Answer
$$A = \frac{{41}}{2}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {x^3}{\text{; }}\left[ { - 1,3} \right] \cr
& {\text{First check if the graph crosses the }}x - {\text{axis in the given interval}} \cr
& {x^3} = 0 \cr
& {\text{solve for }}x \cr
& x = 0 \cr
& {\text{the graph crosses the }}x{\text{ - axis at }}x = 0{\text{ and the given interval }}\left[ { - 1,3} \right],{\text{ }} \cr
& f\left( { - 1} \right) = {\left( { - 1} \right)^3} = - 1,{\text{ }}f\left( x \right){\text{ is negative for }}\left( { - 1,0} \right) \cr
& {\text{then}} \cr
& {\text{the definite integral of the total area will be}}: \cr
& A = \left| {\int_{ - 1}^0 {{x^3}} dx} \right| + \int_0^3 {{x^3}} dx \cr
& A = \left| {\left( {\frac{{{x^4}}}{4}} \right)_{ - 1}^0} \right| + \left( {\frac{{{x^4}}}{4}} \right)_0^3 \cr
& A = \left| {\left( {\frac{{{{\left( 0 \right)}^4}}}{4} - \frac{{{{\left( 1 \right)}^4}}}{4}} \right)} \right| + \left( {\frac{{{{\left( 3 \right)}^4}}}{4} - \frac{{{{\left( 0 \right)}^4}}}{4}} \right) \cr
& {\text{simplifying}} \cr
& A = \left| { - \frac{1}{4}} \right| + \frac{{81}}{4} \cr
& A = \frac{1}{4} + \frac{{81}}{4} \cr
& A = \frac{{41}}{2} \cr} $$
