Answer Key
Calculus II - Final Exam Review, Spring 2013 [Shaw]
1. Integrate:
Z
4x − 1
(a)
dx = 3 ln |(x + 2)(x − 1)| + C [partial fractions]
(x − 1)(x + 2)
Z
π
sec2 (x) tan(x) dx =
(b)
3π/4
Z
(c)
Z
(d)
−1
[substitution with u = tan(x)]
2
1 1 + ex ex
+ C [substitution with u = ex , followed by partial fractions]
dx = ln 1 − e2x
2
1 − ex e
1
11 − 9e2 + 4e3 [Integration by parts with u = ln(x), and then some
18
1
algebra on v du making it easier to integrate]
(x − 1)2 ln(x) dx =
Z
(e)
x cos(2x) dx =
Z
(f)
Z
1
1
x sin(2x) + cos(2x) + C [Partial fractions with u = x]
2
4
√
√ √
cos( x)
√
dx = 2 sin x + C [Substitution with u = x]
x
∞
e3−x dx = 1 [Improper integral, substitution with u = 3 − x]
(g)
3
Z
−1
xe−x dx DIVERGES [Integrate first using parts, u = x, and then see that
(h)
−∞
lim −xe−x
a→−∞
diverges to ∞.]
Z
(i)
∞
1
dx DIVERGES [Note that the function doesn’t exist at x = 1, so the integral is
− x2
1
improper for both the top and bottom limit. You should end up, after partial fractions, with
∞
x − 1 1 b − 1 1
a − 1 1
ln + = lim ln +
− lim ln −
a→1
b→∞
x x
b b
a a
x3
1
All of these elements converge, except lima→1 ln a−1
a , which diverges to −∞.]
2. Find the arc-length along the curve y =
The arc-length integral,
you get 29
24 .
p
7
x3
+
from x = 1 to x = 2.
21 4x
1 + (f 0 (x))2 , works out to
x2
7
+
7
4x2 .
Once you integrate this from 1 to 2,
3. Use any appropriate
√ method to determine the volume of the solid generated by revolving the segment
of the curve y = 3 x + 1 from (0, 1) to (7, 2) about the y-axis.
1
You could use cylindrical shells or disks to answer this one; I use disks with radius r(y) = y 3 − 1 to
get the integral
Z 2
163π
π(y 3 − 1)2 dy =
14
1
4. Find a solution to the differential equation, and to the initial value problem where applicable; if possible,
write the solution as an explicit function of x:
(a)
dy
dx
= x3 y 4 , y(0) = 1
y = − 34 x4 + 1
(b) xey −
−1/3
dy
dx
=0
2
y = − ln − x2 + C (c)
dy
dx
= y2 − y
y=
(d)
dy
dx
1
1+Cex
= 12x2 y − xy
y = Ce4x
3
−x2 /2
5. The population of a small city in the rural midwest satisfies the exponential growth model. In this
city, the population doubled from 1960 to 2010. What will the population be in 2030, relative to the
1960 population?
70
A(70) = A0 · e 50 ln(2)
6. Determine whether each sequence with the given nth term converges. If so, calculate the limit.
(a) an =
n2
n
− 3n
Converges to 0 [Check the leading terms]
(b) bn =
n(1 − en )
e2n
Converges to 0 [Separate the factors, and use L’Hôpital on each]
(c) cn = n20 · 2−n
Converges to 0 [L’Hôpital]
(d) dn =
ln(ln(n))
ln(n)
Converges to 0 [L’Hôpital]
2
(e) αn = sin
1
· en
n
Diverges [Rewrite as a fraction
(f) βn =
1
sin n
e−n
and use L’Hôpital]
n4 − ln(2n)
1 − n4
Converges to -1 [Use L’Hôpital]
√
(g) γn =
n4 − 1
n2
Converges to 1 [Rewrite the limit as
(h) δn =
p
lim(f (x))2 to simplify the inner limit; you then get
√
1 = 1]
3n
n!
Converges to 0 [Compare 3n with n! to see that n! is eventually much larger.]
(i) εn =
tan−1 (n)
n
Converges to 0 [Can’t use L’Hôpital: in fact, the limit of tan−1 (n) is
π
2]
7. Using any appropriate method, determine whether or not the series converges. You do not have to find
the sum.
(a)
∞
X
2
−n
n=1
Convergent [Use the limit comparison test with
(b)
∞
X
10
2n
5
n=1
Convergent [Equivalent to the geometric series
(c)
1
n]
∞ X
n=2
2
n−1
P
10 ·
1 n
]
25
1.001
Convergent [Limit comparison with
1 1.001
]
n
∞
X
2n − 1
(d)
7n
n=1
Convergent [Rewrite as the difference of two geometric series:
(e)
∞
X
(−1)n+1 ·
n=1
2 n
7
P
−
P
1
7n ]
r !
n 1
1−
5
Convergent [Use the alternating series test and note that since
the series approach 0.]
∞ √
X
n
(f)
n
n=1
3
q
n
1
5
approaches 1, the terms of
Divergent [Use the comparison test;
(g)
√n
n
1
n,
>
which diverges.]
∞
X
ln(5n)
(n + 2)2
n=1
Convergent [Use the limit comparison test with n−3/2 , a convergent geometric series.]
(h)
∞
X
n + 6n
n!
n=1
Convergent [Use the ratio test.]
(i)
∞
X
1
2+1
n
n=1
Convergent [Use the comparison test with
(j)
∞
X
(−1)n ·
n=2
1
n2 ]
n−1
ln(n)
Divergent [Use the alternating series test or the nth term test and see that the terms of the series
do not approach 0.]
(k)
∞
X
2n · n!
(2n)!
n=1
Convergent [Ratio test.]
(l)
∞
X
(−1)n ·
n=1
n3 − 5
n4 + 2n − 1
Convergent [Alternating series test]
(m)
∞
X
(n!)e−n
n=1
Divergent [Either the nth term test or the ratio test, which will give you a ratio approaching ∞.]
8. Find the sum:
n
∞ X
2 4
8
−2
1
3
(a) 1 − + −
+ ···=
=
=
3 9 27
3
1
+
2/3
5
n=0
(b)
∞
X
1
1
1
√ −√
This is a telescoping series with nth partial sum sn = −1 − √12 − √13 + √n−2
+
n
n
−
3
i=4
√1
+ √1n , and thus converges to −1 − √12 − √13
n−1
9. For each of the series below, find the radius and interval of convergence for x.
(a)
∞
X
(x − 3)n
n=0
R = 1; the interval of convergence is (2, 4).
4
(b)
∞
X
2n n
(x)
n!
n=0
R = ∞; the series converges on all of R.
(c)
n
∞ 2
X
x +5
n=0
5
This
a power series, in fact: when you perform the ratio test you get that x must satisfy
2 isn’t
x +5 5 < 1, which is impossible; thus the radius of convergence is 0 and the interval of convergence
is ∅.
(d)
∞
X
(−1)n+1 x2n+1
n=0
R = 1; the interval of convergence is (−1, 1).
(e)
∞
X
n
(ex − 4)
n=0
√
R = ln( 15); the interval of convergence is (ln(3), ln(5)).
5