Cho \(f\left( x \right) = \left\{ \begin{array}{l}1\\rm{khi}}\,\,x \ge 1\\2x - 1\,\,\,\,{\rm{khi}}\,x < 1

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Cho \(f\left( x \right) = \left\{ \begin{array}{l}1\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{khi}}\,\,x \ge 1\\2x - 1\,\,\,\,{\rm{khi}}\,x < 1\end{array} \right.\). Tính \(J = \int\limits_{ - 1}^2 {f\left( x \right){\rm{d}}x} \).

\( - 1\).

\(\frac{1}{2}\).

\(4\).

\(5\).

Giải thích

\(J = \int\limits_{ - 1}^2 {f\left( x \right){\mathop{\rm d}\nolimits} x} = \int\limits_{ - 1}^1 {f\left( x \right){\mathop{\rm d}\nolimits} x} + \int\limits_1^2 {f\left( x \right){\mathop{\rm d}\nolimits} x} \).

Tính \({J_1} = \int\limits_{ - 1}^1 {f\left( x \right){\mathop{\rm d}\nolimits} x} = \int\limits_{ - 1}^1 {\left( {2x - 1} \right){\mathop{\rm d}\nolimits} x} = \left. {\left( {{x^2} - x} \right)} \right|_{ - 1}^1 = 0 - 2 = - 2\).

Tính \({J_2} = \int\limits_1^2 {f\left( x \right){\mathop{\rm d}\nolimits} x} = \int\limits_1^2 {1{\mathop{\rm d}\nolimits} x} = 2 - 1 = 1\).

Vậy \(J = - 2 + 1 = - 1\). Chọn A.