Answer
$$A = \frac{1}{4}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {x^3} - x,{\text{ }}\left[ { - 1,0} \right] \cr
& {x^3} - x \geqslant 0{\text{ for the interval }}\left[ { - 1,0} \right],{\text{ then the area is given by}} \cr
& A = \int_{ - 1}^0 {\left( {{x^3} - x} \right)} dx \cr
& {\text{Integrating}} \cr
& A = \left[ {\frac{{{x^4}}}{4} - \frac{{{x^2}}}{2}} \right]_{ - 1}^0 \cr
& {\text{Evaluating}} \cr
& A = \left[ {\frac{{{{\left( 0 \right)}^4}}}{4} - \frac{{{{\left( 0 \right)}^2}}}{2}} \right] - \left[ {\frac{{{{\left( { - 1} \right)}^4}}}{4} - \frac{{{{\left( { - 1} \right)}^2}}}{2}} \right] \cr
& {\text{Simplifying}} \cr
& A = - \left( {\frac{1}{4} - \frac{1}{2}} \right) \cr
& A = \frac{1}{4} \cr} $$
