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```C H A P T E R
Integration
Section 4.1
4
Antiderivatives and Indefinite Integration
Solutions to Odd-Numbered Exercises
1.
d 3
d
9
C 3x3 C 9x4 4
dx x3
dx
x
3.
d 1 3
x 4x C x2 4 x 2x 2
dx 3
5.
dy
3t2
dt
7.
dy
x3
dx
y t3 C
9.
11.
3
x dx
x1 3 dx
1
x3 2 dx
x x
13.
1
dx
2x3
15.
x 3dx Check:
19.
Integrate
Simplify
x4 3
3 4
x
4
4 3
3
1 x2
C
2 2
1 3
x dx
2
x2
3x C
2
x
17.
21.
d 3 5
x
dx 5
3
3
x5 3
C x5
5 3
5
C x2
3
3 x2
C
2
C x3
2
2x 3x 2dx x 2 x3 C
25.
d 2
x x3 C 2x 3x 2
dx
2
x3 2 2x 1 dx x5 2 x2 x C
5
Check:
3
d 2 5
x
dx 5
C
Check:
d 1 4
x 2x C x3 2
dx 4
x2 3 dx C
1
C
4x2
2
C
3
2
d x2
3x C x 3
dx 2
x2 dx Check:
C
x1 2
C
1 2
1
x3 2 dx x 4 2x C
4
Check:
23.
Check:
Rewrite
dx
2
2
y x5
5
d
Check: t3 C 3t2
dt
Given
d 2 5
x
dx 5
1
dx x3
Check:
2
x3 dx x2 x C x3
2
2x 1
x2
1
C 2C
2
2x
d
1
1
C 3
dx 2x2
x
177
178
27.
Chapter 4
x2 x 1
dx x
Check:
29.
Integration
d 2 5
x
dx 5
2
2
2
2
x3 2 x1 2 x1 2 dx x5 2 x3 2 2x1 2 C x1 23x2 5x 15 C
5
3
15
2
x3
3
x 13x 2 dx 2
2x1 2 C x3 2 x1
2
x1
3x2 x 2 dx
2
2
y5 2 dy y7 2 C
7
y2 y dy 31.
1
x3 x2 2x C
2
Check:
x2 x 1
x
d 2 7
y
dy 7
Check:
2
C y5
2
y2 y
d 3 1 2
x x 2x C 3x2 x 2
dx
2
x 13x 2
33.
37.
dx 1 dx x C
Check:
d
x C 1
dx
41.
sec2 sin d tan cos C
39.
d
t csc t C 1 csc t cot t
dt
tan2 y 1 dy d
2 cos x 3 sin x C 2 sin x 3 cos x
dx
Check:
1 csc t cot t dt t csc t C
Check:
2 sin x 3 cos x dx 2 cos x 3 sin x C
35.
d
tan cos C sec2 sin d
Check:
43. f x cos x
sec2 y dy tan y C
y
d
Check: tan y C sec2 y tan2 y 1
dy
3
2
C
3
2C
0
x
2
2
3
45. f x 2
47. fx 1 x2
f x 2x C
f x x f )x)
2x
2
y
x3
3
f )x)
5
f )x)
f )x)
2x
x
3
2
x
3
2
1
2
3
x
3
2
3
1
f
2
2x 1 dx x2 x C
1 12 1 C ⇒ C 1
x3
3
4
3
dy
2x 1, 1, 1
dx
y
x
y
2
49.
x3
C
3
4
f′
C
3
y x2 x 1
Section 4.1
51.
dy
cos x, 0, 4
dx
y
Antiderivatives and Indefinite Integration
179
y
5
cos x dx sin x C
4 sin 0 C ⇒ C 4
y sin x 4
x
−3
5
−3
(b)
dy 1
x 1, 4, 2
dx 2
6
x2
xC
4
y
−4
42
2 4C
4
8
−2
2C
x2
x2
4
y
55. fx 4x, f 0 6
f x 4x dx 2x 2 C
57. ht 8t3 5, h1 4
ht 8t3 5dt 2t4 5t C
f0 6 202 C ⇒ C 6
h1 4 2 5 C ⇒ C 11
f x 2x 2 6
ht 2t4 5t 11
61. f x x3
59. f x 2
f2 5
f4 2
f 2 10
f 0 0
fx fx 2 dx 2x C1
f2 4 C1 5 ⇒ C1 1
fx 2x 1
f x 2x 1 dx x2 x C2
f 2 6 C2 10 ⇒ C2 4
f x x x 4
2
x3 2 dx 2x1
2
C1 fx f x 2
x
2x1 2 3 dx 4x1 2 3x C2
f x 4x1
h0 0 0 C 12 ⇒ C 12
ht 0.75t2 5t 12
(b) h6 0.7562 56 12 69 cm
C1
3
f 0 0 0 C2 0 ⇒ C2 0
1.5t 5 dt 0.75t 2 5t C
x
2
f4 C1 2 ⇒ C1 3
2
2
63. (a) ht 2
2
3x 4 x 3x
180
Chapter 4
Integration
65. f 0 4. Graph of f is given.
(a) f4
1.0
(f) f is a minimum at x 3.
(b) No. The slopes of the tangent lines are greater than 2
on 0, 2. Therefore, f must increase more than 4 units
on 0, 4.
(g)
y
6
4
(c) No, f 5 &lt; f 4 because f is decreasing on 4, 5.
(d) f is an maximum at x 3.5 because f3.5
the first derivative test.
