Download Chapter 8 Solutions

Chapter 8
1.
1
Given the region bounded by the graphs of y = ln x , y = 0 , and x = e 2 .
a.
Use integration by parts to determine the area of the region.
b.
Find the volume of the solid generated by rotating the region about the
line x = e 2 .
c.
Find the volume of the solid generated by rotating the region about the
horizontal line y = −2 . Show your work, using integration by parts.
Chapter 8
2
Given the region bounded by the graphs of y = ln x , y = 0 , and x = e 2 .
1.
a. Use integration by parts to determine the area of the
region.
+2: integral
e2
Area =
∫ ln x dx = x ln x − ∫ 1dx , using integration by
+1: answer
1
parts with u = ln x & dv = dx .
Area = (x ln x − x ) 1 = e 2 + 1
e2
b. Find the volume of the solid generated by rotating
the region about the line x = e 2 .
Volume =
2
+1: limits, constant, answer
2
π ∫ (e 2 − e y ) dy = π ∫ (e 4 − 2e 2 e y + e 2 y )dy =
2
0
+2: integral
0
2
1
1⎞
⎛
⎞
⎛1
π ⎜ e 4 y − 2e 2 e y + e 2 y ⎟ = π ⎜ e 4 + 2e 2 − ⎟ = 41.577π
2
2⎠
⎝
⎠0
⎝2
c.
Find the volume of the solid generated by rotating
the region about the horizontal line y = −2 .
e
2
[
]
π ∫ (ln x + 2)2 − (2) 2 dx =
1
e2
[
+1: limits, constant, answer
]
Volume = π ∫ (ln x ) + 4 ln x dx =
1
[
2
π x(ln x )2 + 2 x ln x − 2 x
π (6e 2 + 2)
]
e2
1
+2: integrand
=
Chapter 8
3
3
2. Given the region bounded by the graphs of y = e − x , x = 0 , y = 0 , and x = b
(b > 0) .
a.
Find the area of the region if b = 1 .
b.
Find the volume of the solid generated if the region is rotated about the xaxis, and b = 1 .
c.
Find the volume of the solid generated when the region is rotated about
the x-axis and b = 10 .
Chapter 8
4
3
2. Given the region bounded by the graphs of y = e − x , x = 0 , y = 0 , and x = b
a. Find the area of the region if b = 1 .
1
Area = ∫ e − x dx = .8075
3
(b > 0) .
+2: integral
+1: answer
0
b. Find the volume of the solid generated if the region is
rotated about the x-axis and b = 1 .
+1: limits, constant, answer
Volume =
1
( )
π ∫ e− x
0
3
2
+2: integrand
1
dx = π ∫ e −2 x dx = 0.6907π ≈ 2.170
3
0
c. Find the volume of the solid generated when the
region is rotated about the x-axis and b = 10 .
10
( ) dx = π ∫ e
Volume = π ∫ e − x
0
3
2
10
0
−2 x3
dx = .7088π ≈ 2.227
+2: integrand
+1: limits, constant, answer
Chapter 8
3.
5
Consider the graph of y = xe
−x
3
.
a.
Set up the integral used to find the area under the graph of this curve
between x = 0 and x = 2 .
b.
Identify u and dv for finding the integral using integration by parts.
c.
Evaluate the integral using integration by parts.
Chapter 8
3.
6
Consider the graph of y = xe
−x
3
.
a. Set up the integral used to find the area under the
graph of this curve between x = 0 and x = 2 .
2
Area =
∫ xe
−x
3
+1: integrand
+1: limits
dx
0
b. Identify u and dv for finding the integral using
integration by parts.
−x
c.
u=x
and dv = e 3 dx
Evaluate the integral using integration by parts.
Let
u=x
dv = e
−x
du = dx v = −3e
2
Then,
∫ xe
−x
3
3
−x
dx = −3 xe
+1: dv = e
−x
3
dx
+2: uv − ∫ vdu
dx
3
−x
+1: u = x
+2: antiderivatives
3
0
2
− ∫ − 3e
−x
3
dx =
⎡− 3 xe − x 3 − 9e − x 3 ⎤ = −15e − 2 3 + 9 ≈ 1.299
⎢⎣
⎥⎦ 0
+1: answer
Chapter 8
4.
7
Consider the graph of y = ln 2 x .
a.
Set up the integral used to find the area under the graph of this curve
between x = 1 and x = 3 .
b.
Identify u and dv for finding the integral using integration by parts.
c.
Evaluate the integral using integration by parts.
Chapter 8
8
Consider the graph of y = ln 2 x .
4.
a.
Set up the integral used to find the area under the
graph of this curve between x = 1 and x = 3 .
+1: integrand
+1: limits
3
Area = ∫ ln 2 x dx
1
b. Identify u and dv for finding the integral using
integration by parts.
c.
