The Fundamental Theorem of Calculus

Two questions look completely unrelated at first: "given a rate of change, how much total accumulation happened?" (integration — e.g., total P&L from a stream of instantaneous returns) and "given a total accumulated amount as a function of time, what's the rate right now?" (differentiation — e.g., a cumulative P&L curve's current slope). It's a genuinely surprising fact, not an obvious one, that these two questions are exact inverses of each other. The fundamental theorem of calculus is the proof of that link, and it's what makes computing an integral by finding an antiderivative valid at all.

An analogy: odometer and speedometer, reconciled

A car's odometer accumulates total distance traveled; its speedometer reads the instantaneous rate of that accumulation. It feels obvious that they're related — of course distance and speed connect — but making that connection precise and computable is nontrivial: it says that if you know the speedometer reading at every instant (the derivative), you can recover the exact total distance (the integral) just by "undoing" the derivative, and conversely, differentiating the accumulated distance function gives back the speedometer reading exactly. The fundamental theorem of calculus is the rigorous version of "the odometer is just the speedometer, accumulated."

The math, one symbol at a time

The theorem has two parts. Part 1: if $F(x) = \int_a^x f(t)\,dt$ (the accumulated total of $f$ from a fixed start $a$ up to $x$), then

$$F'(x) = f(x).$$

In words: differentiating an accumulation function gets you back the original rate-of-accumulation function — accumulation and differentiation undo each other. Part 2, the one used for actual computation: if $F$ is any antiderivative of $f$ (meaning $F'=f$), then

$$\int_a^b f(x)\,dx = F(b) - F(a).$$

In words: to compute a definite integral (a total accumulated amount, or an area), find any function whose derivative is the integrand, and just subtract its value at the two endpoints — no need to add up infinitely many infinitesimal slices directly.

Worked example 1: verifying Part 1 with a known accumulation

Let $f(t) = 2t$ (a linearly rising rate). $F(x) = \int_0^x 2t\,dt = x^2$ (a standard antiderivative, computed via the power rule in reverse). Part 1 says $F'(x)$ should equal $f(x) = 2x$. Differentiating $F(x)=x^2$ directly gives $F'(x) = 2x$ — matching exactly. Accumulating $f$ up to $x$ and then differentiating that accumulation returns $f$ itself, precisely as the theorem promises.

Worked example 2: computing a definite integral with Part 2

A trading strategy's instantaneous P&L rate is $f(t) = 3t^2$ dollars per day, for $t$ from 0 to 4 days. Total P&L over the 4 days is $\int_0^4 3t^2\,dt$. An antiderivative of $3t^2$ is $F(t) = t^3$ (check: $F'(t) = 3t^2 = f(t)$, confirmed). By Part 2, the total is $F(4) - F(0) = 4^3 - 0^3 = 64$. So the strategy earns exactly $64 total over the 4 days — computed by evaluating one function at two points, rather than summing up an infinite number of infinitesimal P&L slivers directly.

The area swept out under this curve up to any point $x$ is the accumulation function $F(x)$; drag along the curve and notice how the rate the shaded area grows at any point matches the curve's own height there — that's Part 1 in action.

What this means in practice

Every closed-form pricing formula that starts life as "the expected value of a payoff is an integral" ultimately relies on Part 2 to turn that integral into an evaluable expression, and every sensitivity ("Greek") computed by differentiating a price is implicitly relying on Part 1's guarantee that differentiation correctly undoes the accumulation. The theorem is also the reason integration techniques like substitution and integration by parts — which work by finding antiderivatives — are valid ways to compute a definite integral at all, rather than just an unrelated algebraic trick.