integral ( 1/x, x, 0,1)
Ok. I said , this implies that the graph of is symmetric with respect to the graph of . It means that if , then . In other words, f has its own bijection equals to itself.
Now to prove it, you just have to show that .
is equals to . It is demonstrated. If we take a precise example maybe you'll understand better. If , then . That means that to get 5 with our function, we need to evaluated it in , which works. Maybe you will understand even better seeing the graph. It's trivial that the area under the curve of from to is equal to the area from 1 to . To be rigorous, just say f is its own bijection (and show it as I did for you), that should be enough.