Math 142, Final Exam. 5/2/11.
Name:
• No notes, calculator, or text.
• There are 200 points total. Partial credit may be given.
• Write your full name in the upper right corner of page 1.
• Number the pages in the upper right corner.
• Do problem 1 on page 1, problem 2 on page 2, etc.
• Circle or otherwise clearly identify your final answer.
Some identities:
1 − cos 2θ
1 + cos 2θ
, sin2 θ =
, sin 2θ = 2 sin θ cos θ,
cos θ =
2
2
2
Z
1
1
−1 x
tan
+ C.
dx
=
x 2 + a2
a
a
1. (64 points): Evaluate the following integrals:
Z
(a)
ex cos 3x dx.
Solution: We integrate by parts twice. To begin, we have
u = ex , du = ex dx;
dv = cos 3x dx, v =
sin 3x
.
3
We obtain:
ex sin 3x 1
I=
−
3
3
Z
ex sin 3x dx.
For the second application of parts, we have
u = ex , du = ex dx;
dv = sin 3x dx, v = −
cos 3x
.
3
We compute:
x
Z
ex sin 3x 1
e cos 3x 1
x
I=
−
−
+
e cos 3x dx
3
3
3
3
ex sin 3x ex cos 3x 1
=
+
− · I.
3
9
9
Adding (1/9) · I to both sides gives
10
ex
·I =
9
3
cos 3x
sin 3x +
3
To conclude, we multiply both sides by 9/10:
3ex
cos 3x
ex
I=
sin 3x +
+ C = (3 sin 3x + cos 3x) + C.
10
3
10
Z
π/2
sin2 θ cos5 θ dθ.
(b) (16 points):
0
Solution: This is a trigonometric integral. We look for a u-substitution, writing
Z
π/2
2
I=
Z
4
π/2
sin θ cos θ(cos θ dθ) =
0
sin2 θ(1 − sin2 θ)2 (cos θ dθ).
0
Hence, we set u = sin θ. It follows that du = cos θ dθ. We therefore obtain
Z 1
Z 1
Z 1
2
2
4
2
2 2
(u2 − 2u4 + u6 ) du
u (1 − 2u + u ) du =
u (1 − u ) du =
I=
0
0
0
3
5
7 1
u
u
u
8
1 2 1
−2·
+
.
= − + =
3
5
7 0 3 5 7
105
Z
1
dt.
t2 − 9
(Some of the identities at the top of the page may be useful; if you make a substitution,
convert the anti-derivative back to the variable t.)
√
Solution: We require a trigonometric substitution. Since the integral contains t2 − 9,
a suitable substitution is t = 3 sec θ. (To see this, one can draw for√reference a right
triangle with base angle θ, hypotenuse t, side opposite of length t2 − 9, and side
adjacent of length 3.) We then compute dt = 3 sec θ tan θ dθ and
√
p
p
√
t2 − 9 = (3 sec θ)2 − 9 = 9(sec2 θ − 1) = 9 tan2 θ = 3 tan θ.
√
opposite
t2 − 9
=
. We substitute
Alternatively, the reference triangle gives tan θ =
adjacent
3
in the integral to obtain
Z
Z
Z
Z 1
1
3 sec θ tan θ
1
1
1 + cos 2θ
2
I=
dθ =
dθ =
cos θ dθ =
dθ
(3 sec θ)3 (3 tan θ)
27
sec2 θ
27
27
2
Z
1
1
sin 2θ
1
=
(1 + cos 2θ) dθ =
θ+
+C =
(θ + sin θ cos θ) + C
54
54
2
54
√
t
3
1
t2 − 9
−1
sec
+
+ C.
=
54
3
t2
(c) (16 points):
t3
√
We note that in the above calculation, we used:
1 + cos 2θ
3
cos θ =
, sin 2θ = 2 sin θ cos θ, cos θ = , sin θ =
2
t
t
t = 3 sec θ =⇒ sec−1
= θ.
3
2
2
√
t2 − 9
,
t
x3 + x2 + x + 2
dx.
(x2 + 1)(x2 + 2)
Solution: We use partial fractions to evaluate the integral. We have
Z
(d) (16 points):
Ax + B Cx + D
x3 + x2 + x + 2
= 2
+ 2
.
