1. No category

MATH 104 SAMPLE FINAL SOLUTIONS
CLAY SHONKWILER
(1) Evaluate the integral
Z
e−x/2 cos xdx.
Answer: We integrate by parts. Let u = e−x/2 and dv = cos xdx.
Then du = − 12 e−x/2 dx and v = sin x. Then the above integral is
equal to
Z
1
e−x/2 sin x +
e−x/2 sin xdx.
2
To compute this integral, again let u = e−x/2 and let dv = sin xdx.
Then du = − 12 e−x/2 and v = − cos x. Hence
Z
Z
1
−x/2
−x/2
e
cos xdx = e
sin x +
e−x/2 sin xdx
2
Z
1
1
−x/2
−x/2
−x/2
−e
cos x −
e
cos xdx
=e
sin x +
2
2
Z
1 −x/2
1
−x/2
=e
sin x − e
cos x −
e−x/2 cos xdx.
2
4
Therefore,
Z
5
1
e−x/2 cos xdx = e−x/2 sin x − e−x/2 cos x,
4
2
so
Z
4
2
e−x/2 cos xdx = e−x/2 sin x − e−x/2 cos x.
5
5
(2) Consider the series
∞
X
n=1
12n
.
6n2 + ln n
Does this series converge or diverge?
Answer: Since ln n grows significantly slower than both n and
n2 , we suspect that it won’t much affect whether the series converges
or
P diverges.
P 2Therefore, let’s do a limit comparison with the series
12n
=
n , which we know diverges:
6n2
12n
6n2 +ln n
2
n→∞
n
lim
12n2
.
n→∞ 12n2 + 2 ln n
= lim
1
2
CLAY SHONKWILER
By two applications of L’Hˆopital’s Rule, this limit is equal to
24n
24
lim
= lim
= 1.
n→∞ 24n + 2
n→∞ 24 − 22
n
n
P2
P 12n
also diverges.
Therefore, since
n diverges, the series
6n2 +ln n
(3) Consider the sequence {an } with
n(n+1)
(−1) 2 5n
an =
.
π 2n
Does this sequence converge or diverge? If it converges, what does
it converge to?
Answer: Note that
n
(−1) n(n+1)
2
5n 5n
5
=
.
=
π 2n
π 2n
π2
n
Since π52 < 1, the sequence { π52 } converges to zero. Therefore, by
the sandwich theorem, the
√ given sequence converges to zero.
(4) Consider the curve y = x. Find the surface area of the surface
obtained by rotating the portion of the curve between x = 3/4 and
x = 15/4 about the x-axis.
Answer: Recall the surface area formula
s
2
Z 15/4
dy
2πy 1 +
dx.
SA =
dx
3/4
2
dy
dy
1
Now, dx
= 2√1 x , so dx
. Hence,
= 4x
r
Z 15/4
Z 15/4 r
√
1
1
SA = 2π
x 1+
dx = 2π
x + dx.
4x
4
3/4
3/4
Letting u = x + 14 , du = dx and so the above integral becomes
Z 4
√
2 3/2 4
2
2
28π
= 2π (8) − (1) =
.
2π
udu = 2π u
3
3
3
3
1
1
(5) Suppose you invest $1000 in an account earning 8% interest compounded continuously. After how many years will you quadruple
your money (use the approximation ln 4 ≈ 1.6).
Answer Remember that the amount of money in your account is
modeled by
A(t) = P ert .
Since the initial amount in the account is $1000, P = 1000. Also,
r = 0.08. You will quadruple your money when
4000 = A(t) = 1000e0.08t ,
so
4 = e0.08t .
MATH 104 SAMPLE FINAL SOLUTIONS
3
Solving for t,
ln 4 = 0.08t,
so
ln 4
1.6
160
≈
=
= 20.
0.08
0.08
8
Therefore, you will quadruple your money in approximately 20 years.
(6) Consider the series
∞
X
4n
.
e2n+1
t=
n=1
Does this series converge or diverge? If it converges, to what sum
does it converge?
Answer: Note that
n−1
∞
∞
∞
∞
X
X
X 4n
1 4n
1 4 n X 4
4
=
=
=
.
2n+1
2n
2
3
e
ee
e e
e
e2
n=1
n=1
n=1
n=1
This is a geometric series with a =
series converges to
4
e3
1−
4
e2
=
4
e3
e2 −4
e2
=
4
e3
and r =
4
e2
< 1. Hence, the
4
.
e3 − 4e
(7) Does the series
∞
X
(2n)!
n=1
2n n!
converge or diverge?
Answer: Use the Ratio Test:
(2(n+1))!
2n+1 (n+1)!
lim
(2n)!
n→∞
2n n!
(2n + 2)(2n + 1)
4n2 + 5n + 2
(2n + 2)! 2n n!
·
=
lim
=
lim
.
n→∞
n→∞
n→∞ 2n+1 (n + 1)! (2n)!
2(n + 1)
2n + 1
= lim
By L’Hˆopital’s Rule, this limit is equal to
8n + 5
lim
= ∞,
n→∞
2
so the series diverges.
(8) Find the interval of convergence of the power series
∞
X
(2x)n
n=0
n2
.
For what values of x does the series converge absolutely? For what
values does it converge conditionally?
Answer: Using the Ratio Test,
(2x)n+1 (n+1)2 (2x)n+1
2xn2 n2 .
lim
·
= lim n = lim n→∞ (2x)
n→∞ (n + 1)2 (2x)n n→∞ (n + 1)2 2
n
4
CLAY SHONKWILER
By two applications of L’Hˆopital’s Rule, this limit is equal to
4xn = lim 4x = |2x|.
lim n→∞ 2(n + 1)
n→∞ 2 Now, |2x| < 1 when |x| < 12 , so the radius of convergence of the
power series is 12 . Now, we need to check the endpoints, x = ± 12 .
