Answer
$$A = \frac{{32}}{3}$$
Work Step by Step
$$\eqalign{
& {\text{Let }}y = {x^2}{\text{ and }}y = 2{x^2} - 4x \cr
& {\text{Calculate the intersection points}} \cr
& {x^2} = 2{x^2} - 4x \cr
& {x^2} - 4x = 0 \cr
& x\left( {x - 4} \right) = 0 \cr
& {x_1} = 0{\text{ and }}{x_2} = 4 \cr
& {\text{Where }}{x^2} \geqslant 2{x^2} - 4x{\text{ on the interval }}\left[ {0,4} \right],{\text{ then}} \cr
& {\text{the enclosed area is given by }} \cr
& A = \int_0^4 {\left( {{x^2} - 2{x^2} + 4x} \right)} dx \cr
& A = \int_0^4 {\left( {4x - {x^2}} \right)} dx \cr
& {\text{Integrating}} \cr
& A = \left[ {2{x^2} - \frac{1}{3}{x^3}} \right]_0^4 \cr
& A = \left[ {2{{\left( 4 \right)}^2} - \frac{1}{3}{{\left( 4 \right)}^3}} \right] - \left[ {2{{\left( 0 \right)}^2} - \frac{1}{3}{{\left( 0 \right)}^3}} \right] \cr
& A = 32 - \frac{{64}}{3} \cr
& A = \frac{{32}}{3} \cr} $$
