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When deriving Simpson's 1/3 Rule, I used a second order polynomial $P(x) = Ax^2 + Bx + C$, and integrated over the region $[-h,h]$

Integrating gave me: $ \ \dfrac{h}{3}(2Ah^2 +6C)$

I evaluated $P(x)$ at values $-h, 0, h$ adding the results together to get:

$ (2Ah^2 +6C) = P(-h) + 4P(0) + P(h)$

And I could carry on from there to get Simpson's 1/3 rule

However, when I am deriving Simpson's 3/8 Rule, I am using a third order polynomial $P(x) = Ax^3 + Bx^2 + Cx + D$

I integrate over the region $[-h,h]$ which gives me:

$\dfrac{2h}{3}(Bh^2 +3D)$

I know I am now meant to evaluate P(x) at values $-h, \frac{-h}{3}, \frac{-h}{3}, h $, but I am unsure why - is it simply because I have a cubic polynomial, thus I have four points and so I need evenly spaced points, increasing by $\frac{2h}{3}$ each time?

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The short answer is yes, if you are doing $n$'th order interpolation, you need $n+1$ points. Simpson's rule works for any function $f$. It sounds like you're doing cubic interpolation, which means you need four points. In general, the error in Simpson's rule is proportional to $f^{(n+1)}(\xi)$, where $\xi$ is some point in your interval $[a,b]$. Thus, if you're integrating a cubic polynomial via Simpson's rule, it's oppertunistic to use the third order (3/8'ths) rule because $f^{(4)}(\xi)$ for a cubic polynomial is zero for any $\xi$, hence the answer is exact.

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