MATH 162
Solutions: QUIZ IV
Friday The13th
Answer any five of the following six questions. You may answer all six to obtain extra credit.
1. Explain why the following improper integral diverges:


e
1
dx
ln x
Solution:
First note that x > ln x for all x  e. Hence:
x 
ln x
and so:
0
1
x

1
ln x
for x  e
Recalling that


e
1
dx
1/ 2
x
diverges by the p-test, we now invoke the Comparison Test to obtain the desired result.
2. Compute the value of the following convergent improper integral. Assume
that b is a positive constant.

 bx
e
 dx
0
Solution:
Using the definition of improper integral, we find:
2

e
 bx
n
dx  lim  e
n 
0
 bx
0

 1 bx  n
dx  lim   e  
n 
 b
0

1
1  1
 1
lim    e bn  e  0  lim 1  bn  
n 
n  b
 b
 e  b
The last limit uses the fact that b > 0.
3. Evaluate (i.e. find the exact value) the following convergent improper integral.
Show your work! Calculator solutions are not acceptable.

 x
x
2
0

 13
3/ 2
dx
Solution:
Using the definition of improper integral:

 x
x
2
0
3
n

 xx
3/ 2
n
lim
n 
2
dx  lim
n 

3

3
2
 x
0
x
2
3

3/ 2
dx 
dx 
0

lim   x 2  3
n 



1

2
n


1 / 2
1

 31 / 2 
  lim  n 2  3
n 
3

0

3
4. Evaluate (i.e. find the exact value) the following convergent improper
integral. Show your work! Calculator solutions are not acceptable.

arctan x
0 1  x 2 dx
Solution:
Using the definition of improper integral:

arctan x
1
2 n
arctan x  
0 1  x 2 dx  nlim
 2
0


1
1  
2
2
lim arctan n   0    
2 n 
2 2 
8
5.
2
Determine convergence or divergence. Justify your answer! (That is, if you use the
comparison test, exhibit the function that you choose to use for comparison and show why the
appropriate inequality holds.) Calculator solutions are not acceptable.


13
13  x  x 2
dx
4
1313  x 
Solution:
To apply the comparison test, observe that, for all x ≥ 13:
13  x  x 2 13x 2  x 2  x 2 15 x 2
1
0



15
x4
x4
x2
1313  x 4
Applying the p-test, the improper integral
4


13
1
dx
2
x
converges, and hence, invoking the Comparison Test, the original improper integral must
converge.
6. Determine convergence or divergence. Justify your answer! (That is, if you use the
comparison test, exhibit the function that you choose to use for comparison and show why the
appropriate inequality holds.) Calculator solutions are not acceptable.

13  x  e x
13 1313  x 5  13e x dx
Solution:
To apply the comparison test, observe that, for all x ≥ 13:
13  x  e x
ex
1


0
5
x
x
x
x
1315
1313  x  13e
1313e  e  e
Since, the improper integral

1
13 1315 dx
clearly diverges, the original improper integral must diverge as well.
5
Extra Extra Credit:
1
1

1 x
0
4
dx (Hint: Try using the Comparison Test.)
Solution:
Since x2 > x4 for 0 ≤ x < 1, 1– x2 < 1 – x4, and thus
0
1
Now

0
1
1 x2
1
1 x4
b
dx  lim
b1

0

1
1 x2
1
1 x2
for 0 ≤ x < 1.
dx  lim (arcsin b  arcsin 0) 
b1

2
.
Thus, invoking the Comparison Test, the original integral converges also.