HW 16
34 – 7
Find the function f given that f 0 (x) = sin x and f (0) = 3.
Solution:
of sin x, so
The equation f 0 (x) = sin x says that the desired function f is an antiderivative
Z
f (x) = sin x dx = − cos x + C,
that is, f (x) = − cos x + C. Using the condition f (0) = 3, we get
3
=
↑
given
f (0)
=
↑
− cos 0 + C.
evaluate
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The preceding equations reveal that 3 = −1+C, so that C = 4. Therefore, f (x) = 4−cos x.
34 – 8
A ball dropped from the top of a tall building has velocity at time t seconds given by
v(t) = −10t m/s. Find the height of the building given that the ball strikes the ground
after 6 seconds.
Solution: Let f (t) be the ball’s height off the ground at time t. Since velocity is the
rate at which position changes, we have v(t) = f 0 (t), which says that f is an antiderivative
of v, so
2
Z
Z
t
f (t) = v(t) dt = (−10t) dt = −10
+ C = −5t2 + C,
2
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that is, f (t) = −5t2 + C. Since the ball hits the ground after 6 seconds, its height at time
t = 6 is 0, that is, f (6) = 0. Therefore,
0
=
↑
f (6)
given
=
↑
−5(6)2 + C.
evaluate
The preceding equations reveal that 0 = −5(6)2 + C, so that C = 180. Therefore, f (t) =
−5t2 + 180. The height of the building is the height of the ball at time t = 0, which is
f (0) = 180 m.
Z
35 – 1
Find
cos(8x − 3) dx.
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Solution: The expression cos(8x − 3) is the composition of 8x − 3 (inside function) and
cos x (outside function).
Let u = 8x − 3, so that du = 8 dx.
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We have
Z
Z
cos(8x − 3) 18 · 8 dx
Z
8dx
= 81 cos(8x − 3) |{z}
|
{z
}
du
cos u
Z
= 18 cos u du
cos(8x − 3) dx =
=
=
1
8
1
8
sin u + C
sin(8x − 3) + C.
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Z
35 – 2
Find
x4 ex
5
+3
dx.
Solution: The expression ex
(outside function).
5
+3
is the composition of x5 + 3 (inside function) and ex
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5
4
Let u = x + 3, so that du = 5x dx.
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We have
Z
4 x5 +3
x e
Z
dx =
ex
5
+3 4
x dx
Z
5
ex +3 51 · 5 x4 dx
Z
5
4
= 15 e|x{z+3} 5x
| {zdx}
=
eu
=
1
5
Z
du
eu du
= 15 eu + C
= 15 ex
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35 – 3
Find
4x2
p
5
+3
+ C.
1 − x3 dx.
Solution:
The expression
√
x (outside function).
√
1 − x3 is the composition of 1 − x3 (inside function) and
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Let u = 1 − x3 , so that du = −3x2 dx.
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We have
Z
4x2
Z p
p
1 − x3 dx = 4
1 − x3 x2 dx
Z p
1
1 − x3 −3
=4
· −3 x2 dx
Z
1/2
1 − x3
(−3x2 )dx
= − 43
|
{z
} | {z }
u1/2
du
Z
u1/2 du
3/2 u
+C
= − 43
3/2
3/2
8 1 − x3
=−
+ C.
9
= − 43
Z
35 – 4
Find
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tan θ dθ.
Hint: Write tan θ in terms of sine and cosine.
Solution:
Following the hint, we have
Z
Z
Z
sin θ
tan θ dθ =
dθ = (cos θ)−1 sin θ dθ.
cos θ
The expression (cos θ)−1 is the composition of cos θ (inside function) and x−1 (outside
function).
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Let u = cos θ, so that du = − sin θ dθ.
We have
Z
Z
(cos θ)−1 sin θ dθ
Z
= − (cos θ)−1 (− sin θ)dθ
Z
= − u−1 du
tan θ dθ =
= − ln |u| + C
= − ln | cos θ| + C.
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Z
35 – 5
Find
x3
√
dx.
x2 + 9
Solution:
Rewriting the integrand, we get
Z
Z
−1/2 3
x3
√
dx =
x2 + 9
x dx.
x2 + 9
The expression x2 + 9
function).
−1/2
is the composition of x2 +9 (inside function) and x−1/2 (outside
Let u = x2 + 9, so that du = 2x dx.
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In order to help make up the du we break the x3 up into x2 x and associate the x with the
dx. Since we need to change everything into u’s (no x’s), we use the substitution to write
the leftover factor x2 as u − 9.
Z
Z
−1/2 2
x3
√
dx =
x2 + 9
x xdx
x2 + 9
Z
−1/2 2
x 2xdx
= 12
x2 + 9
|
{z
}|{z} | {z }
u−1/2
=
1
2
Z
u−9
du
u−1/2 (u − 9) du
Z u1/2 − 9u−1/2 du
3/2
u
u1/2
1
=2
−9
+C
3/2
1/2
3/2
p
x2 + 9
− 9 x2 + 9 + C.
=
3
=
1
2
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Z
35 – 6
Find
3x + 2
dx
x2 + 1
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Hint: First rewrite the integrand as a sum of fractions.
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Solution:
Following the hint, we have
Z
Z 3x + 2
3x
2
dx =
+
dx
x2 + 1
x2 + 1 x2 + 1
Z
Z
x
1
=3
dx + 2
dx
x2 + 1
x2 + 1
We recognize the second integrand as the derivative of tan−1 x; the first integral requires a
substitution:
Let u = x2 + 1, so that du = 2x dx.
We have
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Z
3x + 2
dx =
x2 + 1
Z
Z
1
1
2xdx + 2
dx
2
+1
x +1
Z
Z
1
= 23 u−1 du + 2
dx
x2 + 1
= 32 ln |u| + 2 tan−1 x + C
= 32 ln x2 + 1 + 2 tan−1 x + C.
3
2
x2
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(The absolute value sign was omitted in the final answer since x + 1 is always positive.)
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