[![enter image description here](https://i.stack.imgur.com/i5Gdh.png)](https://i.stack.imgur.com/i5Gdh.png)

Using figure 8, let $x\_0 = -h, x\_1=0, x\_2=h$

So the area under the curve is exemplified below:
$$\int\_{-h}^h(Ax^2+Bx+C)dx=2\int\_0^h(Ax^2+C)dx$$
$$=2\left[A\frac{x^3}{3}+Cx\right]\_0^h$$
$$=2\left(A\frac{h^3}{3}+Ch\right)=\frac{h}{3}(2Ah^2+6C)$$

But since the parabola passes through $P\_0(-h,y\_0),P\_0(0,y\_1),$ and $P\_2(h,y\_2)$, we have
$$y\_0=A(-h)^2+B(-h)+c=AH^2-Bh+C$$
$$y\_1=C$$
$$y\_2=Ah^2+Bh+C$$

and therefore $y\_0+4y\_1+y\_2 = 2Ah^2+6C$

Where does the $4y\_1$ come from?