Answer
\[1\]
Work Step by Step
\[\begin{align}
& \int_{0}^{2}{\int_{x/2}^{1}{dy}dx} \\
& \text{Find the integral} \\
& \int_{0}^{2}{\int_{x/2}^{1}{dy}dx}=\int_{0}^{2}{\left[ y \right]_{x/2}^{1}dx} \\
& =\int_{0}^{2}{\left[ 1-\frac{x}{2} \right]dx} \\
& =\left[ x-\frac{{{x}^{2}}}{4} \right]_{0}^{2} \\
& =\left[ \left( 2 \right)-\frac{{{\left( 2 \right)}^{2}}}{4} \right]-\left[ \left( 0 \right)-\frac{{{\left( 0 \right)}^{2}}}{4} \right] \\
& =2-\frac{4}{4} \\
& =1 \\
& \text{Using the graph to switch the order of integration} \\
& \int_{0}^{2}{\int_{x/2}^{1}{dy}dx}=\int_{0}^{1}{\int_{0}^{2y}{dx}dy} \\
& =\int_{0}^{1}{\left[ x \right]_{0}^{2y}dy} \\
& =\int_{0}^{1}{2ydy} \\
& =\left[ {{y}^{2}} \right]_{0}^{1} \\
& ={{\left( 1 \right)}^{2}}-{{\left( 0 \right)}^{2}} \\
& =1 \\
\end{align}\]
