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 discontin-
uous at
x
= 1
.
(b) This is an improper integral of type 1 where the interval of integration
is unbounded.
(c) This is an improper integral of type 1 where the interval of integration
is unbounded.
(d) This is an improper integral of type 2 where the integrand is discontin-
uous at
x
= 0
Exercise 5
First, letting
u
=
x
-
2
,
we find
Z
dx
(
x
-
2)
3
2
=
Z
u
-
3
2
du
=
-
2
√
u
=
-
2
√
x
-
2
+
C.
Thus,
Z
∞
3
dx
(
x
-
2)
3
2
= lim
b
→∞
2
-
2
√
b
-
2
= 2
so that the improper integral is convergent
Exercise 7
First, letting
u
= 3
-
4
x,
we find
Z
dx
3
-
4
x
=
-
1
4
Z
du
u
=
-
1
4
ln
|
u
|
=
-
1
4
ln
|
3
-
4
x
|
+
C.
Thus,
Z
0
-∞
dx
3
-
4
x
=
lim
a
→-∞
-
ln 3
4
+
1
4
ln
|
3
-
4
a
|
=
∞
so that the improper integral is divvergent
1
Exercise 13
First, letting
u
=
x
2
,
we find
Z
xe
-
x
2
dx
=
1
2
Z
e
-
u
du
=
-
1
2
e
-
u
=
-
1
2
e
-
x
2
+
C.
