Simple integration question help

**aonin** · February 19th 2011, 01:22 AM

Hey guys, after differentiating $y=x^{2}e^{-x^{2}}$ by the product rule to get $y=2xe^{-x^{2}}-2x^{3}e^{-x^{2}}<br />$
how do I use this result when integrating $\displaystyle\int x^{3}e^{-x^{2}}\,dx$

thanks a lot

**Prove It** · February 19th 2011, 01:35 AM

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

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

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

Now solve for your integral.

**aonin** · February 19th 2011, 02:00 AM

Originally Posted by Prove It

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

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

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

Now solve for your integral.

so this is right?
$\displaystyle\int x^3e^{-x^2}\,dx=\frac {-(x^{2}e^{-x^2}+e^{-x^2})}{2}$

**Prove It** · February 19th 2011, 02:02 AM

Originally Posted by aonin

so this is right?
$\displaystyle\int x^3e^{-x^2}\,dx=\frac {-(x^{2}e^{-x^2}+e^{-x^2})}{2}$

Yes

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