APPM 1360
Exam 2 Solutions
Fall 2014
1. Consider the region R in the first quadrant bounded by y = e−2x , y = e−2 , and the y-axis. Set up but
do not evaluate the integrals to calculate the following:
(a) (7 pts) The volume of a solid with R as the base and cross-sections perpendicular to the x-axis
that are squares.
(b) (7 pts) The volume generated by rotating R about the line y = −1 using the shell method.
(c) (7 pts) The volume generated by rotating R about the x-axis using the disk/washer method.
(d) (7 pts) The area of the surface generated by rotating the upper curve about the x-axis.
Solution:
Z 1
2
(a)
e−2x − e−2 dx
0
Z 1
(b)
−π(y + 1) ln y dy
e−2
Z 1
(c)
π e−4x − e−4 dx
0
Z 1
Z
p
−2x
−4x
(d)
2πe
1 + 4e
dx or
r
2πy
e−2
0
2.
1
1+
1
dy
4y 2
(a) (9 pts) A trapezoidal region has vertices at (0, 0), (a, 0), (a, 12), and (0, 6). If the x-coordinate
of the centroid of the region is located at x = 5/2, find the value of a.
p
√
(b) (7 pts) Find the particular solution to the differential equation y 1 − x2 y 0 − x 1 − y 2 = 0
that solves the initial condition y(0) = 1.
Solution:
(a) The area of the trapezoid is A = h(b1 + b2 )/2 = a(6 + 12)/2 = 9a. An equation of the line
connecting points (0, 6) and (a, 12) is y = 6x/a + 6. It follows that the x-coordinate of the
centroid equals
Z
Z a Z a
My
1 a
1
6
1
6 2
x=
xf (x) dx =
x
=
x + 6 dx =
x + 6x dx
m
A 0
9a 0
a
9a 0
a
a
5
1
1 2 3
=
x + 3x2 =
2a2 + 3a2 = a.
9a a
9a
9
0
We are given that x = 5/2 ⇒ a = 9/2 .
Alternate Solution:
The trapezoidal region consists of a rectangle topped by a triangle. The rectangle has area
m1 = 6a and centroid at (x1 , y1 ) = (a/2, 3). The triangle has area m2 = 3a and centroid at
(x2 , y2 ) = (2a/3, 8). (The centroid of a triangle is located 2/3 of the way along the median
from a vertex to the opposite side.) Using additivity of moments, the centroid of the trapezoidal
region has an x-coordinate of
x=
m1 x1 + m2 x2
(6a)(a/2) + (3a)(2a/3)
5
=
= a.
m1 + m2
6a + 3a
9
x = 5a/9 = 5/2 ⇒ a = 9/2 .
(b)
y
p
p
1 − x2 y 0 − x 1 − y 2 = 0
p
p
dy
y 1 − x2
= x 1 − y2
dx Z
Z
y(1 − y 2 )−1/2 dy = x(1 − x2 )−1/2 dx
Let u = 1 − y 2 , du = −2y dy and v = 1 − x2 , dv = −2x dx.
Z
Z
1
1
−
u−1/2 du = −
v −1/2 dv
2
2
−(1 − y 2 )1/2 = −(1 − x2 )1/2 + C
Use the initial value to find C: y(0) = 1 ⇒ 0 = −1 + C ⇒ C = 1
p
p
− 1 − y 2 = − 1 − x2 + 1
p
1 − y 2 = 1 − x2 − 2 1 − x2 + 1
q
p
y=
x2 + 2 1 − x2 − 1
3.
