Calculus II , Final (practice test)
9:00–12:00 noon, Friday, Dec.15
Calculators are not allowed.
Problem
1 Evaluate the following integrals
R
dx
x2 +2x+5
Solution:
1
2
tan−1 ( x+1
2 ) + C.
Problem
2 Evaluate the following integrals
R 5
x cos(x3 )dx
Solution:
1 3
3
3 x sin(x )
+ 13 cos(x3 ) + C.
Problem 3 Find the volume of the solid generated by revolving the region
bounded by the curves y 2 = x and y = x3 about x-axis.
Solution: Find the intersection of these curves: (0, 0) and (1, 1).
Then
Z 1
q
5π
V =
π(( (x))2 − (x3 )2 )dx =
.
14
0
Problem 4 Find the length of the curve given by
x = a(cos t + t sin t), y = a(sin t − t cos t), t ∈ [0, π].
Solution:
Z
π
q
π
√
L =
0
Z
(x0 (t))2 + (y 0 (t))2 dt
=
a2 t2 dt
0
=
Problem 5Evaluate
|a|π 2
.
2
∞
X
1
.
2n(2n + 4)
n=1
Solution:
1
∞
X
1
2n(2n + 4)
n=1
=
=
=
∞
1
1
1X
( −
)
8 n=1 n n + 2
1
(1 − 1/3 + 1/2 − 1/4 + 1/3 − 1/5 + 1/4 − 1/6 + 1/5 − 1/7 + ...)
8
1
(1 + 1/2) = 3/16.
8
Problem 6 Decide the convergence or divergence of
∞
X
nn
n=1
n!
.
Solution: Apply the ratio test,
an+1
n + 1 nn + 1
1
=(
)
= (1 + )n → e > 1.
an
n
n+1
n
Therefore it diverges.
x
−x
Problem 7Let sinh x = e −e
. Find its power series at 0 along with its
2
convergence interval. Show that sinh x is equal to its power series.
x
n+1 −x
Solution: (sinh)(n) x = e +(−1)2 e . So (sinh)(n) (0) = 1 if n is odd and
(sinh)(n) (0) = 0 if n is even.
Therefore formerly the Taylor expansion of sinh x is
∞
X
x2n+1
.
(2n + 1)!
n=0
2n+1
y
Rn+1 (x) = (2n+1)!
→ 0 for some y between 0 and x. Therefore sinh x is
equal to its power series.
Problem 8 Find
ex − 1 − x
.
x→0
x sin x
lim
Solution:
ex − 1 − x
1 + x + x2 /2 + ... − 1 − x
= lim
= 1/2.
x→0
x→0
x sin x
x2
lim
Problem 9 Which value for p > 0 makes
X
sin(
n=1
2
1
)
np
convergent? Prove it. Solution: Apply the limit comparison test for
P
and n=1 n1p .
The answer is
p > 1.
1
1
n=1 sin( np )
P
Problem 10 Find the Taylor Polynomial for f (x) = e− x2 at x = 0, where
f (0) is defined to be 0.
Solution: 0.
3