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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 $F\left( 2 \right)+G\left( 2 \right)=8$ và $F\left( 0 \right)+G\left( 0 \right)=-2$. Khi đó $\int\limits_{0}^{16}{f}\left( \dfrac{x}{8} \right)\text{d}x$ bằng
A. -40.
B. 40.
C. $5$.
D. $-5$.
A. -40.
B. 40.
C. $5$.
D. $-5$.
Ta có: $G\left( x \right)=F\left( x \right)+C\Rightarrow \left\{ \begin{aligned}
& G\left( 2 \right)=F\left( 2 \right)+C \\
& G\left( 0 \right)=F\left( 0 \right)+C \\
\end{aligned} \right.$
$\left\{ \begin{aligned}
& F\left( 2 \right)+G\left( 2 \right)=8 \\
& F(0)+G(0)=-2 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& 2F(2)+C=8 \\
& 2F(0)+C=-2 \\
\end{aligned} \right.\Leftrightarrow F(2)-F(0)=5.$
Vậy: $\int\limits_{0}^{16}{f}\left( \dfrac{x}{8} \right)\text{d}x=8\int\limits_{0}^{2}{f(t)dt=8\left( F(2)-F(0) \right)=40}$.
& G\left( 2 \right)=F\left( 2 \right)+C \\
& G\left( 0 \right)=F\left( 0 \right)+C \\
\end{aligned} \right.$
$\left\{ \begin{aligned}
& F\left( 2 \right)+G\left( 2 \right)=8 \\
& F(0)+G(0)=-2 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& 2F(2)+C=8 \\
& 2F(0)+C=-2 \\
\end{aligned} \right.\Leftrightarrow F(2)-F(0)=5.$
Vậy: $\int\limits_{0}^{16}{f}\left( \dfrac{x}{8} \right)\text{d}x=8\int\limits_{0}^{2}{f(t)dt=8\left( F(2)-F(0) \right)=40}$.
Đáp án B.
