Answer
$$A = \frac{{142}}{3}$$
Work Step by Step
$$\eqalign{
& {\text{From the given graph we have}} \cr
& f\left( x \right) = - {x^2} + 3x + 11{\text{ and }}g\left( x \right) = {x^2} - 3x + 3 \cr
& g\left( x \right) \geqslant f\left( x \right){\text{ on the interval }} - 2 \leqslant x \leqslant - 1 \cr
& f\left( x \right) \geqslant g\left( x \right){\text{ on the interval }} - 1 \leqslant x \leqslant 4 \cr
& {\text{The area is given by}} \cr
& A = \int_{ - 2}^{ - 1} {\left[ {\left( {{x^2} - 3x + 3} \right) - \left( { - {x^2} + 3x + 11} \right)} \right]} dx \cr
& {\text{ }} + \int_{ - 1}^4 {\left[ {\left( { - {x^2} + 3x + 11} \right) - \left( {{x^2} - 3x + 3} \right)} \right]} dx \cr
& A = \int_{ - 2}^{ - 1} {\left( {{x^2} - 3x + 3 + {x^2} - 3x - 11} \right)} dx \cr
& + \int_{ - 1}^4 {\left( { - {x^2} + 3x - 3 - {x^2} + 3x + 11} \right)} dx \cr
& A = \int_{ - 2}^{ - 1} {\left( {2{x^2} - 6x - 8} \right)} dx \cr
& + \int_{ - 1}^4 {\left( {8 + 6x - 2{x^2}} \right)} dx \cr
& {\text{Integrate}} \cr
& A = \left[ {\frac{{2{x^3}}}{3} - 3{x^2} - 8x} \right]_{ - 2}^{ - 1} + \left[ {8x + 3{x^2} - \frac{{2{x^3}}}{3}} \right]_{ - 1}^4 \cr
& A = \left[ {\frac{{2{{\left( { - 1} \right)}^3}}}{3} - 3{{\left( { - 1} \right)}^2} - 8\left( { - 1} \right)} \right] - \left[ {\frac{{2{{\left( { - 2} \right)}^3}}}{3} - 3{{\left( { - 2} \right)}^2} - 8\left( { - 2} \right)} \right] \cr
& + \left[ {8\left( 4 \right) + 3{{\left( 4 \right)}^2} - \frac{{2{{\left( 4 \right)}^3}}}{3}} \right] - \left[ {8\left( { - 1} \right) + 3{{\left( { - 1} \right)}^2} - \frac{{2{{\left( { - 1} \right)}^3}}}{3}} \right] \cr
& {\text{Simplifying}} \cr
& A = \frac{{13}}{3} + \frac{4}{3} + \frac{{112}}{3} + \frac{{13}}{3} \cr
& A = \frac{{142}}{3} \cr} $$
