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Area between two curves: top minus bottom, split where they cross

The area between two curves is one integral: the top curve minus the bottom curve, between the points where they meet. Getting it right comes down to three checks: which curve is on top, where the curves cross, and whether the region is easier to slice sideways.

Short answer

Find where the curves meet by setting them equal. Those x-values are your limits. Test one point in between to see which curve is on top, then integrate ∫ (top − bottom) dx. If the curves cross between your limits, split the integral at each crossing and add the pieces. If the region is bounded by curves written as x = …, integrate in y instead: ∫ (right − left) dy. An area is never negative.

Why it works

Slice the region into thin vertical strips. A strip at position x has height top(x) − bottom(x) and width dx, so its area is (top − bottom) dx. The integral adds up the strips from the left edge of the region to the right edge, and those edges are where the curves meet.

The subtraction only gives a positive height if the top curve really is on top. Where the curves cross, the order flips, and the same formula starts giving negative heights that cancel the area you've already counted. That's why the integral has to be split at every crossing.

Slicing sideways works the same way. A horizontal strip at height y has length right(y) − left(y) and thickness dy. It's the better choice when the top or bottom of the region changes formula partway across, but the left and right sides don't.

Example 1: a parabola and a line

Find the area of the region enclosed by y = x + 2 and y = x2.

  1. x2 = x + 2, so (x − 2)(x + 1) = 0Set the curves equal to find where they meet: x² − x − 2 = 0.
  2. x = −1 and x = 2These are the limits.
  3. At x = 0: line 2, parabola 0Test a point in between. The line is on top.
  4. ∫−12 (x + 2 − x2) dx = [x2/2 + 2x − x3/3]−12Top minus bottom.
  5. = 103 − (−76)At 2: 2 + 4 − 8/3 = 10/3. At −1: 1/2 − 2 + 1/3 = −7/6.
  6. Area = 9220/6 + 7/6 = 27/6.

Example 2: curves that cross in the middle

Find the area between y = 2x and y = x2 from x = 0 to x = 3.

  1. 2x = x2 at x = 0 and x = 2x² − 2x = x(x − 2). The crossing at x = 2 is inside the interval.
  2. ∫02 (2x − x2) dx = 4 − 83 = 43On (0, 2) the line is on top: at x = 1, 2 > 1.
  3. ∫23 (x2 − 2x) dx = (9 − 9) − (83 − 4) = 43On (2, 3) the parabola is on top: at x = 2.5, 6.25 > 5.
  4. Area = 43 + 43 = 83Add the pieces. One integral straight from 0 to 3 gives 0 (mistake 2).

Example 3: integrating in y

Find the area of the region enclosed by x = y2 and x = 6 − y.

  1. y2 = 6 − y, so (y + 3)(y − 2) = 0Both curves are written as x = …, so set them equal and solve for y.
  2. y = −3 and y = 2The limits, as heights. The region runs from (9, −3) up to (4, 2).
  3. At y = 0: line 6, parabola 0The line x = 6 − y is on the right.
  4. ∫−32 (6 − y − y2) dy = [6y − y2/2 − y3/3]−32Right minus left.
  5. = 223 − (−272)At 2: 12 − 2 − 8/3 = 22/3. At −3: −18 − 9/2 + 9 = −27/2.
  6. Area = 1256 ≈ 20.8344/6 + 81/6. In x this region needs two integrals, because its top edge changes from the parabola to the line at x = 4.

Example 4: a curve below the x-axis

Find the area between y = x2 − 4 and the x-axis.

  1. x2 − 4 = 0 at x = ±2The x-axis is the curve y = 0.
  2. ∫−22 (0 − (x2 − 4)) dx = ∫−22 (4 − x2) dxBetween −2 and 2 the parabola is below the axis, so the axis is the top.
  3. = 16 − 163 = 323[4x − x³/3] from −2 to 2: (8 − 8/3) − (−8 + 8/3).

Common mistakes, and the exact line they happen on

1. Bottom minus top

Find the area enclosed by y = x + 2 and y = x2 (Example 1).

  1. Limits x = −1 and x = 2Right so far.
  2. ∫−12 (x2 − (x + 2)) dx = −92This is the mistake. The curves were subtracted in the order they were listed, not top minus bottom. A negative area is the giveaway. Test a point first: at x = 0 the line is higher, so it goes first, and the area is 9/2.

2. One integral across a crossing

Find the area between y = 2x and y = x2 from 0 to 3 (Example 2).

  1. ∫03 (2x − x2) dx = 9 − 9 = 0This is the mistake. The line is on top only until x = 2. After that the parabola is, so the strips from 2 to 3 count as negative and cancel the area from 0 to 2 exactly. Two curves with space between them can't have zero area between them. Split at x = 2: 4/3 + 4/3 = 8/3.

3. Limits from the picture instead of the intersections

Find the area enclosed by y = x + 2 and y = x2 (Example 1).

  1. ∫02 (x + 2 − x2) dx = 103This is the mistake. The lower limit 0 came from where the graph seemed to start, not from solving x² = x + 2. The region begins at x = −1, and this misses the piece from −1 to 0, which has area 7/6. Always get the limits by setting the curves equal: the area is 10/3 + 7/6 = 9/2.

Practice

  1. Find the area enclosed by y = 2x and y = x2.

    Show answer

    They meet at x = 0 and x = 2, with the line on top: ∫02 (2x − x2) dx = 4 − 8/3 = 4/3.

  2. Find the area enclosed by y = x3 and y = 4x for x ≥ 0.

    Show answer

    x3 = 4x at x = 0 and x = 2, and at x = 1, 4 > 1. ∫02 (4x − x3) dx = 8 − 4 = 4.

  3. Find the area between y = x2 − 1 and the x-axis from x = 0 to x = 2.

    Show answer

    The curve crosses the axis at x = 1. ∫01 (1 − x2) dx = 2/3 and ∫12 (x2 − 1) dx = 4/3, so the area is 2. (One integral from 0 to 2 gives 2/3, which isn't the area.)

  4. Find the area enclosed by x = y2 and x = 4.

    Show answer

    In y: they meet at y = ±2, and x = 4 is on the right. ∫−22 (4 − y2) dy = 16 − 16/3 = 32/3.

  5. Find the area between y = ex and y = 1 from x = 0 to x = 1.

    Show answer

    ex ≥ 1 for x ≥ 0, so ∫01 (ex − 1) dx = (e − 1) − 1 = e − 2 ≈ 0.718.

Practice this

Printable PDF: area-between-curves.pdf

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