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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 $F\left( 8 \right)+G\left( 8 \right)=17$ và $F\left( 0 \right)+G\left( 0 \right)=1$. Khi đó $\int\limits_{0}^{\dfrac{\pi }{2}}{\sin x.f}\left( 8\cos x \right)\text{d}x$ bằng
A. $-1$.
B. $1$.
C. $8$.
D. $-8$.
A. $-1$.
B. $1$.
C. $8$.
D. $-8$.
Ta có: $G\left( x \right)=F\left( x \right)+C\Rightarrow \left\{ \begin{aligned}
& G\left( 8 \right)=F\left( 8 \right)+C \\
& G\left( 0 \right)=F\left( 0 \right)+C \\
\end{aligned} \right.$
$\left\{ \begin{aligned}
& F\left( 8 \right)+G\left( 8 \right)=18 \\
& F(0)+G(0)=2 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& 2F(8)+C=18 \\
& 2F(0)+C=2 \\
\end{aligned} \right.\Leftrightarrow F\left( 8 \right)-F\left( 0 \right)=8.$
Vậy: $\int\limits_{0}^{\dfrac{\pi }{2}}{\sin x.f}\left( 8\cos x \right)\text{d}x=\dfrac{1}{8}\int\limits_{0}^{8}{f(t)dt=\dfrac{1}{8}\left( F(8)-F(0) \right)=1.}$
& G\left( 8 \right)=F\left( 8 \right)+C \\
& G\left( 0 \right)=F\left( 0 \right)+C \\
\end{aligned} \right.$
$\left\{ \begin{aligned}
& F\left( 8 \right)+G\left( 8 \right)=18 \\
& F(0)+G(0)=2 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& 2F(8)+C=18 \\
& 2F(0)+C=2 \\
\end{aligned} \right.\Leftrightarrow F\left( 8 \right)-F\left( 0 \right)=8.$
Vậy: $\int\limits_{0}^{\dfrac{\pi }{2}}{\sin x.f}\left( 8\cos x \right)\text{d}x=\dfrac{1}{8}\int\limits_{0}^{8}{f(t)dt=\dfrac{1}{8}\left( F(8)-F(0) \right)=1.}$
Đáp án B.