Area Between Two Curves: Step-by-Step Method and Examples

The area between two curves \( y = f(x) \) (top) and \( y = g(x) \) (bottom) from \( x = a \) to \( x = b \) is

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

“Top minus bottom,” integrated over the interval. This guide shows you why the formula works, how to set it up every time, what to do when the curves cross, and when it’s smarter to integrate with respect to \( y \).

Why it works

Slice the region into thin vertical strips. A strip at position \( x \) has width \( \Delta x \) and height equal to the gap between the curves, \( f(x) - g(x) \). Its area is about \( \big(f(x) - g(x)\big)\Delta x \).

Adding up all the strips gives a Riemann sum. As the strips get thinner, the sum becomes the definite integral, and the Fundamental Theorem of Calculus lets you evaluate it with an antiderivative.

Notice that only the gap matters, not where the curves sit. If you shift both curves down by 10 units, every strip keeps the same height. That’s why the formula still works when part of the region is below the x-axis: top minus bottom is positive as long as you have the order right.

The familiar “area under a curve” is just a special case. When the bottom curve is the x-axis, \( g(x) = 0 \), and the formula reduces to \( \int_a^b f(x)\,dx \). The difference is that a plain definite integral counts area below the axis as negative, while the area between curves never does, because you always subtract the lower curve from the higher one. Keeping that distinction clear avoids most sign errors in this topic.

Step-by-step method

1. Sketch the curves (or use the visual lab).
2. Find the intersection points by solving \( f(x) = g(x) \). These are often the limits of integration.
3. Decide which curve is on top on each interval (test a point).
4. Integrate top minus bottom on each interval and add the results.

Example 1: y = x and y = x²

Intersections: \( x = x^2 \Rightarrow x = 0, 1 \). On \( (0,1) \), \( x > x^2 \) (test \( x = \frac12 \)).

$$A = \int_0^1(x - x^2)\,dx = \frac12 - \frac13 = \frac16$$

Example 2: a parabola and a line

\( y = 2 - x^2 \) and \( y = x \). Solve \( 2 - x^2 = x \): \( x^2 + x - 2 = 0 \), so \( x = -2, 1 \). The parabola is on top (test \( x = 0 \): \( 2 > 0 \)).

$$\begin{aligned} A &= \int_{-2}^{1}(2 - x - x^2)\,dx \\ &= \left[2x - \frac{x^2}{2} - \frac{x^3}{3}\right]_{-2}^{1} \\ &= \frac76 - \left(-\frac{10}{3}\right) = \frac92 \end{aligned}$$

Example 3: curves that cross

\( y = x^3 \) and \( y = x \) on \( [-1, 1] \). They cross at \( x = 0 \), and the top curve switches:

• On \( [-1, 0] \): \( x^3 \ge x \), area \( \int_{-1}^0(x^3 - x)\,dx = \frac14 \)
• On \( [0, 1] \): \( x \ge x^3 \), area \( \int_0^1(x - x^3)\,dx = \frac14 \)

Total area \( \frac12 \). If you integrate \( x^3 - x \) straight across \( [-1, 1] \) you get 0: the two pieces cancel. Area requires splitting at crossings (or integrating the absolute value).

Example 4: trig curves

\( \sin x \) and \( \cos x \) on \( [0, \pi] \) cross where \( \tan x = 1 \), at \( x = \frac\pi4 \). Cosine is on top before the crossing (test \( x = 0 \)) and sine is on top after it (test \( x = \frac\pi2 \)):

$$\int_0^{\pi/4}(\cos x - \sin x)\,dx = \sqrt2 - 1$$

$$\int_{\pi/4}^{\pi}(\sin x - \cos x)\,dx = \sqrt2 + 1$$

The total area is \( 2\sqrt2 \approx 2.828 \).

Example 5: a region partly below the axis

Find the area between \( y = x^2 - 4 \) and \( y = 2x - 1 \).

Set them equal: \( x^2 - 4 = 2x - 1 \), so \( x^2 - 2x - 3 = 0 \) and \( x = -1, 3 \). Test \( x = 0 \): the line gives \( -1 \) and the parabola gives \( -4 \), so the line is on top.

$$\begin{aligned} A &= \int_{-1}^{3}\big((2x - 1) - (x^2 - 4)\big)\,dx \\ &= \int_{-1}^{3}(-x^2 + 2x + 3)\,dx \\ &= 9 - \left(-\frac53\right) = \frac{32}{3} \end{aligned}$$

Most of this region lies below the x-axis, yet you never needed to think about that. Top minus bottom handles it automatically.

