A General Quadrature Formula

Let $f(x_k)=y_k$ be the nodal value at the tabular point $x_k$ for $k = 0, 1, \cdots,x_n,$ where $x_0 = a$ and $x_n = x_0 + nh =b.$ Now, a general quadrature formula is obtained by replacing the integrand by Newton's forward difference interpolating polynomial. Thus, we get,

$$\int\limits^{b}_{a}f(x)dx = \int\limits^{b}_{a}\left[y_0+\frac{\Delta y_0}{h}(x-x_0) + \frac{... ...0}{2!h^2}(x-x_0)(x-x_1) +\frac{\Delta^3 y_0}{3!h^3}(x-x_0)(x-x_1)(x-x_2)\right.$$

$$+ \left.\frac{\Delta^4 y_0}{4!h^4}(x-x_0)(x-x_1)(x-x_2)(x-x_3)+ \cdots \right]dx$$

This on using the transformation $x= x_0 + hu$ gives:

$$\int\limits^{b}_{a}f(x)dx = h \int\limits^{n}_{0}\left[y_0+ u \Delta y_0 + \frac{\Delta^2 y_0}{2!}u(u-1) +\frac{\Delta^3 y_0}{3!}u(u-1)(u-2)\right.$$

$$+ \left.\frac{\Delta^4 y_0}{4!}u(u-1)(u-2)(u-3)+ \cdots \right]du$$

which on term by term integration gives,

$$\int\limits^{b}_{a}f(x)dx = h \left[n y_0+ \frac{n^2}{2} \Delta y_0 + \frac{\Delta^2 y_0}{2!}... ...{2}\right) +\frac{\Delta^3 y_0}{3!}\left(\frac{n^4}{4}-{n^3}+n^2 \right)\right.$$

$$+ \left.\frac{\Delta^4 y_0}{4!}\left(\frac{n^5}{5}-\frac{3n^4}{2}+\frac{11n^3}{3}-3n^2\right)+ \cdots \right]$$ (13.3.1)

For $n = 1,$ i.e., when linear interpolating polynomial is used then, we have

$$\int\limits^{b}_{a}f(x)dx = h\left[y_0+\frac{\Delta y_0}{2}\right]=\frac{h}{2} \left[y_0 +y_1\right].$$ (13.3.2)

Similarly, using interpolating polynomial of degree 2 (i.e. $n = 2$ ), we obtain,

$$\int\limits^{b}_{a}f(x)dx = h\left[2y_0 + 2\Delta y_0 + \left(\frac{8}{3}-\frac{4}{2}\right)\frac{\Delta^2 y_0}{2}\right]$$

$$= 2h\left[y_0 +(y_1- y_0)+\frac{1}{3}\times\frac{y_2 -2y_1 +y_0}{2}\right] =\frac{h}{3}\left[y_0 + 4 y_1 + y_2 \right].$$ (13.3.3)

In the above we have replaced the integrand by an interpolating polynomial over the whole interval $[a, b]$ and then integrated it term by term. However, this process is not very useful. More useful Numerical integral formulae are obtained by dividing the interval $[a, b]$ in $n$ sub-intervals $[x_k, x_{k+1}],$ where, $x_k = x_0 + kh$ for $k = 0, 1, \cdots, n$ with $x_0 = a, x_n = x_0 + nh =b.$