Why Repeated Crossings Create Separate Areas
When two curves intersect several times, their intersections can divide the space between them into multiple bounded regions. Each region has its own horizontal interval, and the curve that lies above may change from one interval to the next.
The key is to treat the regions separately. Find the intersections, put them in order, and check the curves’ relative heights between consecutive intersections. Then integrate the positive difference, upper curve minus lower curve, on each interval.
A frequent error is to find all the intersection points but use the same subtraction order on every interval. If the curves cross, that can make an integral negative. A negative signed integral is not a negative geometric area: it signals that the subtraction order needs attention.
Intersections and Individual Bounds
Suppose the curves are graphs of continuous functions \(y=f(x)\) and \(y=g(x)\). Their intersections occur at the \(x\)-values satisfying \(f(x)=g(x)\). List those values in increasing order. Consecutive intersection values are candidates for the horizontal bounds of separate regions.
On each candidate interval, test the difference \(f(x)-g(x)\) at an interior point. If it is positive, \(f\) is above \(g\); if it is negative, \(g\) is above \(f\). Testing each interval matters: the curves can change order at a crossing, but a contact that does not cross need not change their order.
Equivalently, the area on any such interval is \(\int_a^b |f(x)-g(x)|\,dx\). In practice, checking which curve is above lets us write the integrand without an absolute value. If there are several bounded regions, calculate each on its own interval rather than treating their union as one integral with an unexamined sign.
Worked Example: A Sine Curve Crossing the Axis Twice
Worked Example: A Sine Curve Crossing the Axis Twice
Find the areas of the regions bounded by \(y=\sin(\pi x)\) and \(y=0\) for \(0\leq x\leq 2\).
Step 1: Find the intersections. Set the functions equal: $$ \sin(\pi x)=0. $$ For \(0\leq x\leq 2\), the solutions are \(x=0,1,2\). Thus the two candidate regions have bounds \([0,1]\) and \([1,2]\).
Step 2: Determine the order on each interval. At \(x=\frac12\), \(\sin(\pi/2)=1>0\), so the sine curve is above the axis on \((0,1)\). At \(x=\frac32\), \(\sin(3\pi/2)=-1<0\), so the axis is above the sine curve on \((1,2)\).
Step 3: Calculate each area. An antiderivative of \(\sin(\pi x)\) is \(F(x)=-\frac{\cos(\pi x)}{\pi}\), since \(F'(x)=\sin(\pi x)\).
On the second interval, the axis is the upper curve, so subtract the sine function from zero. The signed integral of the sine function itself is negative there:
Therefore the second geometric area is
Check. The areas are positive, as geometric areas must be. Differentiating \(F(x)=-\cos(\pi x)/\pi\) gives \(F'(x)=\sin(\pi x)\), and the evaluated endpoint differences give signed integrals \(2/\pi\) and \(-2/\pi\). Reversing the subtraction on the second interval makes its area \(2/\pi\). The two-region total is \(4/\pi\).
Why the Interval-by-Interval Method Works
The method follows from how the area between graphs is accumulated. On a short interval of width \(\Delta x\), the vertical distance between the curves is approximately \(|f(x)-g(x)|\). Adding these thin-strip areas and taking the limit gives the integral of that nonnegative distance. If the order is fixed on an interval, the absolute difference there is simply upper function minus lower function.
Here is the argument for a set of consecutive intersection bounds \(x_0<x_1<\cdots<x_n\). On each interval \([x_{i-1},x_i]\), suppose the continuous curves have a fixed order except possibly at the endpoints. If \(f\) is above \(g\) there, the region’s area is \(\int_{x_{i-1}}^{x_i}(f-g)\,dx\). If \(g\) is above \(f\), it is \(\int_{x_{i-1}}^{x_i}(g-f)\,dx\). This is the area-between-curves formula applied to that interval.
Because the curves are continuous, their difference cannot change sign inside an interval without becoming zero somewhere inside it. That would give another intersection, contrary to the bounds being consecutive. Thus one test point determines the ordering throughout the interval. This explains why finding every intersection before integrating is essential.
Set the functions equal, solve, and retain the solutions in the relevant interval.
Consecutive values provide candidate bounds for individual regions.
Choose an interior point in each interval and compare the function values.
Evaluate one definite integral for each bounded region and check that every area is nonnegative.
Worked Example: Two Regions from a Cubic Curve
Worked Example: Two Regions from a Cubic Curve
Find the individual areas bounded by \(y=x^3-3x\) and the \(x\)-axis.
Step 1: Find the intersection bounds. Factor the function: $$ x^3-3x=x(x^2-3)=x(x-\sqrt{3})(x+\sqrt{3}). $$ It meets the axis at \(x=-\sqrt{3},0,\sqrt{3}\). These bounds give two bounded regions.
Step 2: Check the sign on each interval. At \(x=-1\), the function value is \(-1+3=2>0\), so the curve is above the axis on \((-\sqrt{3},0)\). At \(x=1\), its value is \(1-3=-2<0\), so the axis is above the curve on \((0,\sqrt{3})\).
