Finding the Lowest Order Error Term in the Trapezium Rule?

Question

I'm looking at the integral of a function f(x) between x0 and x0+delta_x.

The trapezium rule is that this integral is equal to 1/2 * the sum between i=1 and i= N-1 of( f(x0) + f(x0 + i*delta_x)) * delta_x.

Thanks.

Answer 1

Wikipedia is almost always your friend:
http://en.wikipedia.org/wiki/Newton-Cotes_formulas

One way to think about this is that you are approximating the function by a Taylor series because a polynomial is easy to integrate. In general:

T(f,a,x) = f(a) + f'(a)x + (1/2!)f''(a)x^2 + ...

Suppose we take only the first term and integrate over one interval. That gives us f(a) times delta-x. The error begins at the second term, the one in f'(a).

When we are using trapezoids, we are using the first two terms so the error is in the f''(a)x^2 term.

With Simpson's rule, we re using the first three terms, etc.

For more detail, check out:
http://math.fullerton.edu/mathews/n2003/TrapezoidalRuleMod.html

or
http://www.krellinst.org/UCES/archive/classes/CNA/dir2.3/uces2.3.html