substitution

Printable View

November 14th 2009, 06:39 PM
gralla55

1 Attachment(s)

substitution

Thanks!
November 14th 2009, 06:46 PM
Prove It

Quote:

Originally Posted by gralla55

Thanks!

Is this

$\int{2\pi x \sqrt{4 - x^2}\,dx}$ ?
November 14th 2009, 06:46 PM
gralla55

Yep!
November 14th 2009, 06:50 PM
Prove It

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

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

You can rewrite your integral as

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

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

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

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

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

All times are GMT -8. The time now is 11:00 AM.

Copyright © 2005-2013 Math Help Forum. All rights reserved.

Copyright © 2005-2013 Math Help Forum. All rights reserved.