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Câu hỏi: Nếu $\int\limits_{1}^{2}{\left[ 2x-3f\left( x \right) \right]dx}=4$ thì $\int\limits_{3}^{6}{f\left( \dfrac{x}{3} \right)dx}$ bằng:
A. 4
B. 1
C. $\dfrac{1}{3}$
D. $-1$
A. 4
B. 1
C. $\dfrac{1}{3}$
D. $-1$
Cách giải:
Ta có:
$\int\limits_{1}^{2}{\left[ 2x-3f\left( x \right) \right]dx}=4\Leftrightarrow \int\limits_{1}^{2}{2xdx}-3\int\limits_{1}^{2}{f\left( x \right)dx}=4$
$\Leftrightarrow {{x}^{2}}\left| \begin{aligned}
& 2 \\
& 1 \\
\end{aligned} \right.-3\int\limits_{1}^{2}{f\left( x \right)dx}=4\Leftrightarrow 4-1-3\int\limits_{1}^{2}{f\left( x \right)dx}=4$
$\Leftrightarrow 3-3\int\limits_{1}^{2}{f\left( x \right)dx}=4\Leftrightarrow \int\limits_{1}^{2}{f\left( x \right)dx}=-\dfrac{1}{3}$
Khi đó ta có: $\int\limits_{3}^{6}{f\left( \dfrac{x}{3} \right)dx}=3\int\limits_{3}^{6}{f\left( \dfrac{x}{3} \right)d\left( \dfrac{x}{3} \right)}=3\int\limits_{1}^{2}{f\left( x \right)dx}=3.\dfrac{-1}{3}=-1.$
Ta có:
$\int\limits_{1}^{2}{\left[ 2x-3f\left( x \right) \right]dx}=4\Leftrightarrow \int\limits_{1}^{2}{2xdx}-3\int\limits_{1}^{2}{f\left( x \right)dx}=4$
$\Leftrightarrow {{x}^{2}}\left| \begin{aligned}
& 2 \\
& 1 \\
\end{aligned} \right.-3\int\limits_{1}^{2}{f\left( x \right)dx}=4\Leftrightarrow 4-1-3\int\limits_{1}^{2}{f\left( x \right)dx}=4$
$\Leftrightarrow 3-3\int\limits_{1}^{2}{f\left( x \right)dx}=4\Leftrightarrow \int\limits_{1}^{2}{f\left( x \right)dx}=-\dfrac{1}{3}$
Khi đó ta có: $\int\limits_{3}^{6}{f\left( \dfrac{x}{3} \right)dx}=3\int\limits_{3}^{6}{f\left( \dfrac{x}{3} \right)d\left( \dfrac{x}{3} \right)}=3\int\limits_{1}^{2}{f\left( x \right)dx}=3.\dfrac{-1}{3}=-1.$
Đáp án D.