Why Compare the Integrands?
Two quantities can accumulate over the same interval while changing at different rates. Examples include two objects moving along a line, two tanks filling, or two populations changing. If their rates are \(f\) and \(g\), comparing the rates at a single moment does not by itself tell us which quantity accumulated more over an interval. The accumulated difference depends on the rates throughout that interval.
The central idea is to subtract the rates first and then accumulate that difference. This turns a comparison of two accumulation functions into one integral.
Definitions and the Comparison Identity
Let \(f\) and \(g\) be integrable functions on \([a,b]\). Suppose two quantities have starting values \(A(a)\) and \(B(a)\), and accumulate according to the rates \(f\) and \(g\). Their accumulation functions are
Here, \(t\) is the variable of integration and \(x\) is the endpoint at which we evaluate the accumulated quantity. The functions \(f\) and \(g\) need only be integrable for the identities below. Continuity is a common sufficient condition in AP Calculus and also permits the usual derivative conclusion from the Fundamental Theorem of Calculus.
The first identity compares the changes in the two quantities over \([u,v]\). The second compares their values at \(x\), so it includes any difference in their starting values. When \(A(a)=B(a)\), the starting-value term is zero, and the difference in the values is just the integral of \(f-g\).
Why the Identity Works
The result follows from the definitions of the accumulation functions and linearity of the definite integral. The steps below establish both the interval-change formula and the endpoint-value formula.
By additivity of definite integrals, \(A(v)-A(u)=\int_u^v f(t)\,dt\), and \(B(v)-B(u)=\int_u^v g(t)\,dt\).
Subtracting the second equation from the first gives \(\int_u^v f(t)\,dt-\int_u^v g(t)\,dt\).
By linearity, the result is \(\int_u^v(f(t)-g(t))\,dt\), as claimed.
From the definitions, \(A(x)-B(x)=A(a)-B(a)+\int_a^x f(t)\,dt-\int_a^x g(t)\,dt\). Linearity gives the endpoint-value formula.
This proof also explains why the same interval matters: both changes must be measured between the same endpoints before they can be combined into one integral. The identity is not an approximation. It follows exactly from the definitions and the properties of definite integrals.
What the Sign of the Integral Tells You
The integral of \(f-g\) gives a signed comparison. If it is positive, the change produced by \(f\) exceeds the change produced by \(g\) over the interval. If it is negative, the change produced by \(f\) is smaller. If it is zero, their net changes are equal—even if the rates differ at many points.
A useful sufficient condition is pointwise comparison. If \(f(t)\geq g(t)\) throughout \([u,v]\), then \(f(t)-g(t)\geq 0\) there, so
Thus \(A\) changes by at least as much as \(B\) on that interval. If their values agree at \(u\), then \(A(v)\geq B(v)\). For continuous integrands, if \(f-g\) is positive throughout an interval of positive length, its integral over that interval is positive, giving a strict comparison of the changes.
For a comparison of values at an endpoint, remember to include the starting-value difference. Even if the integral of \(f-g\) is positive, \(A(x)\) need not exceed \(B(x)\) if \(A(a)\) started sufficiently far below \(B(a)\).
Worked Comparisons
Worked Example: Equal Accumulations Despite Different Rates
Let \(A(0)=B(0)=2\), with rates \(f(t)=2t+1\) and \(g(t)=t+2\). Compare the accumulation values at \(x=2\).
First subtract the rates:
Therefore the difference in the accumulation values is
Check by calculating each value directly:
The check gives \(A(2)-B(2)=8-8=0\), matching the integral of the rate difference. The rates are not identical, but their signed difference over the whole interval is zero.
Worked Example: Compare Changes on an Interval with Nonzero Endpoints
Suppose \(A(1)=B(1)=7\), and on \([1,3]\) the rates are \(f(t)=5-t\) and \(g(t)=2\). Compare the accumulation values at \(3\).
Subtract the rates and integrate over the shared interval:
At the upper endpoint, \(3(3)-\frac{3^2}{2}=9-\frac{9}{2}=\frac{9}{2}\). At the lower endpoint, \(3(1)-\frac{1^2}{2}=3-\frac{1}{2}=\frac{5}{2}\). Thus
Check by finding each value separately:
The direct comparison is \(13-11=2\), which agrees with the integral of \(f-g\). Since the starting values were equal, the difference in the endpoint values equals the difference in the accumulated changes.
Worked Example: Add Signed Contributions Across Subintervals
Let \(A(0)=B(0)=0\). On \([0,4]\), suppose \(g(t)=1\), while \(f\) is given by
Compare the accumulated values at \(4\). Since \(f-g\) is respectively \(2\), \(-1\), and \(3\) on the three pieces, split the integral at the changes in its formula:
Check by computing the accumulations separately. The value of \(A\) is \(3\) over the first unit, \(0\) over the next two units, and \(4\) over the last unit:
The rate \(g\) is \(1\) across an interval of length \(4\), so
Hence \(A(4)-B(4)=7-4=3\), as the rate-difference calculation gives. The negative contribution on \([1,3]\) cancels the positive contribution on \([0,1]\); the final interval supplies the remaining net difference.
A Common Mistake: Comparing Only the Current Rates
A tempting but incorrect shortcut is to compare the integrands at the right endpoint and infer which accumulation is larger. Accumulation records the entire history from the starting point, not just the rates at the endpoint.
The correction is to integrate the rate difference over the full interval:
So \(A(2)\) is actually less than \(B(2)\), even though its rate is greater at the endpoint. The earlier negative contribution is larger than the later positive contribution. To compare accumulated values, compare the starting values and the integral of the rate difference across the whole interval.
How to Use the Comparison
When a problem asks which accumulation is greater, or by how much, organize the information before integrating:
- Identify the common interval and the two integrands.
- Determine whether the question compares changes or endpoint values.
- For changes, integrate \(f-g\) over the shared interval.
- For endpoint values, include the difference between the starting values.
- Interpret the sign of the result: positive, negative, or zero.
This method is useful because it avoids calculating two accumulated quantities separately when only their difference is needed. It also makes cancellation visible: the individual rates may vary substantially, while their net difference over the interval may be small or zero.
Check Your Understanding
For each question, identify the relevant interval and account for starting values when needed.
- If \(A(a)=B(a)\), what integral gives \(A(x)-B(x)\)?
- If \(f(t)\geq g(t)\) throughout \([u,v]\), what can you conclude about the changes in \(A\) and \(B\) over that interval?
- If \(\int_u^v(f(t)-g(t))\,dt=0\), must \(f(t)=g(t)\) at every point? Explain.
- If \(A(a)\neq B(a)\), what additional term must be included when comparing \(A(x)\) and \(B(x)\)?
- Why is comparing \(f(v)\) and \(g(v)\) alone not enough to determine which accumulation value is larger at \(v\)?
