Answer
\[A=\frac{140}{3}\]
Work Step by Step
\[\begin{align}
& \text{From the graph} \\
& {{R}_{1}}+{{R}_{2}}=\left\{ \left( x,y \right):6-x\le y\le 5x+6,\text{ }0\le x\le 2\text{ } \right\} \\
& +\left\{ \left( x,y \right):{{x}^{2}}\le y\le 5x+6,\text{ }2\le x\le 6\text{ } \right\} \\
& A=\int_{0}^{2}{\int_{6-x}^{5x+6}{dy}dx}+\int_{2}^{6}{\int_{{{x}^{2}}}^{5x+6}{dy}dx} \\
& A=\int_{0}^{2}{\left( 5x+6-6+x \right)dx}+\int_{2}^{6}{\left( 5x+6-{{x}^{2}} \right)dx} \\
& A=\int_{0}^{2}{6xdx}+\int_{2}^{6}{\left( 5x+6-{{x}^{2}} \right)dx} \\
& A=\left[ 3{{x}^{2}} \right]_{0}^{2}+\left[ \frac{5}{2}{{x}^{2}}+6x-\frac{1}{3}{{x}^{3}} \right]_{2}^{6} \\
& A=3{{\left( 6 \right)}^{2}}+\left[ \frac{5}{2}{{\left( 6 \right)}^{2}}+6\left( 6 \right)-\frac{1}{3}{{\left( 6 \right)}^{3}} \right] \\
& -\left[ \frac{5}{2}{{\left( 2 \right)}^{2}}+6\left( 2 \right)-\frac{1}{3}{{\left( 2 \right)}^{3}} \right] \\
& A=12+54-\frac{58}{3} \\
& A=\frac{140}{3} \\
\end{align}\]
