[tex]\int{xe^{-x^2}}dx[/tex]
Let u=-x^2
du/dx=-2x
-du/2=x dx
[tex]\int{xe^{-x^2}}dx=-1/2\int{e^u}du =- \frac{e^u}{2}+C=-\frac{e^{-x^2}}{2}+C[/tex]
When x approaches infinity (or minus infinity, since it is squared), e^(-x^2) approaches zero. So the answer is zero.
