Integrate improper integral from 3 to 2 of 1/square root of 3-x

Question

Does it converge or diverge? I got 2 and therefore converges.

Answer 1

∫_{(from 3 to 2)}[1/√(3-x)]dx  (the integral is improper at x = 3 because if

                                    x = 3, the integrand is undefined)

 

= -∫_{(from 2 to 3)}[1/√(3-x)]dx

 

= -lim_b→3^- [∫_{(from 2 to b)}[1/√(3-x)]dx     Let u = 3-x  then du = -dx

                                                                        So, dx = -du

 

                                                      When x = 2, u = 3-2 = 1

                                                      When x = b, u = 3-b

 

= -lim_b→3^- [∫_{(from 1 to 3-b)} u^-1/2(-du)]

 

= lim_b→3^- [2√u]_{(from 1 to 3-b)}

 

= lim_b→3^- [2√(3-b) - 2] = 0 - 2 = -2

 

The integral converges to -2.

 

NOTE:  If you meant the integral "from 2 to 3", then the integral                      converges to 2.  

 

Answer 2

Usually one states the lower limit (a) first, and the upper limit (b) second. Let's assume that the problem is a=2, b = 3.

 

Antiderivative of (3-x)^-½ is -2(3-x)^½

and substituting upper b=3 now, we get 0, and substituting and subtracting with a=2 next, we get

-[-1] = 2.