Simpson's rule (simpson) in SciPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Simpson's rule integration with simple function

What is the output of this code that uses Simpson's rule to integrate f(x) = x^2 from 0 to 2 with 5 points?

SciPy

import numpy as np
from scipy.integrate import simpson
x = np.linspace(0, 2, 5)
y = x**2
result = simpson(y, x)
print(round(result, 4))

A 2.6668

B 2.6667

C 2.6665

D 2.6666

💡 Hint

Recall the exact integral of x^2 from 0 to 2 is 8/3 ≈ 2.6667. Simpson's rule should approximate closely.

Simpson's rule approximates the integral of x^2 from 0 to 2 as about 2.6667, which matches the exact value 8/3.

❓ data_output

intermediate

Number of intervals used in Simpson's rule with uneven spacing

Given unevenly spaced points x and function values y, how many intervals does scipy.integrate.simpson use internally for integration?

SciPy

import numpy as np
from scipy.integrate import simpson
x = np.array([0, 1, 1.5, 2.5, 3])
y = np.sin(x)
result = simpson(y, x)
intervals = len(x) - 1
print(intervals)

A 5

B 3

C 4

D 2

💡 Hint

Intervals are the spaces between points, so count points minus one.

Simpson's rule uses intervals between points, so with 5 points, there are 4 intervals.

🔧 Debug

advanced

Identify the error in Simpson's rule usage

What error does this code raise when trying to integrate y = x^3 over 0 to 3 with 4 points using simpson?

SciPy

import numpy as np
from scipy.integrate import simpson
x = np.linspace(0, 3, 4)
y = x**3
result = simpson(y, x)
print(result)

A IndexError: index out of range

B TypeError: unsupported operand type(s) for ** or pow()

C No error, outputs 20.25

D ValueError: Simpson's rule requires an odd number of samples.

💡 Hint

Simpson's rule needs an odd number of points to work correctly.

Simpson's rule in scipy requires an odd number of samples; 4 points is even, so it raises a ValueError.

🚀 Application

advanced

Choosing the correct Simpson's rule call for equally spaced data

You have y values of a function sampled at equal intervals of 0.1 seconds. Which simpson call correctly computes the integral over time?

A simpson(y, dx=0.1)

B simpson(y, x=np.arange(len(y))*0.1)

C simpson(y)

D simpson(y, dx=1)

💡 Hint

When spacing is uniform, use dx parameter for efficiency.

Using dx=0.1 tells simpson the spacing between points, which is correct for equally spaced data.

🧠 Conceptual

expert

Why does Simpson's rule require an odd number of points?

Why must the number of sample points be odd when using Simpson's rule for numerical integration?

A Because Simpson's rule fits parabolas over pairs of intervals, needing an odd number of points.

B Because the function must be sampled at prime number points only.

C Because the algorithm uses trapezoids which require odd points.

D Because the integral is only defined for odd number of points.

💡 Hint

Think about how parabolas are fit between points.

Simpson's rule approximates the function by fitting parabolas over pairs of intervals, which requires an odd number of points.