Answer
$$A = e - \frac{3}{2}$$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( x \right) = {e^x},{\text{ }}g\left( x \right) = x \cr
& {\text{From the graph shown below}} \cr
& f\left( x \right) \geqslant g\left( x \right){\text{ on the interval }}0 \leqslant x \leqslant 1 \cr
& {\text{The area is given by}} \cr
& A = \int_0^1 {\left( {{e^x} - x} \right)} dx \cr
& {\text{ Integrate}} \cr
& A = \left[ {{e^x} - \frac{1}{2}{x^2}} \right]_0^1 \cr
& A = \left[ {{e^1} - \frac{1}{2}{{\left( 1 \right)}^2}} \right] - \left[ {{e^0} - \frac{1}{2}{{\left( 0 \right)}^2}} \right] \cr
& {\text{Simplifying}} \cr
& A = e - \frac{1}{2} - 1 \cr
& A = e - \frac{3}{2} \cr} $$
