Answer
\[ = - \frac{{32}}{3}\]
Work Step by Step
\[\begin{gathered}
\int_0^2 {{x^3}\sqrt {16 - {x^4}} \,dx} \hfill \\
\hfill \\
\,set\,\,u = 16 - {x^4}\,\,\,\,\,then\,\,\,\, - \frac{{du}}{4} = {x^3}dx \hfill \\
\hfill \\
{\text{Changing limits of integration}} \hfill \\
\hfill \\
x = 0\,\,\,\,\,implies\,\,\,\,u = 16 - {0^4} = 16 \hfill \\
x = 2\,\,\,\,\,implies\,\,\,u = 16 - {2^4} = 0 \hfill \\
\hfill \\
apply\,\,the\,\,\,substitution \hfill \\
\hfill \\
= - \frac{1}{4}\int_{16}^0 {{u^{\frac{1}{2}}}} du \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= - \frac{1}{6}\,\,\left[ {{u^{\frac{3}{2}}}} \right]_{16}^0 \hfill \\
\hfill \\
Fundamental\,\,theorem \hfill \\
\hfill \\
= - \frac{1}{6}\,\,\left[ {{0^{\frac{3}{2}}} - {{16}^{\frac{3}{2}}}} \right] \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
= - \frac{1}{6}\,\left( {64} \right) \hfill \\
\hfill \\
= - \frac{{32}}{3} \hfill \\
\end{gathered} \]
