Câu hỏi 40

Ta có:  \(\int\limits_{ - 1}^1 {f\left( {\left| {2x - 1} \right|} \right)} dx = \int\limits_{ - 1}^{\dfrac{1}{2}} {f\left( { - 2x - 1} \right)dx}  + \int\limits_{\dfrac{1}{2}}^1 {f\left( {2x - 1} \right)dx} \)

Đặt \({I_1} = \int\limits_{ - 1}^{\dfrac{1}{2}} {f\left( { - 2x - 1} \right)dx} ;\,\,\,\,{I_2} = \int\limits_{\dfrac{1}{2}}^2 {f\left( {2x - 1} \right)dx} \)

Tính \({I_1} = \int\limits_{ - 1}^{\dfrac{1}{2}} {f\left( { - 2x - 1} \right)dx} \)

Đặt \( - 2x - 1 = t \Rightarrow dt =  - 2dx \Rightarrow dx =  - \dfrac{1}{2}dt\)

Đổi cận: \(\left\{ \begin{array}{l}x =  - 1 \Rightarrow t = 3\\x = \dfrac{1}{2} \Rightarrow t = 0\end{array} \right..\)

\( \Rightarrow {I_1} =  - \dfrac{1}{2}\int\limits_3^0 {f\left( t \right)dt}  = \dfrac{1}{2}\int\limits_0^3 {f\left( t \right)dt}  = \dfrac{1}{2}\int\limits_0^3 {f\left( x \right)dx}  = \dfrac{1}{2}.6 = 3.\)

Tính \({I_2} = \int\limits_{\dfrac{1}{2}}^1 {f\left( {2x + 1} \right)dx} \)

Đặt \(2x - 1 = t \Rightarrow dt = 2dx \Rightarrow dx = \dfrac{1}{2}dt\)

Đổi cận: \(\left\{ \begin{array}{l}x = 1 \Rightarrow t = 1\\x = \dfrac{1}{2} \Rightarrow t = 0\end{array} \right..\)

\(\begin{array}{l} \Rightarrow {I_2} = \dfrac{1}{2}\int\limits_0^1 {f\left( t \right)dt}  = \dfrac{1}{2}\int\limits_0^1 {f\left( x \right)dx}  = \dfrac{1}{2}.2 = 1.\\ \Rightarrow I = {I_1} + {I_2} = 3 + 1 = 4.\end{array}\)

Chọn  B.