I would interpret the problem geometrically, in terms of area. Obviously,

is the area under the function if we assume which we will do for now. Now the graph of the function is just the graph of , but flipped over the line So the integral

is the area under the function If you plot this up for a few functions, you might notice that if you flip the area represented by

over the line , that will just equal the area required to fill the rectangle whose area is

The result follows for positive functions, at least.

I should point out that one assumption here, that is justified by the hypothesis that both and are continuous, is that exists. That means would be 1-1.