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Câu hỏi: Cho số phức $y=f\left( x \right)$ liên tục trên $\mathbb{R}$. Gọi $F\left( x \right)$ là một nguyên hàm của hàm số $f\left( x \right)-x$, $G\left( x \right)$ là một nguyên hàm của hàm số $f\left( x \right)+x$ trên tập hợp $\mathbb{R}$ thỏa mãn $F\left( 4 \right)+G\left( 4 \right)=5$ và $F\left( 1 \right)+G\left( 1 \right)=-1$. Giá trị của $\int\limits_{0}^{1}{f\left( 3x+1 \right)\text{d}x}$ bằng
A. $\dfrac{1}{3}$.
B. $6$.
C. $2$.
D. $1$.
A. $\dfrac{1}{3}$.
B. $6$.
C. $2$.
D. $1$.
Ta có
$\begin{aligned}
& \left\{ \begin{aligned}
& F\left( 4 \right)+G\left( 4 \right)=5 \\
& F\left( 1 \right)+G\left( 1 \right)=-1 \\
\end{aligned} \right.\Rightarrow F\left( 4 \right)-F\left( 1 \right)+G\left( 4 \right)-G\left( 1 \right)=6 \\
& \Leftrightarrow \int\limits_{1}^{4}{\left[ f\left( x \right)-x \right]\text{d}x}+\int\limits_{1}^{4}{\left[ f\left( x \right)+x \right]\text{d}x}=6 \\
& \Leftrightarrow \int\limits_{1}^{4}{f\left( x \right)\text{d}x}-\int\limits_{1}^{4}{x\text{d}x}+\int\limits_{1}^{4}{f\left( x \right)\text{d}x}+\int\limits_{1}^{4}{x\text{d}x}=6 \\
& \Leftrightarrow 2\int\limits_{1}^{4}{f\left( x \right)\text{d}x}=6\Rightarrow \int\limits_{1}^{4}{f\left( x \right)\text{d}x}=3 \\
\end{aligned}$.
Xét $I=\int\limits_{0}^{1}{f\left( 3x+1 \right)\text{d}x}$, đặt $t=3x+1\Rightarrow dt=3dx\Rightarrow dx=\dfrac{dt}{3};x=0\Rightarrow t=1,x=1\Rightarrow t=4$.
Suy ra $I=\dfrac{1}{3}\int\limits_{1}^{4}{f\left( t \right)\text{dt}}=1$.
$\begin{aligned}
& \left\{ \begin{aligned}
& F\left( 4 \right)+G\left( 4 \right)=5 \\
& F\left( 1 \right)+G\left( 1 \right)=-1 \\
\end{aligned} \right.\Rightarrow F\left( 4 \right)-F\left( 1 \right)+G\left( 4 \right)-G\left( 1 \right)=6 \\
& \Leftrightarrow \int\limits_{1}^{4}{\left[ f\left( x \right)-x \right]\text{d}x}+\int\limits_{1}^{4}{\left[ f\left( x \right)+x \right]\text{d}x}=6 \\
& \Leftrightarrow \int\limits_{1}^{4}{f\left( x \right)\text{d}x}-\int\limits_{1}^{4}{x\text{d}x}+\int\limits_{1}^{4}{f\left( x \right)\text{d}x}+\int\limits_{1}^{4}{x\text{d}x}=6 \\
& \Leftrightarrow 2\int\limits_{1}^{4}{f\left( x \right)\text{d}x}=6\Rightarrow \int\limits_{1}^{4}{f\left( x \right)\text{d}x}=3 \\
\end{aligned}$.
Xét $I=\int\limits_{0}^{1}{f\left( 3x+1 \right)\text{d}x}$, đặt $t=3x+1\Rightarrow dt=3dx\Rightarrow dx=\dfrac{dt}{3};x=0\Rightarrow t=1,x=1\Rightarrow t=4$.
Suy ra $I=\dfrac{1}{3}\int\limits_{1}^{4}{f\left( t \right)\text{dt}}=1$.
Đáp án D.