Why is Simpson's rule more accurate than the Trapesium rule when calculating integrals?

Question

Thanks for any answers!

Answer 1

It's not that Simpson's rule is necessarily more accurate for every problem. Picture the graph of a function, under which you are trying to find the area. The Trapesium rule subdivides that area into little trapezoid-shaped slices. So, for functions that have a lot of nearly-straight sections to their graphs, the Trapesium rule is more accurate. By contrast, Simpson's rule subdivides the area into slices that have little curves for their tips, so Simpson's rule is more accurate for functions that have a lot of curvy sections.

Answer 2

Simpson's rule uses a weighted combination of the Trapezium rule and the Midpoint Rule so as to eliminate part of the error generated by both.

There's an explanation here:

http://en.wikipedia.org/wiki/Simpson%27s_rule#Averaging_the_midpoint_and_the_trapezium_rules

Answer 3

If a function does not integrate you can integrate function’s polynomial approximation y(x) = a0 + a1*(x-x0) +a2*(x-x0)^2 +a3*(x-x0)^3 +++ the higher power the better approximation. Thus y(x) = a0 + a1*(x-x0) +a2*(x-x0)^2 is more accurate than just y(x) = a0 + a1*(x-x0);

Answer 4

D'oh!