2
2
(x + 1)(x + 2)
x +1
x +2
We multiply by (x2 + 1)(x2 + 2) to clear denominators, obtaining
x3 + x2 + x + 2 = (Ax + B)(x2 + 2) + (Cx + D)(x2 + 1)
= Ax3 + Bx2 + 2Ax + 2B + Cx3 + Dx2 + Cx + D
= (A + C)x3 + (B + D)x2 + (2A + C)x + (2B + D)
To find the constants, we compare coefficients :
x3 : A + C = 1, x2 : B + D = 1, x1 : 2A + C = 1, x0 : 2B + D = 2.
Subtracting first equation from the third yields A = 0; subtracting the second equation
from the fourth yields B = 1. Hence, we also find that C = 1 and D = 0. To conclude,
we compute
Z
Z
Z
1
x
1
1
1
−1
−1
I=
dx
+
dx
=
tan
x
+
du
=
tan
x
+
ln(x2 + 2) + C.
2
2
x +1
x +2
2
u
2
We note that in the second integral, we made the substitution u = x2 + 2.
Z ∞
x2
dx is convergent or divergent.
2. (16 points): Determine whether the integral
(x3 + 2)3
0
If it converges, find its value. (At some point in the calculation, you will need to make a
substitution.)
Solution: This is an improper integral. We compute
t3 +2
Z t
Z t3 +2
x2
1
1
1
u−2
I = lim
dx = lim
du = lim
t→∞ 0 (x3 + 2)3
3 t→∞ 2
u3
3 t→∞ −2 2
1
1
1
1
= − lim
−
=
.
6 t→∞ (t3 + 2)2 4
24
The integral converges to the value 1/24.
3. (15 points): Use a comparison test (your choice) to determine whether the series
∞
X
n=1
7n + 5
√
3
5
n + 2n2 + 1
converges or diverges.
7n
7
Solution: For large n, we have an ∼ 5/3 = 2/3 . Therefore, we apply the limit comparin
n
1
son test (LCT) with bn = 2/3 . We have
n
7n+5
√
3 5
n +2n2 +1
lim
1
n→∞
n2/3
n2/3 (7n + 5)
= lim √
·
n→∞ 3 n5 + 2n2 + 1
3
1
n5/3
1
n5/3
7 + n5
= lim q
n→∞ 3
1 + n23 +
= 7 > 0.
1
n3
X
X 1
P
Since
bn =
(it is a p-series with p = 2/3 < 1), the LCT implies that
an
2/3
n
diverges.
4. (20 points): Determine whether the series
∞
X
n=1
(−1)n
2n n!
converges absolutely,
3 · 6 · 9 · · · (3n)
converges conditionally, or diverges.
Solution: We apply the ratio test. We have
an+1 3 · 6 · 9 · · · (3n)
2n+1 (n + 1)!
= lim
L = lim ·
n→∞
n→∞ 3 · 6 · 9 · · · (3n) · (3(n + 1))
an
2n n!
2(n + 1)
2
= lim
= < 1.
n→∞ 3(n + 1)
3
Hence, the series converges absolutely.
5. (20 points): Find the radius of convergence and the interval of convergence of the power
∞
X
(2x − 1)n
√ . Test the endpoints of the interval, if necessary.
series
(−3)n n
n=1
Solution: We apply the ratio test:
√ √
(2x − 1)n+1
(−3)n n |2x − 1|
n
√
√
·
=
lim
·
L = lim n
n→∞
n→∞ (−3)n+1 n + 1 (2x − 1)
3
n+1
=
|2x − 1|
1
lim q
n→∞
3
1+
=
1
n
√1
n
√1
n
|2x − 1|
<1
3
⇐⇒ |2x − 1| < 3 ⇐⇒ −3 < 2x − 1 < 3 ⇐⇒ −2 < 2x < 4 ⇐⇒ −1 < x < 2.
Therefore, the radius is R = 3/2 (it is half of the length of the interval of convergence). It
remains to test the endpoints.
X (−3)n
X (−1)n X 1
√
√
√ . It is a p-series with
=
=
(−3)n n
n
n
p = 1/2 < 1; therefore, it diverges.
X
X (−1)n
3n
√
√ . We observe that the sequence of
• x = 2. The series is
=
(−3)n n
n
absolute values of the terms is decreasing and tends to zero. Therefore, the alternating
series test (AST) implies that this series converges.
