# Derivations of the trapezoid rule

I know the general method to derive the trapezoid rule is with Taylor series, or, you know, to just look at the trapezoids and figure out the rule. However, I feel that for such a simple rule, there must exist some other, perhaps simpler, derivations. I can't seem to find any online, however. Are there indeed more ways to derive the trapezoid rule? By simpler, I mean a simpler algebraic method.

• It is not clear from your question if you are referring to the rule or to the error formula? The rule can be derived geometrically or by integrating the interpolating polynomial of degree at most 1. Would you add a few words as to how the rule can be derived using Taylor polynomials? – Carl Christian Apr 10 at 17:01

Adding up the areas of the trapezia is very simple and natural... How much simpler could it be? You can also think of this rule (or any other numerical quadrature) as substituting $$f$$ by a constant function. The value of that constant function is a weighed average of values of $$f$$ (nodes with one neighbor have weight $$\frac 1n$$ and nodes with a single neighbor have weight $$\frac{1}{2n}$$).

• Could you explain more of what you mean by weighted constant functions? I’ll accept this answer if you do. – H Huang Apr 10 at 15:39
• The trapezoidal rule is given by $$Q(f)= \frac{h}{2}(f_0 + 2 f_1 + \cdots + 2 f_{n-1} + f_n)=(b-a)\times \frac{1}{n}(\frac{1}{2n} f_0 + \frac 1n f_1 + \cdots + \frac 1n f_{n-1} + \frac{1}{2n} f_n)$$ Apart from the $(b-a)$ term, what you have is an average value of $f$ on the integration nodes. This is a weighted average in the sense that the nodes don't have all the same weight. When you multiply $(b-a)$ by this average you get the area of a rectangle, that you can identify with the integral of a constant function. – PierreCarre Apr 10 at 17:02
• Thank you! I understand now. – H Huang Apr 10 at 17:59

The trapezoidal rule is derived from the simple area of a trapezium formula. See this video: