SciPy Integration
Integration calculates the total area under a curve, which often means adding up a changing quantity.
Why Integration Matters
If you know a car's speed at every moment, integration gives the total distance. If you know the power used each hour, it gives the total energy. You do not need to do the calculus by hand: scipy.integrate does it numerically.
| Function | Use it when |
|---|---|
quad(f, a, b) | You have a formula f(x) and limits a to b |
trapezoid(y, x) | You only have measured data points |
solve_ivp(f, t_span, y0) | You have a differential equation (a rate of change) |
Integrating a Function With quad
Find the area under y = x² from 0 to 3. The exact answer is 9.
Example
from scipy.integrate import quad
area, error = quad(lambda x: x ** 2, 0, 3)
print(round(area, 6))
print(error < 1e-10)9.0 True
quad returns two values: the result and an estimate of the error, which is tiny here.
A Practical Example
A water tank fills at a rate of r(t) = 20 + 4t litres per minute. How much water flows in during the first 10 minutes?
Example
from scipy.integrate import quad
def rate(t):
return 20 + 4 * t
litres, _ = quad(rate, 0, 10)
print(litres)400.0
Infinite Limits
quad also works with infinity, using np.inf. The area under the standard bell curve is exactly 1:
Example
import numpy as np
from scipy.integrate import quad
bell = lambda x: np.exp(-x ** 2 / 2) / np.sqrt(2 * np.pi)
total, _ = quad(bell, -np.inf, np.inf)
print(round(total, 6))1.0
Integrating Measured Data
Often you have readings, not a formula. A motorbike's speed (metres per second) was recorded every 10 seconds:
Example
import numpy as np
from scipy.integrate import trapezoid
time = np.array([0, 10, 20, 30, 40])
speed = np.array([0, 8, 12, 12, 6])
distance = trapezoid(speed, time)
print(distance, "metres")350.0 metres
The trapezoid rule joins the points with straight lines and adds up the areas of the shapes under them.
The more often you take readings, the more accurate trapezoid becomes. With only a few points, the result is a rough estimate.
Solving a Differential Equation
A differential equation describes how something changes. Your savings grow by 10% per year, continuously: dy/dt = 0.1·y. Starting with Rs 100,000, how much after 5 years?
Example
from scipy.integrate import solve_ivp
def growth(t, y):
return 0.1 * y
sol = solve_ivp(growth, t_span=(0, 5), y0=[100000], t_eval=[5], rtol=1e-8)
print(round(sol.y[0][-1]))164872
The exact answer is 100000 × e^0.5 = 164,872, so SciPy matches it.
Exercise
Exercise
Which function should you use to integrate a list of measured data points?