Answer
\[\frac{8}{5}\,{\left( {8 - r} \right)^{5/2}} - \frac{{64}}{3}\,{\left( {8 - r} \right)^{3/2}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {4r\sqrt {8 - r} dr} \hfill \\
Let\,\,u = 8 - r\,\,Then\,\,r = 8 - u \hfill \\
So\,\,that\,\,dr = - du \hfill \\
\int_{}^{} {4r\sqrt {8 - r} \,dr = 4\int_{}^{} {\,\left( {8 - u} \right){u^{\frac{1}{2}}}\,\left( { - 1} \right)du} } \hfill \\
= 4\int_{}^{} {\,\left( {u - 8} \right){u^{\frac{1}{2}}}du} \hfill \\
= 4\int_{}^{} {\,\left( {{u^{3/2}} - 8{u^{\frac{1}{2}}}} \right)du} \hfill \\
Power\,\,rule \hfill \\
4\,\left( {\frac{{{u^{5/2}}}}{{5/2}}} \right) - 32\,\left( {\frac{{{u^{\frac{3}{2}}}}}{{3/2}}} \right) + C \hfill \\
Substituting\,\,u = 8 - r\,\,gives \hfill \\
\frac{8}{5}\,{\left( {8 - r} \right)^{5/2}} - \frac{{64}}{3}\,{\left( {8 - r} \right)^{3/2}} + C \hfill \\
\end{gathered} \]
