Math 2260 Final Exam Solutions
1. Consider the region bounded above by the graph of y = sin(x/2) and below by the x-axis. What is
the volume of the solid obtained by revolving this region around the x-axis?
Answer: Using the disk method, the volume is given by
Z 2π
Z 2π
2
π (sin(x/2)) dx = π
sin2 (x/2) dx
0
0
2π
1 − cos(x)
dx
2
0
Z 2π 1 cos x
=π
−
dx
2
2
0
2π
x sin x
=π
−
2
2 0
Z
=π
= π [(π − 0) − (0 − 0)]
= π2 .
2. Find the surface area of the cone given by rotating the line y = 2x for 0 ≤ x ≤ 4 around the x-axis.
Answer: Since
dy
dx
= 2, the surface area integral becomes
Z
4
Z
q
2
2π2x 1 + (2) dx = 4π
0
4
√
x 1 + 4 dx
0
√ Z
= 4π 5
4
x dx
0
4
x2
= 4π 5
2 0
√
= 4π 5 (8 − 0)
√
= 32π 5.
√
3. Solve the initial-value problem
dy
− xy = x,
dx
1
y(0) = 3.
Answer: Adding xy to both sides yields
dy
= x + xy = x(1 + y),
dx
and hence separating and integrating gives that
Z
Z
dy
= x dx,
1+y
so
x2
+ C.
2
ln(1 + y) =
Now exponentiate both sides to get
1 + y = ex
2
/2+C
= eC ex
2
/2
= Aex
2
/2
,
where A = eC . Therefore,
y = Aex
2
/2
−1
and then we can use the initial value to determine A:
y(0) = Ae0
2
/2
−1
3 = A − 1,
so A = 4. Therefore, the solution to the initial-value problem is
y = 4ex
2
/2
− 1.
4. Evaluate the integral
Z
e3
1
ln x
√
dx.
3
x2
Answer: First, integrate by parts with
dv = x−2/3 dx
u = ln x
1
du = dx
x
v = 3x1/3
Then the above integral becomes
h
i e3 Z
1/3
3x ln x
−
1
1
e3
Z e3
3x1/3
dx = (9e − 0) − 3
x−2/3 dx
x
1
h
i e3
= 9e − 3 3x1/3
1
= 9e − 3 (3e − 3)
= 9.
5. Evaluate the integral
Z
2
√
0
1
dx.
4 − x2
Answer: First, notice that the denominator equals zero when x = 2, so this is an improper integral.
By definition, it is equal to
Z b
1
√
lim
dx.
−
b→2
4 − x2
0
2
Now, make the trig substitution x = 2 sin θ. Then dx = 2 cos θ dθ and the above integral becomes
Z arcsin(b/2)
Z arcsin(b/2)
1
1
p
2 cos θ dθ
lim
2 cos θ dθ = lim
−
−
2
2 cos θ
b→2
b→2
0
0
4 − 4 sin θ
Z arcsin(b/2)
dθ
= lim−
b→2
0
h iarcsin(b/2)
= lim− θ
b→2
0
= lim (arcsin(b/2) − 0)
b→2−
= arcsin(1)
π
= .
2
6. (a) Does the series
∞
X
(−1)n
n=1
n2 + π
√
n3 − 2/2
converge absolutely, converge conditionally, or diverge?
Answer: The series certainly converges, since it satisfies the hypotheses of the alternating series
test. Now, let’s check the series of absolute values,
∞ ∞
2
X
π X n2 + π
(−1)n n +
√
√
=
.
n3 − 2/2 n=1 n3 − 2/2
n=1
This series looks similar to the series
n2√
+π
n3 − 2/2
lim
n2
n→∞
n3
P n2
n3 ,
so let’s do a limit comparison to that series:
n2 + π n3
n2 + π
n3
√
√
=
lim
= 1.
n→∞ n3 −
n→∞
n2 n3 − 2/2
2/2 n2
= lim
Therefore, the series either both converge or both diverge. Since
∞
∞
X
X
1
n2
=
3
n
n
n=1
n=1
diverges, we see that the series of absolute values diverges, and so the given series only converges
conditionally.
(b) Does the series
∞
X
(n + 3)3
(n + 1)!
n=1
converge absolutely, converge conditionally, or diverge?
Answer: Since all of the terms are positive, each term is equal to its absolute value, so the series
either converges absolutely or it doesn’t converge at all. To determine convergence, use the Ratio
Test:
(n+4)3
(n+2)!
lim
3
n→∞ (n+3)
(n+1)!
