Area of parabola using "weighted" average? I was watching an online lecture of calculus when the professor was going over 3 ways of preforming numerical integration; these were Riemann Sums, the Trapezoid Rule, and the Simpson's Rule. When explaining the Simpson's Rule he used a formula for the area of the parabola with three points $y_1,y_2,$ and $y_3$ where $y_1$ and $y_3$ were the ends of it and $y_2$ would be there midpoint. The formula was
$$2\Delta x(\frac{y_1+4y_2+y_3}{6})$$
where $2\Delta x$ was the width and the rest of the equation was the "weighted" average height. I understand why he used $2 \Delta x$ as his base but cannot seem to find how to derive this "weighted" average without using any integration (It kinda cause a loop where you define an integral using the area of parabola defined by an integral ...). I have seen  Archimedes' Quadrature of the Parabola but cannot seem to to find a way to use it to derive this utterly strange and "weighted" average height.
 A: This can be proved by using Archimedes' Theorem, stating that the area of parabolic segment $P_1P_2P_3$ is $4/3$ the area of triangle $P_1P_2P_3$ (see picture below). Notice first of all that the area $S_{tot}$ comprised between parabolic arc $P_1P_2P_3$ and the $x$ axis, by Archimedes' theorem can be written as:
$$
S_{tot}=S_{ACP_3P_1}+{4\over3}S_{P_3P_1P_2},
$$
where I denote by $S_{P_3P_1P_2}$ the area of polygon ${P_3P_1P_2}$ and so on, that is
$$
S_{tot}=S_{ACP_3P_1}+{4\over3}(S_{ABP_2P_1}+S_{BCP_3P_2}-S_{ACP_3P_1})
={1\over3}(4S_{ABP_2P_1}+4S_{BCP_3P_2}-S_{ACP_3P_1}).
$$
Substitute now here $S_{ABP_2P_1}=\Delta x(y_2+y_1)/2$, 
$S_{BCP_3P_2}=\Delta x(y_2+y_3)/2$ and $S_{ACP_3P_1}=2\Delta x(y_3+y_1)/2$
to obtain your formula.

A: If $y=f(x)=ax^2+bx+c$ is the equation of the parabola, let us compute
$$f(x_1)+4f\left(\dfrac{x_1+x_2}{2}\right)+f(x_2)=$$
$$=a \left(x_1^2+4f\left(\dfrac{x_1+x_2}{2}\right)^2+x_2^2 \right)+b(x_1+4(\dfrac{x_1+x_2}{2})+x_2)+6c$$
When you expand this expression, you get:
$$a(2x_1^2+2x_2^2+2x_1x_2)+b(3x_1+3x_2)+6c$$
Now divide it by $6$ and multiply it by $\Delta x := x_2-x_1$:
(note: my $\Delta x$ is your $2\Delta x$).
$$(x_2-x_1) \ \left[\dfrac{1}{3}a(x_1^2+x_2^2+x_1x_2)+\dfrac{1}{2}b(x_1+x_2)+c \right] $$
you obtain
$$ \dfrac{1}{3}a(x_2^3-x_1^3)+\dfrac{1}{2}b(x_2^2-x_1^2)+c(x_2-x_1)$$
which coincides with 
$$\int_{x=x_1}^{x=x_2}(ax^2+bx+c)dx=\left[a \dfrac{x^3}{3} + b \dfrac{x^2}{2} + cx \right]_{x=x_1}^{x=x_2}$$
Edit: following a discussion with Ziad Fakhoury, there is no true intuitive rationale behind the 1/6, 4/6, 1/6 coefficients. 
Besides, it should be known that there is a more general theory that includes this list of weights as a particular case: it is the theory of Newton-Cotes formulas: for example, instead of choosing one point $m$ in the middle between $x_1$ and $x_2$, you can choose for example two: $m_1=(2x_1+x_2)/3$ and $m_2=(x_1+2x_2)/3$, splitting interval $(x_1,x_2)$ into three equal-length intervals, instead of two ; in this case, the corresponding Newton-Cotes formula is $$(x_2-x_1) \dfrac{f(x_1)+3f(m_1)+3f(m_2)+f(x_2)}{8}$$ (weights $1/8, 3/8, 3/8, 1/8$) giving an exact value of the integral $\int_{x_1}^{x_2}f(x)dx$ for any polynomial function $f$ of degree at most 4.
