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