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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên $\mathbb{R}.$ Gọi $F(x),G(x)$ là hai nguyên hàm của $f\left( x \right)$ trên $\mathbb{R}$ thỏa mãn $F\left( 6 \right)+G\left( 6 \right)=6$ và $F\left( 0 \right)+G\left( 0 \right)=2$. Khi đó $\int\limits_{0}^{2}{f\left( 3x \right) \text{d}x}$ bằng
A. $\dfrac{2}{3}$.
B. $\dfrac{1}{3}$.
C. $\dfrac{1}{4}$.
D. $\dfrac{3}{4}$.
A. $\dfrac{2}{3}$.
B. $\dfrac{1}{3}$.
C. $\dfrac{1}{4}$.
D. $\dfrac{3}{4}$.
Đặt $t=3x$. Khi đó $I=\int\limits_{0}^{2}{f\left( 3x \right)\text{d}x}=\int\limits_{0}^{6}{f\left( t \right)\dfrac{\text{d}t}{3}}=\dfrac{1}{3}\int\limits_{0}^{6}{f\left( x \right)\text{d}x}=\dfrac{1}{3}\left[ F\left( 6 \right)-F\left( 0 \right) \right]$
Ta có: $G\left( x \right)=F\left( x \right)+C$.
Theo đề: $\left\{ \begin{aligned}
& F\left( 6 \right)+G\left( 6 \right)=6 \\
& F\left( 0 \right)+G\left( 0 \right)=2 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& 2F\left( 6 \right)+C=6 \\
& 2F\left( 0 \right)+C=2 \\
\end{aligned} \right.\Rightarrow F\left( 6 \right)-F\left( 0 \right)=2$.
Vậy, $I=\dfrac{2}{3}$.
Ta có: $G\left( x \right)=F\left( x \right)+C$.
Theo đề: $\left\{ \begin{aligned}
& F\left( 6 \right)+G\left( 6 \right)=6 \\
& F\left( 0 \right)+G\left( 0 \right)=2 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& 2F\left( 6 \right)+C=6 \\
& 2F\left( 0 \right)+C=2 \\
\end{aligned} \right.\Rightarrow F\left( 6 \right)-F\left( 0 \right)=2$.
Vậy, $I=\dfrac{2}{3}$.
Đáp án A.
