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 :


*

*Simpson's rule is newton-cotes of the 2nd degree (parabolic function)

*Trapezoidal is newton-cotes of the 1st degree (linear function), 

*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.
 A: 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.
