Answer
$A = \frac{7}{4}$
Work Step by Step
$$\eqalign{
& {\text{Let the functions}} \cr
& y = 1 + {x^3},{\text{ }}y = 2 - x,{\text{ }}x = - 1,{\text{ }}x = 0 \cr
& {\text{Graph the curves using Geogebra }}\left( {{\text{Shown Below}}} \right) \cr
& \cr
& {\text{We can find the area integrating with respect to }}x \cr
& {\text{Use }}A = \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx,{\text{ with }}f\left( x \right) \geqslant g\left( x \right) \cr
& {\text{Let }} \cr
& f\left( x \right) = 2 - x{\text{, }}g\left( x \right) = 1 + {x^3} \cr
& x = a = - 1{\text{ and }}x = b = 0 \cr
& {\text{Therefore}}{\text{,}} \cr
& \underbrace {A = \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx}_ \Downarrow \cr
& A = \int_{ - 1}^0 {\left[ {\left( {2 - x} \right) - \left( {1 + {x^3}} \right)} \right]} dx \cr
& A = \int_{ - 1}^0 {\left( {2 - x - 1 - {x^3}} \right)} dx \cr
& A = \int_{ - 1}^0 {\left( {1 - x - {x^3}} \right)} dx \cr
& {\text{Integrating}} \cr
& A = \left[ {x - \frac{1}{2}{x^2} - \frac{1}{4}{x^4}} \right]_{ - 1}^0 \cr
& A = \left[ 0 \right] - \left[ {\left( { - 1} \right) - \frac{1}{2}{{\left( { - 1} \right)}^2} - \frac{1}{4}{{\left( { - 1} \right)}^4}} \right] \cr
& {\text{Simplifying}} \cr
& A = - \left[ { - \frac{7}{4}} \right] \cr
& A = \frac{7}{4} \cr} $$
