If $f$ is a function that is continuous strictly increasing and differentiable on $[0, a]$ with $f(0)=0$ and g is the inverse of f, show that for any $x \in [0,a], \int_{0}^{x} f(t) \,dt + \int_{0}^{f(x)} g(t) \,dt - xf(x) = 0$.
Looking at some graphs, I know this is true because the first integral gives the area under the curve and the second gives the area above the curve but below $y=f(x)$ which when added together equals $xf(x)$ the area of the rectangle, but I'm having a hard time turning this intuition into a proof. I'm pretty much stuck after "pick an arbitrary $x \in [0, a].$"
Verify that the derivative of the left side is $0$ and that it vanishes when $x=0$. This implies that it is identically $0$.