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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

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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
Letf:[0,1]→Rbeadifferentiablefunctionsuchthatf(f(x))=xforallx∈[0,1]andf(0)=1.Ifnisapositiveinteger,evaluatethefollowingintegral:∫01(x−f(x))2ndx
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))
fdifferntiabl⇒fcontinusf(0)=1,f(f(0))=0⇒f(1)=0⇒[0,1]⊂f[0,1]f∣[0,1]→[0,1]isbijective∫01(x−f(x))2ndx,x=f(t)∫10(f(t)−t)2nf′(t)dt=−∫01((f(t)−t))2nf′(t)dt2∫01(x−f(x))2ndx=∫01(x−f(x))2ndx−∫01(x−f(x))2nf′(x)dx=∫01(x−f(x))2n(1−f′(x))dx=[12n+1(x−f(x))2n+1]01=12n+1((1−f(1))2n+1+f(0)2n+2)=22n+1⇒∫01(x−f(x))2ndx=12n+1exempleofsuchfunctionx→1−x∫01(x−f(x))2ndx=∫01(2x−1)2n=12(2n+1)[2x−1]01=12n+1

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