Comparing Two Accumulations
An accumulation function records the signed amount contributed by a rate between a fixed basepoint and a variable input. When two rates are ordered at every point, their accumulated values can also be compared—but the direction of that comparison depends on which side of the basepoint the input lies.
The reason is orientation. Moving from the basepoint to an input on its right integrates in the usual left-to-right direction. Moving to an input on its left reverses the direction of integration. A pointwise comparison between rates therefore does not translate into the same inequality on both sides of the basepoint.
Definitions and Assumptions
Let \(I\) be an interval, let \(a\in I\), and let \(f\) and \(g\) be continuous functions on \(I\). Define their accumulation functions using the common basepoint \(a\):
The integration variable \(t\) is a dummy variable: it describes the points along the interval being accumulated, while \(x\) is the input at which the accumulation is evaluated. Continuity is a convenient sufficient condition to ensure that these definite integrals exist on every closed subinterval of \(I\).
- if \(x\ge a\), then \(F(x)\le G(x)\);
- if \(x\le a\), then \(F(x)\ge G(x)\).
The hypothesis needs to hold on the interval between \(a\) and the particular input \(x\) being considered. It need not hold elsewhere in \(I\) to compare \(F(x)\) and \(G(x)\).
Why the Comparison Changes Direction
Subtract the two accumulations. By linearity of accumulation, their difference is the accumulation of the difference of the rates:
The pointwise assumption \(f(t)\le g(t)\) means \(f(t)-g(t)\le 0\). If \(x\ge a\), the integral runs from left to right, so integrating a nonpositive function gives a nonpositive value. If \(x\le a\), the orientation identity gives \(\int_a^x h(t)\,dt=-\int_x^a h(t)\,dt\); the minus sign reverses the resulting inequality.
From \(f(t)\le g(t)\), conclude \(f(t)-g(t)\le 0\) on the interval being considered.
For \(x\ge a\), integrate from \(a\) to \(x\). For \(x\le a\), rewrite the integral as the negative of the integral from \(x\) to \(a\).
The sign of \(F(x)-G(x)\) tells which accumulation value is larger.
Proof of the Comparison Theorem
We first use the elementary order property of definite integrals: if a continuous function \(h\) satisfies \(h(t)\le 0\) throughout \([u,v]\), where \(u\le v\), then \(\int_u^v h(t)\,dt\le 0\). Indeed, each Riemann sum for \(h\) on that interval is a sum of terms \(h(\xi_i)\Delta t_i\), where \(\Delta t_i\ge 0\). Each term is nonpositive, so each such sum is nonpositive; its limit, the definite integral, is also nonpositive.
Now take \(h=f-g\). The ordering assumption gives \(h(t)\le 0\) on \(I\). If \(x\ge a\), the order property and linearity give
Thus \(F(x)\le G(x)\). If \(x\le a\), apply the order property on \([x,a]\), then use orientation:
Thus \(F(x)\ge G(x)\). Finally, when \(x=a\), an accumulation over an interval of zero length is zero, so \(F(a)=G(a)=0\). This proves all parts of the theorem.
Worked Examples
Worked Example: Comparing Accumulations to the Right
Let \(a=0\), \(f(t)=t+1\), and \(g(t)=2t+1\) on \([0,2]\). Compare their accumulation values at \(x=2\).
First check the pointwise ordering on the interval:
Therefore \(f(t)\le g(t)\) throughout \([0,2]\), and the theorem predicts \(F(2)\le G(2)\). Compute the two values:
Thus \(F(2)=4\le 6=G(2)\), as predicted. The antiderivatives check: \(\frac{d}{dt}(\frac{t^2}{2}+t)=t+1\) and \(\frac{d}{dt}(t^2+t)=2t+1\). Also, \(G(2)-F(2)=2=\int_0^2 t\,dt\), which directly confirms the difference of the two accumulations.
Worked Example: The Reversed Comparison to the Left
Let \(a=0\), \(f(t)=t\), and \(g(t)=t+2\) on \([-1,0]\). Compare the accumulation values at \(x=-1\).
