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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên $\mathbb{R}$. Gọi $F\left( x \right),G\left( x \right)$ là hai nguyên hàm của $f\left( x \right)$ trên $\mathbb{R}$ thỏa mãn $3F\left( 8 \right)+G\left( 8 \right)=9$ và $3F\left( 0 \right)+G\left( 0 \right)=3$. Khi đó $\int\limits_{0}^{2}{f}\left( 4x \right)\text{d}x$ bằng
A. 3.
B. $\dfrac{1}{4}$.
C. 6.
D. $\dfrac{3}{8}$.
$\left\{ \begin{aligned}
& 3F\left( 8 \right)+G\left( 8 \right)=9 \\
& 3F\left( 0 \right)+G\left( 0 \right)=3 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& 4F\left( 8 \right)+C=9 \\
& 4F\left( 0 \right)+C=3 \\
\end{aligned} \right.\Leftrightarrow F\left( 8 \right)-F\left( 0 \right)=\dfrac{3}{2}.$
Vậy:
$\int\limits_{0}^{2}{f(4x)dx}=\dfrac{1}{4}\int\limits_{0}^{8}{f(x)dx=\dfrac{F\left( 8 \right)-F\left( 0 \right)}{4}=\dfrac{3}{8}.}$
A. 3.
B. $\dfrac{1}{4}$.
C. 6.
D. $\dfrac{3}{8}$.
Ta có: $G\left( x \right)=F\left( x \right)+C$ $\left\{ \begin{aligned}
& 3F\left( 8 \right)+G\left( 8 \right)=9 \\
& 3F\left( 0 \right)+G\left( 0 \right)=3 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& 4F\left( 8 \right)+C=9 \\
& 4F\left( 0 \right)+C=3 \\
\end{aligned} \right.\Leftrightarrow F\left( 8 \right)-F\left( 0 \right)=\dfrac{3}{2}.$
Vậy:
$\int\limits_{0}^{2}{f(4x)dx}=\dfrac{1}{4}\int\limits_{0}^{8}{f(x)dx=\dfrac{F\left( 8 \right)-F\left( 0 \right)}{4}=\dfrac{3}{8}.}$
Đáp án D.
