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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên $\mathbb{R}$. Gọi $F\left( x \right), G\left( x \right), H\left( x \right)$ là ba nguyên hàm của $f\left( x \right)$ trên $\mathbb{R}$ thỏa mãn $F\left( 3 \right)+G\left( 3 \right)+H\left( 3 \right)=4$ và $F\left( 0 \right)+G\left( 0 \right)+H\left( 0 \right)=1$. Khi đó $\int\limits_{0}^{1}{f\left( 3x \right)\text{d}x}$ bằng
A. $1$.
B. $3$.
C. $\dfrac{5}{3}$.
D. $\dfrac{1}{3}$.
A. $1$.
B. $3$.
C. $\dfrac{5}{3}$.
D. $\dfrac{1}{3}$.
Ta có: $F\left( 3 \right)+G\left( 3 \right)+H\left( 3 \right)-F\left( 0 \right)-G\left( 0 \right)-H\left( 0 \right)=3$
$\Leftrightarrow F\left( 3 \right)-F\left( 0 \right)+G\left( 3 \right)-G\left( 0 \right)+H\left( 3 \right)-H\left( 0 \right)=3$
$\Leftrightarrow \int\limits_{0}^{3}{f\left( x \right)\text{d}x}+\int\limits_{0}^{3}{f\left( x \right)\text{d}x}+\int\limits_{0}^{3}{f\left( x \right)\text{d}x}=3$ $\Leftrightarrow \int\limits_{0}^{3}{f\left( x \right)\text{d}x}=1$
Lại có: $\int\limits_{0}^{1}{f\left( 3x \right)\text{d}x}$ $=\dfrac{1}{3}\int\limits_{0}^{3}{f\left( t \right)\text{d}t}$ $=\dfrac{1}{3}\int\limits_{0}^{3}{f\left( x \right)\text{d}x}$.
Vậy: $\int\limits_{0}^{1}{f\left( 3x \right)\text{d}x}=\dfrac{1}{3}$.
$\Leftrightarrow F\left( 3 \right)-F\left( 0 \right)+G\left( 3 \right)-G\left( 0 \right)+H\left( 3 \right)-H\left( 0 \right)=3$
$\Leftrightarrow \int\limits_{0}^{3}{f\left( x \right)\text{d}x}+\int\limits_{0}^{3}{f\left( x \right)\text{d}x}+\int\limits_{0}^{3}{f\left( x \right)\text{d}x}=3$ $\Leftrightarrow \int\limits_{0}^{3}{f\left( x \right)\text{d}x}=1$
Lại có: $\int\limits_{0}^{1}{f\left( 3x \right)\text{d}x}$ $=\dfrac{1}{3}\int\limits_{0}^{3}{f\left( t \right)\text{d}t}$ $=\dfrac{1}{3}\int\limits_{0}^{3}{f\left( x \right)\text{d}x}$.
Vậy: $\int\limits_{0}^{1}{f\left( 3x \right)\text{d}x}=\dfrac{1}{3}$.
Đáp án D.