Miller, Kierste – Homework 8 – Due: Mar 20 2007, 3:00 am – Inst: Gary Berg
2
On the other hand,
lim
t
→∞
ln

6
t

1

=
∞
.
Consequently, the integral
I
is not convergent
.
keywords: improper integral, limit
003
(part 1 of 1) 10 points
Determine if
I
=
Z
∞
3
x
5
√
6

x
2
dx
converges, and if it does, compute its value.
1.
I
=

5
·
3
4
/
5
8
2.
I
=

3
4
/
5
3.
I
= (

3)
4
/
5
4.
I
does not converge
correct
5.
I
=
5
8
(

3)
4
/
5
Explanation:
As
a
1
/
5
exists for all
a
, both positive and
negative, we see that
5
p
6

x
2
exists and is nonzero for all
x >
3.
This
means that
x
5
√
6

x
2
exists and is nonzero on [3
,
∞
), and so the
only reason
I
is improper is because the in
terval of integration is inFnite.
Thus
I
will
converge if
lim
t
→∞
I
t
,
I
t
=
Z
t
3
x
5
√
6

x
2
dx,
exists.
To evaluate
I
t
we use substitution,
setting
u
= 6

x
2
. ±or then
du
=

2
xdx,
while
x
= 3
=
⇒
u
=

3
,
x
=
t
=
⇒
u
= 6

t
2
.
In this case
Z
t
3
x
5
√
6

x
2
dx
=

1
2
Z
6

t
2

3
1
u
1
/
5
du
=

5
8
h
u
4
/
5
i
6

t
2

3
=
5
8
‡
(

3)
4
/
5

(6

t
2
)
4
/
5
·
.
But
(

3)
4
/
5
= 3
4
/
5
,
(6

t
2
)
4
/
5
= (
t
2

6)
4
/
5
,
so
Z
t
3
x
5
√
6

x
2
dx
=
5
8
‡
3
4
/
5

(
t
2

6)
4
/
5
·
.
However,
lim
t
(
t
2

6)
4
/
5
=
∞
.
Consequently,
I
does not converge
.
keywords: improper integral, divergent, odd
rational power
004
(part 1 of 1) 10 points
Determine if the improper integral
I
=
Z
∞
∞
3
xe

2
x
2
dx
is convergent or divergent, and if it is conver
gent, Fnd its value.