<s>[INST] Area enclosed between the roots of a quadratic Let $f(x)=ax^2+bx+c$
If $f(x)$ has roots $\alpha$ and $\beta$, what is the area enclosed by $f(x)$ and the $x$-axis between $x=\alpha$ and $x=\beta$ in terms of $a,b$ and $c$?
It is also given that $\alpha>\beta.$

If $a=1$, then I thought this might be easier since you get:
$$A=\int_\beta^\alpha{(x^2-(\alpha+\beta)x+({\alpha}{\beta})x) dx}$$
But even after evaluating this, I still wasn't even able to find an answer in terms of the coefficients.
I've been at this problem for a while now, and I would love some help. Any ideas?
 [/INST] Have you ever seen Archemedes' proof of the quadrature of a parabola?
https://en.wikipedia.org/wiki/Quadrature_of_the_Parabola
Archimedes proved that any secant through a parabola intersecting at points $\alpha, \beta$ form a region with area $\frac{a(\alpha - \beta)^3}{6}$
But using calculus...
$y = ax^2 + bx + c$
let's put this parabola into vertex form.
$y = a(x + \frac {b}{2a})^2 + c - \frac {b^2}{4a}$
Save this for later...
$\int_\beta^\alpha ax^2 + bx + c\ dx$
$\beta = \frac {-b - \sqrt {b^2 - 4ac}}{2a}\\
\alpha = \frac {-b + \sqrt {b^2 - 4ac}}{2a}$
let's perform a u-substitution.  $u = x + \frac {b}{2a}, du = dx$
$\displaystyle a\int_{-\frac {\sqrt{b^2 - 4ac}}{2a}}^{\frac {\sqrt{b^2 - 4ac}}{2a}} u^2 - \frac {b^2 - 4ac}{4a^2}\ du$
$a(\frac 13 u^3 - \frac {b^2 - 4ac}{4a^2}u)|_{-\frac {\sqrt{b^2 - 4ac}}{2a}}^{\frac {\sqrt{b^2 - 4ac}}{2a}}$
$a(\frac 23 (\frac {b^2 - 4ac}{4a^2})^\frac 32 - 2(\frac {b^2 - 4ac}{4a^2})^\frac 32) = -\frac {(b^2 - 4ac)^\frac 32}{6a^2}$
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