+1: u = ln 2 x
u = ln 2 x and dv = dx
Evaluate the integral using integration by parts.
+1: dv = dx
dv = dx
u = ln 2 x
Let
1
du =
2dx v = x
2x
+2: uv − ∫ vdu
Then,
+2: antiderivatives
+1: answer
3
1
∫ ln 2 xdx = x ln 2 x − ∫ x ⋅ x dx
1
3
1
= [x ln 2 x − x ] 1 = 3 ln 6 − ln 2 − 2 ≈ 2.682
3
Chapter 8
9
Consider the region bounded by the x-axis, the line x = 5 , and the curve
x2
.
f ( x) =
3
2
2
x +9
5.
(
)
a.
Choose an appropriate trigonometric substitution to find the area of this
region. Explain why you made this choice.
b.
Find the area of this region using the method of trigonometric substitution.
c.
Evaluate lim f ( x) .
x →∞
Chapter 8
10
Consider the region bounded by the x-axis, the line x = 5 , and the curve
x2
.
f ( x) =
3
2
2
x +9
5.
(
)
a. Choose an appropriate trigonometric substitution to find the
area of this region. Explain why you made this
choice.
+1: chooses x = 3 tan θ
Let x = 3 tan θ . Using this substitution will result in a
denominator of the form
(9(tan
2
θ + 1)) 2 = (9 sec 2 θ )
3
3
+1: explanation
2
b. Find the area of this region using the method of
trigonometric substitution.
Note:
x = 3 tan θ
+2: Substitutes
for x and dx
+2: antiderivatives
dx = 3 sec θ dθ
Also:
3 tan θ = 0 ⇒ θ = 0
3 tan θ = 5 ⇒ θ = 1.03038
2
5
∫
0
1.03038
x2
(x
2
1.03038
∫
0
+ 9)
3
2
dx =
∫
0
tan 2 θ
dθ =
sec θ
1.03038
∫
0
(
9 tan 2 θ
9 tan θ + 9
2
+1: answer
)
sec 2 θ − 1
dθ =
sec θ
( ln sec θ + tan θ − sin θ )
c. Evaluate lim f ( x) .
1.03038
0
3
⋅ 3sec 2 θ dθ =
1.03038
∫ ( sec θ − cos θ ) dθ =
0
≈ 0.426
x →∞
Using L’Hopital’s rule,
1
x2
= lim
=0
lim
1
3
2
x →∞ (x 2 + 9 ) 2
x →∞ 3 (x + 9 ) 2
2
+1: correct use of
L’Hopital’s Rule
+1: answer
Chapter 8
11
Consider the region bounded by the x-axis, the line x = 2 , and the curve
x3
.
f ( x) =
4 + x2
6.
a.
Choose an appropriate trigonometric substitution to find the area of this
region. Explain why you made this choice.
b.
Find the area of this region using the method of trigonometric substitution.
c.
Evaluate lim f ( x) .
x →∞
Chapter 8
12
Consider the region bounded by the x-axis, the line x = 2 , and the curve
x3
.
f ( x) =
4 + x2
6.
a. Choose an appropriate trigonometric substitution to
find the area of this region. Explain why you made this
choice.
+1: Chooses x = 2 tan θ
Let x = 2 tan θ . Using this substitution will result in a
+1: explanation
denominator of the form 4(tan θ + 1) = 4 sec θ
b. Find the area of this region using the method of
trigonometric substitution.
2
2
∫
dx =
4 + x2
0
π
π
x3
4
∫
0
8 tan 3 θ
4 + 4 tan 2 θ
π
4
2
+2: Substitutes for x and dx
+2: antiderivative
⋅ 2 sec θ dθ =
4
(
2
+1: answer
)
3
2
∫ 8 tan θ secθ dθ = 8 ∫ sec θ − 1 secθ tan θ dθ =
0
0
⎡2 2
2⎤
8 ∫ u 2 − 1 du = 8⎢
− 2+ ⎥
3⎦
1
⎣ 3
c. Evaluate lim f ( x)
2
(
)
x →∞
Using L’Hopital’s Rule,
x3
3x
= lim
=
lim
−1
x →∞
x →∞ (4 + x 2 ) 2
4 + x2
lim 3x(4 + x )
2
1
2
=∞
x →∞
Therefore, this limit does not exist.
+1: Uses L’Hopital’s Rule
+1: answer
Chapter 8
7.
13
Consider the function h( x) =
a.
x − cos x
.
x
Can you use L’Hopital’s Rule to find
lim h( x) ?
Explain your reasoning.
x →∞
b.
Find
lim h( x) analytically.
x →∞
c.
Set up the integral needed to find the area under the graph of h(x) between
x = 1 and x = π (above the x-axis).
Chapter 8
7.
a.
