Why Compare Accumulations?
Two models can describe how quickly a quantity changes over time. One rate may be larger at some times and smaller at others, so comparing the rates at a single moment does not determine which model produces the greater total over an entire interval. To answer that question, compare the amounts accumulated across the interval.
For example, a production rate that is higher early on may be overtaken later by another rate. The later advantage might make up for the earlier deficit, or it might not. Integrating the difference between the rates keeps track of both effects in one calculation.
Rate Differences and Accumulated Differences
Let \(r_A(t)\) and \(r_B(t)\) be two integrable rate functions on an interval \([a,b]\). Assume both describe changes in the same kind of quantity and use the same time units. The accumulated change predicted by model \(A\) from time \(a\) to time \(t\) is \(\int_a^t r_A(s)\,ds\); model \(B\)'s accumulated change is defined in the same way.
The sign of \(D(t)\) answers the comparison question at time \(t\). If \(D(t)>0\), model \(A\) has accumulated more since \(a\). If \(D(t)<0\), model \(B\) has accumulated more. If \(D(t)=0\), their accumulated changes since \(a\) are equal.
In particular, the difference in their totals over the full interval is
A positive value means that model \(A\) accumulates more over \([a,b]\); a negative value means that model \(B\) accumulates more; and zero means that the totals over the interval are equal. The units of \(D\) are the units of the quantity being accumulated: a rate in units per hour, for instance, integrates over hours to give units of the quantity.
This comparison concerns accumulated changes from a common starting time. If the models also describe amounts already present at time \(a\), and those starting amounts are \(Q_A(a)\) and \(Q_B(a)\), then the difference in their amounts at time \(t\) is
The integral alone compares the changes after \(a\); it does not erase a difference in starting amounts.
Why Integrating the Difference Works
The key fact is the linearity of the definite integral: the integral of a difference is the difference of the integrals. Thus, for every \(t\) in the interval,
The left side combines both rate models into one function, \(r_A-r_B\). The right side shows that its integral is exactly the difference between the two accumulated changes. This identity justifies using one integral to compare two totals.
If the rate functions are continuous, the Fundamental Theorem of Calculus also gives \(D'(t)=r_A(t)-r_B(t)\). This explains how the instantaneous rate difference affects the cumulative difference: when \(r_A(t)>r_B(t)\), \(D\) is increasing at that time; when \(r_A(t)<r_B(t)\), \(D\) is decreasing. But increasing does not necessarily mean positive. The current value of \(D\) depends on the full history of the difference since \(a\).
To justify the result, \(r_A-r_B\ge 0\) everywhere, so its integral cannot be negative. If the difference is continuous and strictly positive on a subinterval of positive length, its integral over that subinterval is positive. On the rest of \([a,b]\), its integral is nonnegative. Adding those contributions gives a strictly positive integral over the full interval. In many comparisons, however, one rate is larger at some times and smaller at others. In that situation, the signed integral—not either rate’s value at just one time—determines the total comparison.
Worked Comparisons
Worked Example: Equal Totals Despite a Rate Crossover
Suppose two production rates, in units per hour, are \(r_A(t)=3t+2\) and \(r_B(t)=t+6\) for \(0\le t\le4\), where \(t\) is measured in hours. Compare their total production over the interval and their accumulated amounts at intermediate times.
First subtract the rates:
The accumulated difference from time \(0\) to \(t\) is
At the endpoint,
So the total productions are equal. Check by integrating each rate separately:
The rates are equal when \(2t-4=0\), or \(t=2\) hours. For \(0<t<4\), \(t(t-4)<0\), so \(D(t)<0\): model \(B\) has accumulated more at every interior time, even though the totals tie at \(t=4\). The endpoint calculation and the separate totals agree, as required.
Worked Example: Finding an Intermediate Cumulative Tie
Let \(r_A(t)=t+1\) and \(r_B(t)=3\), in units per hour, on \(0\le t\le5\). Find when their accumulated changes from time \(0\) are equal, then compare their totals at time \(5\).
