Cho hàm số y=f(x) liên tục trên

* Hướng dẫn giải

Xét  \[\mathop \smallint \limits_{ - 1}^{\frac{3}{2}} f\left( {2x + 1} \right)dx\] Đặt\[2x + 1 = t \Leftrightarrow 2dx = dt \Leftrightarrow dx = \frac{{dt}}{2}\]

Đổi cận:\(\left\{ {\begin{array}{*{20}{c}}{x = - 1 \Rightarrow t = - 1}\\{x = \frac{3}{2} \Rightarrow t = 4}\end{array}} \right.\)

Khi đó ta có\[\mathop \smallint \limits_{ - 1}^{\frac{3}{2}} f\left( {2x + 1} \right)dx = \frac{1}{2}\mathop \smallint \limits_{ - 1}^4 f\left( t \right)dt = \frac{1}{2}\mathop \smallint \limits_{ - 1}^4 f\left( x \right)dx\]

\[ = \frac{1}{2}\left( {\mathop \smallint \limits_{ - 1}^1 f\left( x \right)dx + \mathop \smallint \limits_1^4 f\left( x \right)dx} \right)\]

Từ hình vẽ ta có\[\mathop \smallint \limits_{ - 1}^1 f\left( x \right)dx = \frac{{16}}{3};\,\mathop \smallint \limits_1^4 f\left( x \right)dx = - \frac{{63}}{4}\]

Nên\[\mathop \smallint \limits_{ - 1}^{\frac{3}{2}} f\left( {2x + 1} \right)dx = \frac{1}{2}\left( {\mathop \smallint \limits_{ - 1}^1 f\left( x \right)dx + \mathop \smallint \limits_1^4 f\left( x \right)dx} \right) = \frac{1}{2}\left( {\frac{{16}}{3} - \frac{{63}}{4}} \right) = - \frac{{125}}{{24}}\]

Đáp án cần chọn là: C