I forgot to mention that I am trying to find sigma squared.. given that the mean is r and p(x)= (1/r)(e^(-x/r)) on the interval [0,infinity), where r>0..

Question

note: sigma squared= integral sign (-inf, inf) (x-M)^2 p(x) dx

i got (r^2 - 2r)^(1/2) for sigma, but my answer is wrong...

Previous Posts:
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!!!!!!!!


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