# Calculus

posted by Christina

I tried to integrate this:
intergral sign with an interval from -inifinity to + infinity
function: (x-r)^2 (1/r) (e^(-x/r))
with respect to dx.....

I know that Steve set up the u-substitution but I seem to never arrive at an answer... It seems like I keep integrating and integrating and integrating and never arrive at an answer!!!!!!!!

1. Damon

Steve gave you:
let
u = (x-r)^2
du = 2(x-r) dx

dv = (1/r) e^(-x/r) dx
v = e^(-x/r)
CHECK SIGN

∫ u dv = uv - ∫ v du
= (x-r)^2 e^(-x/r) - 2∫(x-r) e^(-x/r) dx

now the second part of it. He said to do it again.
u = x - r
du = dx
dv = e^-x/r
v = -r e^-x/r
-(x-r)e^-x/r - ∫-re^-x/r dx
now you can do ∫-re^-x/r dx = (1/r)e^-x/r

CHECK THE SIGNS !
then put it back together and do the limits

2. Christina

I got r^2 - 2r.. and it is wrong.... please help....

3. Steve

See whether you can get wolframalpha's result:

http://www.wolframalpha.com/input/?i=%E2%88%AB+%28%28x-r%29^2+%281%2Fr%29+%28e^%28-x%2Fr%29%29%29+dx

The limit would be undefined for x -> -∞

I suspect a typo somewhere. Seems like there should be e^(-x^2/r) or something.

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