Chọn phương án B.

Phương trình hoành độ giao điểm: $$\begin{aligned}x^3-x=x-x^2\Leftrightarrow &\,x^3+x^2-2x=0\\
\Leftrightarrow&\,x\left(x^2+x-2\right)=0\\
\Leftrightarrow&\,\left[\begin{array}{l}x=0\\ x=1\\ x=-2.\end{array}\right.\end{aligned}$$
Bảng xét dấu:

Khi đó: $$\begin{aligned}
S&=\displaystyle\int\limits_{-2}^1\left|\left(x^3-x\right)-\left(x-x^2\right)\right|\mathrm{\,d}x\\
&=\displaystyle\int\limits_{-2}^1\left|x^3+x^2-2x\right|\mathrm{\,d}x\\
&=\displaystyle\int\limits_{-2}^0\left(x^3+x^2-2x\right)\mathrm{\,d}x-\displaystyle\int\limits_0^1\left(x^3+x^2-2x\right)\mathrm{\,d}x\\
&=\left(\dfrac{x^4}{4}+\dfrac{x^3}{3}-x^2\right)\bigg|_{-2}^0-\left(\dfrac{x^4}{4}+\dfrac{x^3}{3}-x^2\right)\bigg|_0^1\\
&=\dfrac{8}{3}+\dfrac{5}{12}=\dfrac{37}{12}.\end{aligned}$$