In my notes i have this equals this

Oct 2008
393
10
But i have no idea how it does im getting a different answer, any help please?

 
Oct 2009
4,261
1,836
But i have no idea how it does im getting a different answer, any help please?


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

Tonio
 
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Apr 2010
384
153
Canada
\(\displaystyle \frac{2}{2\pi}\int^a_0xe^{-x^2}\,dx=-\frac{1}{2\pi}\int^a_0(-2x\,dx)e^{-x^2}=\) \(\displaystyle -\frac{1}{2\pi}\left[e^{-x^2}\right]^a_0=-\frac{1}{2\pi}\left[e^{-a^2}-e^0\right]\) \(\displaystyle =\frac{1-e^{-a^2}}{2\pi}\) , so there's a mistake in your notes.

Tonio
\(\displaystyle \frac{2}{2 \pi } \int_0^a xe^{-x^2} dx \)

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.
 
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Oct 2009
4,261
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\(\displaystyle \frac{2}{2 \pi } \int_0^a xe^{-x^2} dx \)

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?


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

Tonio



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.
.
 
Apr 2010
384
153
Canada
So what I got from this thread was trust the judgment of MFH helpers over my own, always! lol thanks
 
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