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