F'=2x-2∫[0, 1]f(x)dx (note that ∫[0, 1]f(x)dx is a number).
Find another derivative
f″= 2-0
f''=2
Integral once
f'=2x+A
Product again.
f=x^2+Ax+B
Substitute original form
x^2+ax+b=x^2-2x(∫[0, 1]f(x)dx)
[A+2∫[0， 1]f(x)dx]x+B=0
It holds true for any X.
therefore
A+2∫[0， 1]f(x)dx
B=0
that is
A=-2∫[0, 1](x^2+Ax)dx
A=-2[x^3/3+Ax^2/2]|[0, 1]
A=-2[ 1/3+A/2-0-0]
A=-2/3-A
2A=-2/3
A=- 1/3
So f (x) = x 2-(1/3) x.