Answer
\[A=5\]
Work Step by Step
\[\begin{align}
& \text{} \\
& 2x-3y=0,\text{ }x+y=5,\text{ }y=0 \\
& \text{Graph the equations using Geogebra} \\
& \\
& \text{From the equations} \\
& 2x-3y=0\text{ }\Rightarrow \text{ }y=\frac{2}{3}x,\text{ for the interval }\left[ 0,3 \right] \\
& x+y=5\text{ }\Rightarrow \text{ }y=5-x,\text{ for the interval }\left[ 3,5 \right] \\
& \text{The area of the region is given by} \\
& A=\int_{0}^{3}{\frac{2}{3}xdx}+\int_{3}^{5}{\left( 5-x \right)}dx \\
& \text{Solving we obtain} \\
& A=5 \\
& \\
& \text{Using an iterated integral we can define the area as:} \\
& 2x-3y=0\text{ }\Rightarrow \text{ }x=\frac{3}{2}y \\
& x+y=5\text{ }\Rightarrow \text{ }x=5-y \\
& 0\le y\le 2 \\
& \text{Then,} \\
& A=\int_{0}^{2}{\int_{\frac{3}{2}y}^{5-y}{dx}dy} \\
& \text{Integrating} \\
& A=\int_{0}^{2}{\left( 5-y-\frac{3y}{2} \right)dy} \\
& A=\int_{0}^{2}{\left( 5-\frac{5y}{2} \right)dy} \\
& A=\left[ 5y-\frac{5}{4}{{y}^{2}} \right]_{0}^{2} \\
& A=\left[ 5\left( 2 \right)-\frac{5}{4}{{\left( 2 \right)}^{2}} \right]-\left[ 5\left( 0 \right)-\frac{5}{4}{{\left( 0 \right)}^{2}} \right] \\
& A=5 \\
\end{align}\]
