Another approximation for integrals is the Trapezoid Rule: There is a built-in function trapezoid in the package scipy .integrate (see the overview for more information).

Question

(a) Compute the Trapezoid approximation using n = 500 subintervals.

(b) Is the Trapezoid approximation equal to the average of the Left and Right Endpoint

approximations?

Answer 1

It'd be preferable to see what the function you're integrating is, as well as the bounds. However, just in general, the way you'd do this is:

import numpy as np

from scipy.integrate import trapezoid

# Step 1: Define the function you want to integrate

def f(x):

return x**2 # Example: f(x) = x^2

# Step 2: Define the interval [a, b] and the number of subintervals n

a = 0 # Lower limit & Upper limit (example bounds [0, 1])

b = 1

n = 500 # Number of subintervals

# Step 3: Generate n evenly spaced points between a and b

x = np.linspace(a, b, n)

# Step 4: Evaluate the function at each point

y = f(x)

# Step 5: Compute the Trapezoid approximation

result = trapezoid(y, x)

print(f"Trapezoid approximation: {result}")

As for the second question, yes, assuming the endpoints are equally spaced, the trapezoid rule is the same as the mean between the left endpoint and right endpoint Riemann sums.

This can be seen quite easily. Assuming h = (b-a)/(# of subintervals)

Left endpoint = h*(f(a)+f(a+h)+f(a+2h)+...+f(b-h))

Right endpoint = h*(f(a+h)+f(a+2h)+...+f(b-h)+f(b))

Due to the commutative property (as well as the distributive property in application to division by 2), the mean can be expressed as:

(L+R)/2 = h*(f(a)+f(a+h)+...+f(b-h))/2 + h*(f(a+h)+f(a+2h)+...+f(b))/2

= h*( (f(a)+f(a+h))/2 + (f(a+h)+f(a+2h))/2 + ... + (f(b-h)+f(b))/2 )

Meanwhile, the trapezoid rule is computed by adding up areas of trapezoids, each area being the width times the mean between the left and right y values:

T = h*( (f(a)+f(a+h))/2 + (f(a+h)+f(a+2h))/2 + ... + (f(b-h)+f(b))/2 )