Comparing Two Accumulations
Two quantities can both accumulate over time while changing at different rates. Their difference depends on more than which rate is larger at a particular moment: it also depends on the difference already present at the starting point and on how the rate differences add up afterward.
The useful comparison is to subtract the rates first. The accumulated difference between the two quantities is then their initial difference plus the signed area under that rate difference. Positive area increases the gap; negative area decreases it.
Definition and Comparison Formula
Let \(r_A\) and \(r_B\) be integrable rate functions on an interval containing a fixed starting point \(a\). Let \(A\) and \(B\) be the corresponding accumulation functions, with initial values \(A(a)\) and \(B(a)\). For inputs \(x\) in the interval, write
The rate difference is the function \(h=r_A-r_B\). The accumulation difference is \(D=A-B\). Subtracting the two accumulation formulas gives the central comparison rule.
The integral is a signed area. Where \(r_A\) exceeds \(r_B\), the rate difference is positive and contributes positively to \(D\). Where \(r_A\) is below \(r_B\), it contributes negatively. Thus \(A(x)=B(x)\) exactly when the signed area accumulated from \(a\) to \(x\) cancels the initial difference \(D(a)\).
Why the Formula Works
The justification is a direct application of the linearity of the definite integral: the integral of a difference equals the difference of the integrals, provided the functions involved are integrable. Starting from the definitions,
Each term has a distinct role: \(D(a)\) is the gap present initially, while the integral measures how the rates change that gap. In particular, matching rates over an interval means the gap stays constant over that interval; it does not mean the two accumulations are equal unless the gap was already zero.
Worked Example: A Positive Rate Difference
Worked Example: A Positive Rate Difference
Suppose \(A(0)=B(0)=0\), \(r_B(t)=0\), and the rate difference \(h(t)=r_A(t)-r_B(t)\) is \(2\) on \([0,1)\) and \(1\) on \([1,2]\). Find the accumulation difference on \([0,2]\) and determine where the two accumulations agree.
The intervals do not overlap at \(t=1\), so the rate difference has a single assigned value there: \(h(1)=1\). For \(0\le x\le1\), the signed area is a rectangle of height \(2\) and width \(x\):
For \(1\le x\le2\), first accumulate the area on \([0,1]\), then add the area from \(1\) to \(x\):
At \(x=1\), both formulas give \(D(1)=2\). At \(x=2\), \(D(2)=3\). The difference is positive for every \(x\) with \(0<x\le2\), so the accumulations agree only at the starting point \(x=0\). As a check, substituting \(x=1.5\) into the second formula gives \(D(1.5)=1.5+1=2.5\), which also equals the area \(2\cdot1+1\cdot0.5=2.5\).
Signed Area Locates Equality
When the initial values are equal, \(D(a)=0\), equality at \(x\) is equivalent to the signed area of \(h\) from \(a\) to \(x\) being zero. That area may vanish after positive and negative contributions cancel, even if the rate difference is not zero throughout the interval. When the initial values are unequal, the area need not be zero: it must equal the negative of the initial gap.
For a piecewise-constant rate difference, calculate the area one interval at a time. On an interval where \(h=c\), traversing a length \(w\) changes \(D\) by \(cw\). This gives a simple way to track the gap at each breakpoint and then solve for any equality point within an interval.
Worked Example: Equality After Positive and Negative Areas
Suppose \(A(0)=B(0)=0\), \(r_B(t)=0\), and \(h(t)=r_A(t)-r_B(t)\) is \(1\) on \([0,1)\), \(-2\) on \([1,2)\), and \(1\) on \([2,3]\). Find all inputs in \([0,3]\) where the accumulations agree.
On the first interval, the difference grows at rate \(1\), so \(D(x)=x\) for \(0\le x\le1\). In particular, \(D(1)=1\). On the second interval, the rate difference is \(-2\), giving
Setting this expression equal to zero gives \(3-2x=0\), so \(x=1.5\). On the final interval, the difference at \(x=2\) is \(D(2)=3-2(2)=-1\). The rate difference is now \(1\), so
This last expression is zero at \(x=3\). The first expression \(D(x)=x\) is zero at \(x=0\). Therefore the equality points are \(0\), \(1.5\), and \(3\). Check the breakpoint values: the formulas give \(D(1)=1\) from both adjacent intervals and \(D(2)=-1\) from both adjacent intervals. The signed areas also confirm the final return: \(1\cdot1+(-2)\cdot1+1\cdot1=0\), so the equal initial accumulations are equal again at \(x=3\).