(n + 4)3 (n + 1)!
1 (n + 4)3
1
= lim
= lim
= 0.
3
n→∞ (n + 2)! (n + 3)
n→∞ n + 2 (n + 3)3
n→∞ n + 2
= lim
Since 0 < 1, the Ratio Test says the series converges, so we conclude that the given series converges
absolutely.
3
7. What is the interval of convergence of the following power series?
∞
X
(−1)n (x − 2)n
.
n5n
n=1
Answer: Using the Ratio Test,
(−1)n+1 (x−2)n+1 (n+1)5n+1 |x − 2|n+1
n5n
|x − 2| n
|x − 2|
=
lim
= lim
lim =
.
(−1)n (x−2)n n→∞ (n + 1)5n+1 |x − 2|n
n→∞
n→∞
5
n
+
1
5
n5n
Therefore, by the Ratio Test, this series definitely converges absolutely when
|x − 2| < 5.
|x−2|
5
< 1, meaning
Now we check the endpoints. When x − 2 = 5, the series becomes
∞
∞
X
X
(−1)n 5n
(−1)n
=
,
n
n5
n
n=1
n=1
which is a convergent alternating series.
When x − 2 = −5, the series becomes
∞
∞
∞
X
X
X
(−1)n (−5)n
(−1)n (−1)n 5n
1
=
=
,
n
n
n5
n5
n
n=1
n=1
n=1
which is a divergent series.
Therefore, the power series converges for
−5 < x − 2 ≤ 5
or
−3 < x ≤ 7,
so the interval of convergence is (−3, 7].
8. Find an approximation for
Z
0
1
sin(t2 )
dt
t
which is accurate to within 0.01. Feel free to give your answer as a fraction.
[Hint: Think about Taylor series.]
Answer: This isn’t an integral we can evaluate exactly, so instead I will find a Taylor series expression
2
)
for sin(t
and integrate that. Since
t
sin(t) = t −
∞
X
t3
t5
t7
(−1)n−1 t2n−1
+ − + ... =
3! 5! 7!
(2n − 1)!
n=1
is the Taylor series centered at x = 0 for the function sin t, the Taylor series centered at x = 0 for
sin(t2 ) arises by replacing every instance of t in the above by t2 :
sin(t2 ) = t2 −
∞
X
t10
t14
(−1)n−1 t4n−2
t6
+
−
+ ... =
.
3!
5!
7!
(2n − 1)!
n=1
4
Therefore
10
6
14
t2 − t3! + t5! − t7! + . . .
sin(t2 )
=
t
t
t5
t9
t13
=t− + −
+ ...
3! 5!
7!
∞
X
(−1)n−1 t4n−3
=
(2n − 1)!
n=1
Now we can integrate:
Z
0
1
sin(t2 )
dt =
t
Z t−
t9
t13
t5
+ −
+ ...
3! 5!
7!
dt
t2
t6
t7
t10
=
−
+
−
+ ...
2
6 · 3! 7 · 5! 10 · 7!
1
1
1
1
+
−
+ ....
= −
2 6 · 3! 7 · 5! 10 · 7!
1
0
Since this sum is alternating, the error at stage k is no bigger than the (k + 1)st term in the series.
Since 7 · 5! = 7 · 125 is much bigger than 100, we can estimate the integral to two decimal places by
adding just the first two terms, so the estimate is
1
1
17
−
=
.
2 36
36
Therefore,
17
36
approximates the integral to within 0.01.
*
*
*
*
*
9. u = h1, 2i and v = h4, 2i are two vectors in the plane. Find vectors a and b so that a is parallel to
*
*
*
v, b is perpendicular to v, and
*
*
*
u = a + b.
*
u
*
*
v
b
*
a
*
*
*
Answer: The vector a will be the orthogonal projection of u onto v, which is
*
*
a=
*
*
*
u·v *
* v
v·v
*
*
and b = u − a.
Now,
*
*
*
*
u · v = (1)(4) + (2)(2) = 4 + 4 = 8
v · v = 42 + 22 = 20.
5
Therefore
*
*
a=
and
*
*
*
*
8 4
,
5 5
8
4
1 − ,2 −
5
5
u·v*
8
h4, 2i =
*
* v =
20
v·v
b = u − a = h1, 2i −
8 4
,
5 5
6
=
=
3 6
− ,
5 5
.