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Program for finding the Integral of a given function using Boole’s Rule
  • Last Updated : 11 Nov, 2020

Given a function, f(x), tabulated at points 

x_i

equally spaced by 

h=x_{i+1} - x_i

such that 
 



f_1 = f(x_1)  [Tex]f_2=f(x_2)[/Tex]

  • …..and so on

The upper and lower limits a, b correspond to which the integral needs to be found, the task is to find the integral value of the given equation f(x).
Examples:  

Input: a = 0, b = 4, 

f(x)= \frac{1}{1+x}

Output: 1.6178 
Explanation: 
Integral of (1 / (1 + x)) is 4ln(|1 + x|)0 + c. 
On substituting the limits, ln(|5|) + ln(|1|) = 1.6178.
Input: a = 0.2, b = 0.6, 

f(x)= \frac{1}{x^2}

Output: 0.3430

Approach: In this article, Boole’s rule is discussed to compute the approximate integral value of the given function f(x). 
Boole’s rule is a numerical integration technique to find the approximate value of the integral. It is named after a largely self-taught mathematician, philosopher, and logician George Boole. The idea of the Boole’s technique is to approximate the integral using the values of ‘fk‘ at some equally spaced values(h in the given image). The following illustration shows how various fk’s are considered: 



The integral value of Boole’s rule is given by the formula:

\int_{x_{1}}^{x_{5}} f(x)dx = \frac{2}{45}h(7f_{1} + 32f_{2} + 12f_{3} + 32f_{4} + 7f_{5}) + \text{Error term}

  • In the above formula, an error term which comes when integration is of the order 6. The error term is

- \frac{8}{945}h^{7}f_{6} [Tex]f_1, f_2, f_3, f_4, f_5, [/Tex]

  • Are the values of f(x) at their respective intervals of x.
  • Therefore, the following steps can be followed to compute the integral of some function f(x) in the interval (a, b): 
    1. The value of n=6, which is the number of parts the interval is divided into.
    2. Calculate the width, h = (b – a)/4.
    3. Calculate the values of x1 to x5 as

x_1 = a, x_2 = x_1 + h, x_3 = x_1 + 2h, ...x_5=x_1 +4h

  • Consider y = f(x). Now find the values 

y(y_1, y_2, .. y_5)

  • for the corresponding

x(x_1, x_2, x_3... x_5)

  • values.
  • Substitute all the values in Boole’s rule to calculate the integral value.

Below is the implementation of the above approach: 
 

C++

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// C++ program to implement Boole's Rule
// on the given function
#include <bits/stdc++.h>
using namespace std;
 
// In order to represent the implementation,
// a function f(x) = 1/(1 + x) is considered
// in this program
 
// Function to return the value of f(x)
// for the given value of x
float y(float x)
{
    return (1 / (1 + x));
}
 
// Function to computes the integrand of y
// at the given intervals of x with
// step size h and the initial limit a
// and final limit b
float BooleRule(float a, float b)
{
 
    // Number of intervals
    int n = 4;
    int h;
 
    // Computing the step size
    h = ((b - a) / n);
    float sum = 0;
 
    // Substituing a = 0, b = 4 and h = 1
    float bl = ((7 * y(a) +
                32 * y(a + h) +
                12 * y(a + 2 * h) +
                32 * y(a + 3 * h) +
                 7 * y(a + 4 * h)) *
                 2 * h / 45);
 
    sum = sum + bl;
    return sum;
}
 
// Driver code
int main()
{
    float lowlimit = 0;
    float upplimit = 4;
     
    cout << fixed << setprecision(4) <<
        "f(x) = " << BooleRule(0, 4);
         
    return 0;
}
 
// This code is contributed by shivanisinghss2110

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C

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// C program to implement Boole's Rule
// on the given function
 
#include <math.h>
#include <stdio.h>
 
// In order to represent the implementation,
// a function f(x) = 1/(1 + x) is considered
// in this program
 
// Function to return the value of f(x)
// for the given value of x
float y(float x)
{
    return (1 / (1 + x));
}
 
// Function to computes the integrand of y
// at the given intervals of x with
// step size h and the initial limit a
// and final limit b
float BooleRule(float a, float b)
{
    // Number of intervals
 
    int n = 4;
    int h;
 
    // Computing the step size
    h = ((b - a) / n);
    float sum = 0;
 
    // Substituing a = 0, b = 4 and h = 1
    float bl = (7 * y(a) + 32 * y(a + h)
                + 12 * y(a + 2 * h)
                + 32 * y(a + 3 * h)
                + 7 * y(a + 4 * h))
               * 2 * h / 45;
 
    sum = sum + bl;
    return sum;
}
 
// Driver code
int main()
{
    float lowlimit = 0;
    float upplimit = 4;
    printf("f(x) = %.4f",
           BooleRule(0, 4));
    return 0;
}

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Java

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// Java program to implement Boole's Rule
// on the given function
class GFG{
     
// In order to represent the implementation,
// a function f(x) = 1/(1 + x) is considered
// in this program
  
// Function to return the value of f(x)
// for the given value of x
static float y(float x)
{
    return (1 / (1 + x));
}
  
// Function to computes the integrand of y
// at the given intervals of x with
// step size h and the initial limit a
// and final limit b
static float BooleRule(float a, float b)
{
    // Number of intervals
  
    int n = 4;
    int h;
  
    // Computing the step size
    h = (int) ((b - a) / n);
    float sum = 0;
  
    // Substituing a = 0, b = 4 and h = 1
    float bl = (7 * y(a) + 32 * y(a + h)
                + 12 * y(a + 2 * h)
                + 32 * y(a + 3 * h)
                + 7 * y(a + 4 * h))
               * 2 * h / 45;
  
    sum = sum + bl;
    return sum;
}
  
// Driver code
public static void main(String[] args)
{
    System.out.printf(("f(x) = %.4f"),
           BooleRule(0, 4));
}
}
 
// This code is contributed by 29AjayKumar

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Python3

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# Python3 program to implement Boole's Rule
# on the given function
 
# In order to represent the implementation,
# a function f(x) = 1/(1 + x) is considered
# in this program
 
# Function to return the value of f(x)
# for the given value of x
def y(x):
    return (1 / (1 + x))
 
# Function to computes the integrand of y
# at the given intervals of x with
# step size h and the initial limit a
# and final limit b
def BooleRule(a, b):
     
    # Number of intervals
    n = 4
 
    # Computing the step size
    h = ((b - a) / n)
    sum = 0
 
    # Substituing a = 0, b = 4 and h = 1
    bl = (7 * y(a) + 32 * y(a + h) + 12 *
        y(a + 2 * h)+32 * y(a + 3 * h)+7 *
        y(a + 4 * h))* 2 * h / 45
 
    sum = sum + bl
    return sum
 
# Driver code
if __name__ == '__main__':
    lowlimit = 0
    upplimit = 4
    print("f(x) =",round(BooleRule(0, 4),4))
 
# This code is contributed by Surendra_Gangwar

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C#

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// C# program to implement Boole's
// Rule on the given function
using System;
class GFG{
 
// In order to represent the
// implementation, a function
// f(x) = 1/(1 + x) is considered
// in this program
 
// Function to return the value of
// f(x) for the given value of x
static float y(float x)
{
  return (1 / (1 + x));
}
 
// Function to computes the integrand
// of y at the given intervals of x
// with step size h and the initial
// limit a and final limit b
static float BooleRule(float a,
                       float b)
{
  // Number of intervals
  int n = 4;
  int h;
 
  // Computing the step size
  h = (int)((b - a) / n);
  float sum = 0;
 
  // Substituing a = 0, b = 4
  // and h = 1
  float bl = (7 * y(a) + 32 *
              y(a + h) + 12 *
              y(a + 2 * h) +
              32 * y(a + 3 *
              h) + 7 * y(a +
              4 * h)) * 2 * 
              h / 45;
 
  sum = sum + bl;
  return sum;
}
 
// Driver code
public static void Main(string[] args)
{
  Console.Write(("f(x) = " +
                  System.Math.Round(
                  BooleRule(0, 4), 4)));
}
}
 
// This code is contributed by Chitranayal

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Output: 

f(x) = 1.6178



 

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