Answer
\[\frac{26}{9}\]
Work Step by Step
\[\begin{align}
& \int_{0}^{2}{\int_{x}^{2}{x\sqrt{1+{{y}^{3}}}}dydx} \\
& \text{Switch the order of integration using the region shown below} \\
& \int_{0}^{2}{\int_{x}^{2}{x\sqrt{1+{{y}^{3}}}}dydx}=\int_{0}^{2}{\int_{0}^{y}{x\sqrt{1+{{y}^{3}}}}dxdy} \\
& \text{Integrating} \\
& =\int_{0}^{2}{\left[ \frac{1}{2}{{x}^{2}}\sqrt{1+{{y}^{3}}} \right]_{0}^{y}dy} \\
& =\int_{0}^{2}{\left[ \frac{1}{2}{{\left( y \right)}^{2}}\sqrt{1+{{y}^{3}}} \right]dy} \\
& =\frac{1}{2}\int_{0}^{2}{\sqrt{1+{{y}^{3}}}\left( {{y}^{2}} \right)dy} \\
& =\frac{1}{6}\int_{0}^{2}{\sqrt{1+{{y}^{3}}}\left( 3{{y}^{2}} \right)dy} \\
& =\frac{1}{6}\left[ \frac{{{\left( 1+{{y}^{3}} \right)}^{3/2}}}{3/2} \right]_{0}^{2} \\
& =\frac{1}{9}\left[ {{\left( 1+{{y}^{3}} \right)}^{3/2}} \right]_{0}^{2} \\
& =\frac{1}{9}\left[ {{\left( 1+{{2}^{3}} \right)}^{3/2}}-{{\left( 1+{{0}^{3}} \right)}^{3/2}} \right] \\
& =\frac{1}{9}\left[ 27-1 \right] \\
& =\frac{26}{9} \\
\end{align}\]
