Program for Simpson’s 1/3 Rule

In numerical analysis, Simpson’s 1/3 rule is a method for numerical approximation of definite integrals. Specifically, it is the following approximation: 
[Tex]$$\int_{a}^{b} f(x) dx \approx \frac{(b-a)}{6} \bigg(f(a) + 4f \frac{(a+b)}{2} + f(b)\bigg)$$      [/Tex]
In Simpson’s 1/3 Rule, we use parabolas to approximate each part of the curve.We divide 
the area into n equal segments of width Δx. 
Simpson’s rule can be derived by approximating the integrand f (x) (in blue) 
by the quadratic interpolant P(x) (in red). 

In order to integrate any function f(x) in the interval (a, b), follow the steps given below:
1.Select a value for n, which is the number of parts the interval is divided into. 
2.Calculate the width, h = (b-a)/n 
3.Calculate the values of x0 to xn as x0 = a, x1 = x0 + h, …..xn-1 = xn-2 + h, xn = b. 
Consider y = f(x). Now find the values of y(y0 to yn) for the corresponding x(x0 to xn) values. 
4.Substitute all the above found values in the Simpson’s Rule Formula to calculate the integral value.
Approximate value of the integral can be given by Simpson’s Rule: 
[Tex]$$\int_{a}^{b} f(x) dx \approx \frac{h}{3} \bigg(f_0 + f_n + 4 * \sum_{i=1,3,5}^{n-1}f_i + 2* \sum_{i=2,4,6}^{n-2}f_i \bigg)$$      [/Tex]
Note : In this rule, n must be EVEN.
Application :
It is used when it is very difficult to solve the given integral mathematically. 
This rule gives approximation easily without actually knowing the integration rules.
Example : 

Evaluate logx dx within limit 4 to 5.2. First we will divide interval into six equal parts as number of interval should be even. x : 4 4.2 4.4 4.6 4.8 5.0 5.2 logx : 1.38 1.43 1.48 1.52 1.56 1.60 1.64 Now we can calculate approximate value of integral using above formula: = h/3[( 1.38 + 1.64) + 4 * (1.43 + 1.52 + 1.60 ) +2 *(1.48 + 1.56)] = 1.84 Hence the approximation of above integral is 1.827 using Simpson's 1/3 rule.

C++

// CPP program for simpson's 1/3 rule
#include <iostream>
#include <math.h>
using namespace std;

// Function to calculate f(x)
float func(float x)
{
    return log(x);
}

// Function for approximate integral
float simpsons_(float ll, float ul, int n)
{
    // Calculating the value of h
    float h = (ul - ll) / n;

    // Array for storing value of x and f(x)
    float x[10], fx[10];

    // Calculating values of x and f(x)
    for (int i = 0; i <= n; i++) {
        x[i] = ll + i * h;
        fx[i] = func(x[i]);
    }

    // Calculating result
    float res = 0;
    for (int i = 0; i <= n; i++) {
        if (i == 0 || i == n)
            res += fx[i];
        else if (i % 2 != 0)
            res += 4 * fx[i];
        else
            res += 2 * fx[i];
    }
    res = res * (h / 3);
    return res;
}

// Driver program
int main()
{
    float lower_limit = 4; // Lower limit
    float upper_limit = 5.2; // Upper limit
    int n = 6; // Number of interval
    cout << simpsons_(lower_limit, upper_limit, n);
    return 0;
}

Java

// Java program for simpson's 1/3 rule

public class GfG{

    // Function to calculate f(x)
    static float func(float x)
    {
        return (float)Math.log(x);
    }

    // Function for approximate integral
    static float simpsons_(float ll, float ul,
                                       int n)
    {
        // Calculating the value of h
        float h = (ul - ll) / n;

        // Array for storing value of x
        // and f(x)
        float[] x = new float[10];
        float[] fx= new float[10];

        // Calculating values of x and f(x)
        for (int i = 0; i <= n; i++) {
            x[i] = ll + i * h;
            fx[i] = func(x[i]);
        }

        // Calculating result
        float res = 0;
        for (int i = 0; i <= n; i++) {
            if (i == 0 || i == n)
                res += fx[i];
            else if (i % 2 != 0)
                res += 4 * fx[i];
            else
                res += 2 * fx[i];
        }

        res = res * (h / 3);
        return res;
    }

    // Driver Code
    public static void main(String s[])
    {
        // Lower limit
        float lower_limit = 4;

        // Upper limit
        float upper_limit = (float)5.2;

        // Number of interval
        int n = 6;

        System.out.println(simpsons_(lower_limit,
                                upper_limit, n));
    }
}

