How fast does accumulated area grow?

Start at $x=1$ with $a=-2$: the accumulated signed area is $F(1)=9/16$. The orange strip adds $\Delta F=-95/256$ — negative, because $f$ is below the axis there — almost exactly $f(1)\,\Delta x=-3/8$. On the right, the slope of $F$ at $x=1$ is $-3/4$, the height of $f$. Now choose Jump in f: $F$ has a corner, and no single slope exists at the jump.

Intuition: the accumulator’s speedometer

Imagine a tank that fills at rate $f(t)$ (negative means draining). The amount in the tank since time $a$ is $F(x)=\int_a^x f(t)\,dt$. How fast is the amount changing at time $x$? Exactly at the current rate, $f(x)$. Integration accumulates a rate; differentiation recovers it.

Definition

For $f$ integrable on an interval containing $a$, the accumulation function is

Fundamental theorem (first form). If $f$ is continuous at $x$, then $F$ is differentiable at $x$ and $F'(x)=f(x)$.

Second form. If $G$ is any antiderivative of a continuous $f$ ($G'=f$), then $\int_a^b f=G(b)-G(a)$.

Mechanics: what the explorer computes

The integrands are piecewise polynomials, so $F$ is computed exactly as a piecewise polynomial that is continuous by construction. The strip’s area $\Delta F=F(x+\Delta x)-F(x)$ is exact, as is the rectangle estimate $f(x)\,\Delta x$. The slope row is not $f(x)$ copied over: it is the derivative of $F$‘s own exact pieces, including one-sided slopes at piece boundaries. Agreement with $f(x)$ is what the theorem predicts — and the jump preset shows where that agreement fails.

Derivation: why F′ = f

For $\Delta x>0$,

the average of $f$ over the strip. If $f$ is continuous at $x$, every value of $f$ on a short enough strip is within any chosen tolerance of $f(x)$, so the average is too. Letting $\Delta x\to0$ (from either side) gives $F'(x)=f(x)$. If $f$ jumps at $x$, the averages from the left and right approach the two different one-sided values of $f$, and $F$ has a corner.

Worked example

Take $f(t)=t^3/4-t$ and $a=-2$. Then

so $F(-2)=0$, $F(1)=9/16$ and $F(2)=0$. At $x=2$, $f$ changes from negative to positive, and $F$ has a local minimum; at $x=0$ it changes from positive to negative and $F$ has a local maximum $F(0)=1$. With $\Delta x=1/2$ at $x=1$, $\Delta F=F(1.5)-F(1)=-95/256$ while $f(1)\,\Delta x=-3/8=-96/256$.

Sign changes, moving limits and jumps

• Where $f>0$, $F$ increases; where $f<0$, it decreases; where $f$ changes sign, $F$ has a local extremum.
• Moving $a$ shifts $F$ by a constant ($\int_{a'}^{a}f$). The slope $F'(x)=f(x)$ does not depend on $a$.
• With $x<a$, $F(x)=-\int_x^a f$: the accumulation runs backward.
• For a jump in $f$, $F$ is still continuous but has a corner; the first form of the theorem needs continuity at $x$.

Executable reference code

The reference builds $F$ exactly with Fraction antiderivatives and checks $F(x+h)-F(x)$ against $f(x)\,h$ for shrinking $h$. The NumPy version accumulates a sampled $f$ with the cumulative trapezoid rule and differentiates the result numerically, recovering $f$ up to discretisation error.

"""The accumulation function F(x) = integral from a to x of f, built exactly; F'(x) = f(x) (III.13)."""
from fractions import Fraction as F

f = lambda t: t**3 / 4 - t
P = lambda t: t**4 / 16 - t**2 / 2          # an antiderivative of f
a = F(-2)
Fx = lambda x: P(x) - P(a)                  # F(a) = 0 by construction

assert (Fx(a), Fx(F(1)), Fx(F(2)), Fx(F(0))) == (0, F(9, 16), 0, 1)

# A thin strip adds nearly f(x)*h; the ratio tends to f(x) as h shrinks.
x = F(1)
gaps = [abs((Fx(x + h) - Fx(x)) / h - f(x)) for h in (F(1, 2**k) for k in range(4, 14))]   # small h: gap ~ |f'(x)| h/2
assert all(b < a_ for a_, b in zip(gaps, gaps[1:]))
assert Fx(x + F(1, 2)) - Fx(x) == F(-95, 256)

# A jump in f gives a corner in F: one-sided slopes 1 and -1 at t = 1.
step_F = lambda x: x if x < 1 else 2 - x    # integral from 0 of (1 for t < 1, -1 for t >= 1)
h = F(1, 1000)
assert (step_F(F(1)) - step_F(1 - h)) / h == 1 and (step_F(1 + h) - step_F(F(1))) / h == -1

"""Cumulative trapezoid accumulation of a sampled f, then numerical differentiation recovers f (III.13)."""
import numpy as np

t = np.linspace(-2, 3, 2001)
f = t**3 / 4 - t
F = np.concatenate([[0.0], np.cumsum((f[1:] + f[:-1]) / 2 * np.diff(t))])   # F(-2) = 0
exact = t**4 / 16 - t**2 / 2 + 1
assert np.max(np.abs(F - exact)) < 1e-5
recovered = np.gradient(F, t)
assert np.max(np.abs(recovered[1:-1] - f[1:-1])) < 1e-4

Explore

Find the maximum and minimum of $F$ by dragging $x$ to the zeros of $f$. Move $a$ and watch the whole curve $F$ shift vertically while its tangent at $x$ keeps the same slope. On the jump preset, set $\Delta x$ negative to see the left slope.

Common mistakes

Confusing $F(x)$ (an accumulated amount) with $f(x)$ (a rate). Thinking the lower limit affects $F'$. Applying $F'=f$ at a discontinuity of $f$. Forgetting that areas below the axis make $F$ decrease.

Numerical considerations

Cumulative numerical integration followed by numerical differentiation recovers $f$ only approximately: discretisation error from the first step and the step-size trade-off of III.14 in the second. Working with exact antiderivatives, when they exist, avoids both.

Engineering and ML connections

Cumulative quantities are everywhere: position from velocity, charge from current, a cumulative distribution function from a density, whose derivative recovers the density (IV.8). Prefix sums in computing are the discrete accumulation function, and their differences recover the original sequence.

Exercises

Summary and checkpoints

The accumulation function $F(x)=\int_a^x f$ grows at the rate $f(x)$: $F'(x)=f(x)$ wherever $f$ is continuous. Changing $a$ shifts $F$ by a constant; a jump in $f$ makes a corner in $F$. Next, III.14 turns to computing derivatives numerically, where finite precision shapes the answer.