212 Calculus 3 Review 2 
1. Prove that the integral x
0 1  x 4 dx converges and find its value. A
Solution: consider the proper integral x
0 1  x 4 dx and perform the substitution
u  x 2 . Then A
A2
x
1
1
1
dx
du
arctan u


2
0 1  x 4

2 0 1 u
2
A2
0
1

 arctan A2  . 2
A 4
1
2. Prove that the integral ln x
0 3 x dx converges and find its value. 1
ln x
 3 x dx and apply integration by parts 1
3 23
1 3
taking u  ln x and dv  x dx . Then du  dx and v  x whence 2
x
Solution: consider the proper integral 1
ln x
3

dx
ln x  x 2 3
 3 x
2
1
1

3
3
9
  x 1 3dx   ln    2 3  x 2 3
2
2
4
1


3
9 9
  ln    2 3    2 3 .
2
4 4
We have to find the limit of the last expression when 
 0  . The limit of the last term is obviously 0 and we will prove that the limit of the first tem is also 0 . Indeed, lim
 0 
ln    2 3  lim
 0 
ln 

2 3
. The last limit is an indeterminate form 
3
 1
3
13
and we apply L’Hopital’s rule obtaining  lim  4 3   lim   0 
2  0  
2  0 
Finally we see that the integral converges and its value is 
9
. 4
3. Use the limit comparison test to decide whether the integral 
x3  3x 2  1
2 x 4 ln x dx converges or diverges. x3  3x 2  1
x3
1
Solution: let f ( x ) 
and
. Then g
(
x
)


4
4
x ln x
x ln x x ln x


f ( x)
lim
 1and therefore the integrals  f ( x)dx and  g ( x) dx either x  g ( x )
2
2
both converge or both diverge. Next consider Performing the substitution u
A
A
A
2
2
 g ( x)dx  
1
dx . x ln x
 ln x we see that ln A
1
1

dx   du  ln u
x ln x
u
2
ln 2
ln A
ln 2
 ln ln A  ln ln 2   . Thus the A
original integral diverges. 4. Use the limit comparison test to decide whether the integral 1

csc xdx converges or diverges. 0
Solution: Let 1
1
and g ( x ) 
. We know that sin x
x
f ( x)  csc x 
sin x
f ( x)
 1and therefore lim
 1. Consider x 0
x 0  g ( x )
x
lim
1
1
 g ( x)dx  
1
1
dx  2 x
x
1

 2  2   2 . Thus  g ( x) dx
 0 
0
1
converges and so does
 f ( x)dx . 0

e x
dx converges or diverges. 5. Decide whether the integral 
x
0
2
Solution: We represent this integral as the sum of two improper integrals. 
e x
e x
I1  
dx and I 2  
dx x
x
0
1
1
2
2
If we prove that the integrals I1 and I 2 converge then the original integral converges as well. To see that I1 converges notice that e
e x
1
dx

0 x
0 x dx  2 x
1
2
1
1
0
 2   .  x2
 1whence To prove that I 2 converges notice first that if x  1 then x  1whence 
1
 x2
 1and I 2   e dx . Next we can reason this way: x
1
x  1  x  x  x  x  e
2
2
 x2
 e x Whence 
A
I 2   e  x dx  lim  e  x dx  lim  e  x
1
A
1
A
and the original integral converges. A
1
 lim(1  e  A )  1   A