When x = − 21 , the series becomes
∞
X
(−1)n
n=1
n2
,
which converges, by the alternating series test. In fact,
∞
∞ X
(−1)n X
1
=
,
n2 n2
n=1
n=1
which is a convergent p-series. On the other hand, when x = 21 , the
series becomes
∞
X
1n
,
n2
n=1
which is a convergent p-series.
Therefore, the interval of convergence
of the power series is − 12 , 12 . The series converges absolutely for
− 21 ≤ x ≤ 21 .
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(9) Consider the region contained by y = 1+x
2 , the x- and y-axes and
x = 1. What is the volume of the solid obtained by rotating this
region about the y-axis?
Answer: Using the shell method, the volume is equal to
Z 1
2πrhdx.
0
1
,
1+x2
Now, r = x and h =
Z 1
V =
2πx
0
so
1
dx = π
1 + x2
Z
0
1
2xdx
.
1 + x2
x2 ,
Letting u = 1 +
du = 2xdx, so we can re-write this integral as
Z 2
du
= π [ln |u|]21 = π [ln 2 − ln 1] = π ln 2.
π
1 u
(10) Evaluate the integral
Z
2dx
√
.
(2x − 2) 4x2 − 8x + 1
Answer: We complete the square:
4x2 − 8x + 1 = (4x2 − 8x + 4) + 1 − 4 = (2x − 2)2 − 3.
MATH 104 SAMPLE FINAL SOLUTIONS
5
Hence, letting u = 2x − 2, du = 2dx and so we can re-write the
integral as
Z
du
1
1
−1 u −1 2x − 2 √
√
√
√
√
=
sec +C =
sec + C.
3
3
3
3 u u2 − 3
(11) Evaluate the integral
Z
x2
2x + 3
dx.
− 7x + 12
Answer: Note that x2 − 7x + 12 = (x − 3)(x − 4). Thus, we set
2x+3
A
B
up the partial fraction (x−3)(x−4)
= x−3
+ x−4
and solve for A and
B:
2x + 3 = A(x − 4) + B(x − 3).
Letting x = 4,
2(4) + 3 = A(0) + B(1) = B,
so B = 11. Now, letting x = 3,
2(3) + 3) = A(−1) + B(0) = −A,
Z
so A = −9. Therefore,
Z −9
11
2x + 3
dx =
+
dx = −9 ln |x−3|+11 ln |x−4|+C.
x2 − 12x + 7
x−3 x−4
(12) Solve the initial value problem
dy
2
+ y = sin x + cos x, x > 0,
y(π) = .
dx
π
Answer: Re-write in standard form:
1
sin x + cos x
dy
+ y=
,
dx x
x
x
so P (x) =
1
x
x
and Q(x) = sin x+cos
. Hence,
x
Z
Z
1
dx = ln |x| = ln x,
P (x)dx =
x
since x > 0. Thus,
R
v(x) = e
P (x)dx
= eln x = x.
Therefore,
Z
Z
Z
1
1
sin x + cos x
1
y=
v(x)Q(x)dx =
x
dx =
[sin x + cos x] dx
v(x)
x
x
x
1
= [− cos x + sin x + C]
x
sin x − cos x + C
=
.
x
6
CLAY SHONKWILER
Using the initial value,
2
sin π − cos π + C
1+C
= y(π) =
=
,
π
π
π
so C = 1. Therefore,
y=
sin x − cos x + 1
.
x
(13) Evaluate the integral
Z
(x2
x
dx.
+ 3)3/2
√
√
Answer: Let x = 3 tan θ. Then dx = 3 sec2 θdθ, so we can
re-write the integral as
√
Z
Z
√
3 tan θ sec2 θdθ
3 tan θ
2
3
sec
θdθ
=
(3 tan2 θ + 3)3/2
(3 sec2 θ)3/2
Z
3 tan θ sec2 θdθ
√
=
3 3 sec3 θ
Z
1
tan θdθ
=√
sec θ
3
Z
1
sin θdθ
=√
3
1
= − √ cos θ + C
3
Now, θ = tan−1
to
√x ,
3
so cos θ =
√
√ 3 .
3+x2
Hence, the integral is equal
√
3
1
1
−√ √
= −√
.
2
3 3+x
3 + x2
(14) Consider the sequence {an } where
√
3
n
.
an =
ln(n + 1)
Does the sequence converge or diverge? If it converges, what does it
converge to?
Answer: Using L’Hˆopital’s Rule,
√
1
3
n
n+1
2/3
lim
= lim 3n1 = lim
.
n→∞ ln(n + 1)
n→∞
n→∞ 3n2/3
n+1
Using L’Hˆopital again, this is equal to
√
3
1
n
= ∞.
lim
= lim
−1/3
n→∞ 2n
n→∞ 2
Therefore, the sequence diverges.
MATH 104 SAMPLE FINAL SOLUTIONS
(15) Does the series
∞
X
n=1
1
n(ln n)2
converge or diverge?
Answer: Using the Integral Test,
Z ∞
Z b
1
dx
dx = lim
.
2
b→∞ 1 x(ln x)2
x(ln x)
1
Now, letting u = ln x, du = x1 dx, so
Z
Z
dx
du
1
1
=
=− +C =−
+ C.
x(ln x)2
u2
u
ln x
Hence,
Z b
1
1 b
1
dx
= lim −
= lim −
+
=∞
lim
b→∞
b→∞ 1 x(ln x)2
ln x 1 b→∞
ln b ln 1
P 1
since ln 1 = 0. Therefore, by the Integral Test,
diverges.
n(ln n)2
DRL 3E3A, University of Pennsylvania
E-mail address: [email protected]
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