On the other hand,
lim
t
→
0
+
t
2
/
3
= 0
.
Consequently,
I
is convergent and
I
= 3
.
017(part1of3)10.0points
(i) Express the improper integral
I
=
integraldisplay
∞
2
xe

x/
2
dx
as lim
t
→ ∞
I
t
with each
I
t
a proper integral.
1.
I
=
lim
t
→ ∞
integraldisplay
t

t
xe

x/
2
dx
2.
I
=
lim
t
→ ∞
integraldisplay
∞
2+
t
xe

x/
2
dx
3.
I
=
lim
t
→ ∞
integraldisplay
∞
2
xe

x/t
dx
4.
I
=
lim
t
→ ∞
integraldisplay
∞
t
xe

x/
2
dx
5.
I
=
lim
t
→ ∞
integraldisplay
t
2
xe

x/
2
dx
correct
Explanation:
The integral
I
=
integraldisplay
∞
2
xe

x/
2
dx
is improper because of the infinite range of
integration. To overcome this we restrict to a
finite interval of integration and consider the
limit
I
=
lim
t
→ ∞
I
t
,
I
t
=
integraldisplay
t
2
xe

x/
2
dx
.
018(part2of3)10.0points
(ii) Compute the value of
I
t
.
1.
I
t
= 8
e

1

2(
t
+ 2)
e

t/
2
correct
2.
I
t
= 2(2

t
)
{
e

1

e

t/
2
}
3.
I
t
=

8
e

1
+ 2(
t
+ 2)
e

t/
2
white (jew3279) – Homework 03 – staron – (53545)
10
4.
I
t
= 4
e

1

(
t
+ 2)
e

t/
2
5.
I
t
= 8
e

1
+ 2(
t
+ 2)
e

t/
2
020
10.0points
Determine if the improper integral
I
=
integraldisplay
∞
021
10.0points
white (jew3279) – Homework 03 – staron – (53545)
11
Find the total area under the graph of
y
=
8
x
3
for
x
≥
1
.
1.
Area =
∞
2.
Area = 4
correct
3.
Area =
7
2
4.
Area =
9
2
5.
Area = 5
6.
Area = 3
Explanation:
The total area under the graph for
x
≥
1 is
an improper integral whose value is the limit
lim
t
→ ∞
integraldisplay
t
1
8
x
3
dx .
Now
integraldisplay
t
1
8
x

3
dx
=

4
bracketleftBig
x

2
bracketrightBig
t
1
.
Consequently,
Area =
lim
t
→ ∞
4
bracketleftBig
1

t

2
bracketrightBig
= 4
.
022
10.0points
Determine if the improper integral
I
=
integraldisplay
∞
0
2
(2
x
+ 3)
2
dx
is convergent, and if it is, find its value.