Calculating the area underneath a mathematical function

Question

I have a range of data that I have approximated using a polynomial of degree 2 in Python. I want to calculate the area underneath this polynomial between 0 and 1.

Is there a calculus, or similar package from numpy that I can use, or should I just make a simple function to integrate these functions?

I'm a little unclear what the best approach for defining mathematical functions is.

Thanks.

Answer 1

If you're integrating only polynomials, you don't need to represent a general mathematical function, use numpy.poly1d, which has an integ method for integration.

>>> import numpy
>>> p = numpy.poly1d([2, 4, 6])
>>> print p
   2
2 x + 4 x + 6
>>> i = p.integ()
>>> i
poly1d([ 0.66666667,  2.        ,  6.        ,  0.        ])
>>> integrand = i(1) - i(0) # Use call notation to evaluate a poly1d
>>> integrand
8.6666666666666661

For integrating arbitrary numerical functions, you would use scipy.integrate with normal Python functions for functions. For integrating functions analytically, you would use sympy. It doesn't sound like you want either in this case, especially not the latter.

Answer 2

'quad' in scipy.integrate is the general purpose method for integrating functions of a single variable over a definite interval. In a simple case (such as the one described in your question) you pass in your function and the lower and upper limits, respectively. 'quad' returns a tuple comprised of the integral result and an upper bound on the error term.

from scipy import integrate as TG

fnx = lambda x: 3*x**2 + 9*x    # some polynomial of degree two
aoc, err = TG.quad(fnx, 0, 1)

[Note: after i posted this i an answer posted before mine, and which represents polynomials using 'poly1d' in Numpy. My scriptlet just above can also accept a polynomial in this form:

import numpy as NP

px = NP.poly1d([2,4,6])
aoc, err = TG.quad(px, 0, 1)
# returns (8.6666666666666661, 9.6219328800846896e-14)

Answer 3

It might be overkill to resort to general-purpose numeric integration algorithms for your special case...if you work out the algebra, there's a simple expression that gives you the area.

You have a polynomial of degree 2: f(x) = ax² + bx + c

You want to find the area under the curve for x in the range [0,1].

The antiderivative F(x) = ax³/3 + bx²/2 + cx + C

The area under the curve from 0 to 1 is: F(1) - F(0) = a/3 + b/2 + c

So if you're only calculating the area for the interval [0,1], you might consider using this simple expression rather than resorting to the general-purpose methods.

Answer 4

Look, Ma, no imports!

>>> coeffs = [2., 4., 6.]
>>> sum(coeff / (i+1) for i, coeff in enumerate(reversed(coeffs)))
8.6666666666666661
>>>

Our guarantee: Works for a polynomial of any positive degree or your money back!

Update from our research lab: Guarantee extended; s/positive/non-negative/ :-)

Update Here's the industrial-strength version that is robust in the face of stray ints in the coefficients without having a function call in the loop, and uses neither enumerate() nor reversed() in the setup:

>>> icoeffs = [2, 4, 6]
>>> tot = 0.0
>>> divisor = float(len(icoeffs))
>>> for coeff in icoeffs:
...     tot += coeff / divisor
...     divisor -= 1.0
...
>>> tot
8.6666666666666661
>>>