# integrate in terms of area

• Dec 11th 2012, 01:20 PM
pnfuller
integrate in terms of area
integrate from -2 to 2 sqrt(4-x^4) in terms of an area.

• Dec 11th 2012, 02:37 PM
Prove It
Re: integrate in terms of area
I think it means you need to use a Riemann Sum, since at face value I can't see a way to integrate this...
• Dec 11th 2012, 02:45 PM
pnfuller
Re: integrate in terms of area
how many rectangles should i do so i dont get negatives under the square root?
• Dec 11th 2012, 03:39 PM
skeeter
Re: integrate in terms of area
Quote:

Originally Posted by pnfuller
integrate from -2 to 2 sqrt(4-x^4) in terms of an area.

you sure the integrand isn't $\color{red}\sqrt{4 - x^2}$ ? the function you posted is undefined at the limits of integration ... $\sqrt{4 - 2^4} = \sqrt{-12}$ ... ???
btw ... y = $\sqrt{4-x^2}$ is a semicircle