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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm liên tục trên $\left[ 0;6 \right]$ và thỏa mãn $f\left( 0 \right)=2,\int\limits_{0}^{2}{\left( 2x-4 \right){f}'\left( x \right)dx=4.}$ Tính $\int\limits_{0}^{6}{f\left( \dfrac{x}{3} \right)dx}.$
A. $I=18.$
B. $I=-6.$
C. $I=-18.$
D. $I=6.$
A. $I=18.$
B. $I=-6.$
C. $I=-18.$
D. $I=6.$
Đặt $\dfrac{x}{3}=t\Rightarrow dx=3dt.$ Đổi cận: $\left\{ \begin{aligned}
& x=0\Rightarrow t=0 \\
& x=6\Rightarrow t=2 \\
\end{aligned} \right.\Rightarrow \int\limits_{0}^{6}{f\left( \dfrac{x}{3} \right)dx}=3\int\limits_{0}^{2}{f\left( t \right)dt}$
Đặt: $\left\{ \begin{aligned}
& 2x-4=u \\
& {f}'\left( x \right)dx=dv \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& 2dx=du \\
& f\left( x \right)=v \\
\end{aligned} \right.$
$\Rightarrow \int\limits_{0}^{2}{\left( 2x-4 \right){f}'\left( x \right)dx=\left. \left( 2x-4 \right)f\left( x \right) \right|_{0}^{2}-2\int\limits_{0}^{2}{f\left( x \right)dx}=4f\left( 0 \right)-}2\int\limits_{0}^{2}{f\left( x \right)dx}=4\Leftrightarrow \int\limits_{0}^{2}{f\left( x \right)dx}=2$
Nên: $\int\limits_{0}^{6}{f\left( \dfrac{x}{3} \right)}dx=3\int\limits_{0}^{2}{f\left( x \right)dx}=6$
& x=0\Rightarrow t=0 \\
& x=6\Rightarrow t=2 \\
\end{aligned} \right.\Rightarrow \int\limits_{0}^{6}{f\left( \dfrac{x}{3} \right)dx}=3\int\limits_{0}^{2}{f\left( t \right)dt}$
Đặt: $\left\{ \begin{aligned}
& 2x-4=u \\
& {f}'\left( x \right)dx=dv \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& 2dx=du \\
& f\left( x \right)=v \\
\end{aligned} \right.$
$\Rightarrow \int\limits_{0}^{2}{\left( 2x-4 \right){f}'\left( x \right)dx=\left. \left( 2x-4 \right)f\left( x \right) \right|_{0}^{2}-2\int\limits_{0}^{2}{f\left( x \right)dx}=4f\left( 0 \right)-}2\int\limits_{0}^{2}{f\left( x \right)dx}=4\Leftrightarrow \int\limits_{0}^{2}{f\left( x \right)dx}=2$
Nên: $\int\limits_{0}^{6}{f\left( \dfrac{x}{3} \right)}dx=3\int\limits_{0}^{2}{f\left( x \right)dx}=6$
Đáp án D.