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Câu hỏi: Cho hàm số $y=f\left( x \right)$ là hàm chẵn, liên tục trên $\mathbb{R}$ và $\int\limits_{-2}^{2}{\dfrac{f\left( x \right)}{{{2020}^{x}}+1}d\text{x}}=29$. Khi đó $\int\limits_{0}^{2}{f\left( x \right)d\text{x}}$ bằng
A. $\dfrac{29}{2}$
B. 29
C. 58
D. 30
A. $\dfrac{29}{2}$
B. 29
C. 58
D. 30
Ta có: $M=\int\limits_{-2}^{2}{\dfrac{f\left( x \right)}{{{2020}^{x}}+1}d\text{x}}=\int\limits_{-2}^{0}{\dfrac{f\left( x \right)}{{{2020}^{x}}+1}d\text{x}}+\int\limits_{0}^{2}{\dfrac{f\left( x \right)}{{{2020}^{x}}+1}d\text{x}}$
Xét $N=\int\limits_{-2}^{0}{\dfrac{f\left( x \right)}{{{2020}^{x}}+1}d\text{x}}$, đặt $t=-x\Rightarrow x=-t$, suy ra $d\text{x}=-dt$.
Đổi cận: $x=-2\Rightarrow t=2;x=0\Rightarrow t=0$. Khi đó
$N=\int\limits_{2}^{0}{\dfrac{f\left( -t \right)}{{{2020}^{-t}}+1}\left( -dt \right)}=\int\limits_{0}^{2}{\dfrac{f\left( t \right)}{{{2020}^{-t}}+1}dt}=\int\limits_{0}^{2}{\dfrac{{{2020}^{t}}.f\left( t \right)}{{{2020}^{t}}+1}dt}=\int\limits_{0}^{2}{\dfrac{{{2020}^{x}}.f\left( x \right)}{{{2020}^{x}}+1}d\text{x}}$
Do đó: $\int\limits_{0}^{2}{f\left( x \right)d\text{x}}=\int\limits_{0}^{2}{\dfrac{{{2020}^{x}}.f\left( x \right)}{{{2020}^{x}}+1}d\text{x}}+\int\limits_{0}^{2}{\dfrac{f\left( x \right)}{{{2020}^{x}}+1}d\text{x}}=\int\limits_{-2}^{0}{\dfrac{f\left( x \right)}{{{2020}^{x}}+1}d\text{x}}+\int\limits_{0}^{2}{\dfrac{f\left( x \right)}{{{2020}^{x}}+1}d\text{x}}=M=29$.
Xét $N=\int\limits_{-2}^{0}{\dfrac{f\left( x \right)}{{{2020}^{x}}+1}d\text{x}}$, đặt $t=-x\Rightarrow x=-t$, suy ra $d\text{x}=-dt$.
Đổi cận: $x=-2\Rightarrow t=2;x=0\Rightarrow t=0$. Khi đó
$N=\int\limits_{2}^{0}{\dfrac{f\left( -t \right)}{{{2020}^{-t}}+1}\left( -dt \right)}=\int\limits_{0}^{2}{\dfrac{f\left( t \right)}{{{2020}^{-t}}+1}dt}=\int\limits_{0}^{2}{\dfrac{{{2020}^{t}}.f\left( t \right)}{{{2020}^{t}}+1}dt}=\int\limits_{0}^{2}{\dfrac{{{2020}^{x}}.f\left( x \right)}{{{2020}^{x}}+1}d\text{x}}$
Do đó: $\int\limits_{0}^{2}{f\left( x \right)d\text{x}}=\int\limits_{0}^{2}{\dfrac{{{2020}^{x}}.f\left( x \right)}{{{2020}^{x}}+1}d\text{x}}+\int\limits_{0}^{2}{\dfrac{f\left( x \right)}{{{2020}^{x}}+1}d\text{x}}=\int\limits_{-2}^{0}{\dfrac{f\left( x \right)}{{{2020}^{x}}+1}d\text{x}}+\int\limits_{0}^{2}{\dfrac{f\left( x \right)}{{{2020}^{x}}+1}d\text{x}}=M=29$.
Đáp án B.