1 + (0.375)2 + 1
2 1 − e2 + (0.625)2 + 1 + (0.875)2 + 1](0.25)
lim
x→∞ n 1 − e2/n = 1.38125.
1/n
= 2(1 − e2 ) lim 2
x→∞ 1 − e2/n
1
= 2(1 − e2 ) lim
1.5
x→∞ −2e2/n
= e2 − 1. 1
n
2
Thus lim e2i/n = e2 − 1.
x→∞
i=1
n 0.5
37. Distance 0
= 50(2) + 60(1) + 70(1/2) + 60(3) 0 0.2 0.4 0.6
x
0.8 1 1.2
= 375 miles.
38. Distance
= 50(1) + 40(1) + 60(1/2) + 55(3)
= 285 miles. (b) Evaluation points:
0.25, 0.75, 1.25, 1.75.
39. On the time interval [0, 0.25], the estimated ve- Notice that ∆x = 0.5.
120 + 116 A4 = [f (0.25) + f (0.75) + f (1.25)
locity is the average velocity = 118
2 + f (1.75)](0.5)
feet per second. = [(0.25)2 + 1 + (0.75)2 + 1 + (1.25)2
We estimate the distance traveled during the + 1 + (1.75)2 + 1](0.5)
time interval [0, 0.25] to be = 4.625.
(118)(0.25 − 0) = 29.5 feet.
Altogether, the distance traveled is estimated 7
as
6
= (236/2)(0.25) + (229/2)(0.25)
+ (223/2)(0.25) + (218/2)(0.25) 5
+ (214/2)(0.25) + (210/2)(0.25) 4
+ (207/2)(0.25) + (205/2)(0.25) 3
= 217.75 feet. 2
40. On the time interval [0, 0.5], the estimated ve- 1
10 + 14.9 0
locity is the average velocity = 12.45 -0.5 0 0.5 1 1.5 2 2.5
2 x
meters per second. We estimate the distance
fallen during the time interval [0, 0.5] to be
(12.45)(0.5 − 0) = 6.225 meters.
Altogether, the distance fallen (estimated)
= (12.45)(0.5) + (17.35)(0.5)
+ (22.25)(0.5) + (27.15)(0.5)
+ (32.05)(0.5) + (36.95)(0.5)
+ (41.85)(0.5) + (46.75)(0.5) 2. (a) Evaluation points:
= 118.4 meters. 1.125, 1.375, 1.625, 1.875.
Notice that ∆x = 0.25.
A4 = [f (1.125) + f (1.375) + f (1.625)
4.3 Area + f (1.875)](0.25)
= [(1.125)3 − 1 + (1.375)3 − 1
1. (a) Evaluation points: + (1.625)3 − 1 + (1.875)3 − 1](0.25)
0.125, 0.375, 0.625, 0.875. = 2.7265625.

13. 252 CHAPTER 4. INTEGRATION
11π/16, 13π/16, 15π/16.
7
Notice that ∆x = π/8.
6 A4 = [f (π/16) + f (3π/16) + f (5π/16)
5 + f (7π/16) + f (9π/16) + f (11π/16)
4
+ f (13π/16) + f (15π/16)](π/8)
= [sin(π/16) + sin(3π/16) + sin(5π/16)
3
+ sin(7π/16) + sin(9π/16)
2 + sin(11π/16) + sin(13π/16)
1 + sin(15π/16)](π/8)
= 2.0129.
0
1 1.2 1.4 1.6 1.8 2
x 1
(b) Evaluation points: 0.8
1.25, 1.75, 2.25, 2.75.
Notice that ∆x = 0.5. 0.6
A4 = [f (1.25) + f (1.75) + f (2.25)
+ f (2.75)](0.5) 0.4
= [(1.25)3 − 1 + (1.75)3 − 1
+ (2.25)3 − 1 + (2.75)3 − 1](0.5)
0.2
= 17.75. 0
0 0.5 1 1.5 2 2.5 3
x
30
25
20
15
4. (a) Evaluation points:
−0.75, −0.25, 0.25, 0.75.
10
Notice that ∆x = 0.5.
5
A4 = [f (−0.75) + f (−0.25) + f (0.25)
0 + f (0.75)](0.5)
1 1.5 2 2.5 3
x = [4 − (−0.75)2 + 4 − (−0.25)2 + 4
− (0.25)2 + 4 − (0.75)2 ](0.5)
3. (a) Evaluation points: = 7.375.
π/8, 3π/8, 5π/8, 7π/8. 4
Notice that ∆x = π/4.
A4 = [f (π/8) + f (3π/8) + f (5π/8) 3
+ f (7π/8)](π/4)
= [sin(π/8) + sin(3π/8) + sin(5π/8)
2
+ sin(7π/8)](π/4)
= 2.05234.
1
1
0
0.8 -1 -0.5 0 0.5 1
x
0.6
0.4 (b) Evaluation points:
−2.75, −2.25, −1.75, −1.25.
0.2
Notice that ∆x = 0.5.
0
A4 = [f (−2.75) + f (−2.25) + f (−1.75)
0 0.5 1 1.5
x
2 2.5 3
+ f (−1.25)](0.5)
= [4 − (−2.75)2 + 4 − (−2.25)2 + 4
(b) Evaluation points: − (−1.75)2 + 4 − (−1.25)2 ](0.5)
π/16, 3π/16, 5π/16, 7π/16, 9π/16, = −0.625.

