Area Between Curves Calculator

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Integral#11

Area Between Two Curves

\(A = \int_a^b |f(x) - g(x)|\,dx\), split automatically at intersections.

This area between curves calculator finds the area of the region enclosed by two graphs \( y = f(x) \) and \( y = g(x) \) over an interval \( [a, b] \). It locates every point where the curves cross, splits the integral there, and adds up the pieces so that no part of the area cancels out.

Use it to check textbook answers, confirm your intersection points, or get the area when the integral is messy. It is free and needs no sign-up.

How to use the area between curves calculator

  1. Enter the first curve in the \( f(x) \) box and the second in \( g(x) \). The order does not matter, because the calculator uses \( |f - g| \).
  2. Type functions with ^ for powers, sqrt(), sin(), ln() and e^(x). Implicit products such as 2x and x sin(x) are fine.
  3. Enter the left and right limits \( a \) and \( b \). If the problem says “the region bounded by the curves”, these are usually the outermost intersection points; solve \( f(x) = g(x) \) first, or plot the curves in the graphing calculator to see roughly where they cross.
  4. Press Compute area. You get any crossing points strictly inside \( (a, b) \) and the total area.

Formula

The area between two curves is the integral of the vertical distance between them:

$$A = \int_a^b \lvert f(x) - g(x) \rvert\,dx$$

On a stretch where \( f \) stays above \( g \), this is just \( \int (f - g)\,dx \). When the curves cross, “top” and “bottom” swap. The calculator finds the crossing points \( c_1 < c_2 < \dots \), integrates \( f - g \) on each piece, and adds the absolute values:

$$A = \left\lvert \int_a^{c_1} (f - g)\,dx \right\rvert + \left\lvert \int_{c_1}^{c_2} (f - g)\,dx \right\rvert + \cdots$$

Worked examples

Example 1. Find the area enclosed by \( y = x + 6 \) and \( y = x^2 \).

Set them equal: \( x^2 = x + 6 \), so \( x^2 - x - 6 = (x - 3)(x + 2) = 0 \), giving \( x = -2 \) and \( x = 3 \). Testing \( x = 0 \) shows the line (6) is above the parabola (0). The area is

$$\begin{aligned}A &= \int_{-2}^{3} (x + 6 - x^2)\,dx \\ &= \left[\tfrac{x^2}{2} + 6x - \tfrac{x^3}{3}\right]_{-2}^{3} \\ &= \tfrac{27}{2} - \left(-\tfrac{22}{3}\right) = \tfrac{125}{6}\end{aligned}$$

That is about 20.8333. Enter x + 6, x^2, -2 and 3 in the calculator above and you get \( \frac{125}{6} \).

Example 2 (curves that cross). Find the area between \( y = x^3 \) and \( y = x \) on \( [-1, 2] \).

The curves meet where \( x^3 = x \), at \( x = -1, 0, 1 \). Only 0 and 1 are inside the interval, and the calculator reports exactly those. Integrating piece by piece:

  • On \( [-1, 0] \): \( \int (x^3 - x)\,dx = \frac14 \)
  • On \( [0, 1] \): \( \int (x - x^3)\,dx = \frac14 \)
  • On \( [1, 2] \): \( \int (x^3 - x)\,dx = \frac94 \)

Total area: \( \frac14 + \frac14 + \frac94 = \frac{11}{4} = 2.75 \). If you integrated \( x^3 - x \) straight from -1 to 2 without splitting, you would get \( \frac94 \), because the middle piece counts as negative and cancels one of the others.

Common mistakes

  • Forgetting to split at intersections. A single integral of \( f - g \) gives signed area, not total area. This is the most common reason a hand answer is too small.
  • Subtracting in the wrong order. Top minus bottom gives a positive integrand. The calculator’s absolute value protects you, but on paper, check which curve is higher with a test point.
  • Using the wrong limits. For a bounded region, the limits come from solving \( f(x) = g(x) \), not from the graph window.
  • Curves given as \( x = h(y) \). Some regions are easier sliced horizontally. Rewrite each curve as a function of \( x \), or swap the roles of \( x \) and \( y \) and use the calculator with \( y \) renamed to \( x \).

For a longer walkthrough of setting up these integrals, including horizontal slices, see our guide to finding the area between two curves.

Related guides

Further reading

FAQ

Does the calculator find where the curves intersect?

Yes, for crossings strictly between \( a \) and \( b \). It lists those x-values and splits the integral at each one. For a bounded region, find the outer intersection points yourself and use them as \( a \) and \( b \).

Does it matter which function I enter as f and which as g?

No. The area uses \( |f(x) - g(x)| \), so swapping them gives the same answer.

Can the area between curves be negative?

No. Area is always positive. A negative number means a signed integral was used without absolute values or splitting.

How do I find the area between a curve and the x-axis?

Enter your function as \( f(x) \) and 0 as \( g(x) \). The calculator then gives the total area between the graph and the axis, counting parts below the axis as positive. For signed area instead, use the definite integral calculator.

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