Sử dụng tính chất tích phân: \(\int\limits_a^b {\left[ {f\left( x \right) + g\left( x \right)} \right]dx} = \int\limits_a^b {f\left( x \right)dx} + \int\limits_a^b {g\left( x \right)dx} \), \(\int\limits_a^b {kf\left( x \right)dx} = k\int\limits_a^b {f\left( x \right)dx} \,\,\left( {k \ne 0} \right)\), \(\int\limits_a^b {f\left( x \right)dx} = \int\limits_a^c {f\left( x \right)dx} + \int\limits_c^b {f\left( x \right)dx} \).Giải chi tiết:Theo bài ra ta có:\(\left\{ \begin{array}{l}\int\limits_0^2 {\left[ {f\left( x \right) - 3g\left( x \right)} \right]dx} = 4\\\int\limits_0^2 {\left[ {2f\left( x \right) + g\left( x \right)} \right]dx} = 8\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\int\limits_0^2 {f\left( x \right)dx} - 3\int\limits_0^2 {g\left( x \right)dx} = 4\\2\int\limits_0^2 {f\left( x \right)dx} + \int\limits_0^2 {g\left( x \right)dx} = 8\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\int\limits_0^2 {f\left( x \right)dx} = 4\\\int\limits_0^2 {g\left( x \right)dx} = 0\end{array} \right.\)Vậy \(I = \int\limits_1^2 {f\left( x \right)dx} = \int\limits_0^2 {f\left( x \right)dx} - \int\limits_0^1 {f\left( x \right)dx} = 4 - 3 = 1\).Chọn C
