Answer
$2\pi$
Work Step by Step
\[\begin{align}
& \int_{0}^{\sqrt{2}}{4x\sqrt{4-{{x}^{4}}}dx} \\
& \text{Let }2u={{x}^{2}}, \text{ }du=xdx \\
& \text{The new limits of integration are:} \\
& x=0\to u=0 \\
& x=\sqrt{2}\to u=1 \\
& \text{Substituting} \\
& \int_{0}^{\sqrt{2}}{4x\sqrt{4-{{x}^{4}}}dx}=\int_{0}^{1}{4x\sqrt{4-{{\left( {{x}^{2}} \right)}^{2}}}dx} \\
& =4\int_{0}^{1}{\sqrt{4-{{4u}^{2}}}du}=8\int_{0}^{1}{\sqrt{1-{{u}^{2}}}du} \\
&\text{ By the equation of a semicirle} \\
& =8\left( \frac{1}{4}\pi {{\left( 1 \right)}^{2}} \right) \\
& =2\pi \\
\end{align}\]
