From a Rate to a Quantity
A rate describes how quickly a quantity changes. In context, that might be the rate at which water enters or leaves a tank, a population grows or declines, or a substance is added to a mixture. An integral of the rate over time gives the net change in the quantity over that interval.
When a rate is piecewise, the formula may change at specified times. Each part must be integrated over its own time interval, with the results added. If the quantity at the beginning is known, that initial amount is then combined with the net change to find the quantity at the end.
Units help keep these ideas distinct. If \(Q\) is measured in liters and time in minutes, then \(r\) is measured in liters per minute. Multiplying a rate by a time interval, or integrating that rate with respect to time, gives liters. The average rate, by contrast, is still measured in liters per minute.
Setting Up a Piecewise Accumulation
Suppose the rate changes formula at an intermediate time \(c\), where \(a<c<b\). The accumulation over the whole interval is the sum of the accumulations on the two parts:
For more than one change time, split at every boundary where the rate formula changes. Use each formula only on its stated interval. A positive rate contributes a positive net change; a negative rate contributes a negative net change. This is why a rate integral gives net change, not necessarily the total amount that entered or left.
Worked Example: A Tank with a Rate That Changes Sign
Worked Example: Finding a Tank’s Final Volume and Average Rate
A tank initially contains \(120\) liters of water. During the next five minutes, its net rate of change is \(4\) liters per minute for \(0\leq t<2\), then \(-2\) liters per minute for \(2\leq t\leq5\). Find the net change, the volume after five minutes, and the average rate over the five-minute interval.
The rate has units of liters per minute, so its integral with respect to time has units of liters. Split at \(t=2\), where the rate formula changes:
The negative contribution means the tank loses \(6\) liters during the second interval. The final volume is the initial volume plus the net change:
The average rate is net change divided by elapsed time:
Check: The average rate multiplied by five minutes gives \(0.4(5)=2\) liters, the net change found from the two integrals. Adding that change to \(120\) liters gives \(122\) liters, as calculated.
Average Rate and Average Value of a Rate
For a quantity \(Q\), its average rate of change from \(a\) to \(b\) is the net change divided by the length of the interval. If \(Q'=r\) and the rate integral gives the net change, then this is also the average value of the rate function:
This equality connects the contextual question “What was the average rate of change?” to the average value of a function. It does not mean that the rate was constant. It means that a constant rate equal to this average would produce the same net change over the same interval.
Why the Accumulation Formula Works
The justification comes from the Fundamental Theorem of Calculus on each interval where the rate formula applies. It is important that the quantity does not jump at the boundaries: piecewise differentiability alone does not rule out jumps. Here is the precise argument for a finite partition \(a=t_0<t_1<\cdots<t_n=b\).
Suppose \(r\) is piecewise continuous on \([a,b]\), \(Q\) is continuous across the partition points, and on each open piece \(Q'=r\). Assume the usual Fundamental Theorem conditions hold on each piece.
For every \(i\), the Fundamental Theorem gives \(\int_{t_{i-1}}^{t_i}r(t)\,dt=Q(t_i)-Q(t_{i-1})\). Continuity at the boundaries ensures these endpoint values belong to the same accumulated quantity.
Summing the equations gives \(\int_a^b r(t)\,dt=\sum_{i=1}^{n}(Q(t_i)-Q(t_{i-1}))\). Each intermediate value appears once with a plus sign and once with a minus sign.
The sum therefore equals \(Q(t_n)-Q(t_0)=Q(b)-Q(a)\). This proves that the integral of the rate is the net change in the quantity.
Without continuity of \(Q\) across the boundaries, this conclusion need not hold: a jump in \(Q\) contributes to its total change but is not captured by integrating the rate on the open pieces. In a contextual model, check that the quantity is being treated as continuous across rate changes, or account separately for any instantaneous jump.
Worked Example: A Quantity with Linear Rates on Two Intervals
Worked Example: Accumulating a Changing Net Rate
A reservoir contains \(40\) kilograms of a dissolved substance at time \(t=0\). Its net rate of change is \(r(t)=12-2t\) kilograms per hour for \(0\leq t<3\), and \(r(t)=2t-6\) kilograms per hour for \(3\leq t\leq5\). Find the net change, the amount at \(t=5\), and the average net rate over the five hours.
Since the rate formula changes at \(t=3\), integrate the first formula from \(0\) to \(3\) and the second from \(3\) to \(5\):
The final amount is \(40+31=71\) kilograms. Divide the net change by the total elapsed time to get the average net rate:
Check: Multiplying the average net rate by \(5\) hours gives \(6.2(5)=31\) kilograms. Adding this to the initial \(40\) kilograms recovers \(71\) kilograms. The two antiderivatives also differentiate to the stated rates: \(\frac{d}{dt}(12t-t^2)=12-2t\) and \(\frac{d}{dt}(t^2-6t)=2t-6\).
Worked Example: A Piecewise Rate with a Negative Contribution
Worked Example: Interpreting a Negative Rate
A storage system holds \(75\) megabytes of data at the start of a six-minute interval. Its net rate of change is \(5\) megabytes per minute for the first two minutes, \(3\) megabytes per minute for the next two minutes, and \(-4\) megabytes per minute for the final two minutes. Find the final amount and the average net rate.
Use time measured in minutes from the start, so the three intervals are \([0,2]\), \([2,4]\), and \([4,6]\). Integrate the constant rate on each interval:
The final amount is \(75+8=83\) megabytes. The average net rate is \(8/6=4/3\) megabytes per minute. The negative rate represents a loss of \(8\) megabytes during the last two minutes; it does not mean that the accumulated amount over the full interval is negative.
Check: The average rate \(4/3\) megabytes per minute multiplied by \(6\) minutes gives \(8\) megabytes, the same net change as the three interval contributions. Adding \(8\) megabytes to the initial \(75\) megabytes gives \(83\) megabytes.
Common Errors in Contextual Problems
A frequent mistake is to average the rates on different pieces without accounting for how long each rate applies. In the tank example, the rates are \(4\) liters per minute for two minutes and \(-2\) liters per minute for three minutes. A simple, unweighted average would give \(\frac{4+(-2)}{2}=1\) liter per minute. That is wrong because the intervals have different lengths.
The correct average rate weights each rate by its duration through the integral:
Another error is treating the integral as the final quantity. In the same tank problem, the integral is \(2\) liters, which is the net change, not the final volume. The final volume is \(120+2=122\) liters. The units help reveal the distinction: an integral of liters per minute over minutes yields liters of change, while the initial amount is also in liters and must be added separately.
Check Your Understanding
For each question, identify the relevant time intervals and keep track of whether the requested result is a net change, a final quantity, or a rate.
- A quantity starts at \(30\) units. Its rate is \(2\) units per minute for \(0\leq t<3\), then \(-1\) unit per minute for \(3\leq t\leq7\). What is the net change and final quantity?
- For the rate in question 1, what is the average rate over the full interval?
- Why must the integral of a rate be split at the time when its formula changes?
- What continuity condition on \(Q\) is needed at piece boundaries to infer \(Q(b)-Q(a)=\int_a^b r(t)\,dt\) from the piecewise derivative relation?
- A rate is measured in gallons per hour and time in hours. What are the units of its integral and of its average value over a time interval?
