Area enclosed between the roots of a quadratic

Question

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?

Answer 1

As @insipintegrator pointed it out,$$\left|\int_\beta^\alpha (ax^2+bx+c)dx\right|=|\frac{a}{3}(\alpha^3-\beta^3)+\frac{b}{2}(\alpha^2-\beta^2)+c(\alpha-\beta)|,$$ with $\alpha=\alpha(a,b,c)$ and $\beta=\beta(a,b,c)$ are very well known.

Answer 2

For $x$-coordinate of the apex $\mu = -\frac{b}{2a}$, discriminant $\Delta = b^2 - 4ac$, we can use the quadratic formula to obtain $\alpha = \mu + \frac{\sqrt{\Delta}}{2a}$, $\beta= \mu - \frac{\sqrt{\Delta}}{2a}$ (WLOG assuming $a>0$).

Thus, the area contained is the absolute value of the definite integral of the function from $\beta$ to $\alpha$, which is

$$\left| \int_\beta^\alpha{ax^2 +bx +c\,\mathrm{d}x}\right| = \left| \left[\frac{a}{3}x^3 + \frac{b}{2}x^2 +cx\right]_\beta^\alpha\right| = \left|\mu^2\sqrt{\Delta}+\frac{1}{3}\frac{\Delta^{\frac{3}{2}}}{4a^2}+b\mu\frac{\sqrt{\Delta}}{a}+c\frac{\sqrt{\Delta}}{a}\right|$$

$$=\left|\frac{\sqrt{b^2-4ac}}{a}\left(c-\frac{b^2+2ac}{6a}\right)\right|$$

Answer 3

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. $u = x + \frac {b}{2a}, du = dx$

$\displaystyle a\int_{-\frac {\sqrt{b^2 - 4ac}}{2a}}^{\frac {\sqrt{b^2 - 4ac}}{2a}} u^2 - \frac {b^2 - 4ac}{4a^2}\ du$

$a(\frac 13 u^3 - \frac {b^2 - 4ac}{4a^2}u)|_{-\frac {\sqrt{b^2 - 4ac}}{2a}}^{\frac {\sqrt{b^2 - 4ac}}{2a}}$

$a(\frac 23 (\frac {b^2 - 4ac}{4a^2})^\frac 32 - 2(\frac {b^2 - 4ac}{4a^2})^\frac 32) = -\frac {(b^2 - 4ac)^\frac 32}{6a^2}$

Answer 4

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 $$ A \ \ = \ \ \frac{a· \left(\frac{\sqrt{D}}{a} \right)^3}{6} \ \ = \ \ \frac{( \ b^2 - 4ac \ )^{3/2}}{6·a^2} \ \ . $$