Cho hàm số \(f\left( x \right)\) liên tục trên \(\mathbb{R}\) và \(\int\limits_0^2 {f\left( x \right){\rm{d}}x}  = 9;\,\int\limits_0^4 {f\left( x \right){\rm{d}}x}  = 12\). Giá trị của \(\int\limits_{ - 1}^1 {f\left( {\left| {3x - 1} \right|} \right){\rm{d}}x} \) bằng

Đặt \(t = 3x - 1\)\( \Rightarrow \)\(\int\limits_{ - 1}^1 {f\left( {\left| {3x - 1} \right|} \right){\rm{d}}x}  \)\( = \int\limits_{ - 4}^2 {f\left( {\left| t \right|} \right).\dfrac{{{\rm{d}}t}}{3}= \dfrac{1}{3}\int\limits_{ - 4}^2 {f\left( {\left| t \right|} \right){\rm{d}}t} } \) \( = \dfrac{1}{3}\left( {\int_{ - 4}^0 {f\left( { - t} \right){\rm{d}}t}  + \int\limits_0^2 {f\left( t \right){\rm{d}}t} } \right)\)

\( = \dfrac{1}{3}\left( {\int\limits_0^4 {f\left( t \right){\rm{d}}t}  + \int\limits_0^2 {f\left( t \right){\rm{d}}t} } \right)\) \( = \dfrac{1}{3}\left( {\int\limits_0^4 {f\left( x \right){\rm{dx}}}  + \int\limits_0^2 {f\left( x \right){\rm{dx}}} } \right)\)\( = \dfrac{1}{3}\left( {9 + 12} \right) = 7\).