Why Piecewise Rates Require Careful Accumulation
A rate does not always follow one formula for an entire time interval. A pump may change its setting, a vehicle may switch operating modes, or a quantity may grow under one condition and then decline under another. In these situations, the rate is described by different formulas on successive intervals.
The net change is still found by integrating the rate. The important step is to divide the time interval wherever the applicable formula changes. Each resulting definite integral gives the signed change over that piece; adding those changes gives the net change over the full interval.
Set Up the Intervals Before Integrating
Suppose a rate \(r(t)\) is defined by one formula from time \(a\) to time \(c\), and by another formula from \(c\) to time \(b\). The time \(c\) is a breakpoint. To find the net change from \(a\) to \(b\), write
In each integral, use the formula that applies on that subinterval. With several breakpoints, repeat the split until every integral covers an interval on which a single formula applies. If the rate is piecewise continuous, these definite integrals exist, and the net change is the sum of the interval-by-interval signed changes.
A breakpoint is not necessarily a point where the rate is zero or where the quantity stops changing. It is a time where the rule used to calculate the rate changes. The rate formulas may agree at the breakpoint, or they may give a jump. Either way, the integral is split there because that is where the formula changes.
Why Splitting the Integral Works
The justification comes from additivity of definite integrals. For an integrable function, if \(a\le c\le b\), then the integral over the whole interval equals the sum of the integrals over the two pieces. This property can also be seen directly from the Fundamental Theorem of Calculus when each piece has an antiderivative.
Let \(c\) be a time where the rate formula changes, with \(a\le c\le b\).
Let \(F_1\) be an antiderivative of the first formula on \([a,c]\), and let \(F_2\) be an antiderivative of the second formula on \([c,b]\).
The sum is \((F_1(c)-F_1(a))+(F_2(b)-F_2(c))\), the two definite integrals joined at \(c\).
The two antiderivatives need not be the same function: they come from different rate formulas. What matters is that each is used on the interval where its formula applies. The definite integral records each interval’s change, so the pieces can be added to recover the change over the entire interval. If a rate formula has a jump at a breakpoint, the value assigned at that single time does not change the definite integral.
A Reliable Calculation Process
Before calculating, mark the starting time, ending time, and all breakpoints that lie between them. Then make sure no integral crosses a breakpoint without being split. The following process keeps the setup and signs visible:
Include only breakpoints between the requested starting and ending times.
Match the formula to the time interval, not to whether its value is positive or negative.
Substitute the upper endpoint and subtract the value at the lower endpoint.
Keep negative values negative; they represent decreases in the quantity.
Confirm the sign makes sense, attach the appropriate units, and add the net change to an initial quantity if one is given.
Worked Examples
Worked Example: A Pump Changes Its Rate
A pump’s rate \(r(t)\), in liters per minute, is \(3t\) for \(0\le t\le2\), and \(12-3t\) for \(2<t\le4\). Find the net change in the amount of water during the first four minutes.
The formula changes at \(t=2\), so split the integral there:
For the first interval, an antiderivative of \(3t\) is \(\frac{3}{2}t^2\). Thus
For the second interval, an antiderivative of \(12-3t\) is \(12t-\frac{3}{2}t^2\). Therefore
Add the two signed changes: \(6+6=12\). The net change is \(12\) liters. As a check, the rate is \(3t\), which is nonnegative on \(0\le t\le2\) and positive for \(t>0\); the second formula decreases from \(6\) at \(t=2\) to \(0\) at \(t=4\). Both pieces therefore contribute a nonnegative change, consistent with a positive net change. The units are liters per minute multiplied by minutes, giving liters.
Worked Example: A Rate That Produces a Negative Net Change
A quantity has rate \(q(t)=2-t\) for \(0\le t\le3\), followed by \(q(t)=-1\) for \(3<t\le5\). Find its net change from \(t=0\) to \(t=5\).
The breakpoint is \(t=3\). On the first interval, an antiderivative of \(2-t\) is \(2t-\frac{t^2}{2}\):
On the second interval, integrate the constant rate:
Combining the signed changes gives
The net change is \(-\frac{1}{2}\) quantity units. The first interval contributes an increase of \(\frac{3}{2}\), while the second contributes a decrease of \(2\); the decrease is larger, so a negative net change is reasonable. Substituting the interval results back into the split integral confirms the calculation: \(\frac{3}{2}-2=-\frac{1}{2}\).
Worked Example: Use Only the Portions Between the Requested Times
A tank initially contains \(10\) liters. Its rate of change is \(r(t)=2\) liters per minute for \(0\le t\le1\), \(r(t)=1-t\) for \(1<t\le3\), and \(r(t)=-2\) for \(3<t\le4\). Find the amount in the tank at \(t=3.5\).
The requested interval is from \(t=0\) to \(t=3.5\). It includes the breakpoints \(1\) and \(3\), but not the entire final interval: only the portion from \(3\) to \(3.5\) is needed. The net change is
Evaluate the first piece:
An antiderivative of \(1-t\) is \(t-\frac{t^2}{2}\), so the second piece is
The third piece is
Therefore, the net change from \(0\) to \(3.5\) is \(2+(-2)+(-1)=-1\) liter. Add this change to the initial amount: \(10+(-1)=9\) liters. The tank contains \(9\) liters at \(t=3.5\). As a check, the total decrease after the initial increase is \(2+1=3\) liters, giving \(10+2-3=9\) liters, in agreement with the integral calculation.
Common Mistake: Using One Formula Across a Breakpoint
A frequent error is to select the formula that applies at the start of the interval and use it all the way to the end. For the first example, this would incorrectly give
This calculation is wrong because \(3t\) applies only through \(t=2\); from after \(t=2\) through \(t=4\), the rate is \(12-3t\). The correction is to split at \(t=2\): the first piece contributes \(6\), and the second contributes \(6\), for a total net change of \(12\) liters. The incorrect value of \(24\) comes from extending a rate formula beyond the interval where it is defined.
Signs, Units, and Final Checks
Every interval integral is a signed change. A positive rate contributes a positive change, and a negative rate contributes a negative change. When the interval includes both positive and negative rates, the final net change depends on their combined effects. Do not replace a negative interval result with its positive magnitude before adding.
Units provide a useful check. If a rate is measured in liters per minute and time is measured in minutes, its integral has units of liters. If the problem gives an initial quantity, the quantity at the end of the interval equals the initial quantity plus the net change. The net change itself is not the ending quantity unless the initial quantity is zero.
Before accepting a result, check that every breakpoint inside the requested time range was included, that each formula was integrated only where it applies, and that the signed interval results were added correctly. These checks are particularly valuable when one piece increases the quantity and another decreases it.
Check Your Understanding
For each question, split at every relevant breakpoint and preserve the sign of each interval’s contribution.
- A rate is \(r(t)=4\) for \(0\le t\le2\) and \(r(t)=t\) for \(2<t\le4\). Find the net change from \(t=0\) to \(t=4\).
- A rate is \(s(t)=3-t\) for \(0\le t\le2\) and \(s(t)=-1\) for \(2<t\le5\). Find the net change from \(t=0\) to \(t=5\).
- A quantity begins at \(7\). Its rate is \(2t\) for \(0\le t\le1\) and \(3\) for \(1<t\le3\). Find the quantity at \(t=3\).
- Why must an integral be split at a time where the rate’s formula changes?
- If one interval’s change is positive and another interval’s change is negative, what should be done before adding them?
