If f(f(x))=x and f(0) =1f(f(x))=xandf(0)=1 then int_0^1(x-f(x))^2018dx=?10(xf(x))2018dx=?

1 Answer
Jul 14, 2017

1/201912019

Explanation:

Proposing

f(x) = a x + bf(x)=ax+b we have

f(f(x))=a(ax+b)+b = a^2x+ab+bf(f(x))=a(ax+b)+b=a2x+ab+b so

{(a^2=1),(ab+b=0):}

solving we have

a=1,b=-1 and a=-1,b=1

according with the condition f(0)=1 we have

f(x) = 1-x

now

(x-f(x))^2018= (2x-1)^2018 and

int_0^1 (2x-1)^2018 dx = (1/2(2x-1)^2019/2019)_0^1 = 1/2019