Calculus · Grade 12 and undergraduate

Area Between Curves

Quick answer

The area between two curves is the integral of the gap between them. With vertical slices, integrate top minus bottom with respect to x, between the x-values where the curves meet. When the boundaries are easier to write as functions of y, slice horizontally and integrate right minus left with respect to y. Where the curves cross, top and bottom trade places, so split the interval at each crossing and add the areas of the pieces.

What you'll learn

  • Find the area between curves with vertical slices
  • Find the area between curves with horizontal slices
  • Split the region where two curves cross
  • Choose the slicing that needs the fewest integrals

Top minus bottom

The integral ∫abf(x) dx\int_a^b f(x)\,dx measures the area between a curve and the xx-axis. Between two curves, measure the gap instead. If f(x)≥g(x)f(x) \ge g(x) on [a,b][a, b], the area between them is

A=∫ab(f(x)−g(x)) dxA = \int_a^b \big(f(x) - g(x)\big)\,dx

The limits are usually where the curves meet, found by solving f(x)=g(x)f(x) = g(x).

Vertical slices: top minus bottom The parabola y = 2 − x² and the line y = x, crossing at (−2, −2) and (1, 1). The region between them is shaded, and one thin vertical strip is drawn at x = −0.5, from the line up to the parabola; part of the strip lies below the x-axis. -2-11-2-112xy
  • y = 2 − x²
  • y = x
Vertical slices: top minus bottom

Why top minus bottom works anywhere

Cut the region into thin vertical strips. A strip at xx runs from the lower curve up to the upper one, so its height is f(x)−g(x)f(x) - g(x) and its area is about (f(x)−g(x)) Δx\big(f(x) - g(x)\big)\,\Delta x. Adding the strips gives a Riemann sum, and its limit is the integral.

The height of a strip is a difference of two yy-values, and that difference is the strip’s length whatever the signs. The strip drawn in the graph runs from −0.5-0.5 up to 1.751.75, and its length is 1.75−(−0.5)=2.251.75 - (-0.5) = 2.25. So the formula needs no adjustment for a region below or across the axis. Slice the region, write the length of one slice, and integrate.

Slicing horizontally

Some regions are easier to slice sideways. A horizontal strip at height yy runs from the left boundary to the right one, so

A=∫cd(xright(y)−xleft(y)) dyA = \int_c^d \big(x_{\text{right}}(y) - x_{\text{left}}(y)\big)\,dy

with the boundaries written as xx in terms of yy and the limits as yy-values. Slice horizontally when the curves are given as x=…x = \ldots, or when a vertical strip would have to change formulas partway across.

Horizontal slices: right minus left The parabolas x = y² opening to the right and x = 2 − y² opening to the left, crossing at (1, 1) and (1, −1). The lens-shaped region between them is shaded, and one thin horizontal strip runs across it at y = 0.5, from the left parabola to the right one. 12-11xy
  • x = y²
  • x = 2 − y²
Horizontal slices: right minus left

Here a vertical strip would run between the two halves of x=y2x = y^2 for x≤1x \le 1 and between the two halves of x=2−y2x = 2 - y^2 for x≥1x \ge 1: two integrals, each with square roots. Horizontal strips need only one.

Worked examples

Common mistakes

Practice problems

  1. Find the area between y=x2y = x^2 and y=2xy = 2x.

    Answer

    43\tfrac{4}{3}

    Full solution

    They meet where x2=2xx^2 = 2x, at x=0x = 0 and x=2x = 2, with the line on top. ∫02(2x−x2) dx=4−83=43\int_0^2 (2x - x^2)\,dx = 4 - \tfrac{8}{3} = \tfrac{4}{3}.

  2. Find the area between y=xy = \sqrt{x} and y=xy = x.

    Answer

    16\tfrac{1}{6}

    Full solution

    They meet at x=0x = 0 and x=1x = 1, with x\sqrt{x} on top. ∫01(x1/2−x)dx=23−12=16\int_0^1 \left(x^{1/2} - x\right)dx = \tfrac{2}{3} - \tfrac{1}{2} = \tfrac{1}{6}.

  3. Find the area between y=exy = e^x and y=1y = 1 from x=0x = 0 to x=1x = 1.

    Answer

    e−2≈0.718e - 2 \approx 0.718

    Full solution

    ∫01(ex−1) dx=[ex−x]01=(e−1)−1=e−2\int_0^1 (e^x - 1)\,dx = \big[e^x - x\big]_0^1 = (e - 1) - 1 = e - 2.

  4. Find the area between y=cos⁡xy = \cos x and y=sin⁡xy = \sin x from x=0x = 0 to x=π4x = \tfrac{\pi}{4}.

    Answer

    2−1≈0.414\sqrt{2} - 1 \approx 0.414

    Full solution

    On this interval cos⁡x≥sin⁡x\cos x \ge \sin x. ∫0π/4(cos⁡x−sin⁡x) dx=[sin⁡x+cos⁡x]0π/4=2−1\int_0^{\pi/4} (\cos x - \sin x)\,dx = \big[\sin x + \cos x\big]_0^{\pi/4} = \sqrt{2} - 1.

