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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm liên tục trên $\mathbb{R}$. Biết $\int\limits_{1}^{\text{e}}{\dfrac{1-f\left( \ln x \right)}{x}\text{d}x}=2$ và $f\left( 1 \right)=\dfrac{1}{3}$. Tích phân $\int\limits_{0}^{1}{x{f}'\left( x \right)\text{d}x}$ bằng
A. $-\dfrac{2}{3}$.
B. $\dfrac{2\text{e}}{3}$.
C. $\dfrac{4}{3}$.
D. $\dfrac{2}{3}$.
A. $-\dfrac{2}{3}$.
B. $\dfrac{2\text{e}}{3}$.
C. $\dfrac{4}{3}$.
D. $\dfrac{2}{3}$.
Ta có $\int\limits_{1}^{\text{e}}{\dfrac{1-f\left( \ln x \right)}{x}\text{d}x}=\int\limits_{1}^{\text{e}}{\dfrac{1}{x}\text{d}x}-\int\limits_{1}^{\text{e}}{\dfrac{f\left( \ln x \right)}{x}\text{d}x}=\left. \ln x \right|_{1}^{\text{e}}-\int\limits_{1}^{\text{e}}{f\left( \ln x \right)\text{d}\left( \ln x \right)}=1-\int\limits_{0}^{\text{1}}{f\left( x \right)\text{d}x}$.
Mặt khác $\int\limits_{1}^{\text{e}}{\dfrac{1-f\left( \ln x \right)}{x}\text{d}x}=2$ suy ra $\int\limits_{0}^{\text{1}}{f\left( x \right)\text{d}x}=-1$.
Do đó $\int\limits_{0}^{1}{x{f}'\left( x \right)\text{d}x}=\int\limits_{0}^{1}{x\text{d}\left( f\left( x \right) \right)}=\left. xf\left( x \right) \right|_{0}^{1}-\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=f\left( 1 \right)+1=\dfrac{4}{3}$.
Mặt khác $\int\limits_{1}^{\text{e}}{\dfrac{1-f\left( \ln x \right)}{x}\text{d}x}=2$ suy ra $\int\limits_{0}^{\text{1}}{f\left( x \right)\text{d}x}=-1$.
Do đó $\int\limits_{0}^{1}{x{f}'\left( x \right)\text{d}x}=\int\limits_{0}^{1}{x\text{d}\left( f\left( x \right) \right)}=\left. xf\left( x \right) \right|_{0}^{1}-\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=f\left( 1 \right)+1=\dfrac{4}{3}$.
Đáp án C.