Arkansas Tech University
MATH 2924: Calculus II
Dr. Marcel B. Finan
Solutions to Assignment 8.7
Exercise 3
We have
n=0
f (n) (0) n
x =
n!
n=0
(n + 1)! n
x =
n!
Using the Ratio Test with an = (n +
lim
n
1)xn ,
(n + 1)xn .
n=0
we nd
(n + 2)xn+1
= |x|.
(n +
Arkansas Tech University
MATH 2924: Calculus II
Dr. Marcel B. Finan
Solutions to Assignment 7.2
Exercise 1
A cross-section is a disk of radius f (x) = 2 x . See gure below.
2
Thus, the volume of the solid of revolution is
2
5 2
1
x
2
2
2
4 2x +
dx =
1
x3
Arkansas Tech University
MATH 2924: Calculus II
Dr. Marcel B. Finan
Solutions to Assignment 7.1
Exercise 1
The area is found as follows
4
4
(4x x2 )dx
(5x x2 x)dx =
0
0
= 2x2
x3
3
4
=
0
32
3
Exercise 3
The area is found as follows
1
y3
+ 2y
3
10
=e e1 +
Arkansas Tech University
MATH 2924: Calculus II
Dr. Marcel B. Finan
Solutions to Assignment 8.4
Exercise 3
4
4
The given series is the alternating series (1)n1 n+6 . Let an = n+6
n=1
4
4
and f (x) = n+6 . Since f (x) = (x+6)2 < 0, f is decreasing and ther
Arkansas Tech University
MATH 2924: Calculus II
Dr. Marcel B. Finan
Solutions to Assignment 8.6
Exercise 3
We have
1
1
=
=
1+x
1 (x)
(x)n = (1)n xn .
n=0
This geometric series is convergent for | x| = |x| < 1. The interval of
convergence is 1 < x < 1
Exer
Arkansas Tech University
MATH 2924: Calculus II
Dr. Marcel B. Finan
Solutions to Assignment 8.3
Exercise 1
From the gure below, we see that
n=2
1
<
n1.3
1
dx
.
x1.3
Since the integral is convergent (pintegral with p = 1.3 > 1), the series is
convergent by
Arkansas Tech University
MATH 2924: Calculus II
Dr. Marcel B. Finan
Solutions to Assignment 8.1
Exercise 5
First notice that an = 2 an1 . Since a1 = 3, we have a2 = 3
3
2
n1
3 2 , , an = 3 2
3
3
2
3
, a3 =
Exercise 7
n2
The general nth term is an = (1)n+1
Arkansas Tech University
MATH 2924: Calculus II
Dr. Marcel B. Finan
Solutions to Assignment 6.1
Exercise 3
We have
u(x) = x v (x) = cos 5x
u (x) = 1 v(x) = sin55x .
Using the method of integration by parts, we nd
x
x cos 5xdx = sin 5x
5
x
= sin 5x +
5
1
Arkansas Tech University
MATH 2924: Calculus II
Dr. Marcel B. Finan
Solutions to Assignment 6.6
Exercise 1
(a) This is an improper integral of type 2 where the integrand is discontinuous at x = 1.
(b) This is an improper integral of type 1 where the inter
Arkansas Tech University
MATH 2924: Calculus II
Dr. Marcel B. Finan
Solutions to Assignment 7.4
Exercise 3
1 + cos2 xdx 3.8202
L=
0
Exercise 5
For x = y y we have
dx
dy
=
1
2 y
4
L=
1+
1
Exercise 7
3
For y = 1 + 6x 2 we have
dy
dx
1
1
2 y
2
dy 3.6095
1
=
Arkansas Tech University
MATH 2924: Calculus II
Dr. Marcel B. Finan
Solutions to Assignment 7.5
Exercise 1
(i) We have
/3
2 tan x
S=
1 + sec4 xdx 10.5017.
0
(ii) We have
/3
2x 1 + sec4 xdx 7.9353
S=
0
Exercise 4
(i) We have
1
S=
2y
1+
0
4
dy 4.2583.
(2y +
Arkansas Tech University
MATH 2924: Calculus II
Dr. Marcel B. Finan
Solutions to Assignment 8.5
Exercise 3
We have
lim
n
an+1
(1)n+1 (n + 1)xn+1
n+1
= lim
= lim
|x| = |x|.
n
n
an
(1)n nxn
n
By the ratio test, the series converges for |x| < 1. Thus, the ra
Arkansas Tech University
MATH 2924: Calculus II
Dr. Marcel B. Finan
Solutions to Assignment 7.3
Exercise 5
The region to be rotated is shown in the gure below.
Using the substitution u = x2 , we have
1
V =
1
2
2xex dx =
0
= eu
1
0
= (1 e
0
1
eu du
)
Exer
Arkansas Tech University
MATH 2924: Calculus II
Dr. Marcel B. Finan
Solutions to Assignment 7.7
Exercise 1
Clearly y(x) = 0 is a trivial solution. Using the separation of variables
method, we have
dy
dx
dy
y2
dy
y2
1
y
=xy 2
=xdx
=
xdx
=
x2
+C
2
x2
+C
2
y
Arkansas Tech University
MATH 2924: Calculus II
Dr. Marcel B. Finan
Solutions to Assignment 7.6
Exercise 1
We have
10
W =
2
5x
1
Exercise 2
We have
2
W =
cos
1
1
dx = 5
x
10
= 4.5 ft lb
1
3
x dx =
sin
x
3
3
2
= 0J.
1
From x = 1 to x = 3 , F (x) > 0 and t
Arkansas Tech University
MATH 2924: Calculus II
Dr. Marcel B. Finan
Solutions to Assignment 6.2
Exercise 3
Let u = cos x then du = sin xdx. Thus,
/2
/2
(1 cos2 x)3 sin x cos5 xdx
sin7 x cos5 xdx =
0
0
1
1
(1 u2 )3 u5 du =
=
0
(1 3u2 + 3u4 u6 )u5 du
0
u6 3
Arkansas Tech University
MATH 2924: Calculus II
Dr. Marcel B. Finan
Solutions to Assignment 8.8
Exercise 3
We have
f (2)
f (2)
f 3(2)
(x 2) +
(x 2)2 +
(x 2)3
1!
2!
2!
1
1
1 1
= (x 2) + (x 2)2 (x 2)3
2 4
8
16
T3 (x) =f (2) +
Exercise 5
We have
f (/2)
f (/2
Arkansas Tech University
MATH 2924: Calculus II
Dr. Marcel B. Finan
Solutions to Assignment 6.5
Exercise 1
(a) We have
L2 =f (0)(2) + f (2)(2) = (0.5)(2) + (2.5)(2) = 6
R2 =f (2)(2) + f (4)(2) = (2.5)(2) + (3.5)(2) = 12
M3 =f (1)(2) + f (3)(2) = (1.6)(2)