Two Different Questions About a Rate
A rate can describe how quickly a quantity increases or decreases. Its sign matters: a positive rate indicates increase, while a negative rate indicates decrease. As a result, integrating a rate answers different questions depending on whether we keep its sign or count every change positively.
Suppose \(r(t)\) is the rate of change of a quantity \(Q(t)\) on an interval \([a,b]\). The signed integral gives the net change: increases contribute positively and decreases negatively. The total accumulated amount instead counts the size of both increases and decreases as positive amounts.
These answers need not be equal. If a quantity rises and later falls, some of its increase is canceled in the net change, but not in the total amount of change.
Definitions and Conditions
Assume that \(r\) is integrable on \([a,b]\). In applications, \(r\) is often continuous or piecewise continuous, so its integral can be computed using the usual methods of AP Calculus.
When \(r(t)>0\), the rate adds to the quantity. When \(r(t)<0\), it subtracts from the quantity. The absolute value \(|r(t)|\) is nonnegative in either case, so its integral counts the magnitude of change without cancellation. At any point where \(r(t)=0\), neither integral receives a contribution at that instant.
If \(r\) has units of quantity per unit of time and \(t\) is measured in units of time, both integrals have units of quantity. The net change can be positive, negative, or zero; the total accumulated amount cannot be negative.
Why the Absolute Value Gives Total Change
The difference between the two integrals follows directly from what they do on intervals where the rate has a constant sign. If \(r\) is nonnegative throughout an interval, then \(|r|=r\), so the signed integral and the total amount agree there. If \(r\) is nonpositive throughout an interval, then \(|r|=-r\), so the total amount is the negative of the signed integral on that interval.
For a rate that changes sign finitely many times, divide the interval at its zeros or other sign-change points. On each resulting subinterval, use the appropriate sign. The following argument establishes the formula for this situation.
Let \(a=c_0<c_1<\cdots<c_n=b\) divide the interval into subintervals on which \(r\) has one sign, allowing \(r\) to be zero at endpoints.
If \(r\geq 0\) on \([c_{i-1},c_i]\), then \(\int_{c_{i-1}}^{c_i}|r(t)|\,dt=\int_{c_{i-1}}^{c_i}r(t)\,dt\). If \(r\leq 0\), then \(\int_{c_{i-1}}^{c_i}|r(t)|\,dt=-\int_{c_{i-1}}^{c_i}r(t)\,dt\).
The integral over the whole interval is the sum of the integrals over these subintervals. Thus, \(\int_a^b |r(t)|\,dt\) adds each positive contribution and the magnitude of each negative contribution, while \(\int_a^b r(t)\,dt\) subtracts the negative contributions.
This explains why the two quantities answer different questions. The signed integral records the overall effect after increases and decreases offset one another. The absolute-value integral records how much change occurred in total.
Worked Examples
Worked Example: A Rate That Decreases and Then Increases
Let \(r(t)=t-2\) on \([0,4]\), where \(r\) is measured in units per hour and \(t\) in hours. Find the net change and the total accumulated amount.
The rate is negative on \([0,2)\), zero at \(t=2\), and positive on \((2,4]\). An antiderivative is \(F(t)=\frac{t^2}{2}-2t\), since \(F'(t)=t-2\).
The net change is
For total accumulated amount, split at \(t=2\). On \([0,2]\), \(|r(t)|=2-t\), and on \([2,4]\), \(|r(t)|=t-2\). Therefore,
The net change is \(0\) units because the increase and decrease cancel. The total accumulated amount is \(4\) units. As a check, the two pieces have equal magnitudes: the rate decreases the quantity by \(2\) units, then increases it by \(2\) units.
Worked Example: A Piecewise-Constant Rate
A tank’s water amount changes at \(3\) liters per minute for the first \(2\) minutes, then at \(-2\) liters per minute for the next \(3\) minutes. Find the net change and the total amount of change over the \(5\)-minute interval.
