Free lesson · Integral calculus

Definite integrals: adding many small contributions

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Start with the idea

An integral accumulates a quantity continuously.
Start with a finite sum of rate times interval width, then imagine refining the intervals.
Signed area is one useful picture, but the physical meaning comes from the units.

Symbols, units & horizon

∫: integral or continuous accumulation
a,b: start and end of the integration interval
x: variable of integration
dx: infinitesimal input width in the integral notation
n: number of equal subintervals
Δx: their finite width
xᵢ*: chosen sample point in subinterval i
Σ: sum all interval contributions
lim n→∞: limit as the equal partition is refined

When and why to use this

Use a definite integral when a quantity is naturally expressed as contributions over time, distance or another continuous variable. The accumulation lab compares rectangles with an exact total.

If water flows at q(t) litres per minute, q(t) times a short duration approximates the litres added in that interval.
Adding over intervals approximates the total volume.
The integral is the limiting total when the function is integrable and the partition is refined.
A region below the horizontal axis contributes negatively to the signed integral.
Total geometric area instead integrates the absolute value.
Reversing the integration bounds reverses the sign.

\int_{a}^{b} f (x) d x = \lim_{n \to \infty} \sum_{i = 1}^{n} f (x_{i}^{*}) Δ x, Δ x = \frac{b - a}{n}

Calculus: derivation and arithmetic

Definite integrals: adding many small contributions

For $f (x) = 2 x$ on $[0,1]$ , divide the interval into $n$ equal pieces: $Δ x = \frac{1}{n}$ , with right endpoints $x_{i} = \frac{i}{n}$ .
The right-endpoint sum is $S_{n} = \sum_{i = 1}^{n} f (x_{i}) Δ x = \frac{2}{n^{2}} \sum_{i = 1}^{n} i$ . Substitute $\sum_{i = 1}^{n} i = \frac{n (n + 1)}{2}$ to obtain $S_{n} = \frac{(n + 1)}{n}$ .
Take the limit: $\lim_{n \to \infty} (1 + \frac{1}{n}) = 1$ . Left endpoints give $\frac{(n - 1)}{n}$ , whose limit is also 1. The finite sums approximate the area; their common limit gives the integral.

Work it by hand

With four right-end rectangles the heights are .5,1,1.5,2 and width .25. Total=1.25. With four left rectangles the total is .75. The exact integral is 1.

Architecture pseudocode

def riemann(function,a,b,n,side="right"):
    if not isinstance(n,int) or n<1: raise ValueError("Positive integer n required")
    if side not in ("left","right","midpoint"): raise ValueError("Unknown sample convention")
    offset={"left":0,"right":1,"midpoint":.5}[side]
    dx=(b-a)/n
    return sum(function(a+(i+offset)*dx)*dx for i in range(n))

print(riemann(lambda x:2*x,0,1,4),riemann(lambda x:2*x,0,1,4,"left"))

An analogy to remember

Adding the water collected in many short time intervals reconstructs how much passed through a pipe. A finer measuring schedule better tracks a changing flow.

How this becomes a building block

An integral can be the entire quantity of interest, such as total flow. Within finance it can accumulate a funding rate, a deterministic execution-cost rate or discounted continuous income. Specify whether the integrand is currency/time, probability density or something else before interpreting its integral.

Further reading: OpenStax Calculus, Volume 1 — definitions and standard theorems ↗

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.

def riemann(function,a,b,n,side="right"):
    if not isinstance(n,int) or n<1: raise ValueError("Positive integer n required")
    if side not in ("left","right","midpoint"): raise ValueError("Unknown sample convention")
    offset={"left":0,"right":1,"midpoint":.5}[side]
    dx=(b-a)/n
    return sum(function(a+(i+offset)*dx)*dx for i in range(n))

print(riemann(lambda x:2*x,0,1,4),riemann(lambda x:2*x,0,1,4,"left"))

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Integral Calculus: Accumulation & Area — all lessons

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