Finding the Area Between Curves That Intersect at More Than Two Points: Study Guide (Calculus AB 8.6)

Course scope. This topic is shared by Calculus AB and Calculus BC. Everything here is examinable for both.

A note on notation

This page has no equation renderer, so integrals are written in a compact form: ∫ (a to b) |f(x) − g(x)| dx means the definite integral of the absolute value of f(x) − g(x) from x = a to x = b. On paper, write a at the bottom of the integral sign and b at the top. The vertical bars must sit inside the integral, around the difference.

When the top curve changes

In Topics 8.4 and 8.5 one curve stayed on top (or on the right) across the whole region. Now the curves cross in the middle, so the region falls into separate pieces, and in some pieces the “top” curve is f while in others it is g.

The strip idea still works. A vertical strip at x always has height equal to the distance between the curves. Where f is on top, that distance is f(x) − g(x). Where g is on top, it is g(x) − f(x). Both cases are captured by one expression: |f(x) − g(x)|.

Area when curves cross. If f and g are continuous on [a, b], the area of the region between their graphs for a ≤ x ≤ b is ∫ (a to b) |f(x) − g(x)| dx. Without a calculator, split [a, b] at every point where the curves cross and add ∫ [top − bottom] dx over each piece.

The same holds for horizontal strips: the area is ∫ (c to d) |R(y) − L(y)| dy, split wherever the right and left curves swap.

The method, step by step

1. Find every intersection in the interval. Solve f(x) = g(x), or use a calculator. Include the ends if the region is closed off by the curves themselves.
2. Make a sign chart for f − g. Pick one test value in each subinterval between neighbouring intersections. Positive means f is on top there; negative means g is.
3. Write one integral per piece, each as ∫ [top − bottom] dx with that piece’s limits.
4. Evaluate and add. Every piece should be positive.
5. With a calculator, you may instead enter ∫ (a to b) |f(x) − g(x)| dx in one go.

Net value or area?

The integral without the absolute value, ∫ (a to b) [f(x) − g(x)] dx, adds positive pieces (where f is on top) and negative pieces (where g is on top). The pieces cancel, at least partly. This is the same idea you met in Topic 8.2: integrating velocity gives displacement (net), while integrating speed gives distance (total).

• ∫ (a to b) [f(x) − g(x)] dx gives a net value: (area where f is on top) − (area where g is on top). Use it for accumulated-difference questions, like displacement.
• ∫ (a to b) |f(x) − g(x)| dx gives the total area between the curves. Use it for area questions.
• The absolute value of ∫ (a to b) [f(x) − g(x)] dx is only the size of the net value. It equals the area only if the curves do not cross between a and b.

Worked example 1: three crossings, no calculator

Question. Find the total area of the regions enclosed by y = x³ − x² and y = 2x (Figure 1).

1. Intersections. x³ − x² = 2x gives x³ − x² − 2x = 0, so x(x² − x − 2) = 0 and x(x − 2)(x + 1) = 0. The curves cross at x = −1, 0 and 2. The points are (−1, −2), (0, 0) and (2, 4).
2. Sign chart for f − g = x³ − x² − 2x.
   • On (−1, 0), test x = −0.5: (−0.5)³ − (−0.5)² − 2(−0.5) = 5/8 > 0. The cubic is on top.
   • On (0, 2), test x = 1: 1 − 1 − 2 = −2 < 0. The line is on top.
3. One integral per piece. Area = ∫ (−1 to 0) [(x³ − x²) − 2x] dx + ∫ (0 to 2) [2x − (x³ − x²)] dx
4. Evaluate. An antiderivative of x³ − x² − 2x is F(x) = x⁴/4 − x³/3 − x². F(−1) = 1/4 + 1/3 − 1 = −5/12, F(0) = 0 and F(2) = 4 − 8/3 − 4 = −8/3.
   • A₁ = F(0) − F(−1) = 5/12.
   • A₂ = −[F(2) − F(0)] = 8/3.
5. Add. 5/12 + 8/3 = 5/12 + 32/12 = 37/12.

Answer. The total area is 37/12 square units (about 3.083).

What the single integral gives. ∫ (−1 to 2) [(x³ − x²) − 2x] dx = F(2) − F(−1) = −8/3 + 5/12 = −9/4. That is 5/12 − 8/3: the small piece minus the large piece. It is not an area. Even its size, 9/4, is wrong.