Let \(t\) be time in minutes, and write the rate as \(r(t)=3\) on \(0\leq t<2\) and \(r(t)=-2\) on \(2<t\leq5\). The value assigned at the single transition time does not affect either integral.
The net change is the sum of each rate multiplied by the duration of its interval:
For total accumulated amount, the negative rate contributes positively after taking its absolute value:
Check the interpretation: the tank first gains \(6\) liters and then loses \(6\) liters, leaving no net change. It has nevertheless undergone \(6+6=12\) liters of total change.
Worked Example: Three Sign Intervals
Let \(r(t)=(t-1)(3-t)\) on \([0,4]\). Find the net change and total accumulated amount.
The zeros are \(t=1\) and \(t=3\). On \([0,1]\), \(t-1\leq0\) and \(3-t\geq0\), so \(r(t)\leq0\). On \([1,3]\), both factors are nonnegative, so \(r(t)\geq0\). On \([3,4]\), \(t-1\geq0\) and \(3-t\leq0\), so \(r(t)\leq0\). In particular, the rate is zero at both \(1\) and \(3\); the sign statements on the closed intervals are non-strict.
Expand the rate and find an antiderivative:
Checking, \(F'(t)=-t^2+4t-3=r(t)\). Evaluate \(F\) at the required endpoints:
The net change is
For the total amount, split at both zeros. Since the rate is nonpositive on the first and last intervals and nonnegative on the middle one,
The results are consistent: the signed contributions are \(-\frac43\), \(+\frac43\), and \(-\frac43\), which sum to \(-\frac43\). Their magnitudes sum to \(4\). If \(r\) has units of quantity per unit of time, the two answers have units of quantity.
A Common Mistake: Using the Net Change as the Total
A common error is to calculate \(\int_a^b r(t)\,dt\) and report its absolute value as the total accumulated amount. The absolute value of the final net change does not restore changes that canceled during the interval.
For example, in the first worked example the net change is \(0\). A mistaken calculation might say, “The total accumulated amount is \(\left|\int_0^4(t-2)\,dt\right|=|0|=0\).” This is wrong: it takes the absolute value after the negative and positive contributions have already canceled.
The correction is to take the absolute value of the rate before integrating, or equivalently to split at sign changes and add the magnitudes of the separate signed integrals. In general, \(\left|\int_a^b r(t)\,dt\right|\) is not the total accumulated amount.
Choosing the Right Integral
| Question | Quantity to calculate | What negative rates do |
|---|---|---|
| What is the overall change from the start to the end? | \(\int_a^b r(t)\,dt\) | Subtract from positive contributions |
| How much change occurred altogether, counting increases and decreases? | \(\int_a^b |r(t)|\,dt\) | Contribute their magnitudes positively |
When choosing, focus on the wording of the question. “Net change,” “overall change,” or “final amount minus initial amount” calls for the signed integral. “Total amount of change,” when increases and decreases are both to count positively, calls for the absolute-value integral. If a question describes a specific physical process, interpret what is being counted before selecting a formula.
Check Your Understanding
For each question, decide whether it asks for net change or total accumulated amount, and show the calculation where appropriate.
- A quantity has rate \(r(t)=t-1\) on \([0,2]\). Find both the net change and the total accumulated amount.
- A flow rate is \(5\) liters per minute for \(3\) minutes, then \(-4\) liters per minute for \(2\) minutes. Find the net change and total amount of change.
- Why is \(\left|\int_a^b r(t)\,dt\right|\) generally not a formula for total accumulated amount?
- If \(r(t)\leq0\) throughout \([a,b]\), how are \(\int_a^b r(t)\,dt\) and \(\int_a^b |r(t)|\,dt\) related?
- For \(r(t)=(t-2)(4-t)\) on \([0,5]\), identify the intervals of nonpositive and nonnegative rate, taking account of the zeros, and state where to split the absolute-value integral.
