Say what the number means AB & BC

Every applied integral in this topic has the same shape: a rate, integrated against the variable it is a rate in.

$$\int_a^b (\text{units of } f) \, d(\text{units of } x) \;\longrightarrow\; \text{their product}.$$

Water flowing at $R(t)$ gallons per minute, integrated over minutes, gives gallons. Traffic passing at $C(t)$ cars per hour, integrated over hours, gives cars. A population changing at $P'(t)$ people per year, integrated over years, gives people. In every case the "per" cancels, and what is left is the quantity itself rather than its rate.

That cancellation is the whole check, and it is fast enough to run every time. If the answer still carries a "per" in it, an integral has been mistaken for a rate somewhere.

It also settles a question students ask backwards. $R(6) = 20$ gallons per minute is how fast water is entering at the single instant $t = 6$; $\int_0^6 R(t)\,dt = 84$ gallons is how much entered across the whole six minutes. One is a speed, the other is a total, and no arithmetic converts between them without an interval.

The Fundamental Theorem gives a change, so an amount is always built the same way:

$$A(b) = A(a) + \int_a^b A'(t)\,dt.$$

A tank holding $50$ gallons into which a net $12$ gallons flow holds $62$. A population of $1200$ that changes by $-180$ over five years stands at $1020$. The accumulated integral is never the answer to "how much is there now" on its own.

The accumulation function makes the same point in function form. If $G(x) = \int_2^x f(t)\,dt$, then $G$ measures change from $x = 2$, so $G(2) = 0$ by construction: an integral over an interval of zero width accumulates nothing. $G$ is not $f$, it is not $f(2)$, and its value at $x$ is an area rather than a height.

When a problem gives an initial condition, that condition is load-bearing. A stated $A(0)$ that never appears in the work is the clearest possible sign that a change has been reported as an amount.

Most exam versions have something entering and something leaving. Then the net rate is the difference and the accumulated change is the integral of that difference:

$$A(b) - A(a) = \int_a^b \big(R_{\text{in}}(t) - R_{\text{out}}(t)\big)\,dt.$$

Take water entering at $R(t) = 8 + 2t$ and leaving at $W(t) = 4t$ gallons per minute, with $50$ gallons present at $t = 0$. Over $[0, 6]$, $\int_0^6 R = 84$ gallons in and $\int_0^6 W = 72$ gallons out, so the tank gains $12$ and finishes with $62$.

Three separate questions live in that one setup, and they have three different answers:

1. How much entered? $\int R = 84$ gallons. The outflow is irrelevant.
2. By how much did the amount change? $\int (R - W) = 12$ gallons.
3. How much is in the tank at the end? $50 + 12 = 62$ gallons.

Reading the question for which of the three it is asking is most of the work.

The amount $A(t)$ has $A'(t) = R(t) - W(t)$, so the candidates for a maximum are exactly the times the net rate is zero, plus the endpoints. This is Unit 5's first derivative test with a rate supplied by the problem instead of by differentiating.

In the example, $R = W$ when $8 + 2t = 4t$, at $t = 4$. The net rate is positive before that and negative after, so the amount rises to $t = 4$ and falls afterwards: the tank is fullest at $t = 4$, holding $50 + 16 = 66$ gallons. Note that the maximum is not at an endpoint and not where either rate is largest.

Finally, the sentence. An interpretation that earns the point names three things:

1. The quantity. Water, cars, people. Not "the integral".
2. The interval. "From $t = 0$ to $t = 6$ minutes", not just "at $6$".
3. The units. Gallons, and gallons rather than gallons per minute.

So $\int_0^6 R(t)\,dt = 84$ becomes: $84$ gallons of water entered the tank during the first six minutes. Compare "the tank has $84$ gallons", which drops the initial amount and the outflow, and "water enters at $84$ gallons per minute", which reports an accumulation as a rate. Both describe a number that was computed correctly.