Calculus II
Final Exam Key
Instructions
1. Do NOT write your answers on these sheets. Nothing written on the test papers will be graded.
2. Please begin each section of questions on a new sheet of paper.
3. Please do not write answers side by side.
4. Please do not staple your test papers together.
5. Limited credit will be given for incomplete or incorrect justification.
6. All numeric answers must be exact: numeric approximations are not acceptable.
Questions
1. Integration (5 each)
Z
(a)
sin7 α cos3 α dα
Z
Z
sin7 α cos3 α dα
=
sin7 α(1 − sin2 α) cos α dα =
Z
u7 (1 − u2 ) du =
Z
u7 − u9 du =
=
Z
(b)
u =
du =
sinα.
cos α dα.
1 8
1
u − u10 + C
8
10
1
1
8
sin α −
sin10 α + C.
8
10
(3x2 + 5x − 1) sin x dx
Z
(3x2 + 5x − 1) sin x dx =
Z
−(3x2 + 5x − 1) cos x − −(6x + 5) cos x dx =
Z
−(3x2 + 5x − 1) cos x + (6x + 5) cos x dx =
Z
2
−(3x + 5x − 1) cos x + (6x + 5) sin x − 6 sin x dx =
−(3x2 + 5x − 1) cos x + (6x + 5) sin x + 6 cos x + C.
1
u =
du =
3x2 + 5x − 1.
6x + 5 dx.
u =
du =
6x + 5.
6 dx
dv
v
dv
v
=
=
= sin x dx.
= − cos x.
cos x dx.
sin x.
Final Exam
Z
(c)
2
x2
dx
(1 − x2 )3/2
x2
dx
(1 − x2 )3/2
Z
Z
sin2 θ
cos θ dθ
(1 − sin2 θ)3/2
Z
sin2 θ
cos θ dθ
cos3 θ
Z
sin2 θ
dθ
cos2 θ
Z
tan2 θ dθ
Z
sec2 θ − 1 dθ
=
x =
dx =
sin θ.
cos θ dθ.
=
=
=
=
=
tan θ − θ + C
=
√
x
− arcsin x + C.
1 − x2
Z
(d)
arctan y dy
=
u =
du =
arctan y.
dy
1+y 2 .
=
u =
du =
1 + y2 .
2y dy.
Z
arctan y dy
Z
y
dy
1 + y2
Z
1
2y
y arctan y −
dy
2
1 + y2
Z
1
du
y arctan y −
2
u
y arctan y −
Z
(e)
dv
v
=
=
=
=
y arctan y −
1
ln |1 + y 2 | + C.
2
6z − 11
dz
+ z − 12
z2
6z − 11
=
+ z − 12
6z − 11
=
(z + 4)(z − 3)
6z − 11 =
z2
A
B
+
.
z+4 z−3
A(z − 3) + B(z + 4).
z
=
3.
7
=
7B.
B
=
1.
z
=
−4.
−35
=
−7A.
A
=
5.
6z − 11
dz
z 2 + z − 12
Z
5
1
+
dz
z+4 z−3
Z
=
=
5 ln |z + 4| + ln |z − 3| + C.
dy.
y.
Final Exam
3
Z
(f)
x3 + x − 7
dx
x2 + 1
x
x3 +0 +x −7
x3 +0 +x
−7
x2 + 0 + 1
x3 + x − 7
dx
x2 + 1
Z
7
x− 2
dx
x +1
Z
Z
(g)
∞
=
=
1 2
x − 7 arctan x + C.
2
2
re−r dr
0
Z
Z0
∞
2
re−r dr
a
2
=
lim
re−r dr =
a→∞ 0
Z
2
1 a
lim −
−2re−r dr =
a→∞
2 0
2
Z
1 −a u
e du =
lim −
a→∞
2 0
−a2
1 u =
lim − e a→∞
2 0
2
1
lim − (e−a − 1) =
a→∞
2
u = −r2 .
du = −2r dr.
1
.
2
Final Exam
4
2. Series (5 each)
Determine if each of the following converge or diverge. State the test(s) used.
(a)
∞
X
5n + n
53n
n=1
∞
X
5n + n
53n
n=1
=
∞
∞
X
X
n
5n
+
3n
3n
5
5
n=1
n=1
=
∞
∞
X
X
1
n
+
2n
5
53n
n=1
n=1
=
∞
X
∞
X n
1
+
.
2
n
(5 )
53n
n=1
n=1
n 1/n
lim
=
n→∞ 53n
n1/n
=
lim
n→∞ 53
<
(b)
1
125
1.
