Answer
$$A = \frac{{10235}}{{66}}$$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( x \right) = {x^2}{\left( {{x^3} + 1} \right)^{10}},{\text{ }}g\left( x \right) = - x{\left( {{x^2} + 1} \right)^{10}} \cr
& {\text{From the graph shown below}} \cr
& f\left( x \right) \geqslant g\left( x \right){\text{ on the interval }}0 \leqslant x \leqslant 1 \cr
& {\text{The area is given by}} \cr
& A = \int_0^1 {\left[ {{x^2}{{\left( {{x^3} + 1} \right)}^{10}} + x{{\left( {{x^2} + 1} \right)}^{10}}} \right]} dx \cr
& A = \int_0^1 {\left[ {{x^2}{{\left( {{x^3} + 1} \right)}^{10}} + x{{\left( {{x^2} + 1} \right)}^{10}}} \right]} dx \cr
& {\text{ Integrate}} \cr
& A = \left[ {\frac{1}{3}\left[ {\frac{{{{\left( {{x^3} + 1} \right)}^{11}}}}{{11}}} \right] + \frac{1}{2}\left[ {\frac{{{{\left( {{x^2} + 1} \right)}^{11}}}}{{11}}} \right]} \right]_0^1 \cr
& A = \left[ {\frac{{{{\left( {{x^3} + 1} \right)}^{11}}}}{{33}} + \frac{{{{\left( {{x^2} + 1} \right)}^{11}}}}{{22}}} \right]_0^1 \cr
& A = \left[ {\frac{{{{\left( {{1^3} + 1} \right)}^{11}}}}{{33}} + \frac{{{{\left( {{1^2} + 1} \right)}^{11}}}}{{22}}} \right] - \left[ {\frac{{{{\left( {{0^3} + 1} \right)}^{11}}}}{{33}} + \frac{{{{\left( {{0^2} + 1} \right)}^{11}}}}{{22}}} \right] \cr
& {\text{Simplifying}} \cr
& A = \frac{{{2^{11}}}}{{33}} + \frac{{{2^{11}}}}{{22}} - \frac{1}{{33}} - \frac{1}{{22}} \cr
& A = \frac{{10240}}{{66}} - \frac{5}{{66}} \cr
& A = \frac{{10235}}{{66}} \cr} $$
