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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 $2F\left( 0 \right)-G\left( 0 \right)=1$, $F\left( 2 \right)-2G\left( 2 \right)=4$ và $F\left( 1 \right)-G\left( 1 \right)=-1$. Tính $\int\limits_{1}^{{{e}^{2}}}{ \dfrac{f\left( \ln x \right)}{2x}} \text{d}x$.
A. $-2$.
B. $-4$.
C. $-6$.
D. $-8$.
A. $-2$.
B. $-4$.
C. $-6$.
D. $-8$.
Ta có: $G\left( x \right)=F\left( x \right)+C$
$\left\{ \begin{aligned}
& 2F\left( 0 \right)-G\left( 0 \right)=1 \\
& F\left( 2 \right)-2G\left( 2 \right)=4 \\
& F\left( 1 \right)-G\left( 1 \right)=-1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& F(0)-C=1 \\
& -F(2)-2C=4 \\
& C=1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& F(0)=2 \\
& F(2)=-6 \\
& C=1 \\
\end{aligned} \right.$.
Do đó $\int\limits_{0}^{2}{f}\left( x \right)\text{d}x=F\left( 2 \right)-F\left( 0 \right)=-8$.
Vậy $\int\limits_{1}^{{{e}^{2}}}{ \dfrac{f\left( \ln x \right)}{2x}} \text{d}x=\int\limits_{1}^{{{e}^{2}}}{ \dfrac{f\left( \ln x \right)}{2}} \text{d}\left( \ln x \right)=\dfrac{1}{2}\int\limits_{0}^{2}{f\left( u \right)} \text{d}u=-4$.
$\left\{ \begin{aligned}
& 2F\left( 0 \right)-G\left( 0 \right)=1 \\
& F\left( 2 \right)-2G\left( 2 \right)=4 \\
& F\left( 1 \right)-G\left( 1 \right)=-1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& F(0)-C=1 \\
& -F(2)-2C=4 \\
& C=1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& F(0)=2 \\
& F(2)=-6 \\
& C=1 \\
\end{aligned} \right.$.
Do đó $\int\limits_{0}^{2}{f}\left( x \right)\text{d}x=F\left( 2 \right)-F\left( 0 \right)=-8$.
Vậy $\int\limits_{1}^{{{e}^{2}}}{ \dfrac{f\left( \ln x \right)}{2x}} \text{d}x=\int\limits_{1}^{{{e}^{2}}}{ \dfrac{f\left( \ln x \right)}{2}} \text{d}\left( \ln x \right)=\dfrac{1}{2}\int\limits_{0}^{2}{f\left( u \right)} \text{d}u=-4$.
Đáp án B.