Answer
\[2\left( {{e}^{2}}-e-1 \right)\]
Work Step by Step
\[\begin{align}
& \iint\limits_{R}{{{x}^{-1/2}}{{e}^{y}}dA} \\
& \text{The region }R\text{ is given by} \\
& R=\left\{ \left( x,y \right):0\le y\le \sqrt{x},\text{ }1\le x\le 4\text{ } \right\} \\
& \text{Then,} \\
& \iint\limits_{R}{{{x}^{-1/2}}{{e}^{y}}dA}=\int_{1}^{4}{\int_{0}^{\sqrt{x}}{{{x}^{-1/2}}{{e}^{y}}}dy}dx \\
& \text{Integrating} \\
& =\int_{1}^{4}{\left[ {{x}^{-1/2}}{{e}^{y}} \right]_{0}^{\sqrt{x}}}dx \\
& =\int_{1}^{4}{\left[ {{x}^{-1/2}}{{e}^{\sqrt{x}}}-{{x}^{-1/2}}{{e}^{0}} \right]}dx \\
& =\int_{1}^{4}{\left( \frac{{{e}^{\sqrt{x}}}}{\sqrt{x}}-{{x}^{-1/2}} \right)}dx \\
& =\left[ 2{{e}^{\sqrt{x}}}-2\sqrt{x} \right]_{1}^{4} \\
& =\left[ 2{{e}^{\sqrt{4}}}-2\sqrt{4} \right]-\left[ 2{{e}^{\sqrt{1}}}-2\sqrt{1} \right] \\
& =2{{e}^{2}}-4-2e+2 \\
& =2\left( {{e}^{2}}-e-1 \right) \\
\end{align}\]
