Cho hàm số là hàm chẵn, liên tục trên R

Ta có: $M=\underset{-2}{\overset{2}{\int }}\frac{f\left(x\right)}{{2020}^x+1}d\text{x}=\underset{-2}{\overset{0}{\int }}\frac{f\left(x\right)}{{2020}^x+1}d\text{x}+\underset{0}{\overset{2}{\int }}\frac{f\left(x\right)}{{2020}^x+1}d\text{x}$

Xét $N=\underset{-2}{\overset{0}{\int }}\frac{f\left(x\right)}{{2020}^x+1}d\text{x}$ , 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$N=\underset{2}{\overset{0}{\int }}\frac{f\left(-t\right)}{{2020}^{-t}+1}\left(-dt\right)=\underset{0}{\overset{2}{\int }}\frac{f\left(t\right)}{{2020}^{-t}+1}dt=\underset{0}{\overset{2}{\int }}\frac{{2020}^t.f\left(t\right)}{{2020}^t+1}dt=\underset{0}{\overset{2}{\int }}\frac{{2020}^x.f\left(x\right)}{{2020}^x+1}d\text{x}$

Do đó: $\underset{0}{\overset{2}{\int }}f\left(x\right)d\text{x}=\underset{0}{\overset{2}{\int }}\frac{{2020}^x.f\left(x\right)}{{2020}^x+1}d\text{x}+\underset{0}{\overset{2}{\int }}\frac{f\left(x\right)}{{2020}^x+1}d\text{x}=\underset{-2}{\overset{0}{\int }}\frac{f\left(x\right)}{{2020}^x+1}d\text{x}+\underset{0}{\overset{2}{\int }}\frac{f\left(x\right)}{{2020}^x+1}d\text{x}=M=29$ .