Proof

We split the interval into subintervals and apply Simpson's Rule on each pair of consecutive intervals:

> int1:=(h/3)*(f(a)+4*f(a+h)+f(a+2*h))+err(xi[1]);

> int2:=(h/3)*(f(a+2*h)+4*f(a+3*h)+f(a+4*h))+err(xi[2]);

> int3:=(h/3)*(f(a+4*h)+4*f(a+5*h)+f(a+6*h))+err(xi[3]);

and so on until

> intm:=(h/3)*(f(a+(n-2)*h)+4*f(a+(n-1)*h)+f(b))+err(xi[m]);

where each of the error term is of the form

> err(xi[i])=-h^5*(D@@4)(f)(xi[i])/90;

Adding all the integrals together yields:

> inttot:=int1+int2+int3+intm;

which can be rewritten as

>

This yields the composite formula. The error term can be simplified by considering that for all values of ,

,

with the minimal value of the fourth derivative over the interval . Then

, or

.

Similarly, for the maximum over :

,

, or

.

Using the Mean Value Theorem, we can then say that there must exist a value in the interval so that

The error term then becomes:

or

which, with can finally be written as

>