This comes from the comments section of this question. The original question was to show the following identity for some increasing invertible function $f$
$$\int_{a}^{b}f(x)\,dx+\int_{f(a)}^{f(b)}f^{-1}(x)\,dx=bf(b)-af(a)$$
Quite obviously this can be done by first a $u=f(x)$ substitution in the second integral, which leads to $$\int_{a}^{b}(f(x)+xf'(x))\,dx$$
which can be done done with another simple $y=xf(x)$ substitution.
But then the OP has come up with another interesting question of accomplishing this geometrically. I thought it was worthy of a post of its own, so here goes.
To accomplish this geometrically would mean to find the area under the curve, but since the limits are different as well simply adding them to get $g(x)=f(x)+f^{-1}(x)$ will not do. My idea to tackle this problem is two these separately and then add the individual areas. One property that might(?) be useful is that the image of $f$ in the line $y=x$ will be $f^{-1}$. On this graph, the $x$ co-ordinate would vary from $a$ to $b$ for $f$ under the area being calculated, while for $f^{-1}$ the same variation would be in the $y$ co-ordinate.
I admit the last lines aren't as much an insight as it is merely an observation that might not even prove useful. The OP's question while seemingly pointless(if we could calculate all areas then why even have integration?) might yield something interesting, or at least be a fun challenge.
I hope someone will come up with something brilliant soon enough, I am hopelessly stuck on this.
Also I am not sure what to tag this, feel free to re-tag.