Question Number 145318 by ZiYangLee last updated on 04/Jul/21
![Let f:[0,1]→R be a differentiable function such that f(f(x))=x for all x∈[0,1] and f(0)=1. If n is a positive integer, evaluate the following integral: ∫_0 ^( 1) (x−f(x))^(2n) dx](https://www.tinkutara.com/question/Q145318.png)
$$\mathrm{Let}\:{f}:\left[\mathrm{0},\mathrm{1}\right]\rightarrow\mathbb{R}\:\mathrm{be}\:\mathrm{a}\:\mathrm{differentiable}\:\mathrm{function} \\ $$$$\mathrm{such}\:\mathrm{that}\:{f}\left({f}\left({x}\right)\right)={x}\:\mathrm{for}\:\mathrm{all}\:{x}\in\left[\mathrm{0},\mathrm{1}\right]\:\mathrm{and} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{1}. \\ $$$$\mathrm{If}\:{n}\:\mathrm{is}\:\mathrm{a}\:\mathrm{positive}\:\mathrm{integer},\:\mathrm{evaluate}\:\mathrm{the} \\ $$$$\mathrm{following}\:\mathrm{integral}:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \left({x}−{f}\left({x}\right)\right)^{\mathrm{2}{n}} \:{dx} \\ $$
Answered by mindispower last updated on 04/Jul/21
![fdifferntiabl⇒f continus f(0)=1,f(f(0))=0⇒f(1)=0 ⇒[0,1]⊂f[0,1] f∣[0,1]→[0,1] is bijective ∫_0 ^1 (x−f(x))^(2n) dx,x=f(t) ∫_1 ^0 (f(t)−t)^(2n) f′(t)dt =−∫_0 ^1 ((f(t)−t))^(2n) f′(t)dt 2∫_0 ^1 (x−f(x))^(2n) dx=∫_0 ^1 (x−f(x))^(2n) dx−∫_0 ^1 (x−f(x))^(2n) f′(x)dx =∫_0 ^1 (x−f(x))^(2n) (1−f′(x))dx =[(1/(2n+1))(x−f(x))^(2n+1) ]_0 ^1 =(1/(2n+1))((1−f(1))^(2n+1) +f(0)^(2n+2) ) =(2/(2n+1))⇒∫_0 ^1 (x−f(x))^(2n) dx=(1/(2n+1)) exemple of such function x→1−x ∫_0 ^1 (x−f(x))^(2n) dx=∫_0 ^1 (2x−1)^(2n) =(1/(2(2n+1)))[2x−1]_0 ^1 =(1/(2n+1))](https://www.tinkutara.com/question/Q145354.png)
$${fdifferntiabl}\Rightarrow{f}\:{continus} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{1},{f}\left({f}\left(\mathrm{0}\right)\right)=\mathrm{0}\Rightarrow{f}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\left[\mathrm{0},\mathrm{1}\right]\subset{f}\left[\mathrm{0},\mathrm{1}\right] \\ $$$${f}\mid\left[\mathrm{0},\mathrm{1}\right]\rightarrow\left[\mathrm{0},\mathrm{1}\right]\:{is}\:{bijective}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}−{f}\left({x}\right)\right)^{\mathrm{2}{n}} {dx},{x}={f}\left({t}\right) \\ $$$$\int_{\mathrm{1}} ^{\mathrm{0}} \left({f}\left({t}\right)−{t}\right)^{\mathrm{2}{n}} {f}'\left({t}\right){dt} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \left(\left({f}\left({t}\right)−{t}\right)\right)^{\mathrm{2}{n}} {f}'\left({t}\right){dt} \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}−{f}\left({x}\right)\right)^{\mathrm{2}{n}} {dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}−{f}\left({x}\right)\right)^{\mathrm{2}{n}} {dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}−{f}\left({x}\right)\right)^{\mathrm{2}{n}} {f}'\left({x}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}−{f}\left({x}\right)\right)^{\mathrm{2}{n}} \left(\mathrm{1}−{f}'\left({x}\right)\right){dx} \\ $$$$=\left[\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\left({x}−{f}\left({x}\right)\right)^{\mathrm{2}{n}+\mathrm{1}} \right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\left(\left(\mathrm{1}−{f}\left(\mathrm{1}\right)\right)^{\mathrm{2}{n}+\mathrm{1}} +{f}\left(\mathrm{0}\right)^{\mathrm{2}{n}+\mathrm{2}} \right) \\ $$$$=\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}−{f}\left({x}\right)\right)^{\mathrm{2}{n}} {dx}=\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}} \\ $$$${exemple}\:{of}\:{such}\:{function} \\ $$$${x}\rightarrow\mathrm{1}−{x} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}−{f}\left({x}\right)\right)^{\mathrm{2}{n}} {dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}{n}} =\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{n}+\mathrm{1}\right)}\left[\mathrm{2}{x}−\mathrm{1}\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$