Find the exact area of the region bounded by ,
$\displaystyle u^2 = 4-x^2$
$\displaystyle x^2 = 4-u^2$
$\displaystyle 2x \, dx = -2u \, du$
$\displaystyle x \, dx = -u \, du$
$\displaystyle \int \frac{x^3}{\sqrt{4-x^2}} \, dx = \int \frac{x^2}{\sqrt{4-x^2}} \cdot x \, dx $
substitute ...
$\displaystyle \int \frac{4-u^2}{u} \cdot (-u) \, du$
$\displaystyle -\int 4 - u^2 \, du$
lower limit ... $\displaystyle x = 0$ , $\displaystyle u = 2$
upper limit ... $\displaystyle x = \sqrt{2}$ , $\displaystyle u = \sqrt{2}$
$\displaystyle -\int_2^{\sqrt{2}} 4 - u^2 \, du = \int_{\sqrt{2}}^2 4 - u^2 \, du$