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Question Number 145318 by ZiYangLee last updated on 04/Jul/21

$$\mathrm{Let}f:[0,1]\to \mathbb{R}\mathrm{be}\mathrm{a}\mathrm{differentiable}\mathrm{function}$$$$\mathrm{such}\mathrm{that}f\left(f\left(x\right)\right)=x\mathrm{for}\mathrm{all}x\in [0,1]\mathrm{and}$$$$f\left(0\right)=1.$$$$\mathrm{If}n\mathrm{is}\mathrm{a}\mathrm{positive}\mathrm{integer},\mathrm{evaluate}\mathrm{the}$$$$\mathrm{following}\mathrm{integral}:$$$${\int }_0^1{(x-f\left(x\right))}^{2n}dx$$

Answered by mindispower last updated on 04/Jul/21

$$fdifferntiabl\Rightarrow fcontinus$$$$f\left(0\right)=1,f\left(f\left(0\right)\right)=0\Rightarrow f\left(1\right)=0$$$$\Rightarrow [0,1]\subset f[0,1]$$$$f\mid [0,1]\to [0,1]isbijective$$$${\int }_0^1{(x-f\left(x\right))}^{2n}dx,x=f\left(t\right)$$$${\int }_1^0{(f\left(t\right)-t)}^{2n}{f}^{\prime }\left(t\right)dt$$$$=-{\int }_0^1{\left((f\left(t\right)-t)\right)}^{2n}{f}^{\prime }\left(t\right)dt$$$$2{\int }_0^1{(x-f\left(x\right))}^{2n}dx={\int }_0^1{(x-f\left(x\right))}^{2n}dx-{\int }_0^1{(x-f\left(x\right))}^{2n}{f}^{\prime }\left(x\right)dx$$$$={\int }_0^1{(x-f\left(x\right))}^{2n}(1-f^{\prime }\left(x\right))dx$$$$={\left[\frac{1}{2n+1}{(x-f\left(x\right))}^{2n+1}\right]}_0^1=\frac{1}{2n+1}({(1-f\left(1\right))}^{2n+1}+f{\left(0\right)}^{2n+2})$$$$=\frac{2}{2n+1}\Rightarrow {\int }_0^1{(x-f\left(x\right))}^{2n}dx=\frac{1}{2n+1}$$$$exempleofsuchfunction$$$$x\to 1-x$$$${\int }_0^1{(x-f\left(x\right))}^{2n}dx={\int }_0^1{(2x-1)}^{2n}=\frac{1}{2(2n+1)}{[2x-1]}_0^1=\frac{1}{2n+1}$$$$$$$$$$$$$$

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