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Câu hỏi: Cho hàm số $f\left( x \right)$ có $f\left( 2 \right)=4$, $\int\limits_{0}^{2}{xf\left( x \right)dx}=1$. Khi đó $\int\limits_{0}^{2}{{{x}^{2}}{f}'\left( x \right)dx}$ bằng
A. $15$.
B. $6$.
C. $18$.
D. $14$.
A. $15$.
B. $6$.
C. $18$.
D. $14$.
Đặt $\left\{ \begin{aligned}
& u={{x}^{2}} \\
& dv={f}'\left( x \right)dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=2xdx \\
& v=f\left( x \right) \\
\end{aligned} \right.$, ta được:
$\int\limits_{0}^{2}{{{x}^{2}}{f}'\left( x \right)dx}=\left. \left( {{x}^{2}}f\left( x \right) \right) \right|_{0}^{2}-2\int\limits_{0}^{2}{xf\left( x \right)dx}=4f\left( 2 \right)-0-2.1=14$.
& u={{x}^{2}} \\
& dv={f}'\left( x \right)dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=2xdx \\
& v=f\left( x \right) \\
\end{aligned} \right.$, ta được:
$\int\limits_{0}^{2}{{{x}^{2}}{f}'\left( x \right)dx}=\left. \left( {{x}^{2}}f\left( x \right) \right) \right|_{0}^{2}-2\int\limits_{0}^{2}{xf\left( x \right)dx}=4f\left( 2 \right)-0-2.1=14$.
Đáp án D.