Cho hàm số y=f(x) liên tục trên đoạn

* Hướng dẫn giải

Ta có:\[\mathop \smallint \limits_1^3 \left( {4 - x} \right)f\left( x \right)dx = 4\mathop \smallint \limits_1^3 f\left( x \right)dx - \mathop \smallint \limits_1^3 xf\left( x \right)dx\]

Đặt\[t = 4 - x \Rightarrow dt = - dx\]

Đổi cận:\(\left\{ {\begin{array}{*{20}{c}}{x = 1 \Rightarrow t = 3}\\{x = 3 \Rightarrow t = 1}\end{array}} \right.\) khi đó ta có:

\[\mathop \smallint \limits_1^3 \left( {4 - x} \right)f\left( x \right)dx = - \mathop \smallint \limits_3^1 tf\left( {4 - t} \right)dt = \mathop \smallint \limits_1^3 tf\left( {4 - t} \right)dt = \mathop \smallint \limits_1^3 tf\left( t \right)dt = \mathop \smallint \limits_1^3 xf\left( x \right)dx\]

\[\begin{array}{*{20}{l}}{ \Rightarrow \mathop \smallint \limits_1^3 xf\left( x \right)dx = 4\mathop \smallint \limits_1^3 f\left( x \right)dx - \mathop \smallint \limits_1^3 xf\left( x \right)dx}\\{ \Leftrightarrow 2\mathop \smallint \limits_1^3 f\left( x \right)dx = \mathop \smallint \limits_1^3 xf\left( x \right)dx = - 2}\end{array}\]

Đáp án cần chọn là: C