Area Between Curves

The area between two curves is the integral of the top one minus the bottom one, as long as they do not swap places. Where they cross, the difference changes sign and the integral starts subtracting area it should add. The page finds the crossings, splits there, and prints the area and the plain integral side by side.

Area between two curves	$A = \int_{a}^{b} f (x) - g (x) d x$
Where they cross at c	$\int_{a}^{b} f - g d x = \int_{a}^{c} (g - f) d x + \int_{c}^{b} (f - g) d x$

Worked example — Area Between Curves

Area, split at the crossings	0.5
The integral of the difference, unsplit	0
Crossings inside the interval	1

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Area Between Curves

The area between two curves f and g on [a, b] is ∫ₐᵇ |f(x) − g(x)| dx. Where one curve stays on top it is ∫(top − bottom) dx; where they cross, the integral is split at the crossings and each piece taken as top minus bottom.

What each symbol means

f, g: the two curves, functions of x, or of y when the region is sliced the other way

Where the formula comes from

Find where they meet. The crossings are where f − g changes sign inside the interval, so the curve on top changes there. Solve f = g to find them; a point where the curves only touch is not one.
$f (x) = g (x) \Rightarrow x = c_{1}, \dots, c_{k}$
Integrate top minus bottom on each piece. Between neighbouring crossings one curve is on top, so the integrand there is positive and the integral is that piece's area.
$A = i \sum \int_{c_{i}}^{c_{i + 1}} (f (x) - g (x)) d x$

Problems solved in full

A parabola and a line 2 steps
Find the area between y = x² and y = x + 2.

Open this problem in the tool
1. They meet where x² = x + 2:
  
  $x^{2} = x + 2 \Rightarrow (x - 2) (x + 1) = 0 \Rightarrow x = - 1, 2$
2. Between −1 and 2 the line is on top:
  
  $\int_{- 1}^{2} (x + 2 - x^{2}) d x = [\frac{x ^{2}}{2} + 2 x - \frac{x ^{3}}{3}]_{- 1}^{2} = \frac{10}{3} + \frac{7}{6} = \frac{9}{2}$
Answer

9/2 = 4.5. Check the order with one point: at x = 0 the line is at 2 and the parabola at 0.

$A = \frac{9}{2} = 4.5$
Curves that cross inside the interval 2 steps
Find the area between y = x³ and y = x on [−1, 1].

Open this problem in the tool
Show solution
1. They meet at x = −1, 0 and 1, and swap places at 0:
  
  $x^{3} = x \Rightarrow x = - 1, 0, 1$
2. Each piece with the curve on top first:
  
  $\int_{- 1}^{0} (x^{3} - x) d x + \int_{0}^{1} (x - x^{3}) d x = \frac{1}{4} + \frac{1}{4}$
Answer

½. The unsplit integral of x − x³ over [−1, 1] is 0, which is the warning in the insight block on a zero integral.

$A = \frac{1}{2}$
Two trigonometric curves 2 steps
Find the area between y = sin x and y = cos x on [0, π/2].

Open this problem in the tool
Show solution
1. They cross where tan x = 1, at π/4; cos x is on top before it:
  
  $sin x = cos x \Rightarrow x = \frac{π}{4}$
2. Both pieces:
  
  $\int_{0}^{π /4} (cos x - sin x) d x + \int_{π /4}^{π /2} (sin x - cos x) d x = (2 - 1) + (2 - 1)$
Answer

2√2 − 2 ≈ 0.8284. The two pieces are equal, √2 − 1 each, because the region is symmetric about x = π/4.

$A = 22 - 2 \approx 0.8284$
Slicing in y 2 steps
Find the area between x = y² and x = y + 2.

Open this problem in the tool
Show solution
1. They meet where y² = y + 2, at y = −1 and y = 2; the line is to the right:
  
  $y^{2} = y + 2 \Rightarrow y = - 1, 2$
2. One integral in y:
  
  $\int_{- 1}^{2} (y + 2 - y^{2}) d y = \frac{9}{2}$
Answer

9/2 = 4.5. The same number as the first problem, because this region is that one reflected in the line y = x.

