Fundamental theorem of calculus · Calculus × Computing

What is it?

Differentiation and integration are inverse operations. Accumulating a rate of change recovers the total change: $\int_a^b f'(x)\,\dd x = f(b) - f(a)$ , and the derivative of an accumulated quantity is the rate.

Why does it exist?

Areas were computed by exhausting them with thin pieces since Archimedes — slow, one curve at a time. Tangents were a separate problem. Newton and Leibniz (with Barrow before them) discovered they are the same problem run in opposite directions, so a table of derivatives is also a table of areas.

Intuition

Let $A(x) = \int_a^x f(t)\,\dd t$ be the area accumulated up to $x$ . Move $x$ a little to the right by $h$ : the area grows by a thin strip of height about $f(x)$ and width $h$ . So $\frac{A(x+h) - A(x)}{h} \approx f(x)$ — the rate at which area accumulates is the height of the curve. If $v(t)$ is a velocity, accumulating it gives distance travelled.

Statement

Part 1. If $f$ is continuous on $[a,b]$ , then $A(x) = \int_a^x f(t)\,\dd t$ is differentiable and $A'(x) = f(x)$ .

Part 2 (Barrow's rule). If $F$ is an antiderivative of a continuous $f$ on $[a,b]$ , then

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

Idea of the proof

Part 1: $\frac{A(x+h) - A(x)}{h} = \frac1h\int_x^{x+h} f$ is an average of $f$ over a tiny interval, which tends to $f(x)$ by continuity. Part 2: $A$ and $F$ have the same derivative, so they differ by a constant (MVT).

Proof

Part 1. For $h > 0$ , by the mean value theorem for integrals there is $c_h \in [x, x+h]$ with $\int_x^{x+h} f = f(c_h)\,h$ . Hence $\frac{A(x+h) - A(x)}{h} = f(c_h) \to f(x)$ as $h \to 0^+$ , because $c_h \to x$ and $f$ is continuous. The case $h < 0$ is identical.

Part 2. $(A - F)' = f - f = 0$ on $(a,b)$ , so by the MVT $A - F$ is constant: $A(x) = F(x) + C$ . Since $A(a) = 0$ , $C = -F(a)$ and $\int_a^b f = A(b) = F(b) - F(a)$ . $\blacksquare$

Formulas

$\frac{\dd}{\dd x}\int_a^x f(t)\,\dd t = f(x)$; $\int_a^b f(x)\,\dd x = F(b) - F(a), \qquad F' = f$; $s(T) = s(0) + \int_0^T v(t)\,\dd t$
position from velocity

Example

Area under $\sin x$ on $[0, \pi]$ : an antiderivative is $-\cos x$ , so the area is $-\cos\pi + \cos 0 = 2$ . Compare with the Riemann sums in the Riemann sums demo, which need hundreds of rectangles to get three digits.

Why does it matter?

It links the two halves of the subject and explains why physics engines and ODE solvers are "integrators": they turn rates (forces, velocities) into states (positions). In probability it is the link between a density and its cumulative distribution, $F' = f$ .

Where it shows up in computing

Physics engines★★★★★fundamentalPhysics and simulation
Integrating velocity gives position: every simulation step is a tiny application of the FTC.
Symbolic computation (CAS)★★★★★frequentScientific computing and algorithms
CAS evaluate definite integrals by finding an antiderivative (Risch algorithm) and applying Barrow's rule.

Where is it used?

Computing topics reachable from here, through the chain of ideas that leads to them:

What depends on it

Exercises

1Computation

Compute $\frac{\dd}{\dd x}\int_0^{x^2} e^{-t^2}\,\dd t$ .

Solution

By the FTC and the chain rule: $e^{-x^4}\cdot 2x$ .

2Computing

A sensor reports velocity every 0.1 s. How do you estimate position, and which theorem justifies it?

Solution

Position is $s(0) + \int_0^T v$ (FTC). With samples, approximate the integral by a sum, e.g. the trapezoidal rule $\sum \frac{v_k + v_{k+1}}{2}\,0.1$ . Errors accumulate (drift), which is why IMUs are fused with GPS.