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Câu hỏi: Cho hàm số y= ƒ(x) liên tục trên $\mathbb{R}$ và $\int\limits_{-1}^{1}{\dfrac{f\left( x \right)+f\left( -x \right)}{{{2018}^{x}}+1}dx=2018}$. Tích phân $\int\limits_{-1}^{1}{f\left( x \right)dx}$ là:
A. 2017
B. 2018
C. 1009
D. 0
A. 2017
B. 2018
C. 1009
D. 0
Xét tích phân $\int\limits_{-1}^{1}{\dfrac{f\left( -x \right)}{{{2018}^{x}}+1}dx}$. Đặt $t=-x\Leftrightarrow dx=-dt$ và$\left\{ \begin{aligned}
& x=-1\Rightarrow t=1 \\
& x=1\Rightarrow t=-1 \\
\end{aligned} \right.$
Do đó $\int\limits_{-1}^{1}{\dfrac{f\left( -x \right)}{{{2018}^{x}}+1}dx}=\int\limits_{1}^{-1}{\dfrac{f\left( t \right)}{{{2018}^{-t}}+1}\left( -dt \right)=\int\limits_{-1}^{1}{\dfrac{f\left( x \right)}{{{2018}^{-x}}+1}dx=}}\int\limits_{-1}^{1}{\dfrac{{{2018}^{x}}f\left( x \right)}{{{2018}^{x}}+1}dx}$
Suy ra $2018=\int\limits_{-1}^{1}{\dfrac{f\left( x \right)+f\left( -x \right)}{{{2018}^{x}}+1}dx=\int\limits_{-1}^{1}{\dfrac{f\left( x \right)}{{{2018}^{x}}+1}dx}+}\int\limits_{-1}^{1}{\dfrac{{{2018}^{x}}f\left( x \right)}{{{2018}^{x}}+1}dx=\int\limits_{-1}^{1}{f\left( x \right)dx.}}$
& x=-1\Rightarrow t=1 \\
& x=1\Rightarrow t=-1 \\
\end{aligned} \right.$
Do đó $\int\limits_{-1}^{1}{\dfrac{f\left( -x \right)}{{{2018}^{x}}+1}dx}=\int\limits_{1}^{-1}{\dfrac{f\left( t \right)}{{{2018}^{-t}}+1}\left( -dt \right)=\int\limits_{-1}^{1}{\dfrac{f\left( x \right)}{{{2018}^{-x}}+1}dx=}}\int\limits_{-1}^{1}{\dfrac{{{2018}^{x}}f\left( x \right)}{{{2018}^{x}}+1}dx}$
Suy ra $2018=\int\limits_{-1}^{1}{\dfrac{f\left( x \right)+f\left( -x \right)}{{{2018}^{x}}+1}dx=\int\limits_{-1}^{1}{\dfrac{f\left( x \right)}{{{2018}^{x}}+1}dx}+}\int\limits_{-1}^{1}{\dfrac{{{2018}^{x}}f\left( x \right)}{{{2018}^{x}}+1}dx=\int\limits_{-1}^{1}{f\left( x \right)dx.}}$
Đáp án B.