# Is a Riemann Sum considered a Newton-Cotes Formula?

I cannot find a definitive answer for this online but my thinking is that much like :

1. Simpson's rule is newton-cotes of the 2nd degree (parabolic function)
2. Trapezoidal is newton-cotes of the 1st degree (linear function),
3. Then a Riemann Sums is a newton-cotes formula of the 0th degree (constant function)

Sorry if this is simple but its just a matter of how I title one of my mathematics essays and I don't want to preface my writing with an error. Any help is appreciated, thank you.

If you are concerned about the precise historical definition, there are closed and open Newton-Cotes formulas.

The closed formulas approximate the integral over an interval $$[a,b]$$ using the points $$x_k = a + k\frac{b-a}{n}$$ for $$k = 0,1,\ldots,n$$, which includes endpoints. The lowest-order closed formula is the trapezoidal rule, corresponding to $$n=1$$, using step size $$h = (b-a)$$, and points $$x_0 = a$$ and $$x_1 = b$$, providing the approximation

$$\int_a^b f(x) \, dx \approx \frac{h}{2}[f(x_0) + f(x_1)],$$

representing the average of terms arising in left- and right-Riemann sums.

By convention, there is no $$0$$-th degree closed formula.

The open formulas use the points $$x_k = a + k\frac{b-a}{n}$$ for $$k = 1,\ldots,n-1$$, which excludes endpoints. In this case, a trapezoidal approximation arises when $$n = 3$$, using using step size $$h = \frac{b-a}{3}$$, and points $$x_1 = a + \frac{h}{3}$$ and $$x_2 = a + \frac{2h}{3}$$, providing the approximation

$$\int_a^b f(x) \, dx \approx \frac{3h}{2}[f(x_1) + f(x_2)]$$

When $$n = 2$$ we have the midpoint rule using step size $$h = \frac{b-a}{2}$$ and the single point $$x_1 = \frac{a+b}{2}$$, providing the approximation

$$\int_a^b f(x) \, dx \approx 2hf(x_1) = f\left(\frac{a+b}{2} \right)(b-a)$$

In this way, the midpoint rule is the open Newton-Cotes formula of degree $$2$$ and is one term of a particular Riemann sum for an integral over a larger interval where $$[a,b]$$ is a partition subinterval.