Answer
$A = \frac{{44}}{3}$
Work Step by Step
$$\eqalign{
& {\text{We have the functions:}} \cr
& y = {x^2} \cr
& 3y = 2x + 16 \to y = \frac{2}{3}x + \frac{{16}}{3} \cr
& y = - 2x + 8 \cr
& {\text{From the graph we can see that:}} \cr
& y = \frac{2}{3}x + \frac{{16}}{3}{\text{ is above }}y = {x^2},{\text{ on the interval }}\left[ { - 2,1} \right] \cr
& y = - 2x + 8{\text{ is above }}y = {x^2},{\text{ on the interval }}\left[ {1,4} \right] \cr
& {\text{The enclosed area can be represented as}} \cr
& A = \int_{ - 2}^1 {\left( {\frac{2}{3}x + \frac{{16}}{3} - {x^2}} \right)} dx + \int_1^2 {\left( { - 2x + 8 - {x^2}} \right)} dx \cr
& {\text{Integrating}} \cr
& A = \left[ {\frac{1}{3}{x^2} + \frac{{16}}{3}x - \frac{1}{3}{x^3}} \right]_{ - 2}^1 + \left[ { - {x^2} + 8x - \frac{1}{3}{x^3}} \right]_1^2 \cr
& A = \left[ {\frac{1}{3}{{\left( 1 \right)}^2} + \frac{{16}}{3}\left( 1 \right) - \frac{1}{3}{{\left( 1 \right)}^3}} \right] - \left[ {\frac{1}{3}{{\left( { - 2} \right)}^2} + \frac{{16}}{3}\left( { - 2} \right) - \frac{1}{3}{{\left( { - 2} \right)}^3}} \right] \cr
& + \left[ { - {{\left( 2 \right)}^2} + 8\left( 2 \right) - \frac{1}{3}{{\left( 2 \right)}^3}} \right] - \left[ { - {{\left( 1 \right)}^2} + 8\left( 1 \right) - \frac{1}{3}{{\left( 1 \right)}^3}} \right] \cr
& A = \left( {\frac{{16}}{3}} \right) - \left( { - \frac{{20}}{3}} \right) + \left( {\frac{{28}}{3}} \right) - \left( {\frac{{20}}{3}} \right) \cr
& A = \frac{{44}}{3} \cr} $$
