Relation between the integral of t·f(t) and (f⁻¹(t))² geometrized

For the horizontal width of the triangle, it is simply just the distance from the origin to our intersection point, on the x-axis; we can bring in that intersection X-coordinate from earlier, then simplifying:

The height of the right side is obtained by doing the second linear equation minus the first one, then plugging that intersection X-coordinate in for x, then simplifying:

We then can apply the half-product formula; notice the 2 values we calculated above are equal, so we can consolidate them into (g(m))²:

***For the visual, I will integrate starting at m = 0, all the way up to m.

If we plug a differential in for h in the triangle area above, and integrate, our integral here calculates the red area:

We can use the intersection X-coordinate formula for our bounds and integrate t·f(t) with those bounds:

Then we can subtract that from the area of the outlined purple triangle, which will be easy to calculate with the half-product formula, with one point being at the origin, and the other being at the intersection from earlier:

Of course, we can reverse the roles of the functions, since they are inverses, and they have a 'symmetric' relation.