• x = −1. The series is
We conclude that the interval of convergence is I = (−1, 2].
4
6. (30 points):
• Set up an integral or sum of integrals which gives the volume obtained by revolving
the give region R around the given line L. You may use disks, washers, or shells.
• Do not compute the integrals.
• Sketch the given region and a sample rectangle (or rectangles) in the region to help justify your reasoning. Label the sketch as necessary. What are the points of intersection
of the equations giving R? What is the thickness of a sample rectangle?
(a) (15 points):
R : y = 0 (the x-axis), y = x, x + y = 2;
L : y = 0 (the x-axis).
Solution: The intersection of y = x and x + y = 2 occurs at (1, 1). To find the
volume using a single integral, a sample rectangle in R should be oriented horizontally
and have thickness ∆y. Since the axis is also horizontal, we use shells and integrate
with respect to y. The radius is r(y) = y, and the height (length) of the rectangle is
h(y) = (2 − y) − y = 2(1 − y). Therefore, we have
Z 1
Volume =
4πy(1 − y) dy.
0
Alternatively, one can use disks:
Z
1
2
2
Z
π(2 − x)2 dx.
πx dx +
Volume =
1
0
The common value is 2π/3.
(b) (15 points):
R : x = 0 (the y-axis), y = x + 1, y = −x + 3;
L : y = 1.
Solution: The intersection of y = x + 1 and y = −x + 3 occurs at (1, 2). To find the
volume using a single integral, a sample rectangle in R should be oriented vertically
and have thickness ∆x. Since the axis is horizontal, we use washers and integrate with
respect to x. The outer radius is r2 (x) = −x + 3 − 1 = 2 − x; the inner radius is
r1 (x) = x + 1 − 1 = x. Therefore, we have
Z 1
Volume =
π((2 − x)2 − x2 ) dx.
0
Alternatively, one can use shells:
Z 2
Z
2
2π(y − 1) dy +
Volume =
1
2
The common value is 2π.
5
3
2π(y − 1)((3 − y) dy.
7. (20 points): Find a power series representation of the form
∞
X
cn xn for each function.
n=n0
What is the radius of convergence?
(a) (10 points):
1
.
3+x
Solution:
∞
∞
1
1
1 X x n X (−1)n n
=
=
−
=
x , |x| < 3.
3+x
3 n=0
3
3n+1
3 1 − − x3
n=0
x2
.
(3 + x)2
(Differentiate the power series from part (a). Suitably adjust the resulting derivative.)
Solution: We first observe that
1
1
d
=−
.
dx 3 + x
(3 + x)2
(b) (10 points):
Hence, we have
d
x2
= −x2 ·
2
(3 + x)
dx
=
1
3+x
∞
X
(−1)n+1
n=0
3n+1
n+1
nx
∞
∞
X
X
d (−1)n n
(−1)n n−1
2
x
nx
= −x ·
=
−x
·
n+1
n+1
dx
3
3
n=0
n=0
2
=
∞
X
(−1)n
n=1
6
3n
(n − 1)xn , |x| < 3.
8. (15 points): Let f (x) = cos x.
(a) (5 points): Write down the Taylor series for f (x) at a = 0. (This is sometimes called
the MacLaurin series for f (x).) What is its radius of convergence? You do not have
to give justification. You can simply write it down from memory, or you can derive it
from scratch.
Solution:
cos x =
∞
X
(−1)n
n=0
x2 x4
x6
x2n
=1−
+
−
+ · · · , x ∈ R.
(2n)!
2
24 720
cos x − 1 +
(b) (5 points): Use your answer from (a) to compute the limit lim
x→0
x6
You are not allowed to use L’Hôpital’s rule.
Solution: We compute
lim
x→∞
1−
x2
2
+
x4
24
−
x6
720
+ ··· − 1 +
x6
x2
2
−
(c) (5 points): Use part (a) to evaluate the sum
x4
24
6
42n (2n)!
(Give an exact numerical value.)
Solution: From part (a), the sum is cos
7
π 4
−
x4
24
.
x
− 720
+ ···
1
1
= lim
=−
=− .
6
x→∞
x
720
6!
∞
X
(−1)n π 2n
n=0
x2
2
1
=√ .
2
.