 # Simpson's 1/3rd Rule MATLAB Program Examples

## Question

Evaluate the integral x^4 within limits -3 to 3 using Simpson's 1/3 rd 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 1/3rd  rule:

## MATLAB Program For Simpson's 1/3rd 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=(h/3)*((y(1)+y(l))+2*(y(3)+y(5))+4*(y(2)+y(4)+y(6)))`

## Second Example

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

## MATLAB Code for Simpson's 1/3rd 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=(h/3)*((y(1)+y(l))+2*(y(3)+y(5))+4*(y(2)+y(4)+y(6)))`

## Image Format

1. % please I need to help how to integral (sin^3 ).x
% Generally sin^3 (x) is equal 1/4[3sinx - sin 3x]
clc;
clear all;
close all;

f=@(x) (sinX)^3 ; %Change here for different function
a=0;b=pi/2; %Given limits
n=b-a; %Number of intervals
N=6;
h=n/N;
p=0;

for i=a:b
p=p+1;
x(p)=i;
y(p)=(sin i)^3 ; %Change here for different function
end

l=length(x);
x
y
2. 