Does anyone know how to find the integral of abs(x)e^((-x)^2)dx from negative infinity to infinity? I can't quite figure out how to do this...
No you're all right, that's not the same problem at all, which is:
$\displaystyle \int_{-\infty}^{\infty} |x| e^{-x^2} dx$
(sorry but I don't believe $\displaystyle e^{(-x)^2}$ because that won't converge.)
You can split it into two bits: one for x positive, one for x negative. Let's do
$\displaystyle \int_0^{\infty} x e^{-x^2} dx$
Put $\displaystyle z = x^2$ and that gives you an easy integral in $\displaystyle e^z$.
Now do the same for
$\displaystyle \int_{-\infty}^0 -x e^{-x^2} dx$
and add them together.
If I'm wrong about it NOT being $\displaystyle e^{(-x)^2}$ then it could be the answer is supposed to be zero because one half equals minus the other half ...?
e^(-x^2) is always positive
Int e^(-x^2) from -inf to pos inf is sqrt(pi)-- this is one of those well known results - the Gaussian Integral See Gaussian integral - Wikipedia, the free encyclopedia
however i read the original pblm here incorrectly --didn't see the abs|x| out front so for your problem it works out to 1