Here is an informal argument:
First note that your integral is of the form $\int_0^1 f(x)-f^{-1}(x)\,dx$, where $f$ is a decreasing function with $f(0)=1$, $f(1)=0$, and $f^{-1}$ is the inverse of $f$.
Then
- $ \int_0^1 f(x)\,dx$ is the area of the region bounded above by the graph of $f$ over the interval $0\le x\le 1$;
and, noting that the graph of the equation $x=f^{-1}(y)$ is precisely the graph of the equation $y=f(x)$,
- $ \int_0^1 f^{-1}(y)\,dy$ is the area of the region bounded to the right by the graph of $f$, below by the interval $0\le x\le 1$, and to the left by the interval $0\le y\le1$.
The two aforementioned regions coincide; thus, we have $\int_0^1 f(x)\,dx =\int_0^1 f^{-1}(x)\,dx $.
And so, $\int_0^1 f(x)-f^{-1}(x)\,dx=0$.