Chọn phương án A.

Phương trình hoành độ giao điểm: $$\begin{aligned}
&\,x^3-3x^2=x^2+x-4\\
\Leftrightarrow&\,x^3-4x^2-x+4=0\\
\Leftrightarrow&\,\left[\begin{array}{l}
x=4\\ x=1\\ x=-1
\end{array}\right.
\end{aligned}$$
Bảng xét dấu:

Khi đó: $$\begin{eqnarray*}
S&=&\int\limits_{-1}^4\left|x^3-4x^2-x+4\right|\mathrm{d}x\\
&=&\int\limits_{-1}^1\left(x^3-4x^2-x+4\right)\mathrm{d}x+\int\limits_1^4\left(-x^3+4x^2+x-4\right)\mathrm{d}x\\
&=&\left(\dfrac{x^4}{4}-\dfrac{4x^3}{3}-\dfrac{x^2}{2}+4x\right)\bigg|_{-1}^1+\left(-\dfrac{x^4}{4}+\dfrac{4x^3}{3}+\dfrac{x^2}{2}-4x\right)\bigg|_1^4\\
&=&\dfrac{253}{12}.
\end{eqnarray*}$$