Determine weather the improper integral converges or diverges.
Evaluate if convergent.  Please show work.

 7
S  ____1_____ dx
 3  sqrt(x-3)

Hi cousinit!!

First of all the definition:
"Improper integrals are integrals of functions which either go to
infinity at the integrands, between the integrands, or where the
integrands are infinite. To evaluate these integrals, we use a limit
process of the antiderivative. Thus we say that an improper integral
converges and/or diverges if the limit converges or diverges."
http://planetmath.org/encyclopedia/ImproperIntegral.html

This says us how to solve the problem.
I will start by a little change of variables to simplify the calculations:

If t = x - 3, then dt = dx and replacing we have:
 7                  4                  4
S  ____1_____ dx = S  ____1_____ dt = S  t^(-1/2) dt
 3  sqrt(x-3)       0  sqrt(t)         0


The antiderivative of t^(-1/2) is 2.sqrt(t), then:
 4                      4
S  t^(-1/2) dt = lim  (S  t^(-1/2) dt) = 
 0               a->0   a
            
               = lim  [2.sqrt(4) - 2.sqrt(a)] =
                 a->0 
                
               = 4 - 0 = 4.

The limit exist, then the improper integral converges, and it is equal to 4.


Hope this helps.