would i just integrate as normal, the double integral is putting me off.
could someone show me how id do:
$\displaystyle \int\limits_0^2 \int\limits_{x^{2}-2}^x x^{2} dydx $
thanks
The inside integral is constant with respect to y, so
$\displaystyle \int_0^2 \left ( \int_{x^{2}-2}^x x^2 dy \right ) dx $
$\displaystyle = \int_0^2 \left ( \left . x^2y \right |_{x^2 - 2}^x \right ) dx $
$\displaystyle = \int_0^2 \left ( x^2((x^2 - 2) - x ) \right ) dx $
I'm sure you can finish from here.
-Dan
$\displaystyle = \int_0^2 (x^4 - x^3 - 2x^2) dx $