Worked example 2: absolute value on a calculator

Question. Find the total area of the regions enclosed by y = x³ − 3x + 1 and y = cos x. (Calculator allowed; radians.)

1. Intersections. Graph both, or find the zeros of x³ − 3x + 1 − cos x. There are three: x = A ≈ −1.923, x = 0 and x = B ≈ 1.540. The middle one is exact, since 0 − 0 + 1 = 1 = cos 0. Store A and B. Check the window carefully: a crossing near the origin is easy to miss when the graphs are close together.
2. Who is on top. At x = −1: (−1)³ + 3 + 1 = 3 and cos(−1) ≈ 0.540, so the cubic is on top on (A, 0). At x = 1: 1 − 3 + 1 = −1 and cos 1 ≈ 0.540, so cos x is on top on (0, B).
3. One integral with absolute value: Area = ∫ (A to B) |x³ − 3x + 1 − cos x| dx ≈ 4.723 square units.
4. Check by pieces. ∫ (A to 0) [(x³ − 3x + 1) − cos x] dx ≈ 3.113 and ∫ (0 to B) [cos x − (x³ − 3x + 1)] dx ≈ 1.611. The sum is 4.723. Both pieces are positive, as they should be.

The trap. ∫ (A to B) [x³ − 3x + 1 − cos x] dx without the bars gives about 1.502, which is 3.113 − 1.611. Never report this as an area.

Notation on paper. Show the intersection values and the full integral with the absolute value bars, limits and dx, then the decimal: “∫ (−1.923 to 1.540) |x³ − 3x + 1 − cos x| dx = 4.723”.

Worked example 3: reading information from a representation (no calculator)

Sometimes you are not given formulas. You are given a graph or the values of some integrals, and you must decide what they tell you.

Question. f and g are continuous on [0, 7]. Their graphs cross only at x = 2 and x = 5, and

• ∫ (0 to 2) [f(x) − g(x)] dx = 6
• ∫ (2 to 5) [f(x) − g(x)] dx = −10
• ∫ (5 to 7) [f(x) − g(x)] dx = 3

(a) On which interval is g above f? (b) Find the total area between the graphs for 0 ≤ x ≤ 7. (c) Find ∫ (0 to 7) [f(x) − g(x)] dx. (d) Find ∫ (0 to 5) |f(x) − g(x)| dx.

1. (a) Between crossings, f − g keeps one sign. The integral over (2, 5) is negative, so f − g < 0 there: g is above f on (2, 5). On (0, 2) and (5, 7), f is above g.
2. (b) Each piece contributes its size: 6 + 10 + 3 = 19.
3. (c) The net value keeps the signs: 6 + (−10) + 3 = −1.
4. (d) Only the first two pieces: 6 + 10 = 16.

Why the sign tells you who is on top. If f − g kept one sign on (2, 5) and that sign were positive, the integral would be positive. It is −10, so the sign must be negative. This reasoning relies on the curves not crossing inside (2, 5), which the question states.

Common misconceptions

• Integrating f − g over the whole interval. The pieces cancel. You get a net value, not the area.
• Putting the absolute value outside. |∫ (f − g) dx| is not ∫ |f − g| dx. In Worked example 1 the first gives 9/4, the second 37/12.
• Missing an intersection. In Worked example 2 the crossing at x = 0 is easy to overlook. Every missed crossing merges two pieces with opposite signs.
• Testing only one point for the whole interval. You need one test value per piece.
• Splitting at the x-axis. For the area between two curves, split only where the curves cross each other, not where one crosses the axis.
• Assuming equal pieces. The pieces of a region are rarely the same size, even when the crossing points look evenly spaced.
• Rounding intersection values before using them as limits on a calculator.

Where this leads

This topic completes the area work of Unit 8: vertical strips (Topic 8.4), horizontal strips (Topic 8.5) and now regions in several pieces. The same “split where the boundary changes” habit returns in Topics 8.7 to 8.12, where volumes with cross-sections and solids of revolution are built from slices. Continue with Volumes with Cross Sections: Squares and Rectangles, or return to Topic 8.5. Students in either course can return to the Calculus AB roadmap or the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.