When the Initial Values Differ
The initial gap matters even if the two rates are easy to compare. If \(D(a)\ne0\), a zero of the accumulated rate difference alone does not generally give an equality point. Instead, the accumulated signed area must cancel the initial gap. The next example tracks that cancellation across two intervals.
Worked Example: Two Equality Points with a Nonzero Initial Gap
Suppose \(A(0)=1\), \(B(0)=0\), \(r_B(t)=0\), and \(h(t)=r_A(t)-r_B(t)\) is \(-2\) on \([0,1)\) and \(1\) on \([1,3]\). Find all equality points on \([0,3]\).
The initial difference is \(D(0)=1-0=1\). For \(0\le x\le1\),
Solving \(1-2x=0\) gives \(x=0.5\), which belongs to this interval. At the breakpoint, \(D(1)=1-2=-1\). For \(1\le x\le3\),
Solving \(x-2=0\) gives \(x=2\), also in the stated interval. The two accumulations therefore agree at \(x=0.5\) and \(x=2\). Substitution checks both values: \(D(0.5)=1-2(0.5)=0\), and \(D(2)=2-2=0\). At the endpoints, \(D(0)=1\) and \(D(3)=1\), consistent with the initial gap and the net signed area \((-2)(1)+(1)(2)=0\).
Equality of Values Does Not Require Equality of Rates
An equality point concerns accumulated values, not the rates at that input. In the second example, \(A(1.5)=B(1.5)\), although the rate difference on the interval containing \(1.5\) is \(-2\). The past positive area has been exactly canceled by the negative area at that point. The rates can therefore differ when the accumulations agree.
Similarly, equal rates at a particular input do not by themselves imply equal accumulated values. They describe the instantaneous change in the gap there, not the size of the gap itself. To locate equality, use the entire signed area from the chosen starting point and include the initial difference.
A Common Mistake: Ignoring the Initial Gap
Suppose \(D(0)=1\) and the rate difference is \(-2\) on \([0,1)\). A tempting but incorrect calculation is to set the accumulated rate difference \(-2x\) equal to zero and conclude that equality occurs only at \(x=0\). That calculation tracks the change in the gap but omits the gap already present.
The correct expression is
Thus equality occurs when \(1-2x=0\), namely at \(x=0.5\). Substitution verifies the result: \(D(0.5)=1-2(0.5)=0\). The rule is to include both parts of the comparison formula: the initial difference and the signed area added afterward.
A Reliable Comparison Process
For a comparison problem, keep the initial gap visible and update it by signed area. This process works for any integrable rates; for piecewise-constant rates, each update is a rectangle area.
Calculate \(D(a)=A(a)-B(a)\), keeping its sign.
Set \(h(t)=r_A(t)-r_B(t)\), so positive values increase \(D\) and negative values decrease it.
Use \(D(x)=D(a)+\int_a^x h(t)\,dt\), splitting the integral at any rate-change points.
Solve \(D(x)=0\), check that each solution lies in the interval used to obtain its formula, and substitute it back into that formula.
Why This Comparison Is Useful
Comparing rates directly can tell you how the gap changes at each moment, but it does not by itself describe the accumulated gap. The comparison formula combines rate information with the starting values, making it possible to determine whether one accumulation is ahead, when the gap changes sign, and whether the two quantities meet again.
Signed area is the key accounting device. A positive contribution can be undone by later negative area, and vice versa. Consequently, an equality point can occur after the rates have differed for a long time. What matters is whether the initial gap and the net signed area sum to zero.
Check Your Understanding
Use the initial difference plus the integral of the rate difference in each question.
- If \(D(a)=0\) and \(r_A-r_B\) is positive throughout an interval to the right of \(a\), can the accumulations agree again inside that interval? Explain using signed area.
- If \(D(a)=3\), what signed area from \(a\) to \(x\) is required for \(A(x)=B(x)\)?
- Suppose the rate difference is \(2\) for a length of \(1\), followed by \(-1\) for a length of \(2\), and \(D(a)=0\). What is the final accumulation difference?
- In the third worked example, why is it necessary to include \(D(0)=1\) when finding the first equality point?
- Can two accumulations agree at an input where their rates are unequal? Explain in terms of accumulated signed area.