Thus the first term is a convergent geometric series and the second is convergent by the root test. Thus
the sum is convergent.
∞ X
2n2
n=1
3n2 + 1
Using the nth term divergence test
2n2
n→∞ 3n2 + 1
4n
lim
n→∞ 6n
2
lim
n→∞ 3
lim
(c)
∞/∞
=
L’Hôpital’s Rule
=
= 23 .
Thus this series diverges.
∞
X
n!
n=0
2n
Using the ratio test
(n + 1)! 2n
·
n→∞ 2n+1
n!
(n + 1)
lim
n→∞
2
lim
This series diverges.
=
=
∞.
Final Exam
(d)
5
√
∞
X
n
n
+
1
n=1
Using the limit comparison test
√
lim
n→∞
n
1
÷√
n+1
n
n
lim
n→∞ n + 1
lim 1
=
∞/∞
=
=
n→∞
(e)
L’Hôpital’s Rule
1.
This means this series diverges like the p-series with p = 1/2.
√
∞
X
(−1)n n
n=1
n+1
Using the alternating series test
√
n
n+1
1
lim √
n→∞ 2 n
lim
n→∞
∞/∞
=
L’Hôpital’s Rule
=
an
=
a0n
=
0.
√
n
.
n+1
√
(n + 1)n−1/2 /2 − n
(n + 1)2
1−n
√
.
2 n(n + 1)2
=
Thus the series converges.
∞
X
5n
(f)
(−1)n 2n
n
n=1
Using absolute convergence with the root test
lim
n→∞
5n
n2n
1/n
lim
n→∞
Thus this series converges.
=
5
n2
=
0
<
1.
Final Exam
6
3. Power Series
(a) Find the interval of convergence for each of the following. (6 each)
∞
X
x3n
i.
(3n)!
n=0
Using the ratio test
x3(n+1)
(3n)!
· 3n
n→∞ (3[n + 1])!
x
x3n+3
(3n)!
lim
· 3n
n→∞ (3n + 3)!
x
x3
lim
n→∞ (3n + 3)(3n + 2)(3n + 1)
lim
=
=
=
0.
This converges
(−∞, ∞).
ii.
∞
X
xn
2n + 1
n=0
Using the ratio test
xn+1
2n + 1
·
n→∞ 2(n + 1) + 1
xn
2n + 1
lim x
n→∞ 2n + 3
lim x
lim
=
∞/∞
=
n→∞
∞
X
1n
2n + 1
n=0
L’Hôpital’s Rule
=
x
<
1.
∞
X
=
1
.
2n + 1
n=0
=
0.
<
1
.
2n + 1
∞
X
(−1)n
2n + 1
n=0
1
2n + 1
1
2(n + 1) + 1
lim
n→∞
Thus the interval of convergence is
[−1, 1).
Final Exam
7
iii.
∞
X
(7x − 3)n
3
n=0
Using the root test
lim
n→∞
1/n
(7x − 3)n
3
7x − 3
lim
n→∞ 31/n
−1
2
2
7
≤
≤
≤
7x − 3
7x
x
=
=
7x − 3
<
1.
≤ 1.
≤ 4.
≤ 74 .
∞
X
(7(4/7) − 3)n
3
n=0
=
∞
X
1
.
3
n=0
∞
X
(7(2/7) − 3)n
3
n=0
=
∞
X
(−1)n
.
3
n=0
Both ends diverge (nth term divergence). The interval of convergence is
2 4
,
.
7 7
(b) Write a power series representation for each of the following. Be sure to write in correct form. (4 each)
i.
1+x x
x e
ex
1 x
e
x
1+x x
e
x
x2
x3
+
+ ...
2!
3!
1
x
x2
=
+1+ +
+ ...
x
2!
3!
1
x
x2
=
+1+ +
+ ...
x
2!
3!
x3
x2
1+x+
+
+ ...
2!
3!
1
3x 4x2
=
+2+
+
+ ...
x
2!
3!
=
1+x+
ii. sinh x
sinh x
=
ex
=
e−x
=
sinh x
=
=
ex − e−x
.
2
x2
x3
1+x+
+
+ ...
2!
3!
x2
x3
1−x+
−
+ ...
2!
3! 3
1
x
2x + 2 + . . .
2
3!
x3
x5
x+
+
+ ...
3!
5!
Final Exam
8
iii. x cos(x2 )
x4
x6
x2
+
−
+ ...
