Answer
$$A = \frac{4}{3}$$
Work Step by Step
$$\eqalign{
& y = {x^2},\,\,\,\,\,y = 2x \cr
& {\text{Find the points of intersection of the graphs}} \cr
& {x^2} = 2x \cr
& {x^2} - 2x = 0 \cr
& {\text{factor}} \cr
& x\left( {x - 2} \right) = 0 \cr
& x = 0{\text{ and }}x = 2 \cr
& {\text{We have the interval }}\left( {0,2} \right) \cr
& {\text{Evaluate both functions at }}x = 1 \cr
& y\left( 1 \right) = {\left( 1 \right)^2} = 1 \cr
& y\left( 2 \right) = 2\left( 1 \right) = 2 \cr
& {\text{For the interval }}\left( {0,2} \right){\text{ we have that }}2x \geqslant {x^2} \cr
& \cr
& {\text{The area is given by}}: \cr
& A = \int_0^2 {\left( {2x - {x^2}} \right)} dx \cr
& {\text{Integrating}} \cr
& A = \left( {{x^2} - \frac{{{x^3}}}{3}} \right)_0^2 \cr
& A = \left( {{{\left( 2 \right)}^2} - \frac{{{{\left( 2 \right)}^3}}}{3}} \right) - \left( {{{\left( 0 \right)}^2} - \frac{{{{\left( 0 \right)}^3}}}{3}} \right) \cr
& A = \left( {\frac{4}{3}} \right) - \left( 0 \right) \cr
& A = \frac{4}{3} \cr} $$
