As we can see, both the first order ($n=1$), second order ($n=2$), and third order ($n=3$) Gaussian quadrature rules give exact answers for integrating $f(x) = 2x$. On the other hand, only the second and third order rule give an (essentially) exact answer for $f(x) = 3x^2$, while the first order rule isn't even close. All three quadrature rules give errors for integrating $f(x) = (3\pi/2)\cos(\pi x)$, but the second order rule is significantly more accurate than the first order rule while the third order rule is only marginally more accurate than the second order rule. The moral of the story is that, when choosing a quadrature rule, one should only choose a high enough order rule to reach a desired accuracy. Using extremely high order quadrature rules gives minimal improvements to accuracy.
We now know how to implement Gaussian quadrature for a differentiable real-valued function $f$ on an interval $[a,b]$. However, most functions that we encounter in practice may not be differentiable on an entire interval $[a,b]$ that we wish to integrate over.
A common example of such a function is a piecewise linear function. More explicitly, suppose that we are given data points $(t_1,y_1), \dots, (t_m,y_m)$ where $a \leq t_1 < t_2 < \cdots < t_m \leq b$. Then the piecewise linear interpolant $g$ with respect to these data points is the function defined by the following property: on each subinterval $[t_i,t_{i+1}]$, $g$ is the straight line connecting the data point $(t_i,y_i)$ to $(t_{i+1}, y_{i+1})$.
The graph of $g$ usually has sharp corners at the interpolation points $(t_i,y_i)$, and so $g$ is in general not differentiable at those points. As a result, Gaussian quadrature on the entire interval $[a,b]$ is not a good approximation to $$\int_a^bg(t)dt.$$ On the other hand, we make the following two observations:
- If we restrict ourselves to just looking at each subinterval $[t_i,t_{i+1}]$, $g$ is just a linear function on that interval, and is definitely differentiable. As a result, Gaussian quadrature on $[t_i, t_{i+1}]$ is a good approximation to $$\int_{t_i}^{t_{i+1}}g(t)dt.$$
- The integral of $g$ can be split up as integrals over the subintervals: $$\int_a^bg(t)dt = \sum_{i=1}^{m-1}\int_{t_i}^{t_{i+1}}g(t)dt.$$
Therefore, to approximate the integral of $g$ using Gaussian quadrature, we do the following two steps:
First we approximate the integral of $g$ on each subinterval: $$\int_{t_i}^{t_{i+1}}g(t)dt \approx \frac{t_{i+1}-t_i}{2}\sum_{j=1}^nw_jg\left(\frac{t_{i+1}-t_i}{2}s_j + \frac{t_i+t_{i+1}}{2}\right),$$ where the pairs $(w_j, s_j)$ are the weights and points for the $n$-th order Gaussian quadrature rule and the $t_i$ are the nodes for which $g$ is linear on each $[t_i,t_{i+1}]$.
We add up all of our approximations to approximate the entire integral of $g$ on $[a,b]$: $$\int_a^bg(t)dt \approx \sum_{i=1}^{m-1}\frac{t_{i+1}-t_i}{2}\sum_{j=1}^nw_jg\left(\frac{t_{i+1}-t_i}{2}s_j + \frac{t_i+t_{i+1}}{2}\right).$$
This simple modification of first breaking up the integral over $[a,b]$ into integrals over subintervals $[t_i,t_{i+1}]$ and then applying Gaussian quadrature to each piece gives us a method to approximate the integral of piecewise linear functions $g$ on an interval $[a,b]$. In fact, this method works for any function $g$ that is simply differentiable on each subinterval $[t_i,t_{i+1}]$; $g$ does not have to be linear on these intervals.
We now know how to use Gaussian quadrature to approximate differentiable functions as well as piece-wise differentiable functions on an interval $[a,b]$. As a final example, we will compute a second order Gaussian quadrature of a piecewise linear function $g$ on $[0,1]$, first using the "wrong" method that doesn't split up the integral over the subintervals, and then using the "right" method that splits the integral over the subintervals. We define our function $g$ in such a way so that the exact integral should be $0$.
