# Simple integration question help

• Feb 19th 2011, 01:22 AM
aonin
Simple integration question help
Hey guys, after differentiating $\displaystyle y=x^{2}e^{-x^{2}}$by the product rule to get $\displaystyle y=2xe^{-x^{2}}-2x^{3}e^{-x^{2}}$
how do I use this result when integrating $\displaystyle \displaystyle\int x^{3}e^{-x^{2}}\,dx$

thanks a lot :D
• Feb 19th 2011, 01:35 AM
Prove It
$\displaystyle \displaystyle \frac{d}{dx}(x^2e^{-x^2}) = 2x\,e^{-x^2} - 2x^3e^{-x^2}$ is equivalent to

$\displaystyle \displaystyle x^2e^{-x^2} = \int{2x\,e^{-x^2} - 2x^3e^{-x^2}\,dx}$

$\displaystyle \displaystyle x^2e^{-x^2} = -e^{-x^2} - 2\int{x^3e^{-x^2}\,dx}$.

• Feb 19th 2011, 02:00 AM
aonin
Quote:

Originally Posted by Prove It
$\displaystyle \displaystyle \frac{d}{dx}(x^2e^{-x^2}) = 2x\,e^{-x^2} - 2x^3e^{-x^2}$ is equivalent to

$\displaystyle \displaystyle x^2e^{-x^2} = \int{2x\,e^{-x^2} - 2x^3e^{-x^2}\,dx}$

$\displaystyle \displaystyle x^2e^{-x^2} = -e^{-x^2} - 2\int{x^3e^{-x^2}\,dx}$.

$\displaystyle \displaystyle\int x^3e^{-x^2}\,dx=\frac {-(x^{2}e^{-x^2}+e^{-x^2})}{2}$
$\displaystyle \displaystyle\int x^3e^{-x^2}\,dx=\frac {-(x^{2}e^{-x^2}+e^{-x^2})}{2}$