Comparison of Newton-Cotes Quadrature and Gaussian Quadrature formulas Newton-Cotes quadrature formulas are a generalization of trapezoidal and Simpson's rule. The trapezoidal rule involves $2$ points, Simpson's rule involves $3$, and in general Newton-Cotes formulas exist for any number of sample points.  
There are also Gaussian quadrature rules, for any numbers of points. 
What are the main differences between a Newton-Cotes Quadrature formula and a Gaussian Quadrature formula?
 A: Newton-Cotes says "pick evenly spaced points in the interval, draw the interpolating polynomial of minimal degree through them, and integrate the polynomial." From the Lagrange interpolation theorem, a Newton-Cotes rule on $n$ nodes is exact for polynomials of degree at most $n-1$. It is not exact for polynomials of any higher degree, because one can add a polynomial of degree $n$ which vanishes at the $n$ nodes and takes on any desired value at an additional point, and the Newton-Cotes rule will not "detect" this change.
Gaussian quadrature says "find a rule that lets us exactly integrate polynomials of as large a degree as possible using $n$ nodes". This winds up being degree $2n-1$, since we have control of $2n$ parameters in specifying a quadrature rule at $n$ nodes, and the parameters are independent.
Analysis of either method is generally inspired by the Weierstrass approximation theorem, which generates the following type of argument. Here $I$ denotes exact integration, $Q_n$ denotes quadrature integration with a rule on $n$ nodes, and $p$ is a polynomial.
$$| I(f) - Q_n(f) | \leq | I(f) - I(p) | + | I(p) - Q_n(p) | + | Q_n(p) - Q_n(f) | $$
The Weierstrass theorem and the fact that $I$ is a bounded linear functional (in particular, $\| I \| = I(1)$) allow us to make the first term small, possibly at the expense of $p$ having a large degree.
Once the degree of $p$ is fixed, choosing $n$ large enough enables us to make the second term zero, since both Newton-Cotes and Gaussian quadrature can be made exact for polynomials by taking enough nodes.
The problem for the numerical analyst is essentially to control the last term, when $\| p - f \|$ is small and $n$ might be arbitrarily large.
Now $Q_n$, like $I$, is linear, so we can write
$$Q_n(p) - Q_n(f) = Q_n(p-f).$$
Also, $Q_n$ is bounded:
$$|Q_n(f)| \leq \| f \| \sum_{k=1}^n |w_k|,$$
where $w_k$ are the weights. Hence
$$\| Q_n \| \leq \sum_{k=1}^n |w_k|.$$
This inequality is actually an equality, as we can check by integrating $f$ such that $\| f \|=1$ and $f(x_k)=\text{sign}(w_k)$. (You should check for yourself that such a continuous $f$ exists.)
The fact that the $Q_n$ are each bounded is not good enough. Since simultaneously controlling the first two terms takes away our control over $n$, we need $Q_n$ to be uniformly bounded.
It turns out that Newton-Cotes rules do not have $Q_n$ uniformly bounded, whereas Gaussian quadrature does. The idea is to recognize that, because these methods are exact for polynomials of degree $0$, $\sum_{k=1}^n w_k = I(1)$. If the $w_k$ are all positive, then $\| Q_n \| = \sum_{k=1}^n |w_k| = I(1)$, which is independent of $n$, as we want.
One can prove from the property of being exact for polynomials of degree $2n-1$ that all of the weights in Gaussian quadrature are positive. (Proof sketch: integrating $\prod_{k=1,k \neq j}^n (x-x_k)^2$, whose degree is $2n-2$, proves $w_j>0$, now iterate over $j$.) Consequently Gaussian quadrature is convergent. In particular, if there exists $p$ such that $\| p - f \| < \varepsilon/(2I(1))$ and the degree of $p$ is at most $m$, then $|I(f)-Q_n(f)|<\varepsilon$ provided $n \geq (m+1)/2$.
The weights in Newton-Cotes quadrature are not positive. Moreover, $\| Q_n \|$ is unbounded for Newton-Cotes rules. This means that if $f=p+g$, where $p$ is a polynomial of degree at most $n-1$ and $g$ is a continuous function with $g(x_k)=\| g \| \text{sign}(w_k)$, then $|Q_n(p-f)|=\| Q_n \| \| g \| \gg \| g \|$.
Technically the above does not prove that Newton-Cotes rules are not convergent, because the "bad function" we picked depends on $n$. But it at least implies that $Q_n(f)$ for fixed large enough $n$ has a sensitive dependence on $f$, whereas $I(f)$ does not, which at least suggests that something bad is happening.
To give a full proof, you need an example. Runge's famous example with $1/(x^2+1)$ on a large enough interval centered at $0$ does the job. This example shows that a smooth function is not necessarily nice enough for Newton-Cotes rules to be convergent. For smooth integrands, one can develop a Taylor-with-remainder argument to give a sufficient condition for convergence, but the full story is really rooted in Fourier coefficients rather than derivatives.
A: @Jed;
Newton Cotes forms pick equally spaced points in the interval of integration, Gaussian quadrature picks the best points. For this reason Gaussian quadrature is more accurate and uses less panels. This means less function evaluations and therefore less chance of roundoff error and better speed. Also Newton's, includes the endpoints ( although there are forms that do not). Often these are difficult for a computer to handle and must be dealt with by the user. Gaussian does not include the endpoints.
Higher forms of Newton's also tend to have large coefficients. These will act as error multipliers and therefore it is less numerically stable than Gaussian.
On the other hand Newton Cotes' forms have the advantage of being easier to compute the error estimate. Because of this the Trapezoidal rule is used in Romberg integration.
