Lời giải:Xét phương trình \({x^3} = 4x \Leftrightarrow x\left( {{x^2} - 4} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}x = 0\\x = 2\\x = - 2\end{array} \right.\).
Do đó:
\(\begin{array}{l}S = \int\limits_{ - 2}^2 {\left| {{x^3} - 4x} \right|dx} = \int\limits_{ - 2}^0 {\left| {{x^3} - 4x} \right|dx} + \int\limits_0^2 {\left| {{x^3} - 4x} \right|dx} = \int\limits_{ - 2}^0 {\left( {{x^3} - 4x} \right)dx} + \int\limits_0^2 {\left( {{x^3} - 4x} \right)dx} \\ = \left. {\left( {\frac{1}{4}{x^4} - 2{x^2}} \right)} \right|_{ - 2}^0 - \left. {\left( {\frac{1}{4}{x^4} - 2{x^2}} \right)} \right|_0^2 = 4 + 4 = 8\end{array}\)
Chọn B.
