Answer
$2\pi$
Work Step by Step
$\displaystyle \int_{0}^{2}x^{3}\sqrt{4x^{2}-x^{4}}dx=\frac{1}{2}\int_{0}^{2}x^{2}\sqrt{2(2x^{2})-(x^{2})^{2}}\cdot 2xdx=$
substitute $\left[\begin{array}{ll}
u=x^{2} & du=2xdx\\
x=0\rightarrow u=0 & x=2\rightarrow u=4
\end{array}\right], a=2$
$=\displaystyle \frac{1}{2}\int_{0}^{4}u\sqrt{2au-u^{2}}du=$
Table of integrals:
$\color{blue}{ 114. \quad \displaystyle \int u\sqrt{2au-u^{2}}du=\\
\displaystyle=\frac{2u^{2}-\mathrm{a}u-3\mathrm{a}^{2}}{6}\sqrt{2\mathrm{a}u-u^{2}}+\frac{a^{3}}{2}\cos^{-1}(\frac{a-u}{a})+C }$
$=\left[\dfrac{2u^{2}-2u-3(2^{2})}{12}\sqrt{2(2)u-u^{2}}+\dfrac{2^{3}}{4}\cos^{-1}(\dfrac{2-u}{2})\right]_{0}^{4}$
$=[\displaystyle \frac{2u^{2}-2u-12}{12}\sqrt{4u-u^{2}}+\frac{8}{4}\cos^{-1}(\frac{2-u}{2})]_{0}^{4}$
$=[\displaystyle \frac{u^{2}-u-6}{6}\sqrt{4u-u^{2}}+2\cos^{-1}(\frac{2-u}{2})]_{0}^{4}$
$=[0+2\cos^{-1}(-1)]-[0+2\cos^{-1}1]$
$=2\cdot \pi-2\cdot 0$
$=2\pi$
