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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm trên $\mathbb{R}$ thỏa mãn $f\left( 1 \right)=1; f\left( 3x \right)-{{x}^{2}}f\left( {{x}^{3}} \right)=4{{x}^{3}}+2x+1, \left( \forall x\in \mathbb{R} \right)$. Khi đó $\int\limits_{1}^{3}{x{f}'\left( x \right)dx}$ bằng:
A. $14$
B. $-1$
C. $5$
D. $6$
A. $14$
B. $-1$
C. $5$
D. $6$
Ta có $f\left( 1 \right)=1; f\left( 3 \right)=8.$
Lấy tích phân hai về trên đoạn $\left[ 0;1 \right]$ ta được
$\begin{aligned}
& \int\limits_{0}^{1}{\left( f\left( 3x \right)-{{x}^{2}}f\left( {{x}^{3}} \right) \right)}\text{d}x=\int\limits_{0}^{1}{\left( 4{{x}^{3}}+2x+1 \right)}\text{d}x\Leftrightarrow \int\limits_{0}^{1}{f\left( 3x \right)}\text{d}x-\int\limits_{0}^{1}{{{x}^{2}}f\left( {{x}^{3}} \right)}\text{d}x=3 \\
& \Leftrightarrow \dfrac{1}{3}\int\limits_{0}^{1}{f\left( 3x \right)\text{d}\left( 3x \right)-}\dfrac{1}{3}\int\limits_{0}^{1}{f\left( {{x}^{3}} \right)}\text{d}\left( {{x}^{3}} \right)=3 \\
& \Leftrightarrow \int\limits_{0}^{3}{f\left( x \right)\text{d}x-}\int\limits_{0}^{1}{f\left( x \right)}\text{d}x=9 \\
& \Leftrightarrow \int\limits_{1}^{3}{f\left( x \right)\text{d}x=9} \\
\end{aligned}$
Do đó $\int\limits_{1}^{3}{x{f}'\left( x \right)\text{d}x}=\int\limits_{1}^{3}{x\text{d}\left( f\left( x \right) \right)}=\left. \left( xf\left( x \right) \right) \right|_{1}^{3}-\int\limits_{1}^{3}{f\left( x \right)\text{d}x=3f\left( 3 \right)-1f\left( 1 \right)-9=3.8-1-9=14}$.
Lấy tích phân hai về trên đoạn $\left[ 0;1 \right]$ ta được
$\begin{aligned}
& \int\limits_{0}^{1}{\left( f\left( 3x \right)-{{x}^{2}}f\left( {{x}^{3}} \right) \right)}\text{d}x=\int\limits_{0}^{1}{\left( 4{{x}^{3}}+2x+1 \right)}\text{d}x\Leftrightarrow \int\limits_{0}^{1}{f\left( 3x \right)}\text{d}x-\int\limits_{0}^{1}{{{x}^{2}}f\left( {{x}^{3}} \right)}\text{d}x=3 \\
& \Leftrightarrow \dfrac{1}{3}\int\limits_{0}^{1}{f\left( 3x \right)\text{d}\left( 3x \right)-}\dfrac{1}{3}\int\limits_{0}^{1}{f\left( {{x}^{3}} \right)}\text{d}\left( {{x}^{3}} \right)=3 \\
& \Leftrightarrow \int\limits_{0}^{3}{f\left( x \right)\text{d}x-}\int\limits_{0}^{1}{f\left( x \right)}\text{d}x=9 \\
& \Leftrightarrow \int\limits_{1}^{3}{f\left( x \right)\text{d}x=9} \\
\end{aligned}$
Do đó $\int\limits_{1}^{3}{x{f}'\left( x \right)\text{d}x}=\int\limits_{1}^{3}{x\text{d}\left( f\left( x \right) \right)}=\left. \left( xf\left( x \right) \right) \right|_{1}^{3}-\int\limits_{1}^{3}{f\left( x \right)\text{d}x=3f\left( 3 \right)-1f\left( 1 \right)-9=3.8-1-9=14}$.
Đáp án A.
