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Câu hỏi: Cho hàm số $y=f\left( x \right)$ liên tục trên $\mathbb{R}$ thỏa mãn $2f\left( 3-x \right)+f\left( x \right)=8x-6$. Khi đó, $\int\limits_{0}^{1}{f\left( x \right)dx}$ bằng:
A. 10
B. 6
C. 8
D. 14
A. 10
B. 6
C. 8
D. 14
Ta có: $2f\left( 3-x \right)+f\left( x \right)=8x-6$.
Đặt $\begin{aligned}
& t=3-x\Rightarrow 2f\left( t \right)+f\left( 3-t \right)=8\left( 3-t \right)-6\Rightarrow f\left( 3-x \right)+2f\left( x \right)=-8x+18 \\
& \Leftrightarrow \left\{ \begin{aligned}
& 2f\left( 3-x \right)+f\left( x \right)=8x-6 \\
& f\left( 3-x \right)+2f\left( x \right)=-8x+18 \\
\end{aligned} \right.\Rightarrow f\left( x \right)=-8x+14 \\
\end{aligned}$
Ta có: $\int\limits_{0}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{\left( -8x+14 \right)dx=10}$.
Đặt $\begin{aligned}
& t=3-x\Rightarrow 2f\left( t \right)+f\left( 3-t \right)=8\left( 3-t \right)-6\Rightarrow f\left( 3-x \right)+2f\left( x \right)=-8x+18 \\
& \Leftrightarrow \left\{ \begin{aligned}
& 2f\left( 3-x \right)+f\left( x \right)=8x-6 \\
& f\left( 3-x \right)+2f\left( x \right)=-8x+18 \\
\end{aligned} \right.\Rightarrow f\left( x \right)=-8x+14 \\
\end{aligned}$
Ta có: $\int\limits_{0}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{\left( -8x+14 \right)dx=10}$.
Đáp án A.