Step 3: Evaluate the areas. An antiderivative is \(F(x)=\frac{x^4}{4}-\frac{3x^2}{2}\), because \(F'(x)=x^3-3x\). At either \(x=\sqrt{3}\) or \(x=-\sqrt{3}\), \(x^2=3\) and \(x^4=9\), so $$ F(\sqrt{3})=F(-\sqrt{3})=\frac94-\frac92=-\frac94, \qquad F(0)=0. $$ The left region’s area is $$ A_1=F(0)-F(-\sqrt{3})=0-\left(-\frac94\right)=\frac94. $$ On the right, the curve is below the axis, so reverse the signed integral: $$ \begin{aligned} A_2 &=-\bigl(F(\sqrt{3})-F(0)\bigr)\\ &=-\left(-\frac94-0\right)\\ &=\frac94. \end{aligned} $$
Check. Differentiating \(F\) recovers \(x^3-3x\). The signed integrals are \(9/4\) on the left and \(-9/4\) on the right, so the geometric areas are both positive and equal to \(9/4\). Their total is \(9/2\).
A Common Mistake: Keeping One Subtraction Order
Suppose the upper curve changes at an intersection. Using “\(f-g\)” on both sides without checking the order can produce a negative result. For example, on \([1,2]\) in the sine example, the incorrect calculation would be $$ \int_1^2 \sin(\pi x)\,dx=-\frac{2}{\pi}. $$ This is the signed integral, not the area. The sine curve is below \(y=0\) there, so the correction is to subtract it from zero: $$ \int_1^2 \bigl(0-\sin(\pi x)\bigr)\,dx =-\left(-\frac{2}{\pi}\right) =\frac{2}{\pi}. $$ The endpoint substitution is important: \(F(2)=-1/\pi\) and \(F(1)=1/\pi\), so \(F(2)-F(1)=-1/\pi-1/\pi=-2/\pi\). Negating that signed integral gives the positive area.
Worked Example: Two Crossings of a Cubic and a Line
Worked Example: Two Crossings of a Cubic and a Line
Find the areas of the regions between \(y=x^3\) and \(y=x\) from \(x=-1\) to \(x=1\).
Step 1: Find the intersections. Solve \(x^3=x\): $$ x^3-x=x(x-1)(x+1)=0. $$ The intersection values are \(-1,0,1\), so the two regions have bounds \([-1,0]\) and \([0,1]\).
Step 2: Determine the order. The difference is \(x^3-x\). At \(x=-\frac12\), it is \(-\frac18+\frac12=\frac38>0\), so \(x^3\) is above \(x\) on the left interval. At \(x=\frac12\), it is \(\frac18-\frac12=-\frac38<0\), so \(x\) is above \(x^3\) on the right.
Step 3: Calculate both areas. An antiderivative of \(x^3-x\) is \(F(x)=\frac{x^4}{4}-\frac{x^2}{2}\), since \(F'(x)=x^3-x\). Also, \(F(-1)=F(1)=\frac14-\frac12=-\frac14\), and \(F(0)=0\). Thus $$ A_1=F(0)-F(-1)=0-\left(-\frac14\right)=\frac14. $$ On the right, \(x\) is above \(x^3\), so $$ \begin{aligned} A_2 &=-\bigl(F(1)-F(0)\bigr)\\ &=-\left(-\frac14-0\right)\\ &=\frac14. \end{aligned} $$
Check. Differentiation verifies \(F'(x)=x^3-x\). The signed integral on the right is \(-1/4\), and reversing the order gives the positive area \(1/4\). Both regions therefore have area \(1/4\), for a total of \(1/2\).
What to Check Before Reporting Areas
Repeated crossings make the setup as important as the integration. A complete solution should show which intersections are used, how they divide the interval, and why the chosen integrand is nonnegative on each part. If the curves meet at several values, do not assume all the resulting regions have the same area or that the upper curve stays the same.
- Check the bounds: Substitute each proposed intersection value into both functions to confirm that the values agree.
- Check the order: Use an interior test point on every interval, not just one point for the entire domain.
- Check the integral: Verify the antiderivative by differentiating and substitute the upper and lower bounds in the correct order.
- Check the interpretation: Each geometric area must be nonnegative; a negative signed integral means the curves were subtracted in the opposite order.
Check Your Understanding
For each question, identify individual intervals before deciding how to set up the area integral.
- For \(y=x^2-1\) and \(y=0\) on \([-2,2]\), find the intersection values and list the intervals between consecutive intersections.
- Why is it necessary to test the curve order separately on intervals divided by multiple intersections?
- If \(\int_a^b(f(x)-g(x))\,dx=-5\) and \(g\) is above \(f\) throughout \([a,b]\), what is the area between the curves?
- For \(y=x^3-x\) and \(y=0\), identify the intersection values and use test points to determine which curve is above on each bounded interval.
- What does a negative value for an area integral usually indicate about the setup?
