[VA] Short & Sweet Math Challenge #25 "San Valentin's Special: Weird Math"

[Albert Chan]

(02-18-2021 09:28 AM)Werner Wrote:  The key to the value of the integral in #3 is to realize that the integral is symmetric in the integral boundaries. Basically it is

integral(a,b,f(a+b-x)/(f(x) + f(a+b-x)),dx)

Without noticing the symmetry, if we "fold" the integral, we get the same result.



Let f(x) be integrand of above integral.

I = ∫(f(x), x = 1 .. φ)
  = ∫(f(x) + f(1+φ-x), x = 1 .. φ) / 2
  = ∫(f(x) + f(φ²-x), x = 1 .. φ) / 2
  = ∫(1, x = 1 .. φ) / 2
  = (φ-1)/2