Why Several Intersections Require Care
When two polynomial curves meet more than once, the curve that lies above the other can change from one interval to the next. A single integral of “upper minus lower” may therefore use the wrong order on part of its interval, or allow positive and negative signed areas to cancel.
The reliable approach is to analyze the difference of the functions. Its zeros give the intersections, and its sign on each interval between consecutive intersections identifies the upper curve. The area of each bounded region can then be computed separately and added.
From Polynomial Roots to Curve Order
The intersections satisfy \(f(x)=g(x)\), or equivalently \(D(x)=0\). Factoring \(D(x)\), when possible, makes the intersection values and their multiplicities visible. Sort the distinct real roots in the interval of interest. They divide that interval into pieces on which the sign of \(D\) can be tested.
A root’s multiplicity helps explain what happens at an intersection. A factor \((x-r)^m\) changes sign across \(r\) when \(m\) is odd, but not when \(m\) is even, provided the remaining factor is nonzero there. Thus, an odd-multiplicity root usually corresponds to a crossing, while an even-multiplicity root usually corresponds to touching without crossing. A test point on each interval provides a direct check of the curve order.
This rule applies when the regions are represented by vertical slices between the two graphs over the stated \(x\)-intervals. In the examples below, each bounded region lies between consecutive intersection values.
A Justification for Splitting at Intersections
The sign analysis works because polynomials are continuous. Suppose \(D\) has no zero on an interval. If \(D\) had opposite signs at two points in that interval, the Intermediate Value Theorem would guarantee a point between them where \(D=0\). That would contradict the assumption that the interval contains no zero. Therefore, \(D\) has a constant sign on each interval between consecutive distinct roots.
Now consider an intersection \(x=r\) where \(D(x)=(x-r)^m Q(x)\), with \(Q(r)\ne 0\). By continuity, \(Q\) keeps the same sign sufficiently close to \(r\). The factor \((x-r)^m\) changes sign across \(r\) precisely when \(m\) is odd. Consequently, the sign of \(D\), and hence which curve is above, switches at odd-multiplicity roots but not at even-multiplicity roots.
Finally, on any one interval with constant sign, the area is the integral of the positive difference. If \(D>0\), that is \(\int D(x)\,dx\); if \(D<0\), it is \(-\int D(x)\,dx\). Adding these nonnegative areas gives the sum of absolute-difference integrals in the area rule.
Worked Example: Two Bounded Regions from a Cubic
Worked Example: Two Bounded Regions from a Cubic
Find the total area between \(y=f(x)=x^3-4x\) and \(y=g(x)=0\) across their bounded regions.
1. Find the intersections. Set the functions equal by factoring their difference:
The intersection values are \(x=-2,0,2\). Substitution confirms the corresponding function values are zero: \(f(-2)=-8+8=0\), \(f(0)=0\), and \(f(2)=8-8=0\).
2. Determine the upper curve. On \((-2,0)\), test \(x=-1\): \(D(-1)=-1+4=3>0\), so \(f\) is above \(g\). On \((0,2)\), test \(x=1\): \(D(1)=1-4=-3<0\), so \(g\) is above \(f\).
3. Integrate on each interval. An antiderivative of \(D(x)\) is \(F(x)=\frac{x^4}{4}-2x^2\), since \(F'(x)=x^3-4x\). On the first interval, the area is
On the second interval, the upper curve is \(g\), so use \(-D\):
The total bounded area is \(4+4=8\) square units. The endpoint checks above verify the intersections, and differentiating \(F\) verifies the antiderivative used.
Worked Example: Three Intersections and Alternating Curve Order
Worked Example: Three Intersections and Alternating Curve Order
Find the total area between \(y=f(x)=(x+1)(x-1)(x-3)\) and \(y=g(x)=0\) across the bounded regions.
1. List the intersections. The difference is already factored, so its roots are \(-1,1,3\). At each root, \(f(x)=0=g(x)\), confirming all three intersections.
2. Determine the order on each interval. Test \(x=0\) on \((-1,1)\): \(D(0)=(1)(-1)(-3)=3>0\), so \(f\) is above \(g\). Test \(x=2\) on \((1,3)\): \(D(2)=(3)(1)(-1)=-3<0\), so \(g\) is above \(f\).
3. Find an antiderivative. Expanding gives \(D(x)=x^3-3x^2-x+3\). Therefore an antiderivative is \(F(x)=\frac{x^4}{4}-x^3-\frac{x^2}{2}+3x\), and differentiating gives \(F'(x)=x^3-3x^2-x+3=D(x)\).
