In my notes i have this equals this

**adam_leeds** · May 25th 2010, 03:54 AM

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

**tonio** · May 25th 2010, 04:07 AM

Originally Posted by adam_leeds

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

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

Tonio

**AllanCuz** · May 25th 2010, 04:55 AM

Originally Posted by tonio

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

Tonio

$\frac{2}{2 \pi } \int_0^a xe^{-x^2} dx$

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

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

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

But the above is NOT equal to

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

Your professor has messed up the constants somewhere along the way.

**tonio** · May 25th 2010, 05:27 AM

Originally Posted by AllanCuz

$\frac{2}{2 \pi } \int_0^a xe^{-x^2} dx$

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

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

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

$\int e^{-u}\,du=-e^{-u}$ , and not only $e^{-u}$ . You forgot the minus sign here.

Tonio

But the above is NOT equal to

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

Your professor has messed up the constants somewhere along the way.

.

**AllanCuz** · May 25th 2010, 05:31 AM

So what I got from this thread was trust the judgment of MFH helpers over my own, always! lol thanks

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