Cho hàm số f(x) liên tục trên \(\mathbb{R}\) và có \[I = \mathop \smallint \limits_0^1 f\left( x \right)dx = 3\mathop \smallint \limits_0^3 f\left( x \right)dt = 6\]. Giá trị của \[\mathop \smallint \limits_{ - 1}^1 f\left( {\left| {2x - 1} \right|} \right)dx\] bằng:

Ta có\[\mathop \smallint \limits_{ - 1}^1 f\left( {\left| {2x - 1} \right|} \right)dx = \mathop \smallint \limits_{ - 1}^{\frac{1}{2}} f\left( {1 - 2x} \right)dx + \mathop \smallint \limits_{\frac{1}{2}}^1 f\left( {2x - 1} \right)dx\]

\[ \Rightarrow I = - \frac{1}{2}\mathop \smallint \limits_{ - 1}^{\frac{1}{2}} f\left( {1 - 2x} \right)d\left( {1 - 2x} \right) + \frac{1}{2}\mathop \smallint \limits_{\frac{1}{2}}^1 f\left( {2x - 1} \right)d\left( {2x - 1} \right)\]

\[\begin{array}{*{20}{l}}{ \Leftrightarrow I = - \frac{1}{2}\mathop \smallint \limits_3^0 f\left( t \right)dt + \frac{1}{2}\mathop \smallint \limits_0^1 f\left( t \right)dt}\\{ \Leftrightarrow I = \frac{1}{2}\mathop \smallint \limits_0^3 f\left( t \right)dt + \frac{1}{2}\mathop \smallint \limits_0^1 f\left( t \right)dt = \frac{1}{2}\left( {2 + 6} \right) = 4}\end{array}\]

Đáp án cần chọn là: B