At the needed endpoints, \(F(-1)=\frac14+1-\frac12-3=-\frac94\), \(F(1)=\frac14-1-\frac12+3=\frac74\), and \(F(3)=\frac{81}{4}-27-\frac92+9=-\frac94\). Thus the first region has area
The second region has area
The total is \(4+4=8\) square units. The root substitutions give zero at all three intersections, and the derivative check confirms the antiderivative.
Worked Example: Three Regions from a Quartic Difference
Worked Example: Three Regions from a Quartic Difference
Find the total area between \(y=f(x)=(x+2)x(x-1)(x-3)\) and the \(x\)-axis across the bounded regions.
1. Find the intersection values. The roots are \(-2,0,1,3\), and substituting any of these values into the factored expression gives \(f(x)=0\). The three bounded intervals between consecutive roots are \([-2,0]\), \([0,1]\), and \([1,3]\).
2. Check the signs. At \(x=-1\), \(f(-1)=(1)(-1)(-2)(-4)=-8<0\). At \(x=\frac12\), \(f(\frac12)=(\frac52)(\frac12)(-\frac12)(-\frac52)=\frac{25}{16}>0\). At \(x=2\), \(f(2)=(4)(2)(1)(-1)=-8<0\). Thus \(f\) is below the axis on the first and third intervals, and above it on the middle interval.
3. Integrate each positive difference. Expanding the polynomial gives \(f(x)=x^4-2x^3-5x^2+6x\). An antiderivative is \(F(x)=\frac{x^5}{5}-\frac{x^4}{2}-\frac{5x^3}{3}+3x^2\), because differentiation gives \(F'(x)=x^4-2x^3-5x^2+6x\).
The endpoint values are \(F(-2)=-\frac{32}{5}-8+\frac{40}{3}+12=\frac{164}{15}\), \(F(0)=0\), \(F(1)=\frac15-\frac12-\frac53+3=\frac{31}{30}\), and \(F(3)=\frac{243}{5}-\frac{81}{2}-45+27=-\frac{99}{10}\). Since \(f<0\) on \((-2,0)\), that region’s area is
On \((0,1)\), \(f>0\), so the area is
On \((1,3)\), \(f<0\), so the area is
The total area is \(\frac{164}{15}+\frac{31}{30}+\frac{164}{15}=\frac{328}{30}+\frac{31}{30}=\frac{359}{30}\) square units. The roots, sign tests, endpoint evaluations, and derivative check verify the setup and calculation.
A Common Mistake: Adding Signed Integrals
A frequent error is to integrate \(f-g\) across several intersections and treat the result as the total area. A definite integral gives signed accumulation: portions where \(f-g<0\) subtract from portions where \(f-g>0\).
For the cubic in the first example, integrating \(D(x)=x^3-4x\) over the entire interval from \(-2\) to \(2\) gives
That zero is not the area: the two bounded regions each have area \(4\), for a total of \(8\). The correction is to split at \(x=0\), reverse the difference where it is negative, and add the positive areas:
A Reliable Workflow
For two polynomial curves with several intersections, use this sequence of checks:
Set \(D(x)=f(x)-g(x)\); the roots of \(D\) are the intersection values.
Use consecutive roots to divide the region into intervals. Check that each listed root is a genuine intersection by substitution.
Use a test point, and use root multiplicity as a useful check on whether the sign switches.
Choose the difference with positive sign on each interval, then add the resulting areas.
This process matters whenever polynomial curves cross repeatedly: the intersection values identify where the description of the region changes, while the sign of the difference identifies the correct integrand. Factoring, testing, and splitting keep those decisions visible instead of hiding them inside a single signed integral.
Check Your Understanding
For each question, identify the intersections and pay attention to the sign of the difference on the intervals they determine.
- For \(y=x(x-1)(x-2)\) and \(y=0\), which curve is above on each interval between consecutive intersections?
- If \(D(x)=(x+1)^2(x-2)\), does the sign of \(D\) change at each root? Explain using the multiplicities.
- Why can \(\int_a^b(f(x)-g(x))\,dx\) be zero even when the area between the curves is positive?
- For \(D(x)=(x+2)(x-1)(x-3)\), use one test point in each interval between consecutive roots to determine the sign pattern.
- What must be done to convert signed integrals across multiple crossings into total area?