  5. Find the area between y=xy = x and y=1xy = \tfrac{1}{x} from x=1x = 1 to x=2x = 2.

    Answer

    32−ln⁡2≈0.807\tfrac{3}{2} - \ln 2 \approx 0.807

    Full solution

    On [1,2][1, 2], x≥1xx \ge \tfrac{1}{x}. ∫12(x−1x)dx=[x22−ln⁡x]12=(2−ln⁡2)−12\int_1^2 \left(x - \tfrac{1}{x}\right)dx = \big[\tfrac{x^2}{2} - \ln x\big]_1^2 = (2 - \ln 2) - \tfrac{1}{2}.

  6. Find the area between x=y2x = y^2 and x=4x = 4.

    Answer

    323\tfrac{32}{3}

    Full solution

    Slice horizontally. The curves meet at y=±2y = \pm 2, and x=4x = 4 is on the right: ∫−22(4−y2) dy=[4y−y33]−22=163+163\int_{-2}^{2} (4 - y^2)\,dy = \big[4y - \tfrac{y^3}{3}\big]_{-2}^{2} = \tfrac{16}{3} + \tfrac{16}{3}.

  7. Find the area between x=y2x = y^2 and x=y+6x = y + 6.

    Answer

    1256\tfrac{125}{6}

    Full solution

    They meet where y2=y+6y^2 = y + 6, at y=−2y = -2 and y=3y = 3, with the line on the right. ∫−23(y+6−y2) dy=[y22+6y−y33]−23=272−(−223)=1256\int_{-2}^{3} (y + 6 - y^2)\,dy = \big[\tfrac{y^2}{2} + 6y - \tfrac{y^3}{3}\big]_{-2}^{3} = \tfrac{27}{2} - \left(-\tfrac{22}{3}\right) = \tfrac{125}{6}.

  8. Find the area between y=xy = x and y=x3y = x^3 from x=0x = 0 to x=2x = 2.

    Answer

    52\tfrac{5}{2}

    Full solution

    They cross at x=1x = 1. ∫01(x−x3) dx=14\int_0^1 (x - x^3)\,dx = \tfrac{1}{4} and ∫12(x3−x) dx=(4−2)−(14−12)=94\int_1^2 (x^3 - x)\,dx = (4 - 2) - \left(\tfrac{1}{4} - \tfrac{1}{2}\right) = \tfrac{9}{4}. The total is 52\tfrac{5}{2}.

  9. Find the total area enclosed by y=x3−4xy = x^3 - 4x and the xx-axis.

    Answer

    88

    Full solution

    x3−4x=x(x−2)(x+2)x^3 - 4x = x(x - 2)(x + 2) is zero at −2-2, 00 and 22. ∫−20(x3−4x) dx=4\int_{-2}^{0} (x^3 - 4x)\,dx = 4, and by symmetry the piece from 00 to 22, below the axis, also has area 44.

  10. A student finds the area between y=x2y = x^2 and y=x3y = x^3 from x=0x = 0 to x=2x = 2 as ∫02(x3−x2) dx=43\int_0^2 (x^3 - x^2)\,dx = \tfrac{4}{3}. What went wrong?

    Hint

    Which curve is on top when x=12x = \tfrac{1}{2}?

    Answer

    The curves cross at x=1x = 1, and the student did not split there. The area is 32\tfrac{3}{2}.

    Full solution

    On [0,1][0, 1], x2≥x3x^2 \ge x^3: ∫01(x2−x3) dx=112\int_0^1 (x^2 - x^3)\,dx = \tfrac{1}{12}. On [1,2][1, 2], x3≥x2x^3 \ge x^2: ∫12(x3−x2) dx=1712\int_1^2 (x^3 - x^2)\,dx = \tfrac{17}{12}.

    The area is 1812=32\tfrac{18}{12} = \tfrac{3}{2}. The single integral subtracted the first piece instead of adding it: 1712−112=43\tfrac{17}{12} - \tfrac{1}{12} = \tfrac{4}{3}.

Frequently asked questions

How do I find the area between two curves?

Find where they meet, then integrate the top curve minus the bottom curve between those x-values: A = ∫ from a to b of (f(x) − g(x)) dx.

Does it matter if the region is below the x-axis?

No. Top minus bottom is the height of a slice wherever the slice sits, so the formula works above, below or across the axis.

When should I integrate with respect to y?

When the boundaries are given as x in terms of y, or when a vertical slice would switch formulas partway across. Then integrate right minus left over y.

What if the curves cross?

Split the interval at every crossing and integrate top minus bottom on each piece, or integrate |f − g|. A single integral of f − g lets the pieces cancel.

Can an area come out negative?

Not if top minus bottom is set up correctly. A negative result means the curves were subtracted in the wrong order somewhere.

What to learn next

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