scipy.integrate.

# cumulative_simpson#

scipy.integrate.cumulative_simpson(y, *, x=None, dx=1.0, axis=-1, initial=None)[source]#

Cumulatively integrate y(x) using the composite Simpson’s 1/3 rule. The integral of the samples at every point is calculated by assuming a quadratic relationship between each point and the two adjacent points.

Parameters:
yarray_like

Values to integrate. Requires at least one point along axis. If two or fewer points are provided along axis, Simpson’s integration is not possible and the result is calculated with cumulative_trapezoid.

xarray_like, optional

The coordinate to integrate along. Must have the same shape as y or must be 1D with the same length as y along axis. x must also be strictly increasing along axis. If x is None (default), integration is performed using spacing dx between consecutive elements in y.

dxscalar or array_like, optional

Spacing between elements of y. Only used if x is None. Can either be a float, or an array with the same shape as y, but of length one along axis. Default is 1.0.

axisint, optional

Specifies the axis to integrate along. Default is -1 (last axis).

initialscalar or array_like, optional

If given, insert this value at the beginning of the returned result, and add it to the rest of the result. Default is None, which means no value at x[0] is returned and res has one element less than y along the axis of integration. Can either be a float, or an array with the same shape as y, but of length one along axis.

Returns:
resndarray

The result of cumulative integration of y along axis. If initial is None, the shape is such that the axis of integration has one less value than y. If initial is given, the shape is equal to that of y.

numpy.cumsum
cumulative_trapezoid

cumulative integration using the composite trapezoidal rule

simpson

integrator for sampled data using the Composite Simpson’s Rule

Notes

The composite Simpson’s 1/3 method can be used to approximate the definite integral of a sampled input function $$y(x)$$ [1]. The method assumes a quadratic relationship over the interval containing any three consecutive sampled points.

Consider three consecutive points: $$(x_1, y_1), (x_2, y_2), (x_3, y_3)$$.

Assuming a quadratic relationship over the three points, the integral over the subinterval between $$x_1$$ and $$x_2$$ is given by formula (8) of [2]:

$\begin{split}\int_{x_1}^{x_2} y(x) dx\ &= \frac{x_2-x_1}{6}\left[\ \left\{3-\frac{x_2-x_1}{x_3-x_1}\right\} y_1 + \ \left\{3 + \frac{(x_2-x_1)^2}{(x_3-x_2)(x_3-x_1)} + \ \frac{x_2-x_1}{x_3-x_1}\right\} y_2\\ - \frac{(x_2-x_1)^2}{(x_3-x_2)(x_3-x_1)} y_3\right]\end{split}$

The integral between $$x_2$$ and $$x_3$$ is given by swapping appearances of $$x_1$$ and $$x_3$$. The integral is estimated separately for each subinterval and then cumulatively summed to obtain the final result.

For samples that are equally spaced, the result is exact if the function is a polynomial of order three or less [1] and the number of subintervals is even. Otherwise, the integral is exact for polynomials of order two or less.

References

[2]

Cartwright, Kenneth V. Simpson’s Rule Cumulative Integration with MS Excel and Irregularly-spaced Data. Journal of Mathematical Sciences and Mathematics Education. 12 (2): 1-9

Examples

>>> from scipy import integrate
>>> import numpy as np
>>> import matplotlib.pyplot as plt
>>> x = np.linspace(-2, 2, num=20)
>>> y = x**2
>>> y_int = integrate.cumulative_simpson(y, x=x, initial=0)
>>> fig, ax = plt.subplots()
>>> ax.plot(x, y_int, 'ro', x, x**3/3 - (x[0])**3/3, 'b-')
>>> ax.grid()
>>> plt.show()


The output of cumulative_simpson is similar to that of iteratively calling simpson with successively higher upper limits of integration, but not identical.

>>> def cumulative_simpson_reference(y, x):
...     return np.asarray([integrate.simpson(y[:i], x=x[:i])
...                        for i in range(2, len(y) + 1)])
>>>
>>> rng = np.random.default_rng()
>>> x, y = rng.random(size=(2, 10))
>>> x.sort()
>>>
>>> res = integrate.cumulative_simpson(y, x=x)
>>> ref = cumulative_simpson_reference(y, x)
>>> equal = np.abs(res - ref) < 1e-15
>>> equal  # not equal when simpson has even number of subintervals
array([False,  True, False,  True, False,  True, False,  True,  True])


This is expected: because cumulative_simpson has access to more information than simpson, it can typically produce more accurate estimates of the underlying integral over subintervals.