ok, i need someone to explain this to me, im having some problems with this.
integral from 0 to infinity of (x-1)e^-x
thanks
2006-10-31 03:51:25 · 5 answers · asked by The Dude 3 in Science & Mathematics ➔ Mathematics
The anti-derivative is -xe^(-x). At x=0, this is 0. As x->infty, it goes to 0 also. Hence the integral is 0.
2006-10-31 05:55:54 · answer #1 · answered by mathematician 7 · 0⤊ 0⤋
You basically need to take limit as t aproaches infinity of that integral from 0 to t. Then you would need to use INTEGRATION BY PARTS to take the integral (since there's no relation between (x-1) and e^). you surely will use L'hopital rule.
2006-10-31 12:21:23 · answer #2 · answered by toto 1 · 0⤊ 0⤋
all integrals are separable. first step is to multiply in the e^-x to the parenthesis. once you have done that you will be left with
(xe^-x)-(e^-x). since the exponents are negative you can then write (x/e^x)-(1/e^x). Then take the integral each parenthesis separately don't forget about the subtraction sign.
2006-10-31 11:58:23 · answer #3 · answered by steve0stac 2 · 0⤊ 0⤋
Multiply through and re-write it as xe^(-x) - e^(-x).
Then do 2 integrals.
int xe^(-x) - int e^(-x) (both from 0 to infinity)
Remember int [G(x)±H(x)] dx = int G(x)dx ± int H(X)dx
Doug
2006-10-31 12:01:30 · answer #4 · answered by doug_donaghue 7 · 0⤊ 0⤋
it converges by comparison with e^-x to the limit 1/e
2006-10-31 11:59:55 · answer #5 · answered by jared p 2 · 0⤊ 0⤋