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