Comparing Areas and Integrals
A graph can give a useful estimate of a definite integral. If one graph stays above another throughout an interval, the integral of the upper function is at least as large as the integral of the lower one. The key condition is that the graph order holds across the entire interval being compared.
Two related situations require care. The functions might cross, so neither graph encloses the other on the whole interval. Or the intervals might be nested, so one includes an extra piece whose integral may be positive or negative. In both cases, the comparison depends on the signed contributions of the regions involved, not just on which graph looks higher in some places or which interval is longer.
Graph Order on a Shared Interval
Suppose \(f\) and \(g\) are integrable on \([a,b]\), where \(a\le b\), and the graph of \(f\) lies on or below the graph of \(g\) at every point of the interval. In symbols, this means \(f(x)\le g(x)\) for every \(x\in[a,b]\). The integral comparison rule says that their definite integrals have the same order.
The theorem compares signed definite integrals. It does not require either function to be nonnegative: a graph can lie below another graph while both are below the \(x\)-axis. The ordering of the two integrals follows from their vertical ordering, not from either graph's position relative to the axis.
Why the Comparison Rule Works
The difference \(g-f\) is nonnegative throughout the interval. Its integral therefore cannot be negative. By linearity of the definite integral, that integral is the difference between the integral of \(g\) and the integral of \(f\). Here is the argument in full.
Define \(h(x)=g(x)-f(x)\). The pointwise condition \(f(x)\le g(x)\) gives \(h(x)\ge0\) for every \(x\in[a,b]\).
For any partition of \([a,b]\), every infimum of \(h\) on a subinterval is nonnegative. Thus every lower Darboux sum for \(h\) is nonnegative, and the integral of \(h\), which is bounded below by its lower sums, is nonnegative.
Since \(h=g-f\), linearity gives \(\int_a^b h(x)\,dx=\int_a^b g(x)\,dx-\int_a^b f(x)\,dx\). The left side is nonnegative, so \(\int_a^b f(x)\,dx\le\int_a^b g(x)\,dx\), as claimed.
This also explains the geometric idea. The difference of the integrals is the signed area associated with the vertical gap \(g-f\). When that gap is nonnegative everywhere, its integral cannot subtract from the total.
Worked Example: Comparing Two Graphs on the Same Interval
Compare the integrals of \(f(x)=x+1\) and \(g(x)=2x+1\) over \([0,2]\).
First check the graph order on the whole interval. Their difference is
For every \(x\in[0,2]\), \(x\ge0\), so \(f(x)\le g(x)\). The comparison theorem therefore predicts that the integral of \(f\) is no greater than the integral of \(g\).
Compute the first integral:
Compute the second:
Thus \(4\le6\), as predicted. A direct check uses the gap: \(\int_0^2(g-f)\,dx=\int_0^2x\,dx=2\), and the difference \(6-4\) is also \(2\).
Nested Intervals and the Added Portion
To compare integrals over nested intervals, use interval additivity. If \(a\le b\le c\) and \(f\) is integrable on \([a,c]\), then
This identity says that extending the interval from \([a,b]\) to \([a,c]\) adds exactly the integral over the extra segment \([b,c]\). Consequently, the larger interval has a larger integral if the added portion has positive integral, and a smaller integral if the added portion has negative integral. If the added portion has integral zero, the two integrals are equal.
A sufficient condition for a nonnegative addition is \(f(x)\ge0\) throughout \([b,c]\). A sufficient condition for a nonpositive addition is \(f(x)\le0\) throughout \([b,c]\). If \(f\) changes sign on the added segment, its net integral determines the comparison.
Worked Example: Extending an Interval Through a Nonnegative Region
Let \(f(x)=(x-1)^2\), and compare its integrals over \([0,2]\) and \([0,3]\).
Since a square is nonnegative, \(f(x)\ge0\) throughout the added segment \([2,3]\). Additivity therefore predicts that the integral over \([0,3]\) is at least the integral over \([0,2]\).
For the shorter interval, expand the square and integrate:
For the longer interval:
Check the comparison by calculating the added portion:
Additivity gives \(\frac{2}{3}+\frac{7}{3}=3\), matching the integral over \([0,3]\). The positive added contribution accounts for the increase.
When the Graphs Cross
If two graphs cross, neither function is necessarily above the other across the full interval. The comparison theorem cannot be applied to the whole interval using a single graph order. Instead, divide the interval at the crossing points. On each resulting piece, determine which graph is higher; then combine the signed contributions using additivity.
Equivalently, examine the difference \(g-f\). Where \(g-f\ge0\), \(g\) contributes at least as much as \(f\) on that piece. Where \(g-f\le0\), the reverse is true. The overall comparison depends on the integral of the difference across all the pieces, so a region where one graph is higher does not by itself settle the comparison over the full interval.
Worked Example: Comparing Integrals When the Graphs Cross
Let \(f(x)=x\) and \(g(x)=3-x\) on \([0,2]\). Find the crossing point and compare their integrals.
Set the functions equal to locate the crossing:
On \([0,\frac32]\), \(g-f=3-2x\ge0\), so \(g\) is at least as high as \(f\). On \([\frac32,2]\), \(f-g=2x-3\ge0\), so \(f\) is higher. The first region favors \(g\); the second favors \(f\).
Calculate the net difference by splitting at the crossing:
The second contribution is negative because \(g-f\le0\) there. In magnitude, the difference favoring \(f\) on the final segment is \(\int_{3/2}^2(2x-3)\,dx=\frac14\), while the difference favoring \(g\) on the first segment is \(\int_0^{3/2}(3-2x)\,dx=\frac94\). Their net is \(\frac94-\frac14=2\).
Now calculate each integral to check the comparison:
Indeed, \(4-2=2\), in agreement with the integral of \(g-f\). Although \(f\) is higher on the final part of the interval, \(g\) has the larger integral overall.
A Common Mistake: Assuming a Longer Interval Means a Larger Integral
An interval extension does not automatically increase a definite integral. Definite integrals are signed: portions below the \(x\)-axis contribute negatively. The added segment must be examined before drawing a conclusion.
For a concrete example, let \(f(x)=x-2\) and compare \([0,1]\) with the nested interval \([0,2]\). A tempting but incorrect claim is that \(\int_0^2 f(x)\,dx\) must be at least \(\int_0^1 f(x)\,dx\) because \([0,2]\) is longer. The actual calculations are
Thus \(-2<-\frac32\): the integral over the longer interval is smaller. The added piece explains why:
Additivity checks the result: \(-\frac32+(-\frac12)=-2\). The mistake was to compare interval lengths instead of checking the signed integral over the added segment.
A Reliable Comparison Strategy
When using graphs to compare integrals, first specify both the functions and the interval. Then check whether a pointwise graph order holds throughout that interval. If it does, the comparison theorem applies directly. If the graphs cross, split at the crossing points and account for the difference on each piece. If the intervals are nested, use additivity and determine the sign of the integral over the added part.
Check Your Understanding
For each question, identify the relevant graph order or added portion before making a comparison.
- If \(f(x)\le g(x)\) for every \(x\in[1,4]\), what inequality follows between their integrals over \([1,4]\)?
- For \(a\le b\le c\), write the additivity identity that compares the integrals over \([a,b]\) and \([a,c]\).
- If \(f\le0\) throughout the added segment \([b,c]\), what can be concluded about \(\int_a^c f(x)\,dx\) compared with \(\int_a^b f(x)\,dx\)?
- If two graphs cross once inside an interval, what should be done before using graph order to compare their integrals?
- Why does the fact that one interval is longer than another not, by itself, determine which integral is larger?
