Most of you probably know this result, but I was surprised and thought it was neat so I wanna share :)

I teach math in high school in Norway and we have one gifted student who's taking Calculus in her last year of high school. I try to help her once a week in addition to the lectures she takes at the university.

She had the following problem to solve one day:

Let [; f:[0,1] \to \mathbb{R} ;]be a continuous function.

Evaluate:

[; \int_0^1 \frac{f(x)}{f(x) f(1-x)} dx ;]

Turns out the answer is 1/2 for any cont. f. I LOVE it when completely general problems all become a constant. Here's the solution if anyone is interested.

Let [;u=1-x;] Then by substitution you get:

[;-\int_1^0 \frac{f(1-u) }{f(1-u) f(u)} du;]

Let's call our original integral for A. By simply replacing u with x in the integral over and call it B. Then A B=1, and since A=B (because it was only a substitution) then 2A=1 and hence A=1/2.