Area Between Curves
Functions of \( x \)

Learn to compute the exact area enclosed between two curves using definite integrals and the "top minus bottom" approach.

AP Calculus AB

What Does "Area Between Curves" Mean?

When two functions \( f(x) \) and \( g(x) \) are graphed on the same axes, they may enclose a bounded region. The area of that region is what we want to compute: the total space trapped between the two curves from one intersection to the other.

From One Curve to Two Curves

You already know that \( \int_a^b f(x)\,dx \) gives the signed area between \( f(x) \) and the \( x \)-axis. The area between two curves is a natural extension: instead of measuring height from the axis, we measure the vertical distance between the curves.

Area under one curve

\( \displaystyle\int_a^b f(x)\,dx \) measures from the \( x \)-axis up to \( f(x) \).

Area between two curves

\( \displaystyle\int_a^b [f(x) - g(x)]\,dx \) measures from \( g(x) \) up to \( f(x) \).

Why This Matters

Computing area between curves appears throughout mathematics, physics, economics, and engineering. Any time two quantities vary with respect to the same variable, the integral of their difference tells you the net accumulation between them.

Physics

Work done between two force functions over a displacement interval.

Economics

Consumer or producer surplus between supply and demand curves.

AP Exam

A core tested skill in both the AB and BC curricula, often with calculator and non-calculator parts.

The Core Formula

The integrand \( f(x) - g(x) \) represents the vertical height of each thin rectangle. Summing these rectangles from \( a \) to \( b \) gives the total area.

What If Curves Switch Position?

Sometimes the "top" and "bottom" curves switch at a point inside the interval. When this happens, you must split the integral at every crossing point so you always subtract the lower function from the upper one.

Signed Area vs. Geometric Area

	Signed Area (Net)	Geometric Area
Formula	\( \int_a^b [f(x) - g(x)]\,dx \)	\( \int_a^b \|f(x) - g(x)\|\,dx \)
Can be negative?	Yes	Never
Requires splitting?	Not required	Yes, at crossings
AP Exam asks for	Occasionally	Almost always

How to Apply the Formula

1

Sketch (or visualize) both curves. Identify which function is on top and which is on the bottom over the interval.

2

Find the intersection points. Set \( f(x) = g(x) \) and solve. These give your limits \( a \) and \( b \) (or interior split points).

3

Set up the integral: \( \int_a^b [\text{top} - \text{bottom}]\,dx \). If curves switch, split into separate integrals.

4

Evaluate each integral using antiderivatives and the Fundamental Theorem of Calculus.

5

Add the results from each sub-interval to get the total area.

Example 1: Parabola and Line

Find the area enclosed between \( f(x) = x + 2 \) and \( g(x) = x^2 \).

Example 1: Computing the Integral

Step 2: On \([-1, 2]\), test \( x = 0 \): \( f(0) = 2 \) and \( g(0) = 0 \), so \( f(x) \geq g(x) \). Top is \( f \), bottom is \( g \).

Example 2: Curves That Cross

Find the area between \( f(x) = x^3 \) and \( g(x) = x \) from \( x = -1 \) to \( x = 1 \).

Example 2: Evaluating Both Integrals

Example 3: Given Limits

Find the area between \( f(x) = \sin x \) and \( g(x) = 0 \) (the \( x \)-axis) on \([0, \pi]\).

Given Limits vs. Finding Limits

Problems come in two flavors. Recognizing which type you face helps you allocate your time wisely.

Limits given

The problem tells you \([a, b]\). You still need to determine which curve is on top, and whether they cross inside the interval.

Limits not given

You must solve \( f(x) = g(x) \) to find the intersection points. These become your limits of integration.

Common Mistakes

✕ Wrong order of subtraction

Writing \( g(x) - f(x) \) when \( f(x) \) is on top gives a negative value: not a valid area.

✕ Ignoring crossings

Integrating straight through a crossing point without splitting cancels area and yields a number that is too small.

✕ Using the wrong limits

Using \( x \)-intercepts of one function instead of the intersection points of the two functions.

Summary Checklist

Step	Action	Watch Out For
1. Sketch	Graph both curves	Identify top vs. bottom
2. Intersections	Solve \( f(x) = g(x) \)	Interior crossings
3. Set up	\( \int [\text{top} - \text{bottom}]\,dx \)	Split if curves swap
4. Evaluate	Antidifferentiate + FTC	Arithmetic errors
5. Verify	Answer must be positive	Negative = mistake

Worked practice: Section 9, Practice Problem 4 — Applied

Work through this original lesson problem, then reveal the solution and illustrations.

A civil engineer models the cross-section of a drainage channel. The bottom of the channel follows the curve y = x²/4 (in meters) and the water surface is at y = 3 (a horizontal line), where x is the horizontal distance from the center. Set up and evaluate the integral that gives the cross-sectional area of water in the channel.

Show answer and explanation

Point 1 (1 pt): Find intersection points. Set x²/4 = 3, so x² = 12, giving x = −2√3 and x = 2√3. Point 2 (1 pt): Set up the integral correctly. The water surface y = 3 is above the channel bottom y = x²/4 on [−2√3, 2√3], so A = ∫ from −2√3 to 2√3 of (3 − x²/4) dx. Point 3 (1 pt): Evaluate the integral. Using symmetry: A = 2∫₀^{2√3} (3 − x²/4) dx = 2[3x − x³/12] from 0 to 2√3 = 2[3(2√3) − (2√3)³/12] = 2[6√3 − (24√3)/12] = 2[6√3 − 2√3] = 2(4√3) = 8√3 ≈ 13.86 square meters.

Water above the parabolic bed y=x²/4 and below y=3, between x=−2√3 and x=2√3. A vertical slice at x=2 has height 2; total area is 8√3 m². — The water cross-section lies between y=3 and the bed y=x²/4; vertical slice heights integrate to 8√3 m². The water face is an explanatory cutaway, not an unsupported real water wall.

Question 1 of 5

What is the area between \( f(x) = 4 - x^2 \) and the \( x \)-axis on \([-2, 2]\)?

The parabola is above the \( x \)-axis on \([-2,2]\). So \( A = \int_{-2}^{2}(4 - x^2)\,dx = \left[4x - \frac{x^3}{3}\right]_{-2}^{2} = \frac{16}{3} - \left(-\frac{16}{3}\right) = \frac{32}{3} \).

Question 