What are the limits? The following is not clear.
-00
∫ dx/(4-x)(3/2)
0
evaluate the improper integral whenever it is convergent
∫0,-00 1/(4-x)^(3/2) dx
2 answers
Ooh, I get it. its from 0 to -∞ (or -inf if you want)
-∞
∫ dx/(4-x)(3/2)
0
Proceed with normal integration
=[2/sqrt(4-x)] from 0 to -∞
=[2/sqrt(4-(-inf)) - 2/sqrt(4-0)]
=[0 - 2/2)
=-1
-∞
∫ dx/(4-x)(3/2)
0
Proceed with normal integration
=[2/sqrt(4-x)] from 0 to -∞
=[2/sqrt(4-(-inf)) - 2/sqrt(4-0)]
=[0 - 2/2)
=-1