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!!!!!!!!
Steve gave you:
u = (x-r)^2
du = 2(x-r) dx
dv = (1/r) e^(-x/r) dx
v = e^(-x/r)
∫ 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
Calculus- NEED HELP!!! -
I got r^2 - 2r.. and it is wrong.... please help....
See whether you can get wolframalpha's result:
The limit would be undefined for x -> -∞
I suspect a typo somewhere. Seems like there should be e^(-x^2/r) or something.