AP® Calculus AB review sheet from Aim for Five (aimforfive.com/calc-ab/units/8/8-6)
Unit 8 · Topic 8.6
8.6 Finding the Area Between Curves That Intersect at More Than Two Points
When two curves cross more than twice, the curve on top switches at each crossing. Split the integral at every intersection and subtract bottom from top on each piece, or integrate the absolute value of the difference.
Key terms
- intersection points
- splitting an integral
- absolute value
- total area
Why one integral isn't enough
If f is above g on one part of the interval and below on another, then ∫ (f − g) dx counts one region as positive and the other as negative. They partly cancel, and you get the net difference, not the total area. For area, every piece must count as positive.
Method 1: split at every crossing
This is the method you'll use without a calculator.
- Find all intersection points by solving f(x) = g(x).
- On each interval between consecutive intersections, test a point to see which curve is on top.
- Integrate (top − bottom) on each interval.
- Add the results.
- Symmetry can save time: if both functions are odd, or the region is symmetric about the y-axis, you may be able to compute one piece and double it. Check that the pieces really match first.
Method 2: absolute value
Total area = ∫ₐᵇ |f(x) − g(x)| dx. The absolute value flips any negative strip heights to positive, which is exactly what splitting does.
This is ideal on a calculator: you don't need to know which curve is on top or even find the middle intersection points, as long as the outer limits are right. By hand, |f − g| doesn't have a simple antiderivative, so you split anyway.
Finding every intersection
Missing a crossing is the main way these problems go wrong. Set f(x) = g(x), move everything to one side and factor completely; a cubic like x³ − 4x factors as x(x − 2)(x + 2), giving three crossings. With a calculator, graph both curves in a window that shows the whole region and use the intersect feature for each crossing, storing each value.
If the problem gives the interval, like 0 ≤ x ≤ π, only crossings inside that interval matter, and the endpoints are your outer limits even if the curves don't meet there.
Area vs. integral
Keep two questions separate. “Find ∫ₐᵇ [f(x) − g(x)] dx” asks for a signed number, which can be zero or negative. “Find the area of the region between the curves” asks for total area, which is positive. The same split-and-add approach also gives total distance from velocity (8.2), where you integrate |v(t)|.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Three intersection points
Find the total area of the regions enclosed by y = x³ − 3x and y = x.
Show the solutionHide the solution
- Step 1: Intersections: x³ − 3x = x → x³ − 4x = 0 → x(x − 2)(x + 2) = 0, so x = −2, 0, 2.
- Step 2: On (−2, 0), test x = −1: x³ − 3x = −1 + 3 = 2, and x = −1. The cubic is on top.
- Step 3: On (0, 2), test x = 1: x³ − 3x = −2, and x = 1. The line is on top.
- Step 4: Area = ∫ from −2 to 0 of [(x³ − 3x) − x] dx + ∫₀² [x − (x³ − 3x)] dx = ∫ from −2 to 0 of (x³ − 4x) dx + ∫₀² (4x − x³) dx.
- Step 5: ∫₀² (4x − x³) dx = 8 − 4 = 4. The other piece is also 4 by symmetry (both functions are odd), and you can confirm: [x⁴/4 − 2x²] from −2 to 0 = 0 − (4 − 8) = 4.
- Step 6: Total = 8.
Answer: 8
- Example 2
Trap: the net integral is not the area
Find the area between y = sin x and y = cos x on [0, π]. Compare it with ∫ from 0 to π of (cos x − sin x) dx.
Show the solutionHide the solution
- Step 1: Intersection: sin x = cos x at x = π/4 in [0, π].
- Step 2: On (0, π/4), cos x is on top; on (π/4, π), sin x is on top.
- Step 3: ∫ from 0 to π/4 of (cos x − sin x) dx = [sin x + cos x] from 0 to π/4 = √2 − 1.
- Step 4: ∫ from π/4 to π of (sin x − cos x) dx = [−cos x − sin x] from π/4 to π = (1 − 0) − (−√2) = 1 + √2.
- Step 5: Area = (√2 − 1) + (1 + √2) = 2√2 ≈ 2.828.
- Step 6: The single integral ∫ from 0 to π of (cos x − sin x) dx = [sin x + cos x] from 0 to π = −1 − 1 = −2. It's negative, and it isn't the area.
Answer: Area = 2√2 ≈ 2.828; the unsplit integral gives −2.
Common mistakes
- Integrating f − g once across all crossings and calling it the area.
- Missing a middle intersection point, so one piece has the wrong curve on top.
- Taking the absolute value of the final answer instead of each piece: |∫ (f − g) dx| is not the area.
- Assuming symmetry without checking it.
On the exam
- On calculator questions, ∫ₐᵇ |f(x) − g(x)| dx is the fastest correct setup. Write it out before the number.
- Multiple-choice questions like to offer the net integral as a wrong choice for an area. If the curves cross inside the interval, expect a split.
Connected topics
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Check yourself
4 questions on 8.6 Finding the Area Between Curves That Intersect at More Than Two Points. Pick an answer to see if you got it, and why.
What is the total area of the region enclosed by the graphs of y = x³ and y = x?
The graphs of f(x) = x³ − 3x and g(x) = sin x intersect at three points. What is the total area of the regions enclosed by the two graphs?
What is the total area of the regions enclosed by the graphs of y = x³ − 3x² and y = −2x?
The graphs of the continuous functions f and g intersect only at x = 0, x = 2 and x = 5. On the interval 0 < x < 2, f(x) > g(x), and on the interval 2 < x < 5, g(x) > f(x). Which of the following gives the total area of the regions enclosed by the graphs of f and g?
0 of 4 answered