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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm ${f}'\left( x \right)$ liên tục trên $\mathbb{R}$, $f\left( 4 \right)=8$ và $\int\limits_{0}^{4}{f\left( x \right)\text{d}x}=6$. Giá trị của $\int\limits_{0}^{2}{x{f}'\left( 2x \right)\text{d}x}$ bằng
A. $\dfrac{13}{2}$.
B. $\dfrac{13}{4}$.
C. $13$.
D. $10$.
Xét $I=\int\limits_{0}^{2}{x{f}'\left( 2x \right)\text{d}x}$.
Đặt $\left\{ \begin{aligned}
& x=u \\
& {f}'\left( 2x \right)\text{d}x=\text{d}v \\
\end{aligned} \right.$$\Rightarrow \left\{ \begin{aligned}
& \text{d}x=\text{d}u \\
& \dfrac{1}{2}f\left( 2x \right)=v \\
\end{aligned} \right.$, khi đó
$I=\int\limits_{0}^{2}{x{f}'\left( 2x \right)\text{d}x}=\left. \dfrac{1}{2}xf\left( 2x \right) \right|_{0}^{2}-\dfrac{1}{2}\int\limits_{0}^{2}{f\left( 2x \right)\text{d}x}$ $=\dfrac{1}{2}.2.f\left( 4 \right)-\dfrac{1}{4}\int\limits_{0}^{4}{f\left( x \right)\text{d}x}=8-\dfrac{1}{4}.6=\dfrac{13}{2}$.
A. $\dfrac{13}{2}$.
B. $\dfrac{13}{4}$.
C. $13$.
D. $10$.
Xét $I=\int\limits_{0}^{2}{x{f}'\left( 2x \right)\text{d}x}$.
Đặt $\left\{ \begin{aligned}
& x=u \\
& {f}'\left( 2x \right)\text{d}x=\text{d}v \\
\end{aligned} \right.$$\Rightarrow \left\{ \begin{aligned}
& \text{d}x=\text{d}u \\
& \dfrac{1}{2}f\left( 2x \right)=v \\
\end{aligned} \right.$, khi đó
$I=\int\limits_{0}^{2}{x{f}'\left( 2x \right)\text{d}x}=\left. \dfrac{1}{2}xf\left( 2x \right) \right|_{0}^{2}-\dfrac{1}{2}\int\limits_{0}^{2}{f\left( 2x \right)\text{d}x}$ $=\dfrac{1}{2}.2.f\left( 4 \right)-\dfrac{1}{4}\int\limits_{0}^{4}{f\left( x \right)\text{d}x}=8-\dfrac{1}{4}.6=\dfrac{13}{2}$.
Đáp án A.