On the interval from \(-1\) to \(0\),
so the theorem predicts \(F(-1)\ge G(-1)\), because the input is to the left of the basepoint. Compute each value with its orientation intact:
Consequently, \(F(-1)=\frac12\ge-\frac32=G(-1)\). The antiderivatives differentiate back to \(t\) and \(t+2\). As a direct difference check, \(F(-1)-G(-1)=2\), while \(\int_0^{-1}(f(t)-g(t))\,dt=\int_0^{-1}(-2)\,dt=2\).
Worked Example: A Basepoint Away from Zero
Let \(a=1\), \(f(t)=2t\), and \(g(t)=t^2\) on \([1,2]\). Compare their accumulations at \(x=2\).
The rate difference is
Here \(g(t)\le f(t)\). Since the input is to the right of the shared basepoint, the accumulation theorem, applied with the roles ordered as \(g\le f\), predicts \(G(2)\le F(2)\). Evaluate:
Therefore \(3\ge\frac73\). The antiderivatives check because \(\frac{d}{dt}(t^2)=2t\) and \(\frac{d}{dt}(\frac{t^3}{3})=t^2\). The difference also checks:
When Is the Comparison Strict?
The theorem gives a weak inequality, which allows the two accumulation values to be equal. For example, if \(f=g\) throughout the interval between \(a\) and \(x\), then their accumulations at \(x\) are equal. Even if the rates differ at some points, the signed integral of their difference could still be zero if positive and negative differences cancel; however, that cancellation cannot occur when one rate is everywhere below the other.
In particular, if \(f(t)<g(t)\) throughout the interval between \(a\) and \(x\), and \(x\ne a\), then the comparison is strict: \(F(x)<G(x)\) for \(x>a\), and \(F(x)>G(x)\) for \(x<a\). For continuous functions, the strict gap at any interior point persists on a small interval around that point, so its integral contributes a strictly positive amount in the left-to-right direction. Orientation then gives the stated strict comparison on the left.
A Common Orientation Mistake
A frequent error is to use the right-side conclusion for an input that lies to the left of the basepoint. In the second example, the rates satisfy \(f(t)\le g(t)\), but it would be wrong to conclude \(F(-1)\le G(-1)\). That mistaken conclusion overlooks the reversed integration limits.
The calculation shows the issue directly:
The rate difference is negative, but integrating it from \(0\) down to \(-1\) changes the sign. The correction is to check the location of \(x\) before stating the comparison: right of \(a\) gives \(F(x)\le G(x)\), while left of \(a\) gives \(F(x)\ge G(x)\).
What This Comparison Does—and Does Not—Say
This result compares two accumulation functions that share a basepoint. It does not, by itself, say that either accumulation function is increasing or decreasing as its input varies. Such a conclusion concerns how one accumulation changes across inputs; here the question is how two accumulated values compare at the same input. Keeping these questions separate prevents a pointwise rate comparison from being mistaken for a claim about the monotonicity of either function.
The shared basepoint matters as well. If the rates are accumulated from different basepoints, their values include different starting intervals, so the comparison theorem in this form does not directly apply. First express both accumulations using a common basepoint, or account for the difference between their starting contributions.
The practical method is concise: identify the common basepoint, verify the rate inequality on the interval actually traversed, and then check whether the input is to the right or left. That last orientation check determines the direction of the accumulated comparison.
Check Your Understanding
For each question, pay attention to both the pointwise rate ordering and the orientation from the basepoint to the input.
- Suppose \(f(t)\le g(t)\) on \([a,x]\), where \(x>a\). Which inequality holds between \(\int_a^x f(t)\,dt\) and \(\int_a^x g(t)\,dt\)?
- Suppose \(f(t)\le g(t)\) on \([x,a]\), where \(x<a\). Which accumulation is larger at \(x\), when both use basepoint \(a\)?
- If \(f=g\) throughout the interval between \(a\) and \(x\), what can you conclude about their accumulation values at \(x\)?
- Why does the pointwise inequality \(f(t)\le g(t)\) on its own not determine the same comparison direction for inputs on both sides of \(a\)?
- For \(a=0\), \(f(t)=1\), and \(g(t)=t+2\) on \([0,1]\), which accumulation is larger at \(x=1\)? State the rate comparison that justifies your conclusion.
