Quadrature rules

Number of points:

The rectangle rule (with $n+1$ points) says that \[ \int_a^b f(x)\,dx \approx h\sum_{i=0}^{n-1} f(x_i), \] where $h = (b-a)/n$ and $x_i = a + ih$, so $x_0,\dotsc,x_n$ are equally spaced points between $a$ and $b$.

The trapezium rule (with $n+1$ points) says that \[ \int_a^b f(x)\,dx \approx h(\tfrac{1}{2}f(x_0)+f(x_1)+\dotsb+f(x_{n-1})+\tfrac{1}{2}f(x_n)), \] where $h = (b-a)/n$ and $x_i = a + ih$, so $x_0,\dotsc,x_n$ are equally spaced points between $a$ and $b$. Equivalently, for $0\leq i\lt n$ we make a trapezium with vertices $(x_i,0)$, $(x_i,f(x_i))$, $(x_{i+1},f(x_{i+1}))$, $(x_{i+1},0)$, then we add together the areas of these trapezia to get our approximation to the integral.

Simpson's rule (with $2m+1$ points) says that \[ \int_a^b f(x)\,dx \approx \frac{h}{6}(y_0+4y_1+2y_2+4y_3+\dotsb+2y_{2m-2}+4y_{2m-1}+y_{2m}), \] where $h = (b-a)/(2m)$ and $x_i = a + ih$ (so $x_0,\dotsc,x_{2m}$ are equally spaced points between $a$ and $b$) and $y_i=f(x_i)$. Equivalently, for $0\leq i\lt m$ we find a quadratic function $q_i(x)$ that agrees with $f(x)$ at $x_{2i}$, $x_{2i+1}$, $x_{2i+2}$, and then we add together the areas under these quadratics to get our approximation to the integral. It is easy to integrate a quadratic function exactly, and with some algebra we can check that the end result is just given by the formula above. Thus, when using this method, there is no need to actually find the quadratics $q_i(x)$, we can just use the formula.

The Gaussian quadrature rule uses some carefully chosen constants $(g_0,\dotsc,g_6)\approx(-1,-0.91,-0.54,0,0.54,0.91,1)$ and $(w_0,\dotsc,w_6)\approx(0,0.24,0.48,0.57,0.48,0.24,0)$. The number of points is always $n=5m$ for some $m$. We first divide the interval $[a,b]$ into $m$ subintervals $[t_i,t_{i+1}]$, of length $2h$ (where $h=(b-a)/(2m)$) and centre $c_i=(t_i+t_{i+1})/2=a+(2i+1)h$. Within each subinterval, we consider the points $c_i + hg_j$ with weight $w_j$ for $j=1,\dotsc,5$. This gives $5m$ points $x_0,\dotsc,x_{5m-1}$ in total, with weights $v_0,\dotsc,v_{5m-1}$. The integral is then approximated by the sum \[ \int_a^b f(x)\,dx \approx \sum_{k=0}^{5m-1} h v_k f(x_k). \] It can be shown that this is equivalent to the following formula. For each subinterval $[t_i,t_{i+1}]$, there is a unique polynomial $p_i(x)$ of degree at most $6$ that agrees with $f(x)$ at the points $c_i+hg_j$ for $j=0,\dotsc,6$ (i.e. the points used in the integral formula together with the extra points $t_i=c_i+hg_0$ and $t_{i+1}=c_i+hg_6$). It turns out that the above formula is the same as $\sum_i\int_{t_i}^{t_{i+1}}p_i(x)\,dx$. Typically the graph of $p_i(x)$ will be very close to that of $f(x)$, so the approximate integral will be very close to the true integral.