Consider the function h( x) =
14
x − cos x
.
x
Can you use L’Hopital’s Rule to find
Explain your reasoning.
lim h( x) ?
x →∞
∞
Yes; lim h( x) is of the form . Even though cosine
∞
x →∞
oscillates, the x term dominates and the numerator goes
to infinity.
b. Find lim h( x) analytically.
+1: yes
+2: explanation
x →∞
x − cos x
⎛ x cos x ⎞
= lim ⎜ −
⎟=
lim
x
x ⎠
x →∞
x →∞ ⎝ x
cos x
1 − lim
= 1− 0 = 1
lim
x
x →∞
x →∞
c. Set up the integral needed to find the area under the
graph of h(x) between x = 1 and x = π
(above the x-axis).
π
x − cos x
dx
Area = ∫
x
1
+2: separation into fractions
+1: answer
+2: integrand
+1: limits
Chapter 8
8.
15
Consider the function g ( x) =
a.
ex −1
.
x2
Can you use L’Hopital’s Rule to find
lim g ( x) ?
Explain your reasoning.
x →0 +
b.
Find
lim g ( x) analytically.
x→0 +
c.
Describe the type of indeterminate form involved in directly evaluating
2 −x
lim x e . Evaluate this limit using L’Hopital’s Rule.
x →∞
Chapter 8
8.
a.
16
Consider the function g ( x) =
ex −1
.
x2
Can you use L’Hopital’s Rule to find
lim g ( x) ?
x→0 +
Explain your reasoning.
lim g ( x) is of the form
Find lim g ( x) analytically.
Yes;
x→0 +
b.
0
.
0
+1: yes
+2: explanation
x→0 +
ex −1
ex 1
=
= . This limit does not exist.
lim
lim
0
x2
x→0 +
x →0 + 2 x
c.
Describe the type of indeterminate form involved in
directly evaluating lim x 2 e − x . Evaluate this limit
x →∞
using L’Hopital’s Rule.
The indeterminate form is the form ∞ ⋅ 0 .
x2
2x
2
2 −x
=
= lim x = lim x = 0
x
e
lim
lim
x
x →∞
x →∞ e
x →∞ e
x →∞ e
+2: Uses L’Hopital’s rule
correctly
+1: answer
+1: indeterminate form
+2: answer
Chapter 8
9.
17
2
.
x3
Consider the function f ( x) =
1
a.
The evaluation of
2
∫x
3
dx = 0 is incorrect. Explain why.
−1
b.
Determine whether this improper integral converges or diverges. Show
your work to justify your conclusion.
c.
Evaluate
∞
∫ f ( x)dx
1
analytically.
Chapter 8
18
Consider the function f ( x) =
9.
1
a.
2
∫x
The evaluation of
3
2
.
x3
dx = 0 is incorrect. Explain
+2: explanation
−1
why.
2
has an infinite discontinuity at x = 0 .
x3
Determine whether this improper integral converges
or diverges. Show your work to justify
your conclusion.
b.
+1: uses limits
1
2
∫x
−1
3
t
1
2
2
dx + lim ∫ 3 dx =
3
s →0 + s x
−1 x
dx = lim ∫
t →0 −
1
t
−1
−1
+ lim 2
lim
2
t → 0 − x −1
s →0 + x s
Therefore, this improper integral diverges.
+1: antiderivative
+1: diverges
∞
c.
Evaluate
∫ f ( x)dx
analytically.
1
∞
∫
1
t
t
2
−1
f ( x)dx = lim ∫ 3 dx = lim 2 = 1
t →∞ 1 x
t →∞ x 1
+2: uses limit
+1: antiderivative
+1: answer
Chapter 8
10.
19
Consider the function g ( x) =
1
.
( x − 2 )2
3
a.
Explain why
∫ g ( x)dx
is an improper integral.
0
3
b.
Determine whether
∫ g ( x)dx
0
justify your conclusion.
∞
c.
Evaluate
∫ g ( x)dx
4
.
converges or diverges. Show your work to
Chapter 8
10.
20
Consider the function g ( x) =
1
.
( x − 2 )2
3
a.
Explain why
∫ g ( x)dx
is an improper integral.
+2: explanation
0
g (x) has an infinite discontinuity at x = 2 .
3
b.
Determine whether
∫ g ( x)dx
converges or diverges.
0
Show your work to justify your conclusion.
3
t
∫ g ( x)dx = lim ∫ (x − 2)
t →2− 0
0
3
1
2
dx + lim ∫
s →2+ s
t
1
(x − 2)
2
dx =
3
−1
−1
=∞
+ lim
lim
t →2− x − 2 0
s→2+ x − 2 s
+1: uses limits
+1: antiderivative
+1: answer
It diverges.
∞
c.
Evaluate ∫ g ( x)dx
+2: integrand
4
t
1
∫ lim (x − 2 )
t→∞
0+
2
dx = lim
1 1
= .
2 2
t→∞
−1
=
x−2 4
+1: limits
+1: answer