Answer
$\frac{1}{6}$
Work Step by Step
\[\begin{align}
& \text{From the graph we can define the area} \\
& A=\int_{2}^{3}{\left[ \left( x-6 \right)-\left( {{x}^{2}}-4x \right) \right]}dx \\
& A=\int_{2}^{3}{\left( x-6-{{x}^{2}}+4x \right)}dx \\
& A=\int_{2}^{3}{\left( -6-{{x}^{2}}+5x \right)}dx \\
& \text{Integrating} \\
& A=\left[ -6x-\frac{1}{3}{{x}^{3}}+\frac{5}{2}{{x}^{2}} \right]_{2}^{3} \\
& A=\left[ -6\left( 3 \right)-\frac{1}{3}{{\left( 3 \right)}^{3}}+\frac{5}{2}{{\left( 3 \right)}^{2}} \right]-\left[ -6\left( 2 \right)-\frac{1}{3}{{\left( 2 \right)}^{3}}+\frac{5}{2}{{\left( 2 \right)}^{2}} \right] \\
& A=-\frac{9}{2}+\frac{14}{3} \\
& A=\frac{1}{6} \\
\end{align}\]
