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