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Câu hỏi: Cho hàm số $f\left( x \right)$ và $g\left( x \right)$ liên tục trên $\mathbb{R}$. Gọi $F\left( x \right),~G\left( x \right)$ là hai nguyên hàm của $f\left( x \right)$ và $g\left( x \right)$ trên $\mathbb{R}$ thỏa mãn $F\left( 6 \right)+3G\left( 0 \right)=3$ và $F\left( 0 \right)+3G\left( 2 \right)=1$. Khi đó $\mathop{\int }_{0}^{2}\left[ f\left( 3x \right)-g\left( x \right) \right]dx$ bằng
A. $1.$
B. $3.$
C. $\dfrac{2}{3}.$
D. $\dfrac{4}{3}.$
A. $1.$
B. $3.$
C. $\dfrac{2}{3}.$
D. $\dfrac{4}{3}.$
Ta có: $\underset{0}{\overset{2}{\mathop \int }} \left[ f\left( 3x \right)-g\left( x \right) \right]dx=\mathop{\int }_{0}^{2}f\left( 3x \right)dx-\mathop{\int }_{0}^{2}g\left( x \right)dx=\dfrac{1}{3}\mathop{\int }_{0}^{6}f\left( x \right)dx-\mathop{\int }_{0}^{2}g\left( x \right)dx$
$=\dfrac{1}{3}\left[ F\left( 6 \right)-F\left( 0 \right) \right]-\left[ G\left( 2 \right)-G\left( 0 \right) \right]=\dfrac{1}{3}\left[ F\left( 6 \right)+3G\left( 0 \right) \right]-\dfrac{1}{3}\left[ F\left( 0 \right)-3G\left( 2 \right) \right]$
$=\dfrac{1}{3}.3-\dfrac{1}{3}.1=\dfrac{2}{3}$.
$=\dfrac{1}{3}\left[ F\left( 6 \right)-F\left( 0 \right) \right]-\left[ G\left( 2 \right)-G\left( 0 \right) \right]=\dfrac{1}{3}\left[ F\left( 6 \right)+3G\left( 0 \right) \right]-\dfrac{1}{3}\left[ F\left( 0 \right)-3G\left( 2 \right) \right]$
$=\dfrac{1}{3}.3-\dfrac{1}{3}.1=\dfrac{2}{3}$.
Đáp án C.
