This file is not updated—let me know if you see this!

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okay gets, essentially, if it were ${\int }_8^{10}r(t)dt$, then it would be the second, but since it is the $r^{\prime }(t)$ (aka the Derivative of $r$ per $t$) it would be the rate of change, not the current rate of increase of subscribers.

the way i understand it, it follows Velocity and Acceleration logic, where derivatives are the acceleration, the way your $r$ is changing, integrals are the *net change of $r$.

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Expression	What It Represents	Analogy
$r(t)$	Current rate of subscriber gain (e.g., 100 subs/month)	Velocity
$r^{\prime }(t)$	How fast that rate is changing (e.g., +3 subs/month²)	Acceleration
${\int }_{}r(t)dt$	Net change in number of subscribers over time	Total distance traveled
${\int }_{}{r}^{\prime }(t)dt$	Net change in the rate of subscriber gain	Change in velocity