If you’ve kept up on recent blog posts, you’ll notice that in our data one method for integration seemed to work without any error. This method was Simpson’s Rule. This fact sparked our interest as to why does this method work so well? In particular, we noticed that it was almost flawless with cubic functions. Why is this? From the text, we were given the idea to algebraically compute Simpson’s rule approximation and compare it to the actual integral of the special case in which our interval is [-h,h] and n, our number of rectangles, being 2 for $f(x)=ax^{3}+bx^{2}+cx+d$.

We began by computing the actual integration:

$\int\limits_{-h}^{h} (ax^{3}+bx^{2}+cx+d) dx$ $\implies$ ($\frac{ax^{4}}{4}$ + $\frac{bx^{3}}{3}$ + $\frac{cx^{2}}{2}$ + $dx$ + $C$)|$^{h}_{-h}$

[$\frac{ah^{4}}{4}$ + $\frac{bh^{3}}{3}$ + $\frac{ch^{2}}{2}$ + $dh$ + $C$]-[$\frac{a(-h)^{4}}{4}$ + $\frac{b(-h)^{3}}{3}$ + $\frac{c(-h)^{2}}{2}$ + $d(-h)$ + $C$]

$= \frac{2bh^{3}}{3} + 2dh$

The equation for Simpson’s Rule is $\frac{1}{3}[f(x_{i-1})+4f(x_{i})+f(x_{i+1})]h$

In using two rectangles we get: $\frac{h}{3}[f(-h)+4f(0)+f(h)]$ $\implies$ $\frac{h}{3}[(a(-h)^{3}+b(-h)^{2}+c(-h)+d)+4d+(a(h)^{3}+b(h)^{2}+c(h)+d)]$

$= \frac{h}{3}(2bh^{2}+6d)= \frac{2bh^{3}}{3} + 2dh$

And thus we find that in cubic functions Simpson’s rule is the most accurate method for computing the integration.

For further research we can look at what happens when we change the bounds to [a,b]. We can also calculate the actual integral and the Simpson’s rule for different types of functions too see if this method really holds up as being the most accurate method for numerical integration.