Math 31B
Integration and Infinite Series
Midterm 2 Practice
Instructions: You have 50 minutes to complete this exam. There are four questions, worth a
total of 40 points. This test is closed book and closed notes. No calculator is allowed.
For full credit show all of your work legibly. Please write your solutions in the space below the
questions; INDICATE if you go over the page and/or use scrap paper.
Do not forget to write your name, discussion and UID in the space below.
Name:
Student ID number:
Discussion:
Question Points Score
1
0
2
0
3
0
4
0
Total:
0
Problem 1.
Use u-substitutions and integration by parts to calculate the following indefinite integrals. You might find it better to use the letter t for your substitutions to avoid confusion
with your integration by part labels.
R √ √x
xe dx
(a)
R 2x
(b) e cos( x3 ) dx.
Solution:
(a) Let t =
√
x. Then dt =
1
√
2 x
Z
dx =
√
1
2t
√
xe
x
dx. So 2t dt = dx and
Z
dx =
2t2 et dt.
Let u = 2t2 and dv = et dt. Then du = 4t dt and v = et . So
Z
Z
2 t
2 t
2t e dt = 2t e − 4tet dt.
Let u = 4t and dv = et dt. Then du = 4 dt and v = et . So
Z
Z
2 t
2 t
2t e dt = 2t e − 4tet dt
Z
2 t
t
= 2t e − 4te + 4et dt = 2t2 et − 4tet + 4et + c,
and
Z
√
√
xe
x
√
dx = 2xe
x
√
√ √
− 4 xe x + 4e x + c.
(b) Integrating by part twice (u = e2x both times) gives
Z
Z
x
x
x
2x
2x
2x
e cos
dx = 3e sin
− 6e sin
dx
3
3
3
Z
x
x
x
2x
2x
2x
= 3e sin
+ 18e cos
− 36 e cos
dx.
3
3
3
So
Z
1
x
x
x
2x
2x
dx =
3e sin
+ 18e cos
+c
e cos
3
37
3
3
3e2x
x
x
=
sin
+ 6 cos
+ c.
37
3
3
2x
Problem 2.
Use L’Hôpital’s rule and associated tricks to calculate the following limits.
(a) limx→0+ x(ln x)2
(b) limx→0+ xsin x .
Solution:
indeterminate and apply L’Hôpital’s
(a) We turn the 0 · ∞ indeterminate into an ∞
∞
rule twice:
2(ln x) · x1
(ln x)2
2
lim x(ln x) = lim
= lim
1
−1
x→0+
x→0+
= lim
x→0+
x→0+
x
2 ln x
= lim
−1
x
x→0+
x2
2
x
1
x2
= lim 2x = 0.
x→0+
(b) Let y(x) = ln x
sin x
. Then y(x) = (sin x)(ln x). Thus,
ln x
1
x
− cos x
sin2 x
= lim
lim y(x) = lim (sin x)(ln x) = lim
1
x→0+
x→0+
x→0+
sin x
sin x sin x
sin x
sin x
·
= − lim
· lim
= −1 · 0 = 0.
= − lim
x→0+
x→0+ cos x
x→0+
x
cos x
x
x→0+
We conclude that limx→0+ xsin x = e0 = 1.
Problem 3.
Calculate the following improper integrals or show they diverge.
R ∞ 2x
(a) 0 (1+x
2 )2 dx.
R1
(b) 0 ln x dx.
Solution:
(a) We use the u-sub, u = 1 + x2 , so that du = 2xdx, but we’re careful to apply it
to the proper integral that shows up after using the definition of the improper
integral.
Z ∞
Z S
2x
2x
dx = lim
dx
2
2
S→∞ 0 (1 + x2 )2
(1 + x )
0
1+S 2
Z 1+S 2
−1
1
du = lim
= lim
2
S→∞
S→∞ 1
u
u 1
−1
= lim
+ 1 = 1.
S→∞ 1 + S 2
(b) We do integration by parts, but we’re careful to apply it to the proper integral
that shows up after using the definition of the improper integral.
We also use the fact that limS→0+ S ln S = 0 since ln always loses.
Z 1
Z 1
ln x dx = lim
ln x dx
S→0+ S
0
1
= lim x ln x − x
S→0+
S
= lim [−1 − S ln S + S]
S→0+
= −1.
Problem 4.
Calculate the limits of the following sequences or say that they don’t exist.
(a) (an ) where an =
(b) (bn ) where bn =
22n
.
n!
n
e +(−3)n
.
5n
(c) (cn ) where c1 = 1 and, for n ≥ 1, cn+1 =
√
2cn . You can assume (cn ) converges.
Solution:
(a) an =
4n
n!
4
and so we can see that 0 ≤ an ≤ 44! · n4 . Since
4
4 4 4 4
4
1
4
lim
·
=
· 4 · lim
=
·4 ·0=0
n→∞ 4! n
n→∞ n
4!
4!
the squeeze lemma tells us that limn→∞ an = 0.
(b) We have
en +(−3)n
5n
= ( 5e )n + ( −3
)n .
5
Limit laws and the fact that | 5e |, | −3
| < 1 give
5
n
n
en + (−3)n
e
−3
lim bn = lim
= lim
+ lim
= 0 + 0 = 0.
n
n→∞
n→∞
n→∞
n→∞
5
5
5
√
√
(c) Let L = limn→∞ cn . Since cn+1 = 2cn , we can take limits to give L = 2L.
So L(L − 2) = 0 and L = 0 or L = 2. We can see that each an ≥ 1, so L = 2.