Is this elementary formula for the parabolic segment new?

Question

Recently (May 2020) a formula for the area of the parabolic segment (i.e. the region enclosed by a parabola and a line), in terms of the coefficients of the Cartesian equations, has been published by an Italian high school student in MatematicaMente (n. 269).

Let the parabola and the line have equations $y=ax^2+bx+c$ and $mx+q$, respectively, and assume the line does intersect the parabola. Then the published formula is

$$A=\frac{\sqrt{[(b-m)^2-4a(c-q)]^3}}{6a^2}$$

where $A$ is the area of the corresponding parabolic segment.

If the parabola's equation has normal form $y=ax^2$, then the formula reduces to

$$A=\frac{\sqrt{(m^2+4aq)^3}}{6a^2}\;\;.$$

Is the above formula actually new? By this, I mean: can it be found anywhere in the literature (articles, preprints, books,...) prior to, say, 2019? If yes, was it proven by an elementary proof (using Cartesian coordinates geometry) or did it use integral calculus?

The coordinates proof involves a quite tedious computation but no original idea. So, it is quite surprising if indeed nobody, since the time of Decartes, had ever thought of seeing how the classical Archimede's result related with coordinate geometry.

Answer 1

The formula for the area of the parabolic segment is usually written more concisely in terms of the two intersection abscissas $x_1$ and $x_2$,

$$A=\tfrac{1}{6}|a||x_1-x_2|^3.$$

See for example this elementary derivation.

Since $|x_1-x_2|=|a|^{-1}\sqrt{4 a (q-c)+(b-m)^2}$, being the difference of the two solutions of the quadratic equation $ax^2+(b-m)x+(c-q)=0$, the formula in the OP follows immediately.