1. ## Frustrating 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.

2. Originally Posted by angel.white
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.
a substitution of u = 1 - x should take care of this

3. 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

4. Originally Posted by angel.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
so change the 1 + x into a function of u.

$\displaystyle u = 1 - x$

$\displaystyle \Rightarrow x = 1 - u$

$\displaystyle \Rightarrow x + 1 = 2 - u$