2
x
0 and
−2
(e) f is concave upward when f is increasing on , 1
and 5, . f is concave downward on 1, 5. Points
of inflection at x 1, 5.
67. at 32 ft sec2
vt v0
st 32t v0 0 when t time to reach
32
maximum height.
s
s0 6 C2
st 16t2 60t 6 Position function
32 1632
v0
v0
32t 60
15
16 15
18 8
v0
32 550
v0
187.617 ft sec
15
seconds
8
2
60
158 6
62.26 feet
73. From Exercise 71, f t 4.9t2 10t 2.
71. at 9.8
v t 9.8t 10 0 (Maximum height when v 0.)
9.8 dt 9.8t C1
9.8t 10
v0 v0 C1 ⇒ vt 9.8t v0
t
9.8t v0 dt 4.9t v0t C2
2
f 0 s0 C2 ⇒ f t 4.9t 2 v0t s0
75.
2
v02 35,200
vt 32t 60 0
t
v0
v02 v02
550
64
32
The ball reaches its maximim height when
f t 8
−6
32t 60dt 16t 2 60t C2
vt 6
st 16t2 v0t
v0 60 C1
s
4
69. From Exercise 68, we have:
32 dt 32t C1
st 2
f
10
9.8
10
9.8
a 1.6
vt 1.6 dt 1.6t v0 1.6t, since the stone was dropped, v0 0.
st 1.6t dt 0.8t2 s0
s20 0 ⇒ 0.8202 s0 0
s0 320
Thus, the height of the cliff is 320 meters.
vt 1.6t
v20 32 m sec
7.1 m
Section 4.1
0 ≤ t ≤ 5
77. xt t3 6t2 9t 2
Antiderivatives and Indefinite Integration
1
79. vt t
(a) vt xt 3t 12t 9
2
xt 3t2 4t 3 3t 1t 3
vt dt 2t1
xt 2t1
(b) vt &gt; 0 when 0 &lt; t &lt; 1 or 3 &lt; t &lt; 5.
(c) at 6t 2 0 when t 2.
2
t &gt; 0
2
C
2 position function
1
at vt t3
2
v2 311 3
81. (a) v0 25 km hr 25
1000 250
m sec
3600
36
v13 80 km hr 80
1000 800
m sec
3600
36
2
1
acceleration
2t 3 2
83. Truck: vt 30
st 30t Let s0 0.
Automobile: at 6
at a constant acceleration
vt 6t Let v0 0.
vt at C
st 3t2 Let s0 0.
v0 v13 At the point where the automobile overtakes the truck:
250
250
⇒ vt at 36
36
30t 3t2
800
250
13a 36
36
0 3t2 30t
0 3tt 10 when t 10 sec.
550
13a
36
a
st a
s13 85.
2
x1 4 21 C ⇒ C 2
at vt 6t 12 6t 2
(b)
t1
(a) s10 3102 300 ft
550 275
468 234
t 2 250
t
2
36
(b) v10 610 60 ft sec
1.175 m sec2
41 mph
s0 0
275 132 250
13
234 2
36
189.58 m
1 mi hr5280 ft mi 22
ft sec
3600 sec hr
15
(a)
t
0
5
10
15
20
25
30
V1ft sec
0
3.67
10.27
23.47
42.53
66
95.33
V2ft sec
0
30.8
55.73
74.8
88
93.87
95.33
(c) S1t S2t V1t dt 0.1068 3 0.0416 2
t t 0.3679t
3
2
V2t dt 0.1208t3 6.7991t2
0.0707t
3
2
In both cases, the constant of integration is 0 because S10 S20 0
S130
953.5 feet
S230
1970.3 feet
The second car was going faster than the first until the end.
(b) V1t 0.1068t2 0.0416t 0.3679
V2t 0.1208t2 6.7991t 0.0707
181
182
Chapter 4
Integration
87. at k
vt kt
k
st t2 since v0 s0 0.
2
At the time of lift-off, kt 160 and k2t2 0.7. Since k2t2 0.7,
1.4k
1.4
1.4
160
v k
k
k
t
1.4k 1602 ⇒ k 1602
1.4
18,285.714 mihr2
7.45 ftsec2.
89. True
91. True
x x dx 93. False. For example,
95. fx x dx x dx because
x2
x3
C
C1
3
2
x2 C 2
2
1, 0 ≤ x &lt; 2
3x, 2 ≤ x ≤ 5
x C1,
0 ≤ x &lt; 2
f x 3x2
C2, 2 ≤ x ≤ 5
2
f 1 3 ⇒ 1 C1 3 ⇒ C1 2
f is continuous: Values must agree at x 2:
4 6 C2 ⇒ C2 2
0 ≤ x &lt; 2
x 2,
f x 3x 2
2, 2 ≤ x ≤ 5
2
The left and right hand derivatives at x 2 do not agree. Hence f is not differentiable at x 2.
Section 4.2
5
1.
Area
2i 1 2
i1
5
i
i1
4
3.
k0
k2
5
1 21 2 3 4 5 5 35
i1
1
1 1
1
1
158
1 1
2 5 10 17
85
9
7.
8
1
3i
i1
20
15.
i1
9.
5
j1
2i 2
20
i2
i1
2021
420
2
8j 3
4
c c c c c 4c
5.
k1
11.
2 n
n i1
20
17.
2in 2in
3
i 12 i1
13.
19
i2
i1
3 n
3i
2 1
n i1
n
192039
2470
6
2
```