$A = \frac{9}{2} = 4.5$
Curves that meet only at the ends 1 steps
Find the area between y = √x and y = x² on [0, 1].

Open this problem in the tool
Show solution
1. √x is on top between 0 and 1:
  
  $\int_{0}^{1} (x - x^{2}) d x = [\frac{2}{3} x^{3/2} - \frac{x ^{3}}{3}]_{0}^{1} = \frac{2}{3} - \frac{1}{3}$
Answer

1/3. Each curve is the other reflected in y = x, so the line y = x cuts this area into two halves of 1/6.

$A = \frac{1}{3}$
Curves meeting at three points 2 steps
Find the total area enclosed by y = x³ − x and y = 3x.
Show solution
1. They meet where x³ − 4x = 0, at x = −2, 0 and 2:
  
  $x^{3} - x = 3 x \Rightarrow x^{3} - 4 x = 0 \Rightarrow x = - 2, 0, 2$
2. The pieces are the same shape turned half a turn about the origin, so double the right one, where 3x is on top:
  
  $A = 2 \int_{0}^{2} (4 x - x^{3}) d x = 2 [2 x^{2} - \frac{x ^{4}}{4}]_{0}^{2} = 2 (8 - 4)$
Answer

8. The unsplit integral over [−2, 2] would again be 0.

$A = 8$

The integral of the difference can be zero for a region with plenty of area 🖖

Press x³ and x: zero, yet area. On [−1, 1] the two curves cross at 0. Left of it x³ is on top, right of it x is, and the two lobes are the same shape turned half a turn about the origin, each 0.25. The integral of x − x³ counts the right lobe as positive and the left one as negative and arrives at 0. The area is 0.5. Press sin and cos, a full turn for the same thing on a bigger scale: an area of 4√2 ≈ 5.6569 and an integral of exactly 0. The integral of f − g measures how much f exceeds g on balance; the area asks for |f − g|, and to integrate an absolute value you split it where the inside changes sign.

The same region can take one integral or two, depending on which way you slice it 🖖

Press x = y² and x = y + 2. Sliced across y, every strip runs from the parabola on the left to the line on the right, so one integral does it: ∫ from −1 to 2 of (y + 2 − y²) dy = 4.5. Slice the same region across x instead and the bottom of each strip changes partway: below x = 1 both edges are on the parabola, y = ±√x, and from 1 to 4 the bottom is the line. Two integrals, √x-shaped ones, and the same 4.5. Before choosing the variable, look at which boundary each strip would hit.

Archimedes found this area without integrals: four thirds of a triangle 🖖

Press Archimedes' parabola. The line x + 2 cuts the parabola at (−1, 1) and (2, 4), and the area between them is 4.5. Draw the triangle on that chord whose third corner is where the parabola's tangent runs parallel to it, at (½, ¼): its area is 3.375, and 4.5 is exactly 4/3 of it. Archimedes proved this around 250 BC, in The Quadrature of the Parabola, by filling the segment with triangles, each generation a quarter of the one before, and summing 1 + ¼ + 1/16 + … = 4/3. The integral reaches the same number in one line; his method needed no calculus, because there was none.

Learning path

Applications of integration

References (2)

The method as the page uses it: Theorem 6.1 (∫ₐᵇ [f − g] dx when f ≥ g), Theorem 6.2 (∫ₐᵇ |f − g| dx for curves that cross, split at the intersections) and Theorem 6.3 (∫꜀ᵈ [u(y) − v(y)] dy along the y-axis): G. Strang and E. Herman, Calculus Volume 1, OpenStax (2016), section 6.1, “Areas between Curves”.
Insight block 3: Proposition 24, “Every segment bounded by a parabola and a chord Qq is equal to four-thirds of the triangle which has the same base as the segment and equal height” (p. 251), proved through Proposition 23's sum of areas each a quarter of the one before (p. 250): T. L. Heath (ed.), The Works of Archimedes, Cambridge University Press (1897), “Quadrature of the Parabola”, Propositions 23–24, pp. 250–251.

By the 42Math team · Reviewed against published sources. Corrections welcome.