 # Simpson's 3/8th Rule MATLAB Program Example

## Question

Evaluate the integral x^4 within limits -3 to 3 using Simpson's 3/8th Rule.

## Solution

Let y(x)=x^4

here a=-3 and b=3

therefore (b-a)=6

let ‘n’ be the number of intervals. assume n=6 in this case.

also h=(b-a)/n = 6/6 =1

x: -3  -2  -1  0  1  2  3

y: 81  16  1  0  1  16  81

According to Simpson's 3/8th Rule:

## MATLAB Code for Simpson's 3/8th Rule

`%Created by myclassbook.org (Mayuresh)  %Created on 24 May 2013  %Question: Evaluate the integral x^4 within limits -3 to 3clc;  clear all;  close all;f=@(x)x^4; %Change here for different function  a=-3;b=3; %Given limits  n=b-a; %Number of intervals  h=(b-a)/n;  p=0;for i=a:b  p=p+1;  x(p)=i;  y(p)=i^4; %Change here for different function  endl=length(x);  x  y  answer=(3*h/8)*((y(1)+y(l))+3*(y(2)+y(3)+y(5)+y(6))+2*(y(4)))`

## Second Example

Evaluate the integral 1/(1+x) within limits 0 to 6 using Simpson's 3/8th rule.

## Solution

Let y(x)=1/(1+x)

here a=0 and b=6

therefore (b-a)=6

let ‘n’ be the number of intervals. assume n=6 in this case.

also h=(b-a)/n = 6/6 =1

x: 0                  1                    2              3                  4               5               6

y: 1.0000   0.5000   0.3333   0.2500   0.2000   0.1667   0.1429

According to Simpson's 3/8th rule:

## MATLAB Code for Simpson's 3/8th Rule

`%Created by myclassbook.org (Mayuresh)  %Created on 24 May 2013  %Question: Evaluate the integral 1/(1+x) within limits 0 to 6clc;  clear all;  close all;f=@(x)1/(1+x); %Change here for different function  a=0;b=6; %Given limits  n=b-a; %Number of intervals  h=(b-a)/n;  p=0;for i=a:b  p=p+1;  x(p)=i;  y(p)=1/(1+i); %Change here for different function  endl=length(x);  x  y  answer=(3*h/8)*((y(1)+y(l))+3*(y(2)+y(3)+y(5)+y(6))+2*(y(4)))`