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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên $\mathbb{R}$ thỏa mãn $f\left( 2 \right)=16,\int\limits_{0}^{1}{f\left( 2\text{x} \right)d\text{x}}=2$. Tích phân $\int\limits_{0}^{2}{x{f}'\left( x \right)dx}$ bằng
A. 16
B. 28
C. 36
D. 30
A. 16
B. 28
C. 36
D. 30
Đặt $t=2\text{x}\Rightarrow dt=2\text{dx}$
Đổi cận $\left| \begin{aligned}
& x=0\Rightarrow t=1 \\
& x=1\Rightarrow t=2 \\
\end{aligned} \right. $ ta có: $ \int\limits_{0}^{1}{f\left( 2\text{x} \right)d\text{x}}=\int\limits_{0}^{2}{f\left( t \right)\dfrac{dt}{2}}=\dfrac{1}{2}\int\limits_{0}^{2}{f\left( x \right)d\text{x}}=2 $. Suy ra $ \int\limits_{0}^{2}{f\left( x \right)d\text{x}}=4$
Đặt $\left\{ \begin{aligned}
& u=x \\
& dv={f}'\left( x \right)d\text{x} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=d\text{x} \\
& v=f\left( x \right) \\
\end{aligned} \right.\Rightarrow \int\limits_{0}^{2}{x{f}'\left( x \right)d\text{x}}=\left. xf\left( x \right) \right|_{0}^{2}-\int\limits_{0}^{2}{f\left( x \right)d\text{x}}=2f\left( 2 \right)-4=28$.
Đổi cận $\left| \begin{aligned}
& x=0\Rightarrow t=1 \\
& x=1\Rightarrow t=2 \\
\end{aligned} \right. $ ta có: $ \int\limits_{0}^{1}{f\left( 2\text{x} \right)d\text{x}}=\int\limits_{0}^{2}{f\left( t \right)\dfrac{dt}{2}}=\dfrac{1}{2}\int\limits_{0}^{2}{f\left( x \right)d\text{x}}=2 $. Suy ra $ \int\limits_{0}^{2}{f\left( x \right)d\text{x}}=4$
Đặt $\left\{ \begin{aligned}
& u=x \\
& dv={f}'\left( x \right)d\text{x} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=d\text{x} \\
& v=f\left( x \right) \\
\end{aligned} \right.\Rightarrow \int\limits_{0}^{2}{x{f}'\left( x \right)d\text{x}}=\left. xf\left( x \right) \right|_{0}^{2}-\int\limits_{0}^{2}{f\left( x \right)d\text{x}}=2f\left( 2 \right)-4=28$.
Đáp án B.