(a) Determine whether each sequence converges or diverges. If it converges, find the limit.
cos2 n
2 √ ∞
k
(i) (7 pts) an = √
n
(ii) (7 pts) bk = (−1) tanh (3k)
(iii) (7 pts)
ln n
n
n=2
Solution:
√
√
i. Use the Squeeze Theorem. 0 ≤ cos2 n ≤ 1 ⇒ 0 ≤ (cos2 n)/ n ≤ 1/ n. Since
√
√
lim 1/ n = 0, lim (cos2 n)/ n = 0 and the sequence converges to 0 .
n→∞
n→∞
sinh(3k)
e3k − e3k
1 − e−6k
= lim 3k
=
lim
= 1.
k→∞ cosh(3k)
k→∞ e + e−3k
k→∞ 1 + e−6k
ii. lim |bk | = lim tanh(3k) = lim
k→∞
k→∞
Since lim |bk | =
6 0, the alternating sequence {bk } diverges .
k→∞
√
√
2 x
2 n
iii. The sequence an =
corresponds to the function f (x) =
.
ln n
ln x
√
√
x−1/2
2 x L0 H
= lim
lim x = ∞
lim an = lim f (x) = lim
n→∞
x→∞
x→∞ ln x
x→∞ 1/x x→∞
The sequence diverges .
(b) (14 pts) Consider the sequence 5n c−2n , where c is a constant.
i. For what values of c will {an } converge?
ii. If {an } converges to L, what is the value of L?
iii. For what values of c will
∞
X
an converge?
n=1
iv. If
∞
X
n=1
an converges to S, what is the value of S?
Solution:
i. The sequence is geometric with r = 5c−2 > 0. An {rn } sequence converges for −1 < r ≤
√
√
1. Solving 0 < 5c−2 ≤ 1 for c yields c ≥ 5 or c ≤ − 5 .
√
√
√
ii. If c > 5 or c < − 5, then {an } converges to L = 0 . If c = ± 5, then {an } converges
to L = 1 .
√
√
iii. The geometric series converges when |r| = 5c−2 < 1 for c > 5 or c < − 5 .
iv. The sum of the geometric series is S =
4. Determine whether the following series
∞
X
5
a
5c−2
.
=
= 2
−2
1−r
1 − 5c
c −5
an converge or diverge. (You may not use the series
n=2
comparison, root or ratio tests.)
(a) (7 pts) an =
1
1 + n2
(b) (7 pts) an =
1 + e−n
n
(c) (7 pts) an = ln
n+1
n
Solution:
(a) Use the Integral Test. an = f (n) for f (x) = 1/(1 + x2 ) which is a positive, continuous,
decreasing function.
Z ∞
t
π
dx
= lim tan−1 x 2 = lim tan−1 t − tan−1 2 = − tan−1 2.
2
t→∞
t→∞
1+x
2
2
Z
Since
2
∞
∞
X 1
dx
converges,
the
series
also converges .
1 + x2
1 + n2
n=2
Alternate Solution:
Z ∞
Z ∞
1
dx
dx
1
<
and
converges
for
p
=
2
so
2
2
p
1+x
x
x
1 + x2
1
1
∞
X 1
also converges. Therefore
converges by the Integral Test.
1 + n2
Use the Integral Comparison Test:
n=2
1 + e−x
which is positive,
x
−e−x (x + 1) − 1
1 + e−x
1
continuous, and decreasing: f 0 (x) =
<
0
for
x
≥
2.
Since
> for
2
x
x
x
Z ∞
Z ∞
∞
−x
X
1
1+e
1 + e−n
x ≥ 2 and
dx is divergent, then
dx also diverges. Therefore
x
x
n
2
2
(b) Use the Integral Comparison Test: an = f (n) for f (x) =
n=2
diverges by the Integral Test.
(c) This is a telescoping sum:
∞
X
n=2
ln
∞
X
n+1
=
(ln(n + 1) − ln n) . The limit of the nth partial
n
n=2
sum is
+ · · · + (ln(n + 2) − ln(n
lim sn = lim ((
ln
3 − ln 2) + (
ln
4 −
ln
3) + (
ln
5 −
ln
4)
+ 1)))
n→∞
n→∞
= lim (− ln 2 + ln(n + 2)) = ∞.
n→∞
Therefore
∞
X
n=2
ln
n+1
diverges .
n