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