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Câu hỏi: Biết $\int\limits_{-1}^{2}{f\left( x \right)}\text{d}x=2$ và $\int\limits_{1}^{2}{f\left( x \right)}\text{d}x=5$, tích phân $\int\limits_{1}^{2}{f\left( 3-2x \right)}\text{d}x$ bằng
A. $3$.
B. $\dfrac{-3}{2}$.
C. $-3$.
D. $\dfrac{3}{2}$.
A. $3$.
B. $\dfrac{-3}{2}$.
C. $-3$.
D. $\dfrac{3}{2}$.
Xét $\int\limits_{1}^{2}{f\left( 3-2x \right)}\text{dx}$
Đặt $t=3-2x$, $\text{d}t=-2\text{d}x$.
Đổi cận $\left\{ \begin{aligned}
& x=2 \\
& x=1 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& x=-1 \\
& x=1. \\
\end{aligned} \right.$
Khi đó $\int\limits_{1}^{2}{f\left( 3-2x \right)}\text{d}x\text{=}\int\limits_{1}^{-1}{f\left( t \right)}.\dfrac{-\text{d}t}{2}=\dfrac{1}{2}\int\limits_{-1}^{1}{f\left( t \right)}\text{d}t=\dfrac{1}{2}\int\limits_{-1}^{1}{f\left( x \right)}\text{d}x$.
Có $\int\limits_{-1}^{2}{f\left( x \right)}\text{d}x=\int\limits_{-1}^{1}{f\left( x \right)}\text{d}x+\int\limits_{1}^{2}{f\left( x \right)}\text{d}x\Rightarrow \int\limits_{-1}^{1}{f\left( x \right)}\text{d}x=2-5=-3$
Vậy $\int\limits_{1}^{2}{f\left( 3-2x \right)}\text{d}x=\dfrac{1}{2}\int\limits_{-1}^{1}{f\left( x \right)}\text{d}x=\dfrac{-3}{2}$.
Đặt $t=3-2x$, $\text{d}t=-2\text{d}x$.
Đổi cận $\left\{ \begin{aligned}
& x=2 \\
& x=1 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& x=-1 \\
& x=1. \\
\end{aligned} \right.$
Khi đó $\int\limits_{1}^{2}{f\left( 3-2x \right)}\text{d}x\text{=}\int\limits_{1}^{-1}{f\left( t \right)}.\dfrac{-\text{d}t}{2}=\dfrac{1}{2}\int\limits_{-1}^{1}{f\left( t \right)}\text{d}t=\dfrac{1}{2}\int\limits_{-1}^{1}{f\left( x \right)}\text{d}x$.
Có $\int\limits_{-1}^{2}{f\left( x \right)}\text{d}x=\int\limits_{-1}^{1}{f\left( x \right)}\text{d}x+\int\limits_{1}^{2}{f\left( x \right)}\text{d}x\Rightarrow \int\limits_{-1}^{1}{f\left( x \right)}\text{d}x=2-5=-3$
Vậy $\int\limits_{1}^{2}{f\left( 3-2x \right)}\text{d}x=\dfrac{1}{2}\int\limits_{-1}^{1}{f\left( x \right)}\text{d}x=\dfrac{-3}{2}$.
Đáp án B.