Answer
$$e - \frac{3}{2}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\left( {{e^x} - x} \right)} dx \cr
& {\text{Integrating}}{\text{, use }}\int {{e^x}} dx = {e^x} + C,\,\,\int {{x^n}dx = \frac{{{x^{n + 1}}}}{{n + 1}}} dx \cr
& = \left[ {{e^x} - \frac{{{x^{1 + 1}}}}{{1 + 1}}} \right]_0^1 \cr
& = \left[ {{e^x} - \frac{{{x^2}}}{2}} \right]_0^1 \cr
& {\text{evaluating the limits}} \cr
& = \left[ {{e^1} - \frac{{{{\left( 1 \right)}^2}}}{2}} \right] - \left[ {{e^0} - \frac{{{{\left( 0 \right)}^2}}}{2}} \right] \cr
& {\text{Simplify}} \cr
& = \left[ {e - \frac{1}{2}} \right] - \left[ {1 - \frac{0}{2}} \right] \cr
& = e - \frac{1}{2} - 1 \cr
& = e - \frac{3}{2} \cr} $$