nanni (arn437) – Assignment 16 – guntel – (54940)
1
This printout should have 4 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
10.0 points
±ind the total area under the graph oF
y
=
6
x
3
For
x
≥
1
.
1.
Area = 4
2.
Area =
5
2
3.
Area =
7
2
4.
Area =
∞
5.
Area = 2
6.
Area = 3
correct
Explanation:
The total area under the graph For
x
≥
1 is
an improper integral whose value is the limit
lim
t
→∞
i
t
1
6
x
3
dx .
Now
i
t
1
6
x

3
dx
=

3
b
x

2
B
t
1
.
Consequently,
Area = lim
t
3
b
1

t

2
B
= 3
.
002
10.0 points
Determine iF the improper integral
I
=
i
3
0
2 ln
x
x
dx
converges, and iF it does, compute its value.
1.
I
= (ln 3)
2
2.
I
does not converge
correct
3.
I
= 2(ln 3)
2
4.
I
= 2 ln 3
5.
I
= ln 3
Explanation:
The integral is improper because
lim
x
→
0
+
ln
x
x
=
∞
,
so we have to consider
lim
t
→
0
+
I
t
,
I
t
=
i
3
t
2 ln
x
x
dx .
To evaluate
I
t
,
set
u
= ln
x.
Then
du
=
dx
x
,
while
x
=
t
=
⇒
u
= ln
t ,
x
= 3
=
⇒
u
= ln 3
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '10
 Gualdini
 Calculus, lim, dx

Click to edit the document details