# In my notes i have this equals this

• May 25th 2010, 03:54 AM
In 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 AM
tonio
Quote:

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

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

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

Tonio
• May 25th 2010, 04:55 AM
AllanCuz
Quote:

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.
• May 25th 2010, 05:27 AM
tonio
Quote:

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. (Wink)

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.

.
• May 25th 2010, 05:31 AM
AllanCuz
So what I got from this thread was trust the judgment of MFH helpers over my own, always! lol thanks