Xét phương trình hoành độ giao điểm là ${{x}^{3}}-x=x-{{x}^{2}}\Leftrightarrow {{x}^{3}}+{{x}^{2}}-2x=0\Leftrightarrow \left[ \begin{align} & x=0 \\ & x=1 \\ & x=-2 \\ \end{align} \right.$

Do vậy $I=\int\limits_{-2}^{1}{\left| {{x}^{3}}+{{x}^{2}}-2x \right|dx=\int\limits_{-2}^{0}{\left| {{x}^{3}}+{{x}^{2}}-2x \right|dx+\int\limits_{0}^{1}{\left| {{x}^{3}}+{{x}^{2}}-2x \right|dx}}}$

$=\int\limits_{-2}^{0}{({{x}^{3}}+{{x}^{2}}-2x)dx-\int\limits_{0}^{1}{({{x}^{3}}+{{x}^{2}}-2x)dx}}$.

.=$\left( \frac{{{x}^{4}}}{4}+\frac{{{x}^{3}}}{3}-{{x}^{2}} \right)\left| \begin{align} & 0 \\ & -2 \\ \end{align} \right.$ - $\left( \frac{{{x}^{4}}}{4}+\frac{{{x}^{3}}}{3}-{{x}^{2}} \right)\left| \begin{align} & 1 \\ & 0 \\ \end{align} \right.$

$=\frac{8}{3}+\frac{5}{12}=\frac{37}{12}$