<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] To pursue your approach, not restricting the polynomial to $ \ a \ = \ 1 \ \ , $ we have
$$ A \ \ = \ \ a \int_\beta^\alpha \ \   x^2 \ - \ (\alpha+\beta)x \ + \  \alpha \beta   \ \  dx $$
[you have an extra "x" on what should be the "constant term" :  I presume that was a typo]
$$ = \ \ a·\left( \ \frac13x^3 \ - \ \frac{\alpha \ + \ \beta}{2} ·x^2 + \ \alpha \beta x \ \right) |_{\beta}^{\alpha} $$
$$ = \ \ a· \frac{2·(\alpha^3 \ - \ \beta^3) \ - \ 3·(\alpha \ + \ \beta)·(\alpha^2 \ - \ \beta^2) \ + \ 6·\alpha \beta·(\alpha \ - \ \beta)}{6} $$
$$ = \ \ \frac{a}{6}·(\alpha \ - \ \beta) · [ \ 2·(\alpha^2 \ + \ \alpha \beta \ + \ \beta^2) \ - \ 3·(\alpha \ + \ \beta)^2 \ + \ 6·\alpha \beta \ ]  $$
[applying "differences of two cubes" and "two squares"]
$$ = \ \ \frac{a}{6}·(\alpha \ - \ \beta) · ( \ 2  \alpha^2 \ + \ 2\alpha \beta \ + \ 2\beta^2  \ - \ 3 \alpha^2 \ - \ 6 \alpha \beta \ - \ 3\beta^2 \ + \ 6 \alpha \beta \ )  $$
$$ = \ \ \frac{a·(\alpha \ - \ \beta)}{6} · ( \ -  \alpha^2 \ + \ 2\alpha \beta \ - \ \beta^2   \ ) \ \ = \ \ - \frac{a·(\alpha \ - \ \beta)^3}{6} \ \ , $$
which is related to the result from Archimedes' quadrature that Doug M mentions.  The result is negative since your integral evaluates an area "below" the $ \ x-$axis.  So the area enclosed by the $ \ x-$axis and the parabola is the absolute-value of this, $ \ \frac{a·(\alpha \ - \ \beta)^3}{6} \ \ . \ $
As I mention in the comment to that post, the difference between the two real zeroes is $ \ \alpha - \beta \ = \ \frac{\sqrt{D}}{a} \ \ . \ $  With the vertex "below" the $ \ x-$axis, $ \ D \ = \ b^2 - 4ac \ > \ 0 \ \ , \ $ so we may at last write
$$ A \ \ = \ \ \frac{a· \left(\frac{\sqrt{D}}{a} \right)^3}{6} \ \ = \ \ \frac{( \ b^2 - 4ac \ )^{3/2}}{6·a^2} \ \ .  $$
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