Cho hàm số \(y=f(x)\) có đồ thị trên đoạn [- 1;4] ...

C

- Hướng dẫn giải

Ta có: \(f\left( x \right) = \left\{ \begin{array}{l}
2x + 2\,\,\,\,khi\,\, - 1 \le x \le 0\\
2\,\,\,\,\,\,\,\,\,\,\,\,\,\,khi\,\,\,0 \le x \le 1\\
 - 2x + 4\,khi\,\,\,1 \le x \le 2\\
 - x + 2\,\,\,\,khi\,\,\,2 \le x \le 3\\
 - 1\,\,\,\,\,\,\,\,\,\,\,\,khi\,\,\,3 \le x \le 4
\end{array} \right.\) 

\(\begin{array}{l}
 \Rightarrow I = \int\limits_{ - 1}^4 {f\left( x \right)dx = \int\limits_{ - 1}^0 {\left( {2x + 2} \right)dx + \int\limits_0^1 {2dx + \int\limits_1^2 {\left( { - 2x + 4} \right)dx + \int\limits_2^3 {\left( { - x + 2} \right)dx + \int\limits_3^4 {\left( { - 1} \right)dx} } } } } } \\
 = \left( {{x^2} + 2x} \right)\left| \begin{array}{l}
^0\\
_{ - 1}
\end{array} \right. + 2x\left| \begin{array}{l}
^1\\
_0
\end{array} \right. + \left( { - {x^2} + 4x} \right)\left| \begin{array}{l}
^2\\
_1
\end{array} \right. + \left( { - \frac{{{x^2}}}{2} + 2x} \right)\left| \begin{array}{l}
^3\\
_2
\end{array} \right. + \left( { - x} \right)\left| \begin{array}{l}
^4\\
_3
\end{array} \right.\\
 = 1 + 2 + 1 - \frac{1}{2} - 1 = \frac{5}{2}
\end{array}\)