Why Multiple Lobes Must Be Added Separately
When two curves cross more than once, they can enclose several separate lobes. A definite integral over the entire horizontal interval records signed area: portions where one curve is above the other contribute with the opposite sign from portions where their order is reversed. As a result, a single integral across several lobes can cancel some or all of the area.
Geometric area does not cancel. To find the total area, locate the endpoints of each lobe, determine which curve is above on that interval, calculate that lobe’s positive area, and add the results.
Endpoints and Orientation of a Lobe
Suppose two continuous curves are given by \(y=f(x)\) and \(y=g(x)\). Their intersection points occur at the \(x\)-values satisfying \(f(x)=g(x)\). Consecutive intersection points can serve as the endpoints of a lobe. On the interval between them, check which curve is above; the integrand for area must be upper curve minus lower curve.
A convenient way to check the order is to examine the difference \(h(x)=f(x)-g(x)\). If \(h(x)>0\) on an interval, then \(f\) is above \(g\) there. If \(h(x)<0\), then \(g\) is above \(f\). For continuous curves, if there is no crossing within an interval, the order cannot switch there: a change in the sign of \(h\) would require \(h\) to equal zero somewhere in between.
The area-between-curves rule from vertical slices gives the area on an interval \([a,b]\) as the definite integral of the vertical height, upper minus lower. For multiple lobes, apply that rule to each interval separately. If the curves are polynomials, an antiderivative can be used to evaluate each definite integral.
Why the Lobe Areas Add
Consider a region divided into \(n\) lobes, with successive endpoints \(x_0,x_1,\ldots,x_n\). Write \(U_i\) for the curve above and \(L_i\) for the curve below on \([x_{i-1},x_i]\). By the vertical-slice area rule, the area of lobe \(i\) is the integral of its nonnegative height, \(U_i(x)-L_i(x)\), over that interval. The lobes meet only at boundary points, so their interiors do not overlap. Their total geometric area is therefore the sum of their individual areas:
This argument also explains why one signed integral may give the wrong total. If the same ordered difference \(f-g\) is integrated across the whole interval, then additivity of the definite integral combines the contributions from all subintervals. A lobe where \(f\) is below \(g\) contributes negatively to that signed integral, even though its geometric area is positive.
Worked Examples
Worked Example: Two Equal Lobes with Complete Cancellation
Find the total area between \(f(x)=x^3-3x^2-x+5\) and \(g(x)=2\) from their leftmost to rightmost intersection.
First subtract the curves and factor:
The intersections are \(x=-1,1,3\), so there are two lobes on \([-1,1]\) and \([1,3]\). At \(x=0\), the difference is \(3>0\), so \(f\) is above \(g\) on the first lobe. At \(x=2\), the difference is \(-3<0\), so \(g\) is above \(f\) on the second. An antiderivative of \(f-g\) is
For the first lobe:
so its area is \(F(1)-F(-1)=\frac74-(-\frac94)=4\). For the second lobe, the upper-minus-lower integrand is \(g-f=-(f-g)\). Evaluate the endpoint values:
Thus the second lobe’s area is
Adding the lobe areas gives \(4+4=8\) square units. As a check, the signed integral of \(f-g\) across the full interval is \(4-4=0\), not the total geometric area. Differentiating \(F\) gives \(F'(x)=x^3-3x^2-x+3=f(x)-g(x)\), confirming the antiderivative used.
Worked Example: Two Lobes with Unequal Areas
Find the total area between the curves whose difference is \(f(x)-g(x)=(x+2)(x+1)(x-2)\), on the interval from \(x=-2\) to \(x=2\).
Expanding the difference gives
Its zeros are \(-2,-1,2\), giving two lobes. At \(x=-\frac32\), the factored expression is positive, so \(f\) is above \(g\) on \([-2,-1]\). At \(x=0\), it is negative, so \(g\) is above \(f\) on \([-1,2]\). An antiderivative is
Evaluate it at the three endpoints:
The first lobe’s area is
On the second lobe \(f\) is below \(g\), so reverse the difference:
The total is
square units. The signed integral of \(f-g\) over the entire interval is \(\frac7{12}-\frac{45}{4}=-\frac{32}{3}\). It is not the geometric area, because the second lobe contributes negatively to that signed integral. Differentiating \(F\) verifies that it equals the stated curve difference.
Worked Example: Adding Three Lobes
Find the total area between \(f(x)=x^4-5x^2+4\) and \(g(x)=0\) on \([-2,2]\).
Factor the difference to find the intersections:
The intersection values \(-2,-1,1,2\) divide the region into three lobes. The difference is negative on \([-2,-1]\), positive on \([-1,1]\), and negative on \([1,2]\). An antiderivative is
The endpoint values are
On the left lobe, \(g\) is above \(f\), so the area is \(-[F(-1)-F(-2)]=\frac{22}{15}\). On the middle lobe, \(f\) is above \(g\), so the area is \(F(1)-F(-1)=\frac{76}{15}\). On the right lobe, \(g\) is above \(f\), so the area is \(-[F(2)-F(1)]=\frac{22}{15}\). Therefore,
square units. As a check, the signed integral across the full interval is \(-\frac{22}{15}+\frac{76}{15}-\frac{22}{15}=\frac{32}{15}\), which differs from the total area. Differentiating \(F\) recovers \(f-g\), as required.
A Common Mistake: Adding Signed Integrals as Though They Were Areas
A frequent error is to evaluate \(\int_a^b(f-g)\,dx\) across several lobes and report that value as the total geometric area. That integral is signed: whenever \(f\) is below \(g\), its contribution is negative. In the first example, the full-interval signed integral is \(4-4=0\), even though the region has \(8\) square units of area. A zero signed integral does not mean that there is no enclosed region.
Another error is to identify all intersection points but fail to test the curve order between them. Do not assume that the same curve stays on top across every lobe. Check the sign of \(f-g\) on each interval, or compare the curves directly at an interior point. Then use the upper curve minus the lower curve for that lobe.
A Reliable Calculation Procedure
Solve \(f(x)=g(x)\) for the relevant endpoints of the bounded region.
Order the intersection values from left to right and use consecutive values as lobe endpoints.
Check the sign of \(f-g\), or compare the curve values at a point inside the interval.
Integrate upper minus lower for each lobe, evaluate the endpoint values, and sum the nonnegative results.
Verify the antiderivative by differentiation and, if useful, compare the total with the signed integral to identify cancellation.
Check Your Understanding
For each question, keep the distinction between a lobe’s geometric area and a signed integral in view.
- Why can the integral of \(f-g\) across several lobes be zero even when the curves enclose positive area?
- If \(f-g\) is negative throughout one lobe, which curve is above, and which difference should be integrated for its area?
- Two consecutive intersections occur at \(x=0\) and \(x=2\). What should you check before choosing the integrand for that lobe?
- For \(f-g=(x+1)(x-1)\), identify the sign of the difference on \((-1,1)\). Which curve is above there?
- What does a negative value for a computed lobe area suggest you should recheck?