2!
4!
6!
x4
x8
x12
cos(x2 ) = 1 −
+
−
+ ...
2!
4!
6!
5
9
13
x
x
x
x cos(x2 ) = x −
+
−
+ ...
2!
4!
6!
cos x
=
1−
Final Exam
9
4. Modeling
(a)
x =
y =
t ∈
t + sin(2t).
t − 2 sin t.
[0, 4π].
i. (4) Find all times when the curve changes from left to right or vice versa.
x0
=
1 + 2 cos(2t).
1 + 2 cos(2t)
=
0.
= −1.
2 cos(2t)
= −1/2.
2π 4π 8π 10π
,
,
,
,...
2t = . . . ,
3 3 3
3
π 2π 4π 5π
t = ..., ,
,
,
,...
3 3 3 3
cos(2t)
ii. (4) Find all times when the curve changes from up to down or vice versa.
y0
=
1 − 2 cos t =
cos t =
t =
1 − 2 cos t.
0.
1/2.
π 5π 7π 11π
,
,
,
,...
3 3 3
3
iii. (2) Where are the non-differentiable points? (Give t values.)
x0 and y 0 are both zero at . . . , π3 , 5π
3 ,...
iv. (2) Graph the curve. Indicate direction.
14
12
10
8
6
4
2
2
4
6
8
10
12
v. (2) Calculate the slope of the tangent at t = π/2.
dy
dx
=
dy dx t=π/2
=
1 − 2 cos t
.
1 + 2 cos(2t)
1 − 2 cos(π/2)
1 + 2 cos(2[π/2])
= −1.
Final Exam
10
(b) r = sin(θ) + sin(2θ)/2.
i. (4) Determine when the curve is at the origin.
sin(θ) + sin(2θ)/2
=
0.
sin(θ) + sin θ cos θ
=
0.
sin(θ)(1 + cos θ)
=
0.
sin θ
θ
=
=
0.
. . . , 0, π, 2π, . . .
cos θ
=
−1.
θ
=
. . . , π, 3π, 5π, . . .
ii. (4) Determine when the curve changes from going out to coming in or vice versa.
r0
=
cos θ + cos(2θ)
=
cos θ + 2 cos2 θ − 1.
2 cos2 θ + cos θ − 1
=
0.
(2 cos θ − 1)(cos θ + 1)
=
0.
cos θ
=
θ
=
cos θ
=
1/2.
π 5π 7π 11π
,
,
,
,...
3 3 3
3
−1.
θ
=
. . . , π, 3π, 5π, . . .
iii. (2) Graph the curve (you may need a few more points).
1.2
1.0
0.8
0.6
0.4
0.2
-0.5
0.5
Final Exam
11
(c) (6) Find the area inside both r = sin(2θ) and r = − sin θ. Do not evaluate integrals.
0.5
0.5
-0.5
-0.5
-1.0
sin(2θ)
=
0.
2θ
=
. . . , 0, π, 2π, 3π, . . .
θ
=
. . . , 0, π/2, π, 3π/2, . . .
sin(2θ)
= − sin θ.
2 sin θ cos θ
= − sin θ.
sin θ
θ
2 cos θ
cos θ
θ
=
0.
= . . . , 0, π, 2π, . . .
= −1.
= −1/2.
2π 4π
,
,...
= ...,
3 3
Also
− sin(2(π + θ))
= − sin θ.
− sin(2π + 2θ)
= − sin θ.
−[sin(2π) cos(2θ) + sin(2θ) cos(2π)]
= − sin θ.
− sin(2θ)
= − sin θ.
−2 sin θ cos θ
= − sin θ.
2 cos θ
=
1.
cos θ
=
1/2.
θ
"Z
2
0
π/3
1
(− sin θ)2 dθ +
2
Z
= ...,
3π/2
4π/3
π 5π 7π
,
,
,...
3 3 3
1
sin2 (2θ) dθ
2
#
Final Exam
12
5. Regional Data (4 each)
Setup but do not integrate to find the following. The region is enclosed by y = x2 and y = x.
(a) The volume produced by rotating the region around the x-axis
1
Z
π(x2 − (x2 )2 ) dx
0
(b) The volume produced by rotating the region around the y-axis
Z
1
2πx(x − x2 ) dx
0
(c) The volume produced by rotating the region around the y = 2
Z
1
π([2 − x2 ]2 − [2 − x]2 ) dx
0
(d) The perimeter of the region
√
Z
2+
0
1
p
1 + (2x)2 dx