Area Between Curves Calculator

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. As a result, 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:

$$A = leftlvert int_a^{c_1} (f - g),dx rightrvert + leftlvert int_{c_1}^{c_2} (f - g),dx rightrvert + cdots$$

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 ). However, 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 )
• Next, 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. As a result, 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. Therefore, 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.

Integrating with respect to y

Sometimes it is easier to slice horizontally. For the region between \( x = y^2 \) and \( x = y + 2 \), the curves meet where \( y^2 = y + 2 \), at \( y = -1 \) and \( y = 2 \). So, the right curve minus the left curve is \( (y + 2) - y^2 \):

$$A = \int_{-1}^{2} (y + 2 - y^2)\,dy = \left[\frac{y^2}{2} + 2y - \frac{y^3}{3}\right]_{-1}^{2} = \frac{10}{3} + \frac{7}{6} = 4.5$$

Slicing vertically would need two separate integrals here.

Step-by-step method

1. Sketch both curves and find where they intersect.
2. Decide which is on top (or on the right) on each part of the interval.
3. Integrate top minus bottom (or right minus left) over each part.
4. Add the parts; the result must be positive.

Where it is used

In economics, the area between demand and price curves measures consumer surplus, the total benefit buyers get above what they pay; in physics, the area between two velocity graphs gives how far apart two objects end up.

Practice problems

1. Area between \( y = x \) and \( y = x^2 \).
2. Area between \( y = 4 - x^2 \) and the x-axis.

Answers: 1. They meet at 0 and 1: \( \int_0^1 (x - x^2)\,dx = \frac{1}{6} \). 2. \( \int_{-2}^{2} (4 - x^2)\,dx = \frac{32}{3} \approx 10.67 \).

Interpreting Your Calculator Result

The area between curves calculator provides two main pieces of information. It gives the total area as a positive numerical value. It also lists any x-values where the curves cross within your specified interval.

Always confirm the total area is positive. Because area is a physical quantity, it cannot be negative. If you were doing this by hand and got a negative number, you likely subtracted the functions in the wrong order or missed splitting the integral.

The listed intersection points are important. They show where the 'top' and 'bottom' functions swap positions. Therefore, they confirm the calculator correctly split the integral into pieces and applied the absolute value to each.

Related guides

• Area Between Two Curves: Step-by-Step Method and Examples
• Fundamental Theorem of Calculus (Parts 1 and 2) Explained
• Riemann Sums: Left, Right, Midpoint and Simpson’s Rule

Further reading

• Area Between Curves (Paul’s Online Math Notes) — more examples, including regions that are easier to slice horizontally.
• Computing Definite Integrals (Paul’s Online Math Notes) — a refresher on evaluating each piece with the Fundamental Theorem.

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, however, 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. However, 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.