Câu hỏi 6

\(\int\limits_{ - 2}^2 {\frac{{f\left( x \right)}}{{{{2018}^x} + 1}}dx = 2020} \,\,(1) \Rightarrow \int\limits_2^{ - 2} {\frac{{f\left( { - x} \right)}}{{{{2018}^{ - x}} + 1}}\left( { - dx} \right) = 2020 \Leftrightarrow } \int\limits_{ - 2}^2 {\frac{{{{2018}^x}f\left( x \right)}}{{{{2018}^x} + 1}}dx = 2020} \,\,\,(2)\)

(do \(y = f\left( x \right)\) là hàm số chẵn, liên tục trên đoạn \(\left[ { - 2;2} \right]\))

Cộng (1) với (2):

\(\begin{array}{l}\,\,\,\,\,\,\int\limits_{ - 2}^2 {\frac{{f\left( x \right)}}{{{{2018}^x} + 1}}dx + } \,\int\limits_{ - 2}^2 {\frac{{{{2018}^x}f\left( x \right)}}{{{{2018}^x} + 1}}dx = 4040} \\ \Leftrightarrow \int\limits_{ - 2}^2 {\left( {\frac{{f\left( x \right)}}{{{{2018}^x} + 1}} + \frac{{{{2018}^x}f\left( x \right)}}{{{{2018}^x} + 1}}} \right)dx}  = 4040 \Leftrightarrow \int\limits_{ - 2}^2 {f\left( x \right)dx}  = 4040\end{array}\)

Lại do \(y = f\left( x \right)\) là hàm chẵn nên \(\int\limits_{ - 2}^2 {f\left( x \right)dx}  = 2.\int\limits_0^2 {f\left( x \right)dx}  \Rightarrow \int\limits_0^2 {f\left( x \right)dx}  = 2020\)

Ta có: \(\int\limits_0^2 {\left( {1 + f\left( x \right)} \right)dx}  = \int\limits_0^2 {dx}  + \int\limits_0^2 {f\left( x \right)dx}  = 2 + 2020 = 2022\).

Chọn: B