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Câu hỏi: Cho hàm số $y=f\left( x \right)$ liên tục trên $\mathbb{R}$, gọi $F\left( x \right), G\left( x \right)$ là hai nguyên hàm của $f\left( x \right)$ trên $\mathbb{R}$ thỏa mãn $F\left( 8 \right)+G\left( 8 \right)=15$ và $F\left( 2 \right)+G\left( 2 \right)=3$. Khi đó $\int\limits_{1}^{3}{f\left( 3x-1 \right)\text{d}x}$ bằng
A. $\dfrac{1}{3}$.
B. $3$.
C. $1$.
D. $2$.
A. $\dfrac{1}{3}$.
B. $3$.
C. $1$.
D. $2$.
Ta có $I=\int\limits_{1}^{3}{f\left( 3x-1 \right)\text{d}x}=\dfrac{1}{3}\int\limits_{1}^{3}{f\left( 3x-1 \right)\text{d}\left( 3x-1 \right)}=\dfrac{1}{3}\int\limits_{2}^{8}{f\left( x \right)\text{d}x}$.
$\left. \begin{aligned}
& 3I=\left. F\left( x \right) \right|_{2}^{8}=F\left( 8 \right)-F\left( 2 \right) \\
& 3I=\left. G\left( x \right) \right|_{2}^{8}=G\left( 8 \right)-G\left( 2 \right) \\
\end{aligned} \right\}\Rightarrow 6I=\left[ F\left( 8 \right)+G\left( 8 \right) \right]-\left[ F\left( 2 \right)+G\left( 2 \right) \right]=12\Rightarrow I=2$.
$\left. \begin{aligned}
& 3I=\left. F\left( x \right) \right|_{2}^{8}=F\left( 8 \right)-F\left( 2 \right) \\
& 3I=\left. G\left( x \right) \right|_{2}^{8}=G\left( 8 \right)-G\left( 2 \right) \\
\end{aligned} \right\}\Rightarrow 6I=\left[ F\left( 8 \right)+G\left( 8 \right) \right]-\left[ F\left( 2 \right)+G\left( 2 \right) \right]=12\Rightarrow I=2$.
Đáp án D.
