# Boole's Rule in python

This is my first post in this forum so please forgive me if it is not the way it should be.

My problem is about implementing "Boole's rule" into python. I have succesfully implementet trapezoidal and simpsons, but there is a difference in the results between the first 2 and Boole's rule when I calculate an integral. The results differ between 1 and 10. Here it's my code:

 # implementation of 3 different integration techniques
from math import *
from scipy import *


# 1. : Trapezoidal rule

# the trapezodial rule approximates the area under the function as a trapezoid. Mostly it averages the left and right
# sums of 2 sums given by the Riemann-Integral.

# the approximation is as follows:
f= lambda x: e**(-x**2)
g=  lambda x: x**6-3*x**5+10*x**4-20*x**3
h= lambda x:1/(x**2-3*x+5)

def trapezoidal (f,a,b,n):
dx= ((b-a)/2)/n
x=a
sum = 0
while x<b:
sum += f(x)*dx
x += dx
return sum

print("trapezoidal f1:",trapezoidal(f,0,1,10000))
print("trapezoidal f2:",trapezoidal(g,0,4,10000))
print("trapezoidal f3:",trapezoidal(h,0,10,10000))

#2. : Simpson´s rule

def simpson(f, a, b, n):
if n % 2: raise ValueError("n must be even (received n=%d)" % n)

dx = (b - a) / n
sum = f(a) + f(b)

for i in range(1, n, 2):
sum += 4 * f(a + i * dx)
for i in range(2, n-1, 2):
sum += 2 * f(a + i * dx)

return sum * dx / 3

print("simpson f1:",simpson(f,0,1,10000))
print("simpson f2:",simpson(g,0,4,10000))
print("simpson f3:",simpson(h,0,10,10000))

# 3. : Boole´s rule:

def boole (f,a,b,n):

dx = (b - a) / (n-1)
sum = 7*(f(a) + f(b))
sum += 32*(f(a+dx)+f(b-dx))
sum += 12*(f(a+2*dx))
return 2*sum * dx / 45

print("boole f1:",boole(f,0,1,5))
print("boole f2:",boole(g,0,4,5))
print("boole f3:",boole(h,0,10,5))


I used the formula given on Wikipedia but without the error term.

$$\int_{x_1}^{x_5} f(x)\,dx = \frac{2 h}{45}\left( 7f(x_1) + 32 f(x_2) + 12 f(x_3) + 32 f(x_4) + 7f(x_5) \right) + \text{error term}\, .$$

• calculate the result of the formula for simple functions by hand, e,g for f(x)=1. it should be 1 in the interval (0,1] but your python program prints 0. Then debug your program or print out the intermediate results to locate the error. Jun 11, 2018 at 6:51
• I used python 2.7. Jun 11, 2018 at 6:58

If you look at the way you have implemented the Trapezoidal rule and Simpson's rule, you are able to $n$ number of steps. So even tho the Trapezoidal rule is a two point integration and Simpson's rule is a three point integration, they will give better results, because you apply them $n$ number of times.
For your implementation of Boole's rule, $n$ can only be equal to five. In this implementation you integrate the entire interval in one go, instead of splitting it up (like you do with the two other methods). Even tho Boole's rule is a five point integration, it cannot outperform the two other methods, because they are applied $n$ times for a given integral, where Boole's rule is only applied once.