HW8 - nguyen (kdn444) Assignment 8 guntel (55200) This...

nguyen (kdn444) – Assignment 8 – guntel – (55200) 1
This print-out should have 24 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001(part1of3)10.0points
(i) Express the improper integral
I =
integraldisplay
∞
2
2 xe
- x/ 3
dx
as lim
t → ∞
I
t
with each I
t
a proper integral.
1. I = lim
t → ∞
integraldisplay
t
- t
2 xe
- x/ 3
dx
2. I = lim
t → ∞
integraldisplay
t
2
2 xe
- x/ 3
dx correct
3. I = lim
t → ∞
integraldisplay
∞
t
2 xe
- x/ 3
dx
4. I = lim
t → ∞
integraldisplay
∞
2
2 xe
- x/t
dx
5. I = lim
t → ∞
integraldisplay
∞
2+ t
2 xe
- x/ 3
dx
Explanation:
The integral
I =
integraldisplay
∞
2
2 xe
- x/ 3
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
2 xe
- x/ 3
dx .
002(part2of3)10.0points
(ii) Compute the value of I
t
.
4. I
t
= 30 e
- 2
3
− 6( t + 3) e
- t/ 3
correct
5. I
t
= 30 e
- 2
3
+ 6( t + 3) e
- t/ 3
Explanation:
To evaluate
I
t
=
integraldisplay
t
2
2 xe
- x/ 3
dx
we integrate by parts. Then
I
t
=
bracketleftBig
− 6 xe
- x/ 3
bracketrightBig
t
2
+ 6
integraldisplay
t
2
e
- x/ 3
dx
= − 6
bracketleftBig
( x + 3) e
- x/ 3
bracketrightBig
t
2
.
Consequently,
I
t
= 30 e
-
2
3
− 6( t + 3) e
- t/ 3
.
003(part3of3)10.0points
(iii) Determine if lim
t → ∞
I
t
exists, and if it does,
find its value.
1. I = 6 e
2
3
2. I = 30 e
- 2
3
correct
3. I does not exist
4. I = 30 e
2
3
5. I = 6 e
-
2
3
Explanation:
By L’Hospital’s Rule,
lim
x → ∞
x
m
e
- x
= 0
for any power m of x . It follows that
lim
t

nguyen (kdn444) – Assignment 8 – guntel – (55200) 2
004 10.0points
Determine if the improper integral
I =
integraldisplay
- 5
-∞
1
√
4 − v
dv
is convergent or divergent, and if convergent,
find its value.
1. I = 3
2. I = 6
3. I =
9
2
4. I =
3
2
5. I is divergent correct
Explanation:
The integral is improper because of the in-
finite interval of integration. It will converge
if
lim
t → ∞
integraldisplay
- 5
- t
1
√
4 − v
dv
exists. Now
integraldisplay
- 5
- t
1
√
4 − v
dv =
bracketleftBig
− 2
√
4 − v
bracketrightBig
- 5
- t
= − 6 + 2
√
4 + t .
But we know that
lim
t → ∞
√
4 + t = ∞ .
Consequently,
I is divergent .
005 10.0points
Determine if the improper integral
I =
integraldisplay
∞
e
1
x (ln 4 x ) 2
dx
converges, and if it does, compute its value.
1. I does not converge
2. I =
4
ln 4 e
3. I = ln 4 e
4. I = 4
5. I =
1
ln 4 e
correct
6. I =
1
4 e
Explanation:
The integral is improper because of the in-
finite interval of integration, so we write
I = lim
t → ∞
I
t
, I
t
=
integraldisplay
t
e
1
x (ln 4 x )
2
dx,
whenever the limit exists. To evaluate I
t
, first
set u = ln 4 x . Then
integraldisplay
1
x (ln 4 x )
2
dx =
integraldisplay
1
u
2
du = −
1
u
+ C ,
and so
I
t
=
bracketleftBig
−
1
ln 4 x
bracketrightBig
t
e
=
parenleftBig
1
ln 4 e
−
1
ln 4 t
parenrightBig
.
On the other hand,
lim
t → ∞
1
ln 4 t
= 0 .
Consequently, lim
t → ∞
I
t
exists, and
I = lim
t → ∞
parenleftBig
1
ln 4 e
−
1
ln 4 t
parenrightBig
=
1
ln 4 e
.
006 10.0points
Determine if the improper integral
I =
integraldisplay
∞
1
2 tan
- 1
x
1 + x 2
dx
converges, and if it does, find its value.

nguyen (kdn444) – Assignment 8 – guntel – (55200) 3
1. I =
3
16
π
2
correct
2. I =
3
16
3. I =
1
4
4. I