The accumulation function
The top panel shows a rate, $f(t) = \sin t + 1.2$. The bottom panel traces its running total $F$ as a sweep moves right. While the sweep crosses the stretch where $f$ is tall, what does $F$ do?
$F$ starts at $0$ and never falls. From $x = 1$ to $2$ — through the rate's peak of $2.2$ — the total jumps from $1.660$ to $3.816$. From $x = 4$ to $5$, where the rate sags toward $0.2$, it creeps from $6.454$ to $6.716$. Same sweep width, an eighth of the growth.
You predicted how a function behaves — so it is one. Freeze the lower endpoint at $0$, let the upper endpoint slide, and record the area collected so far at each stop:
This is the accumulation function: one output per stopping point $x$. Where $f$ is tall, area piles up fast and $F$ climbs steeply; where $f$ is short, $F$ barely moves; $F(0) = 0$ because a zero-width sweep collects nothing. The letter changed inside the integral: $x$ marks where the sweep stops, so a dummy variable $t$ does the sweeping and vanishes. This rate never goes negative — negative rates are lesson 4's business.
Whatever else changes, the instrument shows $F(0) = 0$. Why?
At $x = 2$ the stats read $f(x)$ rate $= 2.109$ and $F(x)$ total $= 3.816$. What does $3.816$ measure?
Transfer: a tank fills at $r(t) = 3$ liters per minute for the first $2$ minutes, then $1$ liter per minute for the next $2$. What is $\int_0^4 r(t)\,dt$?
An integral with a moving upper endpoint is a function: the running total of a rate. Tall rate, steep total; short rate, shallow total — $F$ echoes the height of $f$ in its slope. Naming that echo is the next lesson.