Frustrating integral

**angel.white** · December 3rd 2007, 10:11 PM

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

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

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

**Jhevon** · December 3rd 2007, 10:33 PM

Originally Posted by angel.white

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

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

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

a substitution of u = 1 - x should take care of this

**angel.white** · December 3rd 2007, 10:48 PM

Perhaps I'm doing it wrong:

Substitution definitions:
$u=1-x$

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

$-du=dx$

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

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

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

**Jhevon** · December 3rd 2007, 10:56 PM

Originally Posted by angel.white

Perhaps I'm doing it wrong:

Substitution definitions:
$u=1-x$

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

$-du=dx$

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

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

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

so change the 1 + x into a function of u.

$u = 1 - x$

$\Rightarrow x = 1 - u$

$\Rightarrow x + 1 = 2 - u$

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