Cho hàm số \(f\left( x \right)\) liên tục trên \(\...

B

- Hướng dẫn giải

Có \(I = \int\limits_{ - 1}^1 {f\left( {\left| {2x - 1} \right|} \right)dx}  = \int\limits_{ - 1}^{\frac{1}{2}} {f\left( {1 - 2x} \right)dx}  + \int\limits_{\frac{1}{2}}^1 {f\left( {2x - 1} \right)dx} \)

\( =  - \frac{1}{2}\int\limits_{ - 1}^{\frac{1}{2}} {f\left( {1 - 2x} \right)d\left( {1 - 2x} \right)} \left| {\mathop {}\limits_{t = 1 - 2x}^{}  + } \right.\frac{1}{2}\int\limits_{\frac{1}{2}}^1 {f\left( {2x - 1} \right)d\left( {2x - 1} \right)} \left| {\mathop {}\limits_{t = 2x - 1}^{} } \right.\)

\( =  - \frac{1}{2}\int\limits_3^0 {f\left( t \right)dt}  + \frac{1}{2}\int\limits_0^1 {f\left( t \right)dt}  =  - \frac{1}{2}\int\limits_3^0 {f\left( x \right)dx}  + \frac{1}{2}\int\limits_0^1 {f\left( x \right)dx}  = \frac{1}{2}.6 + \frac{1}{2}.2 = 4\)