Answer
$$\eqalign{
& \left( {\text{a}} \right)A = \frac{4}{3} \cr
& \left( {\text{b}} \right)A = \frac{4}{3} \cr} $$
Work Step by Step
$$\eqalign{
& \left( {\text{a}} \right){\text{Integrating with respect to }}x \cr
& {\text{The area is given by}} \cr
& A = \int_0^2 {\left( {2x - {x^2}} \right)} dx \cr
& A = \left[ {{x^2} - \frac{1}{3}{x^3}} \right]_0^2 \cr
& A = \left[ {{{\left( 2 \right)}^2} - \frac{1}{3}{{\left( 2 \right)}^3}} \right] - \left[ {{{\left( 0 \right)}^2} - \frac{1}{3}{{\left( 0 \right)}^3}} \right] \cr
& A = \frac{4}{3} \cr
& \cr
& \left( {\text{b}} \right){\text{Integrating with respect to }}y \cr
& y = {x^2} \Rightarrow x = \sqrt y \cr
& y = 2x \Rightarrow x = \frac{1}{2}y \cr
& \sqrt y \geqslant \frac{1}{2}y{\text{ on the interval 0}} \leqslant {\text{y}} \leqslant {\text{4}} \cr
& {\text{The area is given by}} \cr
& A = \int_0^4 {\left( {\sqrt y - \frac{1}{2}y} \right)} dy \cr
& A = \left[ {\frac{2}{3}{x^{3/2}} - \frac{1}{4}{y^2}} \right]_0^4 \cr
& A = \left[ {\frac{2}{3}{{\left( 4 \right)}^{3/2}} - \frac{1}{4}{{\left( 4 \right)}^2}} \right] - \left[ {\frac{2}{3}{{\left( 0 \right)}^{3/2}} - \frac{1}{4}{{\left( 0 \right)}^2}} \right] \cr
& A = \frac{{16}}{3} - 4 + 0 \cr
& A = \frac{4}{3} \cr} $$
