Midterm 4, Problem 1
December 11, 2013
Problem (25pts): Determine whether the following series converge (absolutely/conditionally). If they converge, nd their value.
1.
k=2
(1)k 5k1
32k
2.
1
n ln n
n=2
Solution:
Part 1. 15 points.
(2pts): We have
k=2
(1)k
Solution: Exam 2, Problem 4
18th October 2013
Grading scheme 1:
log2 (log3 x)
x
log5 x
L = lim
ln(log3 x)
ln 5
) xlim
. . . (5 points)
ln 2 ln x
which is indeterminant, so use the LHospitals Rule . . . (5 points)
L=(
As x , L
ln 5
L=(
) lim
ln 2 x
L=(
1
MATH 141, FALL 2013, MIDTERM 2
Problem 3
Part a.
.
The derivative, f (x) = 1 +
1
1
2 x,
(5 pt.)
satises f (x) > 0 on (0, ). Then there is an inverse on the interval (0, ),
which includes x = 2. (5 pt.)
The largest interval in which an inverse exists is [0
Rubric for Midterm 2, Problem 2
October 18, 2013
PART A. 15 POINTS.
x2
1
dx
3x + 3
=
1
32
2)
(x
+
3
4
dx
[4 points]
Make u = x 3 . Then du = dx, so
2
1
(x
32
2)
+
3
4
dx
=
u2
1
du
+3
4
[5 points]
Now, there are two ways to proceed.
(i). Recall that
1
d
MATH 141, Midterm 3, 11/06/2013
Problem 2
e2 x 4
u ex
du dx
( ex 1)2 dx
u
u2 4
u(u 1)2 du
[2+4]
u2 4
A
B
C
2
u(u 1)
u u 1 (u 1)2
[2]
u 2 4 A(u 1)2 Bu(u 1) Cu
[1]
u0 4 A
u 1 5C
u 2 8 A 2 B 2C 14 2 B B 3
e2 x 4
du
du
du
dx 4
3
5
x
2
( e 1)
u
u1
(u 1
Midterm 3, Problem 3
November 13, 2013
Problem (25pts): Determine whether
cos(x)dx
sin(x)
0
converges. If so, eval-
uate.
Solution:
(6pts): Since the integrand is dened on (0, ), and
cos(x)
lim
= +
sin(x)
x0+
and
cos(x)
lim
x
sin(x)
= ,
one must break th
Math 141, Midterm 4
Problem 2 :
2n 2n
nx
n n 3
2 n 1
2n
2n
|x |2 n2
an n | x| , an1 n1
3 ( n 1)
n3
Applying the ratio test :
an1 2
n
|x|2
an
3
n 1
an
2
2
lim an1 3 |x|
n
[1]
[2]
[2]
[3]
Or, applying the root test :
2|x|2
na
[2]
n
3 nn
2
2
[3]
lim n an
Solution: Exam 3, Problem 4
10th November 2013
Problem 4 a
Grading scheme 1:
L = lim
n
n+
2
n2 = lim n n
n+
2
x
Say f (x) = x , then,
2
2
L = lim n n = lim x x = lim e
n+
x+
2 ln x
x
x+
Now, limx+ ln x is of the form , which is indeterminate. Therefore, u
Problem 1
y cos( x )y cos x
Method (i) : P( x ) cos x
(2)
Q( x ) cos( x )
(2)
S( x ) P( x )dx sin x
(4)
y( x ) e S( x ) e S( t )Q(t )dt
e sin x e sin t cos tdt e sin x [ e sin x c ] ce sin x 1
y(0) 2 c 3
y( x ) 3 e sin x 1 .
(10)
(4)
(3)
dy
(4)
cos x(