Answer
$$A = 20$$
Work Step by Step
$$\eqalign{
& x = - 3,\,\,\,\,\,x = 1,\,\,\,\,\,y = {x^3} + 1,\,\,\,\,\,\,y = 0 \cr
& {\text{Check if the graph crosses the }}x{\text{ - axis}} \cr
& \,\,\,\,\,\,\,\,{x^3} + 1 = 0 \cr
& \,\,\,\,\,\,\,x = - 1 \cr
& {\text{The critical point of the graph is }}x = - 1.{\text{ Evaluaing the graph at }}x = - 3 \cr
& \,\,\,\,\,\,\,y\left( { - 3} \right) = {\left( { - 3} \right)^3} + 1 \cr
& \,\,\,\,\,\,\,y\left( { - 3} \right) = - 26 \cr
& {\text{Then}}{\text{, the region of the graph lies below for the interval }}\left( { - 3, - 1} \right). \cr
& {\text{The area is given by}}: \cr
& A = \left| {\int_{ - 3}^{ - 1} {\left( {{x^3} + 1} \right)dx} } \right| + \int_{ - 1}^1 {\left( {{x^3} + 1} \right)} dx \cr
& A = \left| {\left( {\frac{{{x^4}}}{4} + x} \right)_{ - 3}^{ - 1}} \right| + \left( {\frac{{{x^4}}}{4} + x} \right)_{ - 1}^1 \cr
& {\text{Evaluating}} \cr
& A = \left| {\left( {\frac{{{{\left( { - 1} \right)}^4}}}{4} + \left( { - 1} \right)} \right) - \left( {\frac{{{{\left( { - 3} \right)}^4}}}{4} + \left( { - 3} \right)} \right)} \right| + \left( {\frac{{{{\left( 1 \right)}^4}}}{4} + \left( 1 \right)} \right) - \left( {\frac{{{{\left( { - 1} \right)}^4}}}{4} + \left( { - 1} \right)} \right) \cr
& {\text{Simplifying}} \cr
& A = \left| { - \frac{3}{4} - \frac{{69}}{4}} \right| + \frac{5}{4} + \frac{3}{4} \cr
& A = 18 + \frac{5}{4} + \frac{3}{4} \cr
& A = 20 \cr} $$