// This code is contributed by Gitanjali

Python3

# Python code for simpson's 1 / 3 rule
import math

# Function to calculate f(x)
def func( x ):
    return math.log(x)

# Function for approximate integral
def simpsons_( ll, ul, n ):

    # Calculating the value of h
    h = ( ul - ll )/n

    # List for storing value of x and f(x)
    x = list()
    fx = list()

    # Calculating values of x and f(x)
    i = 0
    while i<= n:
        x.append(ll + i * h)
        fx.append(func(x[i]))
        i += 1

    # Calculating result
    res = 0
    i = 0
    while i<= n:
        if i == 0 or i == n:
            res+= fx[i]
        elif i % 2 != 0:
            res+= 4 * fx[i]
        else:
            res+= 2 * fx[i]
        i+= 1
    res = res * (h / 3)
    return res

# Driver code
lower_limit = 4   # Lower limit
upper_limit = 5.2 # Upper limit
n = 6 # Number of interval
print("%.6f"% simpsons_(lower_limit, upper_limit, n))

C#

// C# program for simpson's 1/3 rule
using System;

public class GfG
{

    // Function to calculate f(x)
    static float func(float x)
    {
        return (float)Math.Log(x);
    }

    // Function for approximate integral
    static float simpsons_(float ll, float ul,
                                        int n)
    {
        // Calculating the value of h
        float h = (ul - ll) / n;

        // Array for storing value of x
        // and f(x)
        float[] x = new float[10];
        float[] fx= new float[10];

        // Calculating values of x and f(x)
        for (int i = 0; i <= n; i++) {
            x[i] = ll + i * h;
            fx[i] = func(x[i]);
        }

        // Calculating result
        float res = 0;
        for (int i = 0; i <= n; i++) {
            if (i == 0 || i == n)
                res += fx[i];
            else if (i % 2 != 0)
                res += 4 * fx[i];
            else
                res += 2 * fx[i];
        }

        res = res * (h / 3);
        return res;
    }

    // Driver Code
    public static void Main()
    {
        // Lower limit
        float lower_limit = 4;

        // Upper limit
        float upper_limit = (float)5.2;

        // Number of interval
        int n = 6;

        Console.WriteLine(simpsons_(lower_limit,
                                upper_limit, n));
    }
}

// This code is contributed by vt_m

PHP

<?php
// PhP program for simpson's 1/3 rule

// Function to calculate f(x)
function func($x)
{
    return log($x);
}

// Function for approximate integral
function simpsons_($ll, $ul, $n)
{

    // Calculating the value of h
    $h = ($ul - $ll) / $n;

    // Calculating values of x and f(x)
    for ($i = 0; $i <= $n; $i++)
    {
        $x[$i] = $ll + $i * $h;
        $fx[$i] = func($x[$i]);
    }

    // Calculating result
    $res = 0;
    for ($i = 0; $i <= $n; $i++)
    {
        if ($i == 0 || $i == $n)
            $res += $fx[$i];
        else if ($i % 2 != 0)
            $res += 4 * $fx[$i];
        else
            $res += 2 * $fx[$i];
    }

    $res = $res * ($h / 3);

    return $res;
}

// Driver program
    $lower_limit = 4; // Lower limit
    $upper_limit = 5.2; // Upper limit
    $n = 6; // Number of interval
    echo simpsons_($lower_limit, $upper_limit, $n);

// This code is contributed by ajit.
?>

Javascript

<script>

// JavaScriptprogram for simpson's 1/3 rule

    // Function to calculate f(x)
    function func(x)
    {
        return Math.log(x);
    }

    // Function for approximate integral
    function simpsons_(ll, ul, n)
    {

        // Calculating the value of h
        let h = (ul - ll) / n;

        // Array for storing value of x
        // and f(x)
        let x = [];
        let fx= [];

        // Calculating values of x and f(x)
        for (let i = 0; i <= n; i++) {
            x[i] = ll + i * h;
            fx[i] = func(x[i]);
        }

        // Calculating result
        let res = 0;
        for (let i = 0; i <= n; i++) {
            if (i == 0 || i == n)
                res += fx[i];
            else if (i % 2 != 0)
                res += 4 * fx[i];
            else
                res += 2 * fx[i];
        }

        res = res * (h / 3);
        return res;
    }

// Driver code

        // Lower limit
        let lower_limit = 4;

        // Upper limit
        let upper_limit = 5.2;

        // Number of interval
        let n = 6;

        document.write(simpsons_(lower_limit,
                                upper_limit, n));

        // This code is contributed by code_hunt.
</script>

Output: 

1.827847

Time Complexity: O(n)
Auxiliary Space: O(1)