One Pair of Curves Can Enclose More Than One Region
Two curves can intersect several times and form more than one bounded lobe. Each lobe has its own pair of endpoint intersections and its own area. The total area is found by calculating the lobes separately and adding their nonnegative areas.
The lobes may meet at a single intersection point. Their interiors are still separate, and the shared point contributes no area. In precise topological language, the closed regions then touch, while their interiors are disconnected. In either case, area calculations must account for every lobe without allowing signed contributions to cancel.
Intersection Values Divide the Enclosure
Suppose continuous curves \(y=f(x)\) and \(y=g(x)\) intersect at several values of \(x\). Between consecutive intersection values, the difference \(f(x)-g(x)\) cannot change sign unless the curves intersect again. This lets us treat each interval between intersections as a separate candidate lobe.
Not every interval between intersections must belong to a bounded region: the curve pieces must actually enclose a region there. But when a bounded lobe is formed between two consecutive intersections, its area is the integral of the positive vertical distance between the curves over that interval.
When the same two curves form several lobes, the intersections partition the \(x\)-axis into intervals. Check which intervals enclose regions, and determine the curve order separately on each one. In particular, do not assume that one curve remains above the other across all the lobes.
Worked Example: Two Lobes from a Sine Curve
Worked Example: Two Lobes from a Sine Curve
Find the total area enclosed by \(y=\sin x\) and \(y=0\) on \(0\leq x\leq 2\pi\).
First solve for intersections: \(\sin x=0\) on this interval at \(x=0,\pi,2\pi\). These values bound two lobes: \([0,\pi]\) and \([\pi,2\pi]\).
On \([0,\pi]\), \(\sin x\geq 0\), so the sine curve is above the \(x\)-axis. The area there is
On \([\pi,2\pi]\), \(\sin x\leq 0\), so the \(x\)-axis is above the sine curve. The area of that lobe is
Adding the component areas gives
Check: \(-\cos x\) differentiates to \(\sin x\), and \(\cos x\) differentiates to \(-\sin x\), so both antiderivatives used are correct. Each area is positive, and the total is \(4\) square units.
Why Component Areas Add
Area is additive over regions whose interiors do not overlap. If a region is divided into lobes that meet at most along boundary points, its total area is the sum of the areas of those lobes. A point has zero area, so a shared endpoint does not need to be added or subtracted.
For curves \(y=f(x)\) and \(y=g(x)\), suppose their intersections split the region into bounded component intervals \([a_1,b_1],\ldots,[a_n,b_n]\). The area of each component is the integral of the vertical distance between the curves. If the curve order changes from one interval to another, the distance must be written with the order appropriate to that interval.
This sum is not necessarily the same as one signed integral across the entire interval. A definite integral records signed accumulation: portions where the integrand is negative subtract from portions where it is positive. Geometric area, by contrast, is nonnegative on every component.
Worked Example: Three Lobes of a Polynomial
Worked Example: Three Lobes of a Polynomial
Find the total area between \(y=x^4-5x^2+4\) and the \(x\)-axis over the bounded lobes formed by their intersections.
Factor the polynomial to find its zeros:
The intersections occur at \(x=-2,-1,1,2\). These give three bounded lobes on \([-2,-1]\), \([-1,1]\), and \([1,2]\). At \(x=0\), the polynomial has value \(4\), so it is above the axis on the central interval. At \(x=\tfrac32\), its value is \((\tfrac94-1)(\tfrac94-4)=\tfrac54(-\tfrac74)<0\), so it is below the axis on the right interval. The polynomial is even, so it is also below the axis on the left interval.
An antiderivative is
For the central lobe, the polynomial is above the axis:
On \([1,2]\), the polynomial is below the axis, so the area is the negative of its signed integral:
Evaluate the endpoint values:
Therefore,
Because the polynomial is even, the left lobe has the same area: \(A_{\text{left}}=\tfrac{22}{15}\). Thus
Check: differentiating \(F\) gives \(x^4-5x^2+4\), and the three component areas are all positive. The total area is \(8\) square units.
A General Justification for Splitting at Intersections
Here is why consecutive intersections provide a useful way to separate lobes. Define the difference \(h(x)=f(x)-g(x)\). It is continuous when \(f\) and \(g\) are continuous. If \(h\) changed sign between two consecutive intersection values \(a\) and \(b\), there would be points \(u,v\) in \((a,b)\) with \(h(u)>0\) and \(h(v)<0\). By the Intermediate Value Theorem, \(h\) would equal zero somewhere between \(u\) and \(v\). That would be another intersection inside \((a,b)\), contradicting that \(a\) and \(b\) are consecutive intersections.
Consequently, on each such interval where the curves enclose a lobe, one curve stays above the other. Its vertical distance is nonnegative throughout that interval. Adding the integrals of these nonnegative distances gives the area of the union, because the component interiors do not overlap. Any shared endpoints contribute no area.
Worked Example: Two Unequal Lobes
Worked Example: Two Unequal Lobes
Find the total bounded area between \(y=x(x-1)(x-3)\) and the \(x\)-axis.
The curve intersects the axis at \(x=0,1,3\), which give candidate intervals \([0,1]\) and \([1,3]\). At \(x=\tfrac12\), the polynomial is \(\tfrac12(-\tfrac12)(-\tfrac52)>0\); at \(x=2\), it is \(2(1)(-1)<0\). Thus the curve is above the axis on \([0,1]\) and below it on \([1,3]\).
Expand the polynomial and find an antiderivative:
On the first interval,
On the second interval, the curve is below the axis. Evaluate the endpoint values:
Therefore,
The total is
Check: \(F'(x)=x^3-4x^2+3x\). Both lobe areas are positive, and their sum is \(\tfrac{37}{12}\) square units.
Common Mistake: Adding Signed Integrals Instead of Areas
A frequent error is to integrate across all the lobes at once and treat the result as total area. For the sine example, this calculation gives
Zero is the net signed accumulation, not the geometric area. The positive lobe cancels the negative lobe in the integral, even though both enclose area. The correction is to split at the intersection \(x=\pi\), reverse the difference where the curve is below the axis, and add the two areas: \(2+2=4\).
A Reliable Component-by-Component Method
When a problem describes one enclosure formed by curves that intersect several times, use an organized process:
Solve for the \(x\)-values where the relevant boundary curves meet.
Order the intersection values and identify which consecutive intervals form bounded lobes.
Use the positive vertical distance between its boundary curves on that interval.
Sum the nonnegative values; do not let a negative signed contribution cancel a positive area.
This approach matters whenever a graph or equation includes several enclosed lobes. Counting only one component misses area; integrating the full interval without accounting for sign can produce a net value instead. A component-by-component setup makes both the bounds and the geometric meaning of the result explicit.
Check Your Understanding
For each question, identify the components before deciding how to combine their areas.
- If two boundary curves intersect at four ordered \(x\)-values, what role do the consecutive intervals between those values play when looking for bounded lobes?
- Why does a shared endpoint between adjacent lobes contribute no area?
- For \(y=\sin x\) and the \(x\)-axis on \([0,2\pi]\), which curve is above on each component interval?
- What does a signed integral of zero tell you, and what does it not necessarily tell you about enclosed area?
- State the correct way to combine the areas of three nonoverlapping component interiors.
