- The difference between the Simpson’s 1/3rd and 3/8th Rules is that the former has been derived using quadratic, while the latter has been derived using cubic interpolation (Trapezoidal Rule uses linear interpolation). I hope you are familiar with the concept of interpolation. If not, this is a brief outline:
You can sketch a random function using some finite points alone. For instance take the function x + 5. Now, to represent this equation on a graph, you need to ideally plot infinite points by varying the value of x. However, interpolation lets you approximate the full graph using just a few points. In this case it becomes very easy – you just have to join all the finite points using a straight line (linear fit) and extend it. However, in functions of higher orders, how you join the points becomes very crucial, and interpolation becomes more meaningful. The higher the degree of interpolation, the better will be the approximation to the actual function.
You can look up the derivation of the two methods in Math text books. I think it is beyond the scope and necessity of this paper, since interpolation methods will then need to be covered. Further, yes, they have different formulae, as follows:
Simpson’s 1/3rd Rule:
Simpson’s 3/8th Rule:
For power 4 you can use Boole’s Law. For higher orders, the math involved becomes very complex, which means the whole point of interpolation (which is for convenience) is lost. So generally approximation with higher degrees is not seen. However, a common method followed in such cases uses Newton-Cote’s formulas.
- The math behind it is as follows:
Let u = ln(x), and dv = dx
Using integration by parts, du = (1/x) dx, and v = x
Integral ln(x) dx = Integral u dv which yields uv - integral vdu. Substituting the above values, this becomes ln(x) x – integral (x (1/x) dx) = ln(x) x - integral dx = x lnx – x. Applying limits to this yields the required answer.
- ‘h’ in the formula is the width of each partition. Assume you want to integrate a function between the limits a and b. To do this, you divide the function into n intervals or partitions. Then the size or width of each partition = (b-a)/n.
In this particular formula, n = 3. Therefore, h = (b-a)/3.
