Answer
$\frac{1}{32}\pi$
Work Step by Step
\[\begin{align}
& \int_{0}^{1/2}{x\sqrt{1-16{{x}^{4}}}dx} \\
& \text{Let }u=4{{x}^{2}},\text{ }du=8xdx \\
& \text{The new limits of integration are:} \\
& x=0\to u=0 \\
& x=1/2\to u=1 \\
& \text{Substituting} \\
& \int_{0}^{1/2}{x\sqrt{1-16{{x}^{4}}}dx}=\int_{0}^{1}{x\sqrt{1-{{\left( 4{{x}^{2}} \right)}^{2}}}dx} \\
& =\frac{1}{8}\int_{0}^{1}{\sqrt{1-{{u}^{2}}}du} \\
& \text{Solving }\int_{0}^{1}{\sqrt{1-{{u}^{2}}}du}\text{ by the equation of a semicirle} \\
& =\frac{1}{8}\left( \frac{1}{4}\pi {{\left( 1 \right)}^{2}} \right) \\
& =\frac{1}{32}\pi \\
\end{align}\]
