# In my notes i have this equals this

#### adam_leeds

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

#### tonio

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

adam_leeds

#### AllanCuz

$$\displaystyle \frac{2}{2\pi}\int^a_0xe^{-x^2}\,dx=-\frac{1}{2\pi}\int^a_0(-2x\,dx)e^{-x^2}=$$ $$\displaystyle -\frac{1}{2\pi}\left[e^{-x^2}\right]^a_0=-\frac{1}{2\pi}\left[e^{-a^2}-e^0\right]$$ $$\displaystyle =\frac{1-e^{-a^2}}{2\pi}$$ , 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.

adam_leeds

#### 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?

$$\displaystyle \int e^{-u}\,du=-e^{-u}$$ , 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.
.

#### AllanCuz

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

adam_leeds
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