A rate, times a time, is an amount AB & BC

Water runs into a tank at a constant $4$ liters per minute for $3$ minutes. The tank gains $12$ liters, and nobody needs calculus to say so: the amount is the rate times the elapsed time.

Draw that on a graph of the rate against time and something useful appears. The rate is a horizontal line at height $4$, the elapsed time is a span of $3$ on the horizontal axis, and $4 \times 3$ is the area of the rectangle between the line and the axis. The multiplication and the area are the same arithmetic.

Now let the rate vary. The rectangle stops being a rectangle, and the multiplication stops being one multiplication, but the area is still there and it still measures the amount. That is the whole idea of the unit:

1. The height of a rate graph is a rate.
2. The width is an elapsed amount of the input variable.
3. The area is the accumulated change in the quantity whose rate you drew.

Notice which of the three is the answer. A question about how fast is answered by a height. A question about how much is answered by an area. Reading a height where an area was asked for, or the reverse, accounts for a large share of the marks lost in this unit and it is a reading error, not a calculus error.

A rate can be negative. Water leaves the tank, a particle moves backwards, a population shrinks. On the graph that is a piece of curve below the horizontal axis, and the accumulation over that stretch is a loss.

So the area we mean is signed: pieces above the axis count positive, pieces below count negative, and the accumulated change is the sum. Take a flow rate that runs positive for the first $6$ minutes, enclosing an area of $18$, and negative for the next $4$, enclosing an area of $9$. The tank gained $18$ liters and then lost $9$, so it is up by

$$18 - 9 = 9 \text{ liters}.$$

That number, $9$, is the net change. It is what the integral of the rate will turn out to be, and it is what a question about the change in the amount is asking for.

A different question about the very same picture asks how much water moved through the pipe in total, and the answer is $18 + 9 = 27$ liters, with both pieces counted positively. That number is the total area, and no amount of care with the first calculation produces it.

These two numbers agree only when the rate never changes sign. The moment it does, they part company, and which one is wanted has to be read out of the wording rather than assumed.

Every accumulation question in context can be checked in one step, and the check catches more errors than re-doing the arithmetic does.

An area is a height times a width, so its units are the vertical unit times the horizontal unit. If the vertical axis is liters per minute and the horizontal axis is minutes, the area carries

$$\frac{\text{liters}}{\text{minute}} \times \text{minutes} = \text{liters}.$$

The minutes cancel, which is exactly why the area answers a question about liters. The same cancellation runs everywhere: meters per second against seconds gives meters; bacteria per hour against hours gives bacteria; dollars per unit against units gives dollars.

Two consequences worth keeping.

1. An accumulation is a change, not a level. The area says how much the quantity moved, and it says nothing about where it started. A tank holding $50$ liters that gains $9$ holds $59$, and the area only ever produced the $9$.
2. If the units do not cancel, the picture is not a rate graph. Area under a graph of the amount itself has units of liter-minutes, which measures nothing anyone asked about.

Stop the clock early and you get a smaller amount. Stop it later and you get a larger one, or a smaller one if the rate has gone negative in the meantime. The accumulated change depends on where you stop, which makes it a function of the stopping point.

Call it $g$: the area collected from the start up to $x$. Topic 6.4 gives it a symbol and Topic 6.5 studies it properly. What matters now is how it behaves, and it is governed by the sign of the rate, not by the rate's size:

1. While the rate is positive, $g$ is increasing, however small the rate is.
2. While the rate is negative, $g$ is decreasing, however close to zero it is.
3. $g$ is largest where the rate crosses from positive to negative.

For the tank above, the flow is positive until minute $6$ and negative after, so the tank holds the most water at minute $6$, with $18$ liters gained. The rate itself reached its largest value back at minute two, and at that moment the tank was still filling and nowhere near full. The rate peaking and the amount peaking are different events at different times.

A rate that is positive and falling is the case worth saying out loud, because it sounds like a decrease and is not one. Filling more slowly is still filling. The amount goes up the whole time; it just goes up less steeply.