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# Parabola Proof

We will prove that for any parabola, the area enclosed by the curve and the x-axis is always equal to $$\frac{2}{3}$$ times the width of the parabola (the distance between the zeros) times the height of the parabola (above/below the x-axis). In other words, we will prove:

$$\forall$$ quadratic equations of the form $$y(x) = m(x-x_v)^2 +y_v$$, where $$(x_v,y_v)$$ is the sole extremum of $$y(x)$$ (i.e. the vertex) and $$m \in \mathbb{R}$$:

$$\displaystyle \int_{b}^{a} y(x)\,dx = \frac{2}{3}(\mid a-b\mid )( y_v )$$

Where $$a$$ and $$b$$ satisfy:

$$y(a) = 0$$ and $$y(b) = 0$$, $$b<a$$

Proof:

Let $$y(x) = -m(x-x_v)^2 +y_v$$, where $$(x_v,y_v)$$ is a point in either quadrant I or II of the xy plane. (It is clear that this is a "downward-pointing" parabola which intersects $$y=0$$ at two points. We may just as easily have chosen an "upward pointing" parabola with vertex in either quadrant III or IV. For the sake of ease of visualization, we choose the former.)

Let $$(-b,0)$$ and $$(a,0)$$ be two points on $$y(x)$$ ($$-b<a$$). (It is clear that $$a$$ and $$-b$$ are the roots of $$y(x)$$)

Define the height of the parabola: $$h=y_v$$

Define the width of the parabola: $$w=\mid a-(-b) \mid = \mid a + b \mid$$

Now, the area enclosed by the line $$y=0$$ and $$y(x)$$ is clearly the integral:

$$A = \displaystyle \int_{-b}^{a} y(x)\,dx =\displaystyle \int_{-b}^{a} -m(x-x_v)^2 +y_v\,dx$$

$$\Rightarrow$$ $$A = \frac{-m}{3}(x-x_v)^3 \mid_{-b}^{a} +y_v(a+b)$$

$$\Rightarrow$$ $$A= \frac{-m}{3}\left[(a-x_v)^3 - (-b-x_v)^3 \right] +y_v(a+b)$$

Now, since $$y(x)$$ is a parabola, dist$$(a,x_v) =$$ dist$$(-b,x_v)$$. However, $$-b<a$$ and $$y(-b) = 0$$ imply that $$-b<x_v$$. Then $$-b-x_v<0$$, but dist$$(a,x_v) =$$ dist$$(-b,x_v)$$. Then we may write $$-(a-x_v) = (-b-x_v)$$. It follows that:

$$A = \frac{-m}{3}\left[-2(-b-x_v)^3 \right] +y_v(a+b)$$

$$\Rightarrow$$ $$A = \frac{2m}{3}\left[(-b-x_v)^3 \right] +y_v(a+b)$$

Now, let us expand $$y(x)$$:

$$y(x) = -m(x-x_v)^2 +y_v$$

$$\Rightarrow$$ $$y(x) = -mx^2 +2mx_vx - mx_v^2+y_v$$

Since $$a$$ and $$-b$$ are the roots of $$y(x)$$, we may apply the quadratic formula and set it equal to our roots. We have:

$$a$$,$$-b$$ $$= \frac{-2mx_v \pm \sqrt{4m^2x_v^2 -4m^2x_v^2 +4my_v}}{-2m}$$

$$\Rightarrow$$ $$a$$,$$-b$$ $$=x_v \mp \sqrt{\frac{y_v}{m}}$$

Now, since $$-b<a$$, we must choose the solution with the minus sign for $$-b$$, leading to:

$$-b = x_v-\sqrt{\frac{y_v}{m}}$$

$$\Rightarrow$$ $$-b-x_v = -\sqrt{\frac{y_v}{m}}$$

$$\Rightarrow$$ $$\sqrt{m} = \frac{-\sqrt{y_v}}{-b-x_v}$$

$$\Rightarrow$$ $$m= \frac{y_v}{(-b-x_v)^2}$$

Plugging this into our expression for the area, we have:

$$A=\frac{2y_v}{3(-b-x_v)^2}\left[(-b-x_v)^3 \right] +y_v(a+b)$$

$$\Rightarrow$$ $$A=\frac{2y_v}{3}(-b-x_v) +y_v(a+b)$$

Since $$-(a-x_v) = (-b-x_v)$$:

$$\Rightarrow$$ $$A=\frac{2y_v}{3}(x_v-a) +y_v(a+b)$$

Now, $$y(a)=0$$ and $$a>-b$$ imply that $$a>x_v$$. Then:

$$x_v-a<0$$

But $$x_v-a$$ is just the negative horizontal distance between the vertex of $$y(x)$$ and its greatest root. Then since dist$$(a,x_v) =$$ dist$$(-b,x_v)$$, $$x_v-a$$ must be half of the width of the parabola times negative one, that is:

$$x_v-a = -\frac{1}{2}(a+b)$$

Then we have:

$$A= \frac{-2y_v}{3}(\frac{1}{2}(a+b)) +y_v(a+b)$$

$$\Rightarrow$$ $$A= \frac{-y_v}{3}(a+b) +y_v(a+b)$$

$$\Rightarrow$$ $$A= \frac{2y_v}{3}(a+b)$$

$$\Rightarrow$$ $$A = \frac{2}{3}hw$$

Which was to be proven.

QED

It is clear, geometrically speaking, that if this result is true for "downward opening" parabolas with vertex in quadrant I or II, then it must be true for "upward opening" parabolas with vertex in quadrants III or IV. This can be rigorously proven, but the proof is almost identical to the above and thus I will omit it to avoid redundancy (I actually proved both cases on paper, the only difference is the sign of the areas, which is to be expected).

To me, this is a very interesting result. I came across a statement claiming what we just proved, and didn't believe it. Since the person who said it was not a math professional, I was suspicious. So I began an attempt to prove the statement and to my surprise, it was true. I hope this surprises a few readers as well.

Note by Ethan Robinett
2 years, 9 months ago

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