Example 6: bounded by a vertical line

Find the area between \( y = x \) and \( y = \frac1x \) from where they meet to the line \( x = 3 \).

The curves meet where \( x = \frac1x \), so \( x = 1 \) (for \( x > 0 \)). At \( x = 2 \), \( 2 > \frac12 \), so \( y = x \) is on top:

$$A = \int_1^3\left(x - \frac1x\right)dx = \left[\frac{x^2}{2} - \ln x\right]_1^3 = 4 - \ln 3$$

That’s about 2.901 square units. A vertical line like \( x = 3 \) simply supplies one of the limits of integration.

Integrating with respect to y

Sometimes “right minus left” in \( y \) is easier. Region between \( x = y^2 \) and \( x = y + 2 \): they meet at \( y = -1, 2 \), and the line is on the right:

$$A = \int_{-1}^{2}\big((y + 2) - y^2\big)\,dy = \frac92$$

In \( x \) this same region would need two separate integrals, because the bottom boundary changes formula partway.

Switch to \( y \) when the region has one clear left boundary and one clear right boundary, when the curves are easier to write as \( x = \) something, or when integrating in \( x \) would force you to split the region.

Common mistakes

• Subtracting in the wrong order. Bottom minus top gives a negative number. Always test a point between the intersections to confirm which curve is on top.
• Not splitting at a crossing. If the curves cross inside \( [a, b] \), one integral over the whole interval lets positive and negative pieces cancel, as Example 3 showed. Find every crossing first.
• Using the wrong limits. When the problem says “the region enclosed by the curves,” the limits are the intersection points, not numbers you pick.
• Mixing variables. If you integrate with respect to \( y \), both boundaries and both limits must be in terms of \( y \).
• Forgetting a boundary. Regions can be bounded by the x-axis (\( y = 0 \)) or a vertical line. Include it in the setup.

Where it’s used

The same setup is the first step for many other applications:

• Volumes of revolution. Rotate the region and the top and bottom curves become outer and inner radii in the disk and washer method, or heights in the shell method.
• Average gap between two quantities. Divide the area by the interval length to get the average value of \( f - g \).
• Accumulated difference. If two curves are rates, like income and spending per month, the area between them is the total difference over the period.

Area between curves calculator

The gadget finds the crossing points itself and splits the integral:

For a single integral you’ve already set up, the definite integral calculator evaluates it directly, and the full area between curves calculator page has more examples.

Practice problems

Try these before checking the answers.

1. \( y = x^2 \) and \( y = 2x \)
2. \( y = \sqrt x \) and \( y = x \)
3. \( y = 4 - x^2 \) and the x-axis
4. \( y = x^2 \) and \( y = x^3 \) on \( [0, 1] \)
5. \( y = \sin x \) and the x-axis on \( [0, \pi] \)
6. \( y = \sqrt x \), \( y = x - 2 \) and the x-axis (hint: integrate in \( y \), or split in \( x \))

Answers: (1) \( \frac43 \); (2) \( \frac16 \); (3) \( \frac{32}{3} \); (4) \( \frac{1}{12} \); (5) \( 2 \); (6) \( \frac{10}{3} \).

For problem 6, in \( y \) the region runs from \( y = 0 \) to \( y = 2 \), with left boundary \( x = y^2 \) and right boundary \( x = y + 2 \). In \( x \) you would need two integrals, because the bottom boundary is the x-axis up to \( x = 2 \) and the line after that.

FAQ

Can the area between curves be negative?

No. Area is always positive. A negative result means top and bottom were swapped, or the curves crossed and you didn’t split.

How do I find the limits of integration?

If the problem gives an interval, use it, but still check for crossings inside it. If it says “the region enclosed by,” solve \( f(x) = g(x) \) and use the smallest and largest solutions.

When should I integrate with respect to y?

When the curves are naturally written as \( x = \) something, or when the region’s top or bottom boundary changes formula partway but its left and right boundaries don’t.

What if the curves intersect more than twice?

Split the interval at every intersection point, find the top curve on each piece, and add the areas. Integrating \( |f(x) - g(x)| \) does the same thing in one expression.

How is this related to volume?

Rotate the region and you get a washer or shell volume problem. The same top/bottom functions become the radii or heights.

Further reading

• Area Between Curves (Paul’s Online Math Notes) — seven more worked examples, including regions that need two integrals.
• Computing Definite Integrals (Paul’s Online Math Notes) — a refresher on evaluating the integrals, including absolute values and piecewise functions.