Question Number 192345 by Mingma last updated on 15/May/23
Answered by aleks041103 last updated on 15/May/23
$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \left({f}\left(\mathrm{1}+{x}\right)+{f}\left(\mathrm{1}−{x}\right)\right){dx}= \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left(\mathrm{1}+{x}\right){dx}\:−\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {f}\left(\mathrm{1}−{x}\right){d}\left(−{x}\right)= \\ $$$$=\int_{\mathrm{1}} ^{\:\mathrm{2}} {f}\left({x}\right){dx}−\int_{\mathrm{1}} ^{\:\mathrm{0}} {f}\left({x}\right){dx}= \\ $$$$=\int_{\mathrm{0}} ^{\:\mathrm{2}} {f}\left({x}\right){dx} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\:\mathrm{2}} {f}\left({x}\right){dx}\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \left({f}\left(\mathrm{1}+{x}\right)+{f}\left(\mathrm{1}−{x}\right)\right){dx}= \\ $$$$=\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{\mathrm{2022}} {dx}=\frac{\mathrm{1}}{\mathrm{2023}} \\ $$