Answer
\[\frac{1}{4}\left( {{e}^{8}}-1 \right)\]
Work Step by Step
\[\begin{align}
& \int_{0}^{2}{\int_{0}^{4-{{x}^{2}}}{\frac{x{{e}^{2y}}}{4-y}}dydx} \\
& y=4-{{x}^{2}}\Rightarrow x=-\sqrt{4-y}\text{, }x\le 0.\text{ }x=\sqrt{4-y}\text{, }x>0 \\
& \text{Using the graph to switch the order of integration} \\
& R=\left\{ \left( x,y \right):0\le x\le \sqrt{4-y},\text{ 0}\le y\le 4\text{ } \right\} \\
& \text{Then}, \\
& \int_{0}^{2}{\int_{0}^{4-{{x}^{2}}}{\frac{x{{e}^{2y}}}{4-y}}dydx}=\int_{0}^{4}{\int_{0}^{\sqrt{4-y}}{\frac{x{{e}^{2y}}}{4-y}}dxdy} \\
& \text{Integrating} \\
& =\int_{0}^{4}{\left[ \frac{{{x}^{2}}{{e}^{2y}}}{2\left( 4-y \right)} \right]_{0}^{\sqrt{4-y}}dy} \\
& =\int_{0}^{4}{\left[ \frac{{{\left( \sqrt{4-y} \right)}^{2}}{{e}^{2y}}}{2\left( 4-y \right)}-\frac{{{\left( 0 \right)}^{2}}{{e}^{2y}}}{2\left( 4-y \right)} \right]dy} \\
& =\int_{0}^{4}{\left[ \frac{\left( 4-y \right){{e}^{2y}}}{2\left( 4-y \right)}-\frac{\left( 0 \right){{e}^{2y}}}{2\left( 4-y \right)} \right]dy} \\
& =\int_{0}^{4}{\frac{{{e}^{2y}}}{2}dy} \\
& =\frac{1}{4}\left[ {{e}^{2y}} \right]_{0}^{4} \\
& =\frac{1}{4}\left( {{e}^{8}}-1 \right) \\
\end{align}\]
