Answer
$$A = \frac{{82}}{3}$$
Work Step by Step
$$\eqalign{
& {\text{From the graph we can note that }} \cr
& - {x^2} + 3x + 12 > 12 - 6x{\text{ on the interval }}1 \leqslant x \leqslant 3,{\text{ then}} \cr
& {\text{The area of the shadow region is given by}} \cr
& A = \int_1^3 {\left[ {\left( { - {x^2} + 3x + 12} \right) - \left( {12 - 6x} \right)} \right]dx} \cr
& A = \int_1^3 {\left( { - {x^2} + 3x + 12 - 12 + 6x} \right)dx} \cr
& A = \int_1^3 {\left( { - {x^2} + 9x} \right)dx} \cr
& {\text{Integrating}} \cr
& A = \left[ { - \frac{1}{3}{x^3} + \frac{9}{2}{x^2}} \right]_1^3 \cr
& A = \left[ { - \frac{1}{3}{{\left( 3 \right)}^3} + \frac{9}{2}{{\left( 3 \right)}^2}} \right] - \left[ { - \frac{1}{3}{{\left( 1 \right)}^3} + \frac{9}{2}{{\left( 1 \right)}^2}} \right] \cr
& A = \frac{{63}}{2} - \frac{{25}}{6} \cr
& A = \frac{{82}}{3} \cr} $$