The difference is \(r_A(t)-r_B(t)=t-2\). Therefore, the cumulative difference is
The accumulated changes are equal when \(D(t)=0\). Factoring gives \(t(t-4)/2=0\), so \(t=0\) or \(t=4\). The initial equality at \(t=0\) is automatic because neither model has yet accumulated anything. The later tie occurs at \(t=4\).
Check the amounts at that time:
At the endpoint \(t=5\), the totals are
Thus model \(A\) has accumulated \(35/2-30/2=5/2\) more units by time \(5\). Differentiating \(D(t)=t^2/2-2t\) gives \(D'(t)=t-2\), the original rate difference, which checks the cumulative-difference formula.
Worked Example: Equal Endpoint Totals Do Not Mean Equal Accumulations Throughout
Suppose \(r_A(t)=t^2\) and \(r_B(t)=2t\), in units per hour, for \(0\le t\le3\). Compare their total accumulations and determine whether their cumulative amounts ever tie after the starting time.
The rate difference is
Its integral from \(0\) to \(t\) is
On \(0<t<3\), \(t^2>0\) and \(t/3-1<0\), so \(D(t)<0\). The cumulative amounts do not tie at an interior time: model \(B\) is ahead throughout the interior of the interval. At \(t=3\),
Indeed, each model accumulates \(9\) units over the full interval:
The instantaneous rates are equal at \(t=2\), since \(2^2=2(2)=4\). But their cumulative amounts are not equal then:
Thus a rate crossover is not, by itself, a cumulative tie. The integral tracks the net advantage accumulated over all earlier times.
A Reliable Comparison Procedure
For two rate models on \([a,b]\), the following sequence keeps the comparison focused on the accumulated quantity:
Check that both rates describe the same quantity per the same time unit.
For example, define model \(A\) minus model \(B\); keep that order consistent when interpreting signs.
Use \(D(t)=\int_a^t(r_A-r_B)\) for cumulative comparisons, or evaluate at \(b\) for the full-interval comparison.
A zero of \(D(t)\) is a time when the accumulated changes since \(a\) are equal. The sign of \(D(t)\) identifies which model is ahead.
When useful, calculate the separate integrals or differentiate \(D(t)\) to verify the comparison.
A Common Mistake: Confusing Rates with Cumulative Totals
A rate tells how fast a quantity is changing at one instant. A cumulative amount records the net effect of the rate over an interval. These are different quantities, with different units and meanings.
In the third example, a tempting but incorrect calculation is: “The rates are equal at \(t=2\), so the cumulative amounts are equal at \(t=2\).” The rates are indeed both \(4\) units per hour at that instant, but the cumulative difference is
The correction is to integrate the rate difference from the common starting time and then solve \(D(t)=0\) if the question asks when the cumulative amounts coincide. Setting the rates equal answers a different question: when their instantaneous rates match.
Why This Comparison Matters
Comparing accumulations is useful whenever two rate models describe the same process under different assumptions, plans, or conditions. The signed integral combines periods of advantage and disadvantage, so it can answer questions that a point-by-point comparison alone cannot settle. It also makes the target precise: the total difference over the interval is \(D(b)\), while a time of equal cumulative change is a solution of \(D(t)=0\).
Always interpret the result in context. A zero integral over the full interval says the net accumulated changes agree at the endpoint; it does not say that the rates agree at every time or that the cumulative amounts agree at every intermediate time. If the starting amounts differ, include that initial difference before drawing conclusions about which model has the greater amount.
Check Your Understanding
Use the accumulated difference \(D(t)=\int_a^t(r_A-r_B)\) to answer these questions.
- If \(D(b)<0\), which model has accumulated more over \([a,b]\)?
- What equation should you solve to find a time when the models’ accumulated changes since \(a\) are equal?
- If \(r_A(t)\ge r_B(t)\) throughout an interval, what can you conclude about their integrals over that interval?
- Why does equality of the instantaneous rates at one time not generally imply equality of the cumulative amounts at that time?
- If the models begin with different amounts at time \(a\), what additional term must be included when comparing their amounts later?
