# Composite Simpsons 1/3 matlab code

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1. May 12, 2014

### haz

hey guys,
So I'm working on a matlab function that uses simpsons 1/3 rule to find an integral.
This is what I have done so far, but I'm not 100% confident.. I seem to get double when I use greater segments.
If anyone would be able to have a look at give me some tips that would be very much appreciated.

function [ I ] = simpsons3( func,a,b,n )
%Finds estimate integral from a to b of function using simpson 1/3 rule

% % INPUTS
% % func = function
% % a = lower limit
% % b = upper limit
% % n=number of segments used for integration
% % OUTPUTS
% % I = integral estimate

h = (b-a)/(n-1);
sumE = 0;
for i = a+2*h:h:b-2*h
sumE = sumE + func(i);
end
sumO = 0;
for i = a+h:h:b-h
sumO = sumO + func(i);
end
I = (h/3)*(func(a)+4*sumE+2*sumO+func(b));

2. May 12, 2014

### Staff: Mentor

Always good to test with a function for which you know the analytical result and compare.

Minor point, but if n is the number of intervals, then
Code (Text):

h = (b-a)/n;

First, it is a bad idea to use real values for a for loop counter. Use integers and calculate the independent variable from h at each step.

Second, you are doubly counting the interior points. Each sum should add points that are 2h apart, not h.

Third, such an implementation of the Simpson rule only works if the number of intervals is even (or n odd according to your definition of h), otherwise it will return a result with a greater error.

3. May 13, 2014

### haz

I am however confused as to why it is a bad idea to set up my for loop counters as I have done. Would you be able to tell be me why it is bad and give me examples as too when this would fail.

4. May 13, 2014

### Staff: Mentor

Because of rounding errors, you can end up with unexpected behaviour at the termination of a for loop. Say you have a for loop that counts to 15, but when you add the step sizes h, you end up with the last values being 15.00000001. This being > 15, the loop will exit without the last point being counted.