Ta có: \(\displaystyle {x^3} = 4x \Leftrightarrow x\left( {{x^2} - 4} \right) = 0\) \(\displaystyle \Leftrightarrow \left[ \begin{array}{l}x = 0\\x = - 2\\x = 2\end{array} \right.\).
\(\displaystyle S = \int\limits_{ - 2}^2 {\left| {{x^3} - 4x} \right|dx} \) \(\displaystyle = \int\limits_{ - 2}^0 {\left| {{x^3} - 4x} \right|dx} + \int\limits_0^2 {\left| {{x^3} - 4x} \right|dx} \) \(\displaystyle = \int\limits_{ - 2}^0 {\left( {{x^3} - 4x} \right)dx} - \int\limits_0^2 {\left( {{x^3} - 4x} \right)dx} \)
\(\displaystyle = \left. {\left( {\frac{{{x^4}}}{4} - 2{x^2}} \right)} \right|_{ - 2}^0 - \left. {\left( {\frac{{{x^4}}}{4} - 2{x^2}} \right)} \right|_0^2\) \(\displaystyle = 0 - \frac{{16}}{4} + 2.4 - \frac{{16}}{4} + 2.4 = 8\).
Chọn C.
