Đáp án đúng: C
Giải chi tiết:Ta có: \(f\left( x \right) = \left\{ \begin{array}{l}2x + 2\,\,\,khi\,\,\, - 1 \le x \le 0\\2\,\,\,\,khi\,\,\,0 \le x \le 1\\ - 2x + 4\,\,\,\,khi\,\,\,1 \le x \le 2\\ - x + 2\,\,\,\,khi\,\,\,2 \le x \le 3\\ - 1\,\,\,khi\,\,\,3 \le x \le 4\end{array} \right.\)
\(\begin{array}{l} \Rightarrow I = \int\limits_{ - 1}^4 {f\left( x \right)dx} = \int\limits_{ - 1}^0 {\left( {2x + 2} \right)dx} + \int\limits_0^1 {2dx} + \int\limits_1^2 {\left( { - 2x + 4} \right)dx} + \int\limits_2^3 {\left( { - x + 2} \right)dx} + \int\limits_3^4 {\left( { - 1} \right)dx} \\ = \left. {\left( {{x^2} + 2x} \right)} \right|_{ - 1}^0 + \left. {2x} \right|_0^1 + \left. {\left( { - {x^2} + 4x} \right)} \right|_1^2 + \left. {\left( { - \dfrac{{{x^2}}}{2} + 2x} \right)} \right|_2^3 + \left. {\left( { - x} \right)} \right|_3^4\\ = 1 + 2 + 1 - \dfrac{1}{2} - 1 = \dfrac{5}{2}.\end{array}\)
Chọn C.
