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

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Answer 1 · 2017-07-14T09:14:46

#1/2019#

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#