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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm liên tục trên $\mathbb{R}$. Biết $f\left( 3 \right)=1$ và $\int\limits_{0}^{1}{xf\left( 3x \right)\text{d}x=1}$, khi đó $\int\limits_{0}^{3}{{{x}^{2}}{f}'\left( x \right)\text{d}x}$ bằng
A. $-9$.
B. $\dfrac{25}{3}$.
C. $3$.
D. $7$.
A. $-9$.
B. $\dfrac{25}{3}$.
C. $3$.
D. $7$.
Đặt $t=3x\Rightarrow \text{d}t=3\text{d}x\Rightarrow \text{d}x=\dfrac{1}{3}\text{d}t$.
Suy ra $1=\int\limits_{0}^{1}{xf\left( 3x \right)}\text{d}x=\dfrac{1}{9}\int\limits_{0}^{3}{tf\left( t \right)}\text{d}t\Leftrightarrow \int\limits_{0}^{3}{tf\left( t \right)}dt=9$.
Đặt $\left\{ \begin{matrix}
u=f\left( t \right) \\
\text{d}v=t\text{d}t \\
\end{matrix} \right.\Rightarrow \left\{ \begin{matrix}
\text{d}u={f}'\left( t \right)\text{d}t \\
v=\dfrac{{{t}^{2}}}{2} \\
\end{matrix} \right.$.
$\Rightarrow \int\limits_{0}^{3}{tf\left( t \right)}\text{d}t=\left. \dfrac{{{t}^{2}}}{2}f\left( t \right) \right|_{0}^{3}-\int\limits_{0}^{3}{\dfrac{{{t}^{2}}}{2}{f}'\left( t \right)\text{d}t}=\dfrac{9}{2}f\left( 3 \right)-\dfrac{1}{2}\int\limits_{0}^{3}{{{t}^{2}}{{f}^{'}}\left( t \right)\text{d}t}$.
$\Leftrightarrow 9=\dfrac{9}{2}-\dfrac{1}{2}\int\limits_{0}^{3}{{{t}^{2}}{f}'\left( t \right)\text{d}t}\Leftrightarrow \int\limits_{0}^{3}{{{t}^{2}}{f}'\left( t \right)\text{d}t}=-9$.
Vậy $\int\limits_{0}^{3}{{{x}^{2}}{f}'\left( x \right)\text{d}x}=-9$.
Suy ra $1=\int\limits_{0}^{1}{xf\left( 3x \right)}\text{d}x=\dfrac{1}{9}\int\limits_{0}^{3}{tf\left( t \right)}\text{d}t\Leftrightarrow \int\limits_{0}^{3}{tf\left( t \right)}dt=9$.
Đặt $\left\{ \begin{matrix}
u=f\left( t \right) \\
\text{d}v=t\text{d}t \\
\end{matrix} \right.\Rightarrow \left\{ \begin{matrix}
\text{d}u={f}'\left( t \right)\text{d}t \\
v=\dfrac{{{t}^{2}}}{2} \\
\end{matrix} \right.$.
$\Rightarrow \int\limits_{0}^{3}{tf\left( t \right)}\text{d}t=\left. \dfrac{{{t}^{2}}}{2}f\left( t \right) \right|_{0}^{3}-\int\limits_{0}^{3}{\dfrac{{{t}^{2}}}{2}{f}'\left( t \right)\text{d}t}=\dfrac{9}{2}f\left( 3 \right)-\dfrac{1}{2}\int\limits_{0}^{3}{{{t}^{2}}{{f}^{'}}\left( t \right)\text{d}t}$.
$\Leftrightarrow 9=\dfrac{9}{2}-\dfrac{1}{2}\int\limits_{0}^{3}{{{t}^{2}}{f}'\left( t \right)\text{d}t}\Leftrightarrow \int\limits_{0}^{3}{{{t}^{2}}{f}'\left( t \right)\text{d}t}=-9$.
Vậy $\int\limits_{0}^{3}{{{x}^{2}}{f}'\left( x \right)\text{d}x}=-9$.
Đáp án A.