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Question Number 192345 by Mingma last updated on 15/May/23

Answered by aleks041103 last updated on 15/May/23

$${\int }_0^1(f(1+x)+f(1-x))dx=$$$$={\int }_0^{1}f(1+x)dx-{\int }_0^{1}f(1-x)d(-x)=$$$$={\int }_1^{2}f\left(x\right)dx-{\int }_1^{0}f\left(x\right)dx=$$$$={\int }_0^{2}f\left(x\right)dx$$$$\Rightarrow {\int }_0^{2}f\left(x\right)dx={\int }_0^1(f(1+x)+f(1-x))dx=$$$$={\int }_0^1{x}^{2022}dx=\frac{1}{2023}$$

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