One Rate, Two Different Questions
A rate that changes over time can describe a quantity increasing during some intervals and decreasing during others. Integrating the rate answers a specific question: what is the net change between the start and the end? It does not, in general, tell us how much increase and decrease occurred altogether.
For example, if a quantity rises by 6 units and later falls by 6 units, its net change is zero. Yet it has changed by 12 units in total when both the rise and the fall are counted. The distinction matters whenever the question concerns both the final difference and the total amount of movement.
Suppose \(r(t)\) is the rate of change of a quantity \(Q(t)\) over the interval \([a,b]\). A positive rate means \(Q\) is increasing, and a negative rate means \(Q\) is decreasing. The signed integral adds these changes with their signs. To count the total amount of increase and decrease, add the magnitudes of the changes instead.
Here “gross amount of change” means the total accumulated size of the increases and decreases. It is not the same quantity as the signed net change. Both integrals have the units of \(Q\): the units of the rate are multiplied by the units of time.
Why the Signed Integral Gives Net Change
The connection to net change follows from the Fundamental Theorem of Calculus. If \(Q\) is continuous on \([a,b]\), differentiable on the interior, and \(Q'(t)=r(t)\), then integrating its rate recovers the endpoint difference.
The signs in this result matter. On an interval where \(r(t)\) is positive, \(Q\) increases and the integral contributes a positive change. Where \(r(t)\) is negative, \(Q\) decreases and the integral contributes a negative change. Thus a decrease offsets an earlier increase when we calculate the net result.
For gross amount of change, the rate’s magnitude \(|r(t)|\) is used instead. Since \(|r(t)|\) is nonnegative, decreases are counted by their size rather than used to cancel increases. If the rate changes sign, the integral for gross change is often easiest to calculate by splitting the interval at each sign-change point.
A Derivation by Splitting at Sign Changes
Suppose \(r\) is continuous and has finitely many sign changes on \([a,b]\), at points \(c_1,\ldots,c_n\). Include the endpoints by setting \(c_0=a\) and \(c_{n+1}=b\). On each interval between consecutive points, the rate has one sign or is zero. Therefore the rate’s magnitude on that interval is either \(r\) or \(-r\).
Split \([a,b]\) at the sign-change points so that \(r\) does not switch sign inside any resulting interval.
On \([c_i,c_{i+1}]\), the change is \(\int_{c_i}^{c_{i+1}}r(t)\,dt\), positive for an increase and negative for a decrease.
Because the rate has a constant sign on that interval, its gross amount is \(\left|\int_{c_i}^{c_{i+1}}r(t)\,dt\right|=\int_{c_i}^{c_{i+1}}|r(t)|\,dt\).
The signed interval changes add to the net change; their magnitudes add to the gross amount of change.
To justify the equality in Step 3, first consider an interval where \(r(t)\geq 0\). Then \(|r(t)|=r(t)\), so the integral of \(|r|\) equals the signed integral, which is nonnegative. If \(r(t)\leq 0\), then \(|r(t)|=-r(t)\), so the integral of \(|r|\) is the negative of the signed integral and equals its magnitude. Adding these interval results gives the gross total across the whole interval.
Worked Examples
Worked Example: Alternating Constant Rates
A tank’s water amount changes at a rate of \(3\) liters per hour for the first \(2\) hours, \(-2\) liters per hour for the next \(3\) hours, and \(1\) liter per hour for the final \(2\) hours. Find the net change and the gross amount of change.
The time intervals are \([0,2]\), \([2,5]\), and \([5,7]\). Calculate the signed change on each interval by multiplying its rate by its duration:
Add the signed changes to obtain the net change:
For the gross amount, count each interval’s change by its magnitude:
Check: The water amount rises by \(6\) liters, falls by \(6\) liters, and rises by \(2\) liters. The first rise and fall cancel in the final difference, leaving a net increase of \(2\) liters. Their sizes, together with the last rise, total \(14\) liters. The answer has units of liters because each rate in liters per hour was multiplied by a duration in hours.
Worked Example: A Rate That Changes Sign Once
Let the rate of change of a quantity, in kilograms per minute, be \(r(t)=t-1\) on \(0\leq t\leq 3\), where \(t\) is in minutes. Find the net change and gross amount of change.
