Short on time? This page is the recap. For explanations, the diagram and worked examples, use the full study guide. This topic is shared by Calculus AB and Calculus BC.
Notation: ∫ (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.
Recap
- When the curves cross inside the interval, the top curve changes and the region splits into pieces.
- A strip’s height is always the distance between the curves, |f(x) − g(x)|.
- No calculator: find every crossing, make a sign chart for f − g, and add ∫ [top − bottom] dx over each piece.
- Calculator: enter ∫ (a to b) |f(x) − g(x)| dx in one go. You still need the crossings if they are the region’s ends.
- The same works sideways: ∫ (c to d) |R(y) − L(y)| dy, split wherever right and left swap.
Key relationships
| Situation | Expression |
|---|---|
| f on top on [a, c], g on top on [c, b] | Area = ∫ (a to c) [f − g] dx + ∫ (c to b) [g − f] dx |
| Any number of crossings, calculator allowed | Area = ∫ (a to b) |f − g| dx |
| Net (signed) value | ∫ (a to b) [f − g] dx = (pieces with f on top) − (pieces with g on top) |
| Sign of ∫ [f − g] dx over a piece where the curves do not cross | Positive: f on top. Negative: g on top. |
Assumptions behind the method
- f and g are continuous on [a, b].
- You have found every crossing in [a, b]; between neighbouring crossings, f − g keeps one sign.
- Each test value lies strictly inside its piece, not at a crossing.
Mistakes to avoid
- One integral of f − g across all the pieces, so they cancel.
- Absolute value outside the integral: |∫ (f − g) dx| is not the area.
- Missing a crossing, especially one at x = 0 or near where the curves almost touch.
- One test value used for the whole interval.
- Splitting at the x-axis instead of at the crossings.
- Rounding crossings before using them as limits.
Quick self-check
- Two curves cross at x = 1 and x = 4 inside [0, 6], and nowhere else. Without a calculator, how many integrals do you need for the area on [0, 6]? (Three: on [0, 1], [1, 4] and [4, 6].)
- On three pieces, ∫ (f − g) dx equals 4, −7 and 2. Find the area and the net value. (Area 4 + 7 + 2 = 13; net 4 − 7 + 2 = −1.)
- A student writes the area as |∫ (a to b) (f − g) dx|. When is this correct? (Only when the curves do not cross between a and b.)
Next: practice questions.