acalculator

What is the area between two curves?

Type two curves. Give an interval [a, b], or leave it empty to use the points where the curves cross. The page gives the area between the two curves, exactly and as a decimal.

Your numbers

Area
21.33333333

The area between 9 - (x/2)^2 and 6 - x is 21.33333333.

Exact area
64/3
From x =
-2
To x =
6

Area: 21.33333333. The area between 9 - (x/2)^2 and 6 - x is 21.33333333.

How to calculate

Finds the area between y = f(x) and y = g(x), from a to b or between their crossings.

Example with the default inputs (Curve y = f(x) 9 - (x/2)^2, Curve y = g(x) 6 - x): The area between 9 - (x/2)^2 and 6 - x is 21.33333333.

Method: Area = ∫ |f(x) − g(x)| dx from a to b, split where the curves cross.

  • Crossings searched for in ±10^6.

Machine-readable copies: Markdown, JSON.

Worked examples

Each example is checked against the calculator on every build.

  1. Curve y = f(x) 9 - (x/2)^2, Curve y = g(x) 6 - x gives Area 21.333333, Exact area 64/3, From x = -2, To x = 6.Source: OpenStax (Strang and Herman, 2016), Calculus Volume 1, section 6.1 Areas between Curves, Example 6.2
  2. Curve y = f(x) x + 4, Curve y = g(x) 3 - x/2, From x = a (optional) 1, To x = b (optional) 4 gives Area 14.25, Exact area 57/4.Source: OpenStax (Strang and Herman, 2016), Calculus Volume 1, section 6.1 Areas between Curves, Example 6.1

How it works

The area between y = f(x) and y = g(x) from x = a to x = b is

Area = ∫ from a to b of |f(x) − g(x)| dx

With h(x) = f(x) − g(x):

  1. The ends. If you type a and b, they are the ends (a must be less than b; typing only one gives a message on the other). If both are empty, the ends are the first and the last crossing: the zeros of h between −∞ and ∞, found as described below (a break of h is not a crossing). With fewer than two crossings the page asks for a and b.
  2. The pieces. The interval is split at every crossing strictly between the ends (zeros of h between a and b, found as below).
  3. Each piece is a definite integral of h, worked out by a computer algebra system (nerdamer, open source) as on the integral calculator: F(end) − F(start) with a checked antiderivative F, compared with a numeric integral of h (Gauss-Kronrod quadrature) to 1 part in 10⁷ of the integral of |h|. A pole of h inside a piece, or a part where h is not a real number, gives no value. The ends and crossings go to the algebra exactly when they are exact, and as decimals when they are not.
  4. The area is the sum of the absolute values of the pieces, shown to 10 significant figures (rounded half up).
  5. The exact area is shown when every end and crossing is exact and every piece has an exact value: the pieces' exact values are added with their signs (a piece whose value is negative is subtracted) and simplified by the algebra; it is left out when the simplified form fails its check or holds a rounded number.

The page also shows the ends used, From x = and To x =, as decimals to 10 significant figures.

How the points are found

The zeros and breaks of a function g (a zero: g passes through or touches 0; a break: g jumps, goes to ±∞, or stops being a real number) are found between two ends the ends given in steps 1 and 2 (each end clipped to ±10⁶) in two ways at once.

By the algebra. The computer algebra system solves g(x) = 0. A solution is an exact candidate when it is a number strictly between the ends, is not a rounded number (a whole number past 2⁵³; a fraction p/q, p and q being the numbers that multiply its top and its bottom, such as 288557167/(342919925e), when q after removing its factors 2 and 5 is over 1,000,000, or is over 1 while |p| times it is over 10¹²; or a decimal with more than 12 significant digits), and g is 0 there up to rounding: |g(c)| is at most 10⁻¹² times the sum of the sizes of the terms of g at c (the parts of g joined by + and −). So x² + 10⁻²⁰ has no zero at 0, though the algebra gives one.

On a grid. g is worked out at 8,001 points x₀ to x₈₀₀₀, evenly spaced in asinh x (so they reach ±10⁶ when an end is infinite and bunch up near 0): xᵢ = sinh(u₀ + (u₁ − u₀)(i + 0.3183)/8001), with u₀ and u₁ the asinh of the two ends. For each step from xᵢ to xᵢ₊₁:

  • If g is a number at one end of the step and not a number at the other (as ln or √ of a negative number is not), the point where it stops being one is found by halving the step 80 times: a break. A value past the largest computer number, about 1.8 × 10³⁰⁸, still counts as a number with its sign (x e^(−x) and its derivatives for x below about −703), so an overflow is not a break. But when |g| is over 10³⁰⁰ at the end where g is a number (e^x − e^(2x) near x = 355, where e^(2x) overflows and the difference is not a number), the page cannot tell an overflow from the end of the domain and gives no answer (a value is too large).
  • If g has opposite signs at the two ends, the sign change is found by halving 80 times. It is a zero if |g| there is at most 10⁻⁶ of the larger of |g(xᵢ)| and |g(xᵢ₊₁)|, and a break otherwise (g jumps or goes through ±∞). If g is past the largest computer number at either end, the page gives no answer (a value is too large).
  • If g has one sign at xᵢ, xᵢ₊₁ and xᵢ₊₂ and turns at xᵢ₊₁ (a peak or a dip), the turning point r is found by halving (on the sign of g(x + h) − g(x − h), h = 10⁻⁷ × max(1, |x|)). If g(r) has the other sign, g dips through 0 and back between xᵢ and xᵢ₊₂: two zeros too close together for the grid. Unless the algebra gives at least two exact candidates strictly between xᵢ and xᵢ₊₂ (then they are the zeros), the page gives no answer (two points are too close together). If |g(r)| is at most 10⁻⁹ of the larger of |g(xᵢ)| and |g(xᵢ₊₂)|, g touches 0 there: that is a zero when an exact candidate lies within 10⁻⁶ × max(1, |candidate|) of r, and otherwise the page says a zero could not be confirmed and gives no answer. If |g(r)| is over 10⁶ times that size (or not a number), g peaks through ±∞: a break. A reason to give no answer that turns up on the way (a dip, a touch the algebra does not confirm) is given only after the whole grid, so a function with more than 30 points (sin x) gets "too many points to check".

A zero found on the grid within 10⁻⁶ × max(1, |c|) of an exact candidate c takes its exact value and text; otherwise, when it is within 10⁻⁹ × max(1, |x|) of a fraction p/q with q = 1, 2, 3, 4, 6, 8 or 12 (the first that fits) where g is 0 up to rounding (the rule above), it is shown as p/q; otherwise as ≈ and a decimal to 10 significant figures (rounded half up). A break is shown exactly as p/q when x is within 10⁻⁹ × max(1, |x|) of p/q for q = 1, 2, 3, 4, 6, 8 or 12 (the first that fits) and g is not a real number at p/q or at p/q ± 10⁻⁹ × max(1, |x|) (a pole, a jump, or an end of the domain); otherwise as ≈ and a decimal. An exact candidate that the grid did not find (two zeros inside one step) is added. A zero right next to a pole, inside the same grid step, is found only when the algebra gives it. The same point found twice counts once (a zero before a break). Two different points within 10⁻⁶ × max(1, |x|) of each other give no answer (two points are too close together), and so do two exact candidates that close: x²(x − 10⁻⁷)² has critical numbers at 0, 5 × 10⁻⁸ and 10⁻⁷, which the page does not merge into one. More than 30 points gives no answer (too many points to check).

What you can type

  • A number has at most 15 digits in a row. A computer number keeps only about 16 digits, so a longer one (9007199254740993) would stand for a nearby number (9007199254740992), and the page asks for fewer digits instead. Write very large or very small numbers with a power of ten (1e-20).
  • The curves f and g use the variable x. Numbers can have decimals (2.5) and powers of ten (1e-3).
  • Operations: + − * / and ^ for powers. Brackets group. A number or bracket next to a letter multiplies: 2x, 3(x + 1), x sin(x).
  • Constants: pi (or π) and e.
  • Functions: sqrt, cbrt, ln (and log, the same natural logarithm), log10, exp, abs (or |x|), sin, cos, tan, sec, csc, cot, asin, acos, atan (arcsin, arccos, arctan also work), sinh, cosh, tanh and their inverses. Angles are in radians. sin x without brackets means sin(x); sin x^2 means sin(x²).
  • a and b are numbers or constant expressions (0, 2.5, pi, e), or both empty.

What gets no answer

  • Fewer than two crossings with a and b empty (eˣ and x never meet).
  • A pole of f − g between the ends, more than 30 crossings (sin x and cos x), or a turn of h that touches 0 without the algebra confirming it.
  • A step that fails its check, finds no formula, or takes over 3 seconds.

Worked examples by hand

9 − (x/2)² and 6 − x, between their crossings (OpenStax Calculus Volume 1, section 6.1, Example 6.2). 9 − x²/4 = 6 − x gives x² − 4x − 12 = (x − 6)(x + 2) = 0: the crossings are x = −2 and x = 6, with none between. The area is ∫ from −2 to 6 of (3 + x − x²/4) dx = [3x + x²/2 − x³/12] from −2 to 6 = 18 − (−10/3) = 64/3 = 21.33333333.

x + 4 and 3 − x/2 on [1, 4] (OpenStax Calculus Volume 1, section 6.1, Example 6.1). h(x) = (x + 4) − (3 − x/2) = 3x/2 + 1, positive on [1, 4], so there are no crossings. The area is [3x²/4 + x] from 1 to 4 = 16 − 7/4 = 57/4 = 14.25.

Other questions people ask

How do I find the area between two curves?

Integrate the top curve minus the bottom curve: area = ∫ from a to b of (f(x) − g(x)) dx when f ≥ g on [a, b]. When the curves cross inside [a, b], split the interval at each crossing and integrate |f(x) − g(x)| on each piece, so no part counts as negative area.

How do I find the limits of integration?

When the region is enclosed by the two curves, the limits are the x values where they meet: solve f(x) = g(x). For 9 − (x/2)² and 6 − x, 9 − x²/4 = 6 − x gives x² − 4x − 12 = 0, so x = −2 and x = 6. Leave a and b empty and the page does this.

What if the curves cross more than twice?

With a and b empty, the page uses the first and the last crossing as the limits and splits the interval at the crossings in between. For x³ and x it integrates from −1 to 0 and from 0 to 1 and adds the two areas: 1/4 + 1/4 = 1/2.

Why is the answer not the same as the integral of f − g?

The integral of f − g counts area where g is above f as negative, so the pieces can cancel: from −1 to 1, the integral of x³ − x is 0, while the area between the curves is 1/2. The page adds the size of each piece.

Can I find the area between curves given as x = f(y)?

Swap the letters: the area between x = √y and x = 2 − y is the same as the area between y = √x and y = 2 − x. Type them in x.

How is the answer checked?

Each piece is a definite integral from a computer algebra system, whose value is compared with a numeric integral of f − g. The crossings are found both by the algebra and by a sign scan on a fine grid, so none is missed.