After seeing how cosh(x) has the property of the area under cosh(x) from an interval a to b equals the arc length above that area, I wondered if there was a graph with similar properties.
Arc length, given by:
And obviously, the area under a graph is given by:
Applying both of these to cosh(x), from any interval a to b, we get sinh(b) for both.
I have ended up with this several times, but cant seem to make it work:
Suppose there is a function $f$ such that that the graph of $y = f(x)$ has the property you seek, that is, the area under the curve between any two values of $x$ is the square of the arc length of the curve between those two points.
The property is trivial unless the graph of the function actually has an arc between two points defining the area under that arc. So let's consider some such arc of the graph of this function starting at the point $(x_1,y_1)$ and ending at the point $(x_2,y_2)$. This arc has some length, $L$, and the area under the arc is $L^2$:
$$ \int_{x_1}^{x_2} f(x) \, dx = L^2. $$
Assuming we actually have an arc and have not just named the same point twice, the arc has a positive length, that is, $L > 0.$
There is a point exactly halfway along this arc, that is, a point $(x_m,y_m)$ on the arc from $(x_1,y_1)$ to $(x_2,y_2)$ so that the distance from $(x_1,y_1)$ to $(x_m,y_m)$ is $L/2$ and the distance from $(x_m,y_m)$ to $(x_2,y_2)$ is $L/2.$ So we can integrate the areas under each half of the arc on either side of $(x_m,y_m)$:
$$ \int_{x_1}^{x_m} f(x) \, dx = \left(\frac L2\right)^2 = \int_{x_m}^{x_2} f(x) \, dx. $$
But now we have
$$ \int_{x_1}^{x_m} f(x) \, dx + \int_{x_m}^{x_2} f(x) \, dx = 2\left(\frac L2\right)^2 = \frac12 L^2 \neq L^2 = \int_{x_1}^{x_2} f(x) \, dx, $$
where the inequality arises from the fact that $L \neq 0$; but now if all three integrals are defined it is not possible for the sum of integrals on the left to have a value different from the integral on the right.
That is a reason why you cannot find a function with the property you seek.