Well, I think I'm switching study subjects after this, because I've about had my fill.

$\displaystyle 2\pi \int_{0}^{1} (1+x) \sqrt{1-x} dx$

I've tried about 10 different things, and can't seem to get anywhere.

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- Dec 3rd 2007, 10:11 PMangel.whiteFrustrating integral
Well, I think I'm switching study subjects after this, because I've about had my fill.

$\displaystyle 2\pi \int_{0}^{1} (1+x) \sqrt{1-x} dx$

I've tried about 10 different things, and can't seem to get anywhere. - Dec 3rd 2007, 10:33 PMJhevon
- Dec 3rd 2007, 10:48 PMangel.white
Perhaps I'm doing it wrong:

Substitution definitions:

$\displaystyle u=1-x$

$\displaystyle \frac{du}{dx}=-1$

$\displaystyle -du=dx$

Origional equation:

$\displaystyle 2\pi \int_{0}^{1} (1+x) \sqrt{1-x} dx$

Substitute:

$\displaystyle 2\pi \int_{0}^{1} (1+x) \sqrt{u} (-1)du$

I still have 1+x leftover. But I am pretty tired now, I've been looking at these all day, I'm probably overlooking something obvious (Worried) - Dec 3rd 2007, 10:56 PMJhevon