Why is the integral of a parabola between its two roots always equals to twice the integral between a root and half the root separation?

Question

I came across an interesting relationship today, which I assume to be related to symmetry and the fact that all parabolas are a transformation of $y=x^2$. However, I am unable to prove it for the general case and would be interested to know more information or where I can find it.

Here is the maths (excuse my notation, I'm still new to this!):

Assume a parabola with roots $a$ and $b$.

$\int_a^b \!f(x)$ = $\int_b^y \! f(x)$ where $y=b+\frac{b - a}{2}$ (assuming absolute value, the geometric area, and therefore ignoring any positive/negative signs).

Any ideas? I'm interested to know how this can be proven to be the case and what it shows geometrically.

Answer 1

Note that your integrals will generally have opposite sign. I will show that their absolute values are equal.

Suppose first that $f(x)=x^2-b^2$ with $b>0$. Then

$$ \int_{-b}^b f(x) \, dx = \left[\frac{x^3}{3}-b^2x\right]_{-b}^b=\frac{2b^3}{3}-2b^3=-\frac{4b^3}{3} $$ and $$ \int_b^{2b} f(x) \, dx = \left[\frac{x^3}{3}-b^2x\right]_b^{2b}=\frac{7b^3}{3}-b^3=\frac{4b^3}{3} $$ and so the claim is true for this special case.

In general, if $f$ has roots $a$ and $b$, then $$f(x)=k(x-a)(x-b)=k\left[\left(x-\frac{b+a}{2}\right)^2-\left(\frac{b-a}{2}\right)^2\right]$$

We know by the argument above that the two areas are equal for the polynomial $$ g_1(x)=x^2-\left(\frac{b-a}{2}\right)^2 $$ Since shifting horizontally preserves area, they are also equal for the polynomial $$ g_2(x)=g_1\!\left(x-\frac{b+a}{2}\right)=\left(x-\frac{b+a}{2}\right)^2-\left(\frac{b-a}{2}\right)^2 $$ Since scaling vertically preserves the ratios of areas, this finally means that they are equal for the polynomial $$ f(x)=kg_2(x)=k\left[\left(x-\frac{b+a}{2}\right)^2-\left(\frac{b-a}{2}\right)^2\right] $$ which establishes the claim for general quadratic functions with real roots.

(Of course, you could also compute this integral for general $f$, but it'd be messier...)

Answer 2

Here is an easy and intuitive approach:

A parabola is the locus of points equidistant from a focus, $p$, and a directrix $y=d$. It easily follows that the parabola is symmetric about the line through $p$, orthogonal to $y=d$. The $x$-axis is where any parabola on the real plane vanishes, it follows that the roots of the parabola lie on the same horizontal line. Since the $x$-axis is parallel to the directrix, both $a$ and $b$ must be equidistant from the directrix, meaning that they are equidistant from the focus, and thus enclose the same area between the curve and the $x$-axis.

Answer 3

There is a theorem that says that the area bounded by a parabola and one of its chords is two-thirds of the area of the surrounding parallelogram. W.l.o.g. we can place the parabola so that its axis of symmetry is the $y$-axis and it opens upward (in the positive $y$-direction).

Let the $x$-intercepts of the parabola be $\pm a$ and its vertex be at $(0,-b)$. By the above theorem, the area between the parabola and $y$-axis is $\frac43ab$. The area of the large rectangle is $16ab$, so the area of the colored region below the parabola and to the right of the $y$-axis is one-sixth of this, or $\frac83ab$. We subtract $ab$ from this for the lower-right rectangle and a further $\frac13ab$ for the small wedge below the parabola, leaving $\frac43ab$.

P.S.: A possibly relevant or interesting fact is that the $x$-axis here bisects the area below the curve.