substitution

**gralla55** · November 14th 2009, 06:39 PM

Thanks!

**Prove It** · November 14th 2009, 06:46 PM

Originally Posted by gralla55

Thanks!

Is this

$\int{2\pi x \sqrt{4 - x^2}\,dx}$ ?

**gralla55** · November 14th 2009, 06:46 PM

Yep!

**Prove It** · November 14th 2009, 06:50 PM

$\int{2\pi x \sqrt{4 - x^2}\,dx} = \int{2\pi x (4 - x^2)^{\frac{1}{2}}\,dx}$ .

Let $u = 4 - x^2$ so that $\frac{du}{dx} = -2x$ .

You can rewrite your integral as

$-\pi \int{ (4 - x^2)^{\frac{1}{2}}(-2x)\,dx}$

$= -\pi\int{u^{\frac{1}{2}}\,\frac{du}{dx}\,dx}$

$= -\pi \int{u^{\frac{1}{2}}\,du}$

$= -\frac{2\pi}{3}u^{\frac{3}{2}} + C$

$= C -\frac{2\pi \sqrt{(4 - x^2)^3}}{3}$ .

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