Cho hàm số \(f\left( x \right)\) liên tục trên \(\left[ 0;3 \right]\) và \(\int\limits_{0}^{1}{f\left( x \right)dx}=2,\,\,\int\limits_{0}^{3}{f\left( x \right)dx=8}\). Khi đó \(\in...

* Hướng dẫn giải

Vì \(\left| {2x - 1} \right| = \left\{ \begin{array}{l} - 2x + 1,\,\forall x \le \frac{1}{2}\\ 2x - 1,\,\forall x \ge \frac{1}{2} \end{array} \right.\) nên \(\int\limits_{ - 1}^1 {f\left( {\left| {2x - 1} \right|} \right)} dx = \int\limits_{ - 1}^{\frac{1}{2}} {f\left( { - 2x + 1} \right)} dx + \int\limits_{\frac{1}{2}}^1 {f\left( {2x - 1} \right)} dx = {I_1} + {I_2}\)

\({I_1} = \int\limits_{ - 1}^{\frac{1}{2}} {f\left( { - 2x + 1} \right)} dx =  - \frac{1}{2}\int\limits_{ - 3}^0 {f\left( t \right)dt}  = \frac{1}{2}\int\limits_0^3 {f\left( t \right)dt} \) với t =  - 2x + 1

\({I_2} = \int\limits_{\frac{1}{2}}^1 {f\left( {2x - 1} \right)} dx = \frac{1}{2}\int\limits_0^1 {f\left( t \right)dt} \) với t = 2x + 1

Vậy \(\int\limits_{ - 1}^1 {f\left( {\left| {2x - 1} \right|} \right)} dx = \frac{1}{2}\int\limits_0^3 {f\left( t \right)dt}  + \frac{1}{2}\int\limits_0^1 {f\left( t \right)dt}  = 4 + 1 = 5.\)