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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 $F\left( 40 \right)+G\left( 40 \right)=8$ và $F\left( 0 \right)+G\left( 0 \right)=-2$. Khi đó $\int\limits_{1}^{{{e}^{8}}}{\dfrac{1}{x}f}\left( 5\ln \left( x \right) \right)\text{d}x$ bằng
A. $-1$.
B. $1$.
C. $5$.
D. $-5$.
A. $-1$.
B. $1$.
C. $5$.
D. $-5$.
Ta có: $G\left( x \right)=F\left( x \right)+C\Rightarrow \left\{ \begin{aligned}
& G\left( 40 \right)=F\left( 40 \right)+C \\
& G\left( 0 \right)=F\left( 0 \right)+C \\
\end{aligned} \right.$
$\left\{ \begin{aligned}
& F\left( 40 \right)+G\left( 40 \right)=8 \\
& F(0)+G(0)=-2 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& 2F(40)+C=8 \\
& 2F(0)+C=-2 \\
\end{aligned} \right.\Leftrightarrow F(40)-F(0)=5.$
Vậy: $\int\limits_{1}^{{{e}^{8}}}{\dfrac{1}{x}f}\left( 5\ln \left( x \right) \right)\text{d}x=\dfrac{1}{5}\int\limits_{0}^{40}{f(t)dt=\dfrac{1}{5}\left( F(40)-F(0) \right)=1.}$
& G\left( 40 \right)=F\left( 40 \right)+C \\
& G\left( 0 \right)=F\left( 0 \right)+C \\
\end{aligned} \right.$
$\left\{ \begin{aligned}
& F\left( 40 \right)+G\left( 40 \right)=8 \\
& F(0)+G(0)=-2 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& 2F(40)+C=8 \\
& 2F(0)+C=-2 \\
\end{aligned} \right.\Leftrightarrow F(40)-F(0)=5.$
Vậy: $\int\limits_{1}^{{{e}^{8}}}{\dfrac{1}{x}f}\left( 5\ln \left( x \right) \right)\text{d}x=\dfrac{1}{5}\int\limits_{0}^{40}{f(t)dt=\dfrac{1}{5}\left( F(40)-F(0) \right)=1.}$
Đáp án B.