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# Program for finding the Integral of a given function using Boole’s Rule

• Last Updated : 19 May, 2021

Given a function, f(x), tabulated at points equally spaced by such that [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, 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, 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: • In the above formula, an error term which comes when integration is of order 6. The error term is [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 • Consider y = f(x). Now find the values • for the corresponding • values.
• Substitute all the values in Boole’s rule to calculate the integral value.

Below is the implementation of the above approach:

## C++

 // C++ program to implement Boole's Rule// on the given function#include 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 xfloat 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 bfloat 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 codeint main(){    float lowlimit = 0;    float upplimit = 4;         cout << fixed << setprecision(4) <<        "f(x) = " << BooleRule(0, 4);             return 0;} // This code is contributed by shivanisinghss2110

## C

 // C program to implement Boole's Rule// on the given function #include #include  // 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 xfloat 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 bfloat 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 codeint main(){    float lowlimit = 0;    float upplimit = 4;    printf("f(x) = %.4f",           BooleRule(0, 4));    return 0;}

## Java

 // Java program to implement Boole's Rule// on the given functionclass 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 xstatic 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 bstatic 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 codepublic static void main(String[] args){    System.out.printf(("f(x) = %.4f"),           BooleRule(0, 4));}} // This code is contributed by 29AjayKumar

## Python3

 # 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 xdef 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 bdef 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 codeif __name__ == '__main__':    lowlimit = 0    upplimit = 4    print("f(x) =",round(BooleRule(0, 4),4)) # This code is contributed by Surendra_Gangwar

## C#

 // C# program to implement Boole's// Rule on the given functionusing 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 xstatic 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 bstatic 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 codepublic static void Main(string[] args){  Console.Write(("f(x) = " +                  System.Math.Round(                  BooleRule(0, 4), 4)));}} // This code is contributed by Chitranayal

## Javascript

 
Output:
f(x) = 1.6178

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