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Numerical integration and a continuous cash-flow building block

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Start with the idea

Many integrals do not have a convenient antiderivative.
Numerical quadrature approximates the integrand on small intervals and adds the resulting areas.
The step size controls one source of error, while inaccurate input data create another.

Symbols, units & horizon

I: exact integral being approximated
h=(b−a)/n: uniform interval width
xᵢ=a+ih: sample locations
n: number of intervals
PV: present value in currency
c: constant currency flow per year
r: continuous discount rate per year
T: horizon in years
t: integration time
At r=0: PV=cT by continuity

When and why to use this

Use a known exact integral to validate an integration routine, then examine step convergence on the target integrand. Separate mathematical approximation error from uncertainty in the inputs.

The trapezoidal rule joins neighbouring samples with straight lines and integrates those lines.
For a sufficiently smooth integrand its error decreases quadratically with uniform step size over a fixed interval.
Halving the step is a useful convergence check, not a cure for discontinuities or biased data.
As a separate application, a continuous currency flow c(t) discounted at a constant continuously compounded rate r has present value ∫ c(t)exp(−rt)dt.
This is a model of continuous payments; a discrete coupon schedule should normally be summed at its actual payment dates.

I \approx \frac{h}{2} [f (x_{0}) + 2 \sum_{i = 1}^{n - 1} f (x_{i}) + f (x_{n})], P V = \int_{0}^{T} c e^{- r t} d t = \frac{c (1 - e^{- r T})}{r} (r \neq 0)

Calculus: derivation and arithmetic

Numerical integration and a continuous cash-flow building block

One strip has trapezoidal area $\frac{h [f (x_{i}) + f (x_{i + 1})]}{2}$ . Adding strips counts every interior sample twice and each endpoint once, giving the displayed weighted sum.
For $f (x) = x^{2}$ on $[0,1]$ with $n = 2$ , use $h = 0.5$ : $I_{2} = 0.25 [0 + 2 (0.25) + 1] = 0.375$ . This approximates the exact integral $\frac{1}{3}$ ; it is not an identity for the curved integrand.
For constant flow and $r \neq 0$ , an antiderivative is $\frac{- c e^{- r t}}{r}$ . Subtract endpoints: $P V = [\frac{- c e^{- r t}}{r}]_{0}^{T} = \frac{c (1 - e^{- r T})}{r}$ .
As $r \to 0$ , the ratio $\frac{(1 - e^{- r T})}{r}$ tends to $T$ , giving $P V = c T$ . With $c = 100$ dollars/year, $r = 0.05$ per year and $T = 2$ years, $P V = \frac{100 (1 - e^{- 0.1})}{0.05} \approx 190.325$ dollars.

Work it by hand

A continuous $100/year flow over 2 years at r=.05 has PV=100(1−exp(−.1))/.05≈$190.325. At zero discount rate it is $200.

Architecture pseudocode

from math import expm1

def trapezoid(function,a,b,n):
    if not isinstance(n,int) or n<1: raise ValueError("Positive integer n required")
    h=(b-a)/n
    return h*(.5*function(a)+sum(function(a+i*h) for i in range(1,n))+.5*function(b))

def continuous_flow_pv(flow,rate,years):
    if years<0: raise ValueError("Nonnegative horizon required")
    return flow*years if rate==0 else -flow*expm1(-rate*years)/rate

print(trapezoid(lambda x:x*x,0,1,2),continuous_flow_pv(100,.05,2))

An analogy to remember

Surveyors can approximate uneven terrain with many flat-sided strips. Finer strips help if the terrain is adequately sampled; they cannot reveal a feature absent from the data.

How this becomes a building block

Quadrature is a reusable numerical component for expectations, discounting and some option-pricing integrals. The continuous-flow formula is a complete valuation under stated assumptions and also a term in a broader valuation model.

Python implementation

Self-contained teaching example. Python 3.10+; dependencies and input conventions are shown in the code and notation. Run in your own Python environment.

from math import expm1

def trapezoid(function,a,b,n):
    if not isinstance(n,int) or n<1: raise ValueError("Positive integer n required")
    h=(b-a)/n
    return h*(.5*function(a)+sum(function(a+i*h) for i in range(1,n))+.5*function(b))

def continuous_flow_pv(flow,rate,years):
    if years<0: raise ValueError("Nonnegative horizon required")
    return flow*years if rate==0 else -flow*expm1(-rate*years)/rate

print(trapezoid(lambda x:x*x,0,1,2),continuous_flow_pv(100,.05,2))

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