Answer
$A = 36$
Work Step by Step
$$\eqalign{
& {\text{Let the functions}} \cr
& y = {x^2} - 4x,{\text{ }}y = 2x \cr
& {\text{Graph the curves using Geogebra }}\left( {{\text{Shown Below}}} \right) \cr
& \cr
& {\text{We can find the area integrating with respect to }}x \cr
& {\text{Use }}A = \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx,{\text{ with }}f\left( x \right) \geqslant g\left( x \right) \cr
& {\text{Let }} \cr
& f\left( x \right) = 2x{\text{, }}g\left( x \right) = {x^2} - 4x \cr
& x = a = 0{\text{ and }}x = b = 6 \cr
& {\text{Therefore}}{\text{,}} \cr
& \underbrace {A = \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx}_ \Downarrow \cr
& A = \int_0^6 {\left[ {2x - \left( {{x^2} - 4x} \right)} \right]} dx \cr
& A = \int_0^6 {\left( {2x - {x^2} + 4x} \right)} dx \cr
& A = \int_0^6 {\left( {6x - {x^2}} \right)} dx \cr
& {\text{Integrating}} \cr
& A = \left[ {3{x^2} - \frac{1}{3}{x^3}} \right]_0^6 \cr
& A = \left[ {3{{\left( 6 \right)}^2} - \frac{1}{3}{{\left( 6 \right)}^3}} \right] - \left[ {3{{\left( 0 \right)}^2} - \frac{1}{3}{{\left( 0 \right)}^3}} \right] \cr
& {\text{Simplifying}} \cr
& A = 36 - 0 \cr
& A = 36 \cr} $$
