Answer
$$A = \frac{{19}}{6}$$
Work Step by Step
$$\eqalign{
& {\text{The intersection point between }}y = 4\sqrt {2x} {\text{ and }}y = - 4x + 6,{\text{ is}} \cr
& 4\sqrt {2x} = - 4x + 6 \cr
& 2\sqrt {2x} = - 2x + 3 \cr
& 4\left( {2x} \right) = 9 - 12x + 4{x^2} \cr
& 8x = 9 - 12x + 4{x^2} \cr
& 4{x^2} - 20x + 9 = 0 \cr
& \left( {2x - 9} \right)\left( {2x - 1} \right) = 0 \cr
& {x_1} = \frac{9}{2},{\text{ }}{x_2} = \frac{1}{2} \cr
& {\text{The intersection point between }}y = 2{x^2}{\text{ and }}y = - 4x + 6,{\text{ is}} \cr
& 2{x^2} = - 4x + 6 \cr
& 2{x^2} + 4x - 6 = 0 \cr
& {x^2} + 2x - 3 = 0 \cr
& \left( {x + 3} \right)\left( {x - 1} \right) = 0 \cr
& {x_1} = - 3{\text{ and }}x = 1 \cr
& {\text{The area of the region is given by:}} \cr
& A = \int_{1/2}^1 {\left[ {4\sqrt {2x} - \left( { - 4x + 6} \right)} \right]} dx + \int_1^2 {\left( {4\sqrt {2x} - 2{x^2}} \right)} dx \cr
& A = \int_{1/2}^1 {\left( {4\sqrt {2x} + 4x - 6} \right)} dx + \int_1^2 {\left( {4\sqrt {2x} - 2{x^2}} \right)} dx \cr
& A = \left[ {\frac{{4\sqrt 2 }}{{3/2}}{x^{3/2}} + 2{x^2} - 6x} \right]_{1/2}^1 + \left[ {\frac{{4\sqrt 2 }}{{3/2}}{x^{3/2}} - \frac{{2{x^3}}}{3}} \right]_1^2 \cr
& A = \left[ {\frac{{8\sqrt 2 }}{3}{x^{3/2}} + 2{x^2} - 6x} \right]_{1/2}^1 + \left[ {\frac{{8\sqrt 2 }}{3}{x^{3/2}} - \frac{{2{x^3}}}{3}} \right]_1^2 \cr
& {\text{Evaluating}} \cr
& A = \left( {\frac{{8\sqrt 2 - 12}}{3} + \frac{7}{6}} \right) + \left( {\frac{{16}}{3} - \frac{{8\sqrt 2 - 2}}{3}} \right) \cr
& {\text{Simplifying}} \cr
& A = \frac{{8\sqrt 2 - 12}}{3} + \frac{7}{6} + \frac{{16}}{3} - \frac{{8\sqrt 2 - 2}}{3} \cr
& A = \frac{{19}}{6} \cr} $$
