Answer
\[\int_{0}^{\ln 2}{\int_{1/2}^{{{e}^{-x}}}{f\left( x,y \right)}dy}dx\]
Work Step by Step
\[\begin{align}
& \int_{1/2}^{1}{\int_{0}^{-\ln y}{f\left( x,y \right)}dx}dy \\
& x=-\ln y\to y={{e}^{-x}} \\
& x=0 \\
& \text{Using the graph to switch the order of integration} \\
& R=\left\{ \left( x,y \right):\frac{1}{2}\le y\le {{e}^{-x}},\text{ 0}\le x\le \ln 2\text{ } \right\} \\
& \text{Then} \\
& \int_{1/2}^{1}{\int_{0}^{-\ln y}{f\left( x,y \right)}dx}dy=\int_{0}^{\ln 2}{\int_{1/2}^{{{e}^{-x}}}{f\left( x,y \right)}dy}dx \\
\end{align}\]