The rate is zero when \(t-1=0\), so the sign-change point is \(t=1\). For \(0\leq t<1\), \(r(t)<0\); for \(1<t\leq3\), \(r(t)>0\). An antiderivative of \(t-1\) is \(F(t)=\frac{t^2}{2}-t\), since \(F'(t)=t-1\).
First find the net change on the whole interval:
For the gross amount, split at \(t=1\). On the first interval the rate is negative, and on the second it is positive:
Evaluate the first contribution:
Evaluate the second contribution:
Therefore the gross amount is \(\frac12+2=\frac52\) kilograms.
Check: The quantity first decreases by \(\frac12\) kilogram and then increases by \(2\) kilograms. The net result is \(-\frac12+2=\frac32\) kilograms, matching the signed integral. The total size of those changes is \(\frac12+2=\frac52\) kilograms.
Worked Example: A Rate That Changes Sign Twice
Suppose \(r(t)=(t-1)(t-3)\) is a rate, in units per hour, on \(0\leq t\leq4\). Find the net change and gross amount of change.
Expand the rate to simplify integration:
The rate is zero at \(t=1\) and \(t=3\). It is positive on \([0,1]\), negative on \([1,3]\), and positive on \([3,4]\). An antiderivative is \(F(t)=\frac{t^3}{3}-2t^2+3t\), because \(F'(t)=t^2-4t+3\).
The net change is the signed integral over the full interval:
For the gross amount, calculate each interval’s signed change, then take its magnitude. On \([0,1]\):
On \([1,3]\):
On \([3,4]\):
The gross amount is consequently
Check: The three interval changes are \(\frac43\), \(-\frac43\), and \(\frac43\). Their signed sum is \(\frac43\), as found by integrating over \([0,4]\). Their magnitudes sum to \(4\), the gross amount of change.
A Common Mistake: Treating Net Change as the Total
A common error is to report the signed integral as the total amount of increase and decrease. In the first worked example, this would mean calculating \(6-6+2=2\) liters and claiming that only \(2\) liters of change occurred altogether. The calculation is valid for net change, but it does not count the earlier rise and fall separately.
For the gross amount, take the magnitude of each interval’s change before adding:
Taking the magnitude only after adding does not fix the error: \(|6-6+2|=2\) still gives the magnitude of the net change, not the gross amount. The correction is to locate the intervals where the rate changes sign, find each interval’s change, and add those changes’ magnitudes.
How to Choose the Right Total
Read the question carefully before choosing an integral. If it asks how much the quantity differs at the end compared with the start, use net change. If it asks for the combined amount of increase and decrease over the interval, use gross amount of change. The rate may be the same in both calculations; the question determines whether its signed value or its magnitude is integrated.
| Question | Quantity to calculate | How decreases affect the result |
|---|---|---|
| What is the overall change from \(a\) to \(b\)? | \(\int_a^b r(t)\,dt\) | A decrease offsets an increase. |
| How much increase and decrease occurred altogether? | \(\int_a^b |r(t)|\,dt\) | A decrease contributes its positive size. |
It is also useful to check whether the result is plausible. The gross amount cannot be less than the magnitude of the net change, because adding interval-change magnitudes cannot produce less than the magnitude of their signed sum. In symbols, \(\left|\int_a^b r(t)\,dt\right|\leq\int_a^b|r(t)|\,dt\). The examples satisfy this check: \(2\leq14\), \(\frac32\leq\frac52\), and \(\frac43\leq4\).
The distinction is useful in models of inventory, water levels, balances, and other changing quantities. A final difference can be small even when substantial movement occurred in both directions. Keeping net change and gross amount separate prevents a correct integral from being used to answer the wrong question.
Check Your Understanding
For each question, decide whether it asks for net change or gross amount of change, and show how you would calculate the result.
- A quantity rises by \(8\) units and later falls by \(3\) units. What are its net change and gross amount of change?
- If \(r(t)=t-2\) on \([0,4]\), at what point should the gross-change integral be split?
- Why does \(\left|\int_a^b r(t)\,dt\right|\) generally fail to give the gross amount of change?
- What conditions on \(Q\) and \(r\) justify the equation \(\int_a^b r(t)\,dt=Q(b)-Q(a)\)?
- For a rate that is nonnegative throughout an interval, how do net change and gross amount of change compare?
