Answer
a) $A = \int_0^2 {\left( {2x - {x^2}} \right)} dx $
b) $A = \frac{4}{3}$
Work Step by Step
$$\eqalign{
& {\text{From the graph we can see that }}y = 3x - {x^2}{\text{ is above }}y = x,{\text{ }} \cr
& {\text{Let }}f\left( x \right) = 3x - {x^2}{\text{ and }}g\left( x \right) = x{\text{, on the interval }}\left[ {0,2} \right] \cr
& \cr
& \left( {\text{a}} \right){\text{Use }}A = \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx,{\text{ with }}f\left( x \right) \geqslant g\left( x \right) \cr
& x = a = 0{\text{ and }}x = b = 2 \cr
& {\text{Therefore}}{\text{,}} \cr
& \underbrace {A = \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx}_ \Downarrow \cr
& A = \int_0^2 {\left[ {\left( {3x - {x^2}} \right) - x} \right]} dx \cr
& A = \int_0^2 {\left( {3x - {x^2} - x} \right)} dx \cr
& A = \int_0^2 {\left( {2x - {x^2}} \right)} dx \cr
& \cr
& \left( {\text{b}} \right) \cr
& {\text{Integrate}} \cr
& A = \left[ {{x^2} - \frac{1}{3}{x^3}} \right]_0^2 \cr
& A = \left[ {{{\left( 2 \right)}^2} - \frac{1}{3}{{\left( 2 \right)}^3}} \right] - \left[ {{{\left( 0 \right)}^2} - \frac{1}{3}{{\left( 0 \right)}^3}} \right] \cr
& {\text{Simplify}} \cr
& A = \left( {\frac{4}{3}} \right) - \left( 0 \right) \cr
& A = \frac{4}{3} \cr} $$
