The Trapezoid rule

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The Trapezoid rule#

The next polynomial degree is \(1\); a line.

Given the points \(x_i\), \(x_{i+1}\), \(f_i\), and \(f_{i+1}\),

\[ f(x) = f_i + \frac{f_{i+1} - f_i}{x_{i+1} - x_i} (x - x_i) \]

and integrating,

\[\begin{split} \begin{align} \int_{x_i}^{x_{i+1}} f(x) \, dx &= \int_{x_i}^{x_{i+1}} \left( f_i + \frac{f_{i+1} - f_i}{x_{i+1} - x_i} (x - x_i) \right) dx \\ &= \frac{f_i + f_{i+1}}{2} (x_{i+1} - x_i) \\ &= h \frac{f_i + f_{i+1}}{2} \end{align} \end{split}\]

Graphically:

Trapezoidal_rule_illustration.svg

Error#

The accuracy is calculated from the Taylor series expansion of \(f(x)\) around the midpoint \(y_i = \frac{x_{i+1} + x_i}{2}\), which is the midpoint between \(x_i\) and \(x_{i+1}\).

Calculating the error is a little more complicated since we’ll need the function value at the midpoint. Begin with the Taylor series at \(x_i\) and \(x_{i+1}\), noting that \(x_i - y_i = -\frac{h}{2}\) and \(x_{i+1} - y_i = \frac{h}{2}\):

\[\begin{split} \begin{align} f(x_i) &= f(y_i) - \frac{hf^{\prime}(y_i)}{2} + \frac{h^2f''(y_i)}{8} - \cdots \\ f(x_{i+1}) &= f(y_i) + \frac{hf^{\prime}(y_i)}{2} + \frac{h^2f''(y_i)}{8} + \cdots. \end{align} \end{split}\]

Taking the average

\[\frac{f(x_{i+1})+f(x_i)}{2} = f(y_i) + O(h^2).\]

cancels odd derivatives, and we can now find,

\[f(y_i) = \frac{f(x_{i+1})+f(x_i)}{2} + O(h^2).\]

Now we take the Taylor series expanded about \(y_i\) and integrate / distribute as before:

\[\begin{split} \begin{align} f(x) &= f(y_i) + f^{\prime}(y_i)(x - y_i) + \frac{f''(y_i)(x - y_i)^2}{2!} + \cdots \\ \int_{x_i}^{x_{i+1}} f(x) dx &= \int_{x_i}^{x_{i+1}} \left(f(y_i) + f^{\prime}(y_i)(x - y_i) + \frac{f''(y_i)(x - y_i)^2}{2!} + \cdots\right) dx \\ &= \int_{x_i}^{x_{i+1}} f(y_i) dx + \int_{x_i}^{x_{i+1}} f^{\prime}(y_i)(x - y_i)dx + \int_{x_i}^{x_{i+1}} \frac{f''(y_i)(x - y_i)^2}{2!} dx + \cdots \end{align} \end{split}\]

Once again, since \(x_i\) and \(x_{i+1}\) are symmetric around \(y_i\), odd integrals evaluate to zero. Now we can insert our result from above,

\[\begin{split} \begin{align} \int_{x_i}^{x_{i+1}} f(x) dx &= hf(y_i) + O(h^3) \\ & = h \left(\frac{f(x_{i+1})+f(x_i)}{2} + O(h^2)\right) + O(h^3) \\ &= h \left(\frac{f(x_{i+1})+f(x_i)}{2}\right) + hO(h^2) + O(h^3) \\ &= h \left(\frac{f(x_{i+1})+f(x_i)}{2}\right) + O(h^3) \end{align} \end{split}\]

Therefore the trapezoid rule is is \(O(h^3)\) over subintervals and totals \(O(h^2)\) over the whole interval.

For the same function as the midpoint method (which approximated 2.033), the trapezoid method yields,

9_Numerical_integration_13_0.png

Example#

Integrate \(sin(x)\) from 0 to \(\pi\) with the midpoint and trapezoid rules.

import numpy as np
import scipy as sp

def f(x):
  """The function to integrate."""
  return np.sin(x)

def trapezoid_method(f, a, b, n):
  """Approximates the integral of f from a to b using the trapezoid method."""
  h = (b - a) / n
  x = np.linspace(a, b, n + 1)
  integral = np.trapz(f(x), x, h)
  return integral

def midpoint_method(f, a, b, n):
  """Approximates the integral of f from a to b using the midpoint method."""
  h = (b - a) / n
  x = np.linspace(a + h / 2, b - h / 2, n)
  integral = h * np.sum(f(x))
  return integral

a = 0
b = np.pi
for n in [2,4,8,16,32]:
  trapezoid_result = trapezoid_method(f, a, b, n)
  midpoint_result = midpoint_method(f, a, b, n)
  simps_result = simps_method(f, a, b, n)
  print(f"n = {n}: Trapezoid error = {abs(trapezoid_result-2):.6f}, Midpoint error= {midpoint_result-2:.6f}")

n = 2: Trapezoid error = 0.429204, Midpoint error= 0.221441
n = 4: Trapezoid error = 0.103881, Midpoint error= 0.052344
n = 8: Trapezoid error = 0.025768, Midpoint error= 0.012909
n = 16: Trapezoid error = 0.006430, Midpoint error= 0.003216
n = 32: Trapezoid error = 0.001607, Midpoint error= 0.000803

C:\Users\wellandm\AppData\Local\Temp\ipykernel_20800\3530925064.py:12: DeprecationWarning: `trapz` is deprecated. Use `trapezoid` instead, or one of the numerical integration functions in `scipy.integrate`.
  integral = np.trapz(f(x), x, h)

Notice we get quadratic convergence for both but that the midpoint method is actually more accurate! However, we had to evaluate the midpoint rule at the midpoints this is problematic if we just have tabulated data, whereas the trapezoid rule can be applied directly.

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