Answer
\[{{e}^{4}}-1\]
Work Step by Step
\[\begin{align}
& \int_{0}^{1}{\int_{2x}^{2}{4{{e}^{{{y}^{2}}}}}dydx} \\
& \text{Switch the order of integration using the region shown below} \\
& \int_{0}^{1}{\int_{2x}^{2}{4{{e}^{{{y}^{2}}}}}dydx}=\int_{0}^{2}{\int_{0}^{y/2}{4{{e}^{{{y}^{2}}}}}dxdy} \\
& \text{Integrating} \\
& =\int_{0}^{2}{\left[ 4x{{e}^{{{y}^{2}}}} \right]_{0}^{y/2}dy} \\
& =\int_{0}^{2}{\left[ 4\left( \frac{y}{2} \right){{e}^{{{y}^{2}}}} \right]dy} \\
& =\int_{0}^{2}{2y{{e}^{{{y}^{2}}}}dy} \\
& =\left[ {{e}^{{{y}^{2}}}} \right]_{0}^{2} \\
& ={{e}^{{{\left( 2 \right)}^{2}}}}-{{e}^{{{\left( 0 \right)}^{2}}}} \\
& ={{e}^{4}}-1 \\
\end{align}\]
