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Câu hỏi: Cho $f\left(x \right)$ liên tục trên $\mathbb{R}$ và thỏa mãn $f\left(2 \right)=16, \int\limits_{0}^{1}{f\left(2x \right)\text{d}x=2}$. Tích phân $\int\limits_{0}^{2}{x{f}'\left(x \right)}\text{d}x$ bằng
A. $30$.$$
B. $28$.
C. $36$.
D. $16$.
A. $30$.$$
B. $28$.
C. $36$.
D. $16$.
Ta có: $\int\limits_{0}^{1}{f\left( 2x \right)\text{d}x=2}\Leftrightarrow \frac{1}{2}\int\limits_{0}^{1}{f\left( 2x \right)\text{d}\left( 2x \right)=2}\Leftrightarrow \int\limits_{0}^{2}{f\left( x \right)\text{d}x=4}$.
Đặt $\left\{ \begin{aligned}
& u=x \\
& \text{d}v={f}'\left( x \right)\text{dx} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=\text{d}x \\
& v=f\left( x \right) \\
\end{aligned} \right.$
$\Rightarrow \int\limits_{0}^{2}{x{f}'\left( x \right)}\text{d}x=\left. xf\left( x \right) \right|_{0}^{2}-\int\limits_{0}^{2}{f\left( x \right)}\text{d}x=2f\left( 2 \right)-4=32-4=28$.
Đặt $\left\{ \begin{aligned}
& u=x \\
& \text{d}v={f}'\left( x \right)\text{dx} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=\text{d}x \\
& v=f\left( x \right) \\
\end{aligned} \right.$
$\Rightarrow \int\limits_{0}^{2}{x{f}'\left( x \right)}\text{d}x=\left. xf\left( x \right) \right|_{0}^{2}-\int\limits_{0}^{2}{f\left( x \right)}\text{d}x=2f\left( 2 \right)-4=32-4=28$.
Đáp án B.