Why Rational Curves Require Care
Two rational graphs can intersect at several points, reverse their vertical order, or fail to be defined at values where a denominator is zero. Those domain restrictions matter when deciding whether an interval actually contains a finite region and when choosing an area integral.
The basic area calculation is familiar: on an interval where two continuous curves are defined, the area between them is the integral of the upper function minus the lower function. The work specific to rational curves is to find valid intersections, inspect denominator zeros, and establish the order of the graphs throughout the interval.
Rational Functions and Their Domains
For example, \((x^2-1)/(x-1)\) is equal to \(x+1\) whenever \(x\ne 1\), but the quotient as written is undefined at \(x=1\). Cancelling the factor does not put that input back into the original domain. The simplified expression can define a continuous extension there, but that extension is a separately specified function.
To find intersections of two rational graphs, set their values equal only at inputs where both functions are defined. Clearing denominators can produce candidate solutions, but each candidate must be checked against the original denominators. A candidate at a pole is not an intersection.
From Intersections to an Area Interval
Suppose \(f\) and \(g\) are continuous on a closed interval \([a,b]\), their graphs meet at the endpoints, and their order does not change in the interior. If \(f(x)\ge g(x)\) throughout the interval, the area between their graphs is
If \(g\) is above \(f\), reverse the difference. Equivalently, the area is the integral of \(|f(x)-g(x)|\), but writing upper minus lower makes the geometry explicit.
Continuity of both functions on \([a,b]\) is a sufficient condition for applying this ordinary area formula directly. It is not necessary for every finite area to exist: an integrable singularity or a removable hole can still correspond to finite area. In such a case, however, the ordinary formula cannot simply be applied to the original functions as though they were continuous across the whole interval. The hole or singularity must be treated appropriately.
For rational curves, check the interval as well as its endpoints. If either denominator vanishes at an interior point, the function is not continuous there, and the interval must not be treated as one ordinary continuous piece. A removable hole may be filled by an extension; a nonremovable singularity requires separate analysis and can prevent the area from being finite.
A Reliable Procedure
Set the two expressions equal and clear denominators, keeping track of where those denominators are zero.
Substitute each candidate into the original denominators. Discard any candidate where one of the curves is undefined.
Check for denominator zeros or other domain restrictions inside the interval. Do not integrate across an excluded value as if the functions were continuous there.
Evaluate both functions at a test point between consecutive valid intersections. If there is no change of order in the interval, that comparison identifies the upper and lower curves throughout it.
Integrate upper minus lower, evaluate at the endpoints, and confirm that the result is nonnegative.
The test-point step works when the functions are continuous between the intersections: their difference is continuous there, so it cannot change sign without becoming zero. A zero of the difference would be another intersection. Thus, in an interval containing no additional intersection, one test point determines the order throughout.
Worked Example: A Symmetric Region with a Horizontal Line
Worked Example: A Symmetric Region with a Horizontal Line
Find the area enclosed by \(y=x^2/(x^2+1)\) and \(y=1/2\).
First find intersections. The denominator \(x^2+1\) is positive for every real \(x\), so clearing it introduces no excluded real inputs:
Between the intersections, test \(x=0\). The rational curve has value \(0\), while the line has value \(1/2\); hence the line is above the rational curve on this interval. The area is
Since \(x^2/(x^2+1)=1-1/(x^2+1)\), the integrand becomes \(1/(x^2+1)-1/2\). Using the antiderivative \(\arctan x-x/2\),
This is positive, as an area must be. Differentiating the antiderivative gives \(1/(1+x^2)-1/2\), the original integrand, which checks the integration.
Why the Difference Must Be Ordered Correctly
The definite integral of \(f-g\) measures signed accumulation: wherever \(f\) lies below \(g\), the difference is negative. Geometric area, by contrast, is nonnegative. On an interval where the graphs exchange order, the signed integral over the whole interval can cancel positive and negative parts instead of adding their areas. That is why intersections and a test point are not merely setup details.
The area formula follows by adding thin vertical slices. At a position \(x\), a slice has height equal to the vertical distance between the graphs, namely upper minus lower, and width approximately \(\Delta x\). Its area is approximately \((\text{upper}-\text{lower})\Delta x\). Summing these slice areas and taking the limit as the widths shrink gives the definite integral of that height.
Worked Example: Two Rational Curves Meet Twice
Worked Example: Two Rational Curves Meet Twice
Find the area enclosed by \(y=x/(x^2+1)\) and \(y=x^2/(x^2+1)\) between their intersections.
Both denominators are positive for every real \(x\). Equating the functions and multiplying by their common positive denominator gives
The intersections occur at \(x=0\) and \(x=1\). At \(x=1/2\), the two function values are \(2/5\) and \(1/5\), respectively, so \(x/(x^2+1)\) is above \(x^2/(x^2+1)\) between the intersections. Therefore,
The result is positive: approximately \(0.132\). Differentiating the antiderivative gives \(x/(x^2+1)-1+1/(x^2+1)\), which simplifies to \(x/(x^2+1)-x^2/(x^2+1)\), as required.
Worked Example: A Removable Hole and Finite Area
Worked Example: A Removable Hole and Finite Area
Consider \(y=(x^2-1)/(x-1)\) for \(x\ne1\) and \(y=0\) on the interval \([0,2]\). Determine the area between the graph and the \(x\)-axis, treating the missing point correctly.
For \(x\ne1\), factor the numerator and cancel only on the domain of the original quotient:
The original rational function is undefined at \(x=1\), so it is not continuous on the whole closed interval \([0,2]\). The ordinary area formula cannot be applied to it as written across that point. However, the hole is removable: defining the value at \(x=1\) to be \(2\) extends the function continuously as \(x+1\). This single point does not change the area, so compute using the extension. Since \(x+1\) is nonnegative on \([0,2]\),
As a check, differentiating \(x^2/2+x\) gives \(x+1\), the extended upper curve. The calculation also shows why a missing point must be acknowledged even when it does not make the area infinite.
Common Mistake: Keeping a Forbidden Intersection
Suppose one tries to find intersections of \(y=1/(x-1)\) and \(y=(x+1)/(x^2-1)\) by cross-multiplying:
The resulting equation is true for every \(x\), because both sides equal \(x^2-1\). It would be wrong to conclude that the graphs coincide everywhere. In the original expressions, \(x=1\) is excluded from the first function, and \(x=-1\) and \(x=1\) are excluded from the second. For other \(x\), the second expression simplifies to \(1/(x-1)\), so the graphs do coincide on their common domain; neither excluded value is an intersection.
This example illustrates the pitfall: clearing denominators is an algebraic step, not a replacement for checking the original domains. Always check candidate points in the expressions that defined the curves.
What Makes a Region Finite?
A pair of valid intersections can provide the left and right boundaries of a finite region when both curves are defined and continuous between them and one remains above the other. A denominator zero inside that interval changes the analysis: it is not an ordinary point of either graph. A vertical asymptote may make the area infinite, while a removable hole may allow a continuous extension and finite area. The denominator test therefore comes before treating two intersection values as the endpoints of one standard area integral.
Also check whether the proposed boundary points genuinely meet the curves. If clearing denominators gave a point where one original expression is undefined, it cannot serve as an endpoint intersection. Likewise, a pair of curves meeting just once does not, by itself, establish a bounded region between them; some additional boundary would be needed.
In practice, a careful solution makes the geometry visible: list the valid intersections, state the interval being considered, note any excluded denominator zeros, identify the upper curve, and then write the definite integral. These checks prevent both invalid intervals and negative signed answers from being mistaken for area.
Check Your Understanding
Use domain restrictions and graph order as part of each setup.
- When solving an equation between two rational functions by clearing denominators, why must candidate intersections be checked in the original expressions?
- For \(y=x^2/(x^2+1)\) and \(y=1/2\), which graph is higher on the interval between their intersections?
- What condition makes the ordinary upper-minus-lower area formula directly applicable on a closed interval?
- Why does a removable hole require attention even when the area is finite?
- What can go wrong if the order of two curves changes inside an interval and one integrates their difference without splitting or using an absolute value?