14. 4.3. AREA 253
15
A16 = ∆x f (ci )
2
i=0
15 2
x 1 i 1
-3 -2.5 -2 -1.5 -1 = + + 1 ≈ 4.6640
0
8 i=0
8 16
-2
(c) There are 16 rectangles and the evalua-
tion points are given by ci = i∆x + ∆x
-4
where i is from 0 to 15.
15
-6
A16 = ∆x f (ci )
i=0
15 2
1 i 1
= + + 1 ≈ 4.9219
5. (a) There are 16 rectangles and the evalua- 8 i=0
8 8
tion points are given by ci = i∆x where i
is from 0 to 15. 7. (a) There are 16 rectangles and the evalua-
15
tion points are the left endpoints which
A16 = ∆x f (ci )
are given by
i=0
15 2 ci = 1 + i∆x where i is from 0 to 15.
1 i 15
= + 1 ≈ 1.3027
16 16 A16 = ∆x f (ci )
i=0
i=0
(b) There are 16 rectangles and the evalua- 3
15
3i
∆x = 1+ + 2 ≈ 6.2663
tion points are given by ci = i∆x + 16 16
2 i=0
where i is from 0 to 15.
15 (b) There are 16 rectangles and the evalua-
A16 = ∆x f (ci ) tion points are the midpoints which are
i=0 given by
15 2 ∆x
1 i 1 ci = 1 + i∆x + where i is from 0 to
= + +1 2
16 i=0
16 32 15.
15
≈ 1.3330 A16 = ∆x f (ci )
(c) There are 16 rectangles and the evalua- i=0
15
tion points are given by ci = i∆x + ∆x 3 3i 3
= 1+ + +2
where i is from 0 to 15. 16 16 32
15 i=0
A16 = ∆x f (ci ) ≈ 6.3340
i=0
15 2 (c) There are 16 rectangles and the evalua-
1 i 1
= + +1 tion points are the right endpoints which
16 i=0
16 16 are given by
≈ 1.3652 ci = 1 + i∆x where i is from 1 to 16.
16
A16 = ∆x f (ci )
6. (a) There are 16 rectangles and the evalua-
i=1
tion points are given by ci = i∆x where i 16
3 3i
is from 0 to 15. = 1+ + 2 ≈ 6.4009
15 16 i=1
16
A16 = ∆x f (ci )
i=0
15 2 8. (a) There are 16 rectangles and the evalua-
1 i tion points are the left endpoints which
= + 1 ≈ 4.4219
8 i=0
8 are given by
ci = −1 + i∆x − ∆x
(b) There are 16 rectangles and the evalua-
where i is from 1 to 16.
∆x 16
tion points are given by ci = i∆x +
2 A16 = ∆x f (ci )
where i is from 0 to 15. i=1

15. 254 CHAPTER 4. INTEGRATION
16
1 i 1 given by ci = −1 + i∆x − ∆x where i is
= e−2(−1+ 8 − 8 ) ≈ 4.0991 from 1 to 100.
8 i=1 100
(b) There are 16 rectangles and the evalua- A100 = ∆x f (ci )
tion points are the midpoints which are i=1
100 3
given by 2 2i 2
∆x = −1 + − −1
ci = −1 + i∆x − 100 i=1
100 100
2
where i is from 1 to 16. ≈ −2.02
16
A16 = ∆x f (ci ) (b) There are 100 rectangles and the evalua-
i=1
tion points are midpoints which are given
16 ∆x
1 i 1 by ci = −1 + i∆x − where i is from 1
= e−2(−1+ 8 − 16 ) ≈ 3.6174 2
8 to 100.
i=1 100
(c) There are 16 rectangles and the evalua- A100 = ∆x f (ci )
tion points are the right endpoints which i=1
100 3
are given by 2 2i 1
ci = −1 + i∆x where i is from 1 to 16. = −1 + − −1
100 i=1
100 100
16
A16 = ∆x f (ci ) = −2
i=1
16
(c) There are 100 rectangles and the evalua-
1 i
tion points are right endpoints which are
= e−2(−1+ 8 ) ≈ 3.1924
8 i=1
given by ci = −1 + i∆x where i is from 1
to 100.
9. (a) There are 50 rectangles and the evalua- 100
tion points are given by ci = i∆x where i A100 = ∆x f (ci )
is from 0 to 49. i=1
50 100 3
2 2i
A50 = ∆x f (ci ) = −1 + − 1 ≈ −1.98
100 i=1
100
i=0
50
π πi 1
= cos ≈ 1.0156 11. (a) ∆x = . We will use right endpoints as
100 i=0
100 n
i
(b) There are 50 rectangles and the evalua- evaluation points, xi = .
n
∆x n
tion points are given by ci = + i∆x An = f (xi )∆x
2
where i is from 0 to 49. i=1
50 n 2 n
1 i 1
A50 = ∆x f (ci ) = +1 = 3 i2 + 1
i=0
n i=1 n n i=1
50 1 n(n + 1)(2n + 1)
π π πi = 3 +1
= cos + n 6
100 i=0
200 100
≈ 1.00004 8n2 + 3n + 1
=
6n2
(c) There are 50 rectangles and the evalua- Now to compute the exact area, we take
tion points are given by ci = ∆x + i∆x the limit as n → ∞:
where i is from 0 to 49. 8n2 + 3n + 1
50 A = lim An = lim
n→∞ n→∞ 6n2
A50 = ∆x f (ci )
8 3 1 4
i=0 = lim + + =
50 n→∞ 6 6n 6n2 3
π π πi 2
= cos + (b) ∆x = . We will use right endpoints as
100 i=0
100 100 n
2i
≈ 0.9842 evaluation points, xi = .
n
n
10. (a) There are 100 rectangles and the evalu- An = f (xi )∆x
ation points are left endpoints which are i=1