Quadrature of Parabola

Theorem
Let $T$ be a parabola.

Consider the parabolic segment bounded by an arbitrary chord $AB$.

Let $C$ be the point on $T$ where the tangent to $T$ is parallel to $AB$.

Let

Then the area $S$ of the parabolic segment $ABC$ of $T$ is given by:
 * $S = \dfrac 4 3 \triangle ABC$

Proof
, consider the parabola $y = a x^2$.

Let $A, B, C$ be the points:


 * ParabolaQuadrature2.png

The slope of the tangent at $C$ is given by using:
 * $\dfrac {\d y} {\d x} 2 a x_1$

which is parallel to $AB$.

Thus:
 * $2 a x_1 = \dfrac {a {x_0}^2 - a {x_2}^2} {x_0 - x_2}$

which leads to:
 * $x_1 = \dfrac {x_0 + x_2} 2$

So the vertical line through $C$ is a bisector of $AB$, at point $P$.

Complete the parallelogram $CPBQ$.

Also, find $E$ which is the point where the tangent to $T$ is parallel to $BC$.

By the same reasoning, the vertical line through $E$ is a bisector of $BC$, and so it also bisects $BP$ at $H$.

Next:

At the same time:

So:
 * $QB = 4 FE = FH$

and because $CB$ is the diagonal of a parallelogram:
 * $2 FE = 2 EG = FG$

This implies that:
 * $2 \triangle BEG = \triangle BGH$

and:
 * $2 \triangle CEG = \triangle BGH$

So:
 * $\triangle BCE = \triangle BGH$

and so as $\triangle BCP = 4 \triangle BGH$ we have that:


 * $BCE = \dfrac {\triangle BCP} 4$

A similar relation holds for $\triangle APC$:


 * ParabolaQuadrature1.png

so it can be seen that:
 * $\triangle ABC = 4 \paren {\triangle ADC + \triangle CEB}$

Similarly, we can create four more triangles underneath $\triangle ADC$ and $\triangle CEB$ which are $\dfrac 1 4$ the area of those combined, or $\dfrac 1 {4^2} \triangle ABC$.

This process can continue indefinitely.

So the area $S$ is given as:
 * $S = \triangle ABC \paren {1 + \dfrac 1 4 + \dfrac 1 {4^2} + \cdots}$

But from Sum of Geometric Progression it follows that:


 * $S = \triangle ABC \paren {\dfrac 1 {1 - \dfrac 1 4} } = \dfrac 4 3 \triangle ABC$