Cho hàm số sau \(y = f\left( x \right)\) liên tục trên \(\mathbb{R}\) thỏa mãn \(\int\limits_0^7 {f\left( x \right)dx} {\rm{\;}} = 10\) và \(\int\limits_0^3 {f\left( x \right)dx} {...

* Hướng dẫn giải

Đặt \(t = 3 - 2x \Rightarrow dt = {\rm{\;}} - 2dx\). Đổi cận \(\left\{ {\begin{array}{*{20}{l}}{x = {\rm{\;}} - 2 \Rightarrow t = 7}\\{x = 3 \Rightarrow t = {\rm{\;}} - 3}\end{array}} \right.\).

Khi đó ta có:

\(\begin{array}{*{20}{l}}{I =  - \dfrac{1}{2}\int\limits_7^{ - 3} {f\left( {\left| t \right|} \right)dt}  = \dfrac{1}{2}\int\limits_{ - 3}^7 {f\left( {\left| t \right|} \right)dt} }\\{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu}  = \dfrac{1}{2}\left( {\int\limits_{ - 3}^0 {f\left( { - t} \right)dt}  + \int\limits_0^7 {f\left( t \right)dt} } \right)}\\{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu}  = \dfrac{1}{2}\left( { - \int\limits_3^0 {f\left( x \right)dx}  + \int\limits_0^7 {f\left( x \right)dx} } \right)}\\{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu}  = \dfrac{1}{2}\left( {\int\limits_0^3 {f\left( x \right)dx}  + \int\limits_0^7 {f\left( x \right)dx} } \right)}\\{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu}  = \dfrac{1}{2}\left( {6 + 10} \right) = 8}\end{array}\)

Chọn D.