Question

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

Answer 1

`1/2019`

Explanation:

Proposing

`f(x) = a x + b` we have

`f(f(x))=a(ax+b)+b = a^2x+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`