But i have no idea how it does im getting a different answer, any help please?

http://i50.tinypic.com/33kwymb.jpg

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- May 25th 2010, 03:54 AMadam_leedsIn my notes i have this equals this
But i have no idea how it does im getting a different answer, any help please?

http://i50.tinypic.com/33kwymb.jpg - May 25th 2010, 04:07 AMtonio

$\displaystyle \frac{2}{2\pi}\int^a_0xe^{-x^2}\,dx=-\frac{1}{2\pi}\int^a_0(-2x\,dx)e^{-x^2}=$ $\displaystyle -\frac{1}{2\pi}\left[e^{-x^2}\right]^a_0=-\frac{1}{2\pi}\left[e^{-a^2}-e^0\right]$ $\displaystyle =\frac{1-e^{-a^2}}{2\pi}$ , so there's a mistake in your notes.

Tonio - May 25th 2010, 04:55 AMAllanCuz
$\displaystyle \frac{2}{2 \pi } \int_0^a xe^{-x^2} dx $

Let $\displaystyle u=x^2 $ and $\displaystyle du = 2xdx $

$\displaystyle \frac{1}{2 \pi } \int_0^{a^2} e^{-u} du = \frac{1}{2 \pi } [ e^{-a^2} - 1 ] $

I think you dont carry through your negative on the third step?

But the above is NOT equal to

$\displaystyle \frac{2}{ \sqrt{ \pi }} [ e^{-a^2} - 1 ] $

Your professor has messed up the constants somewhere along the way. - May 25th 2010, 05:27 AMtonio
- May 25th 2010, 05:31 AMAllanCuz
So what I got from this thread was trust the judgment of MFH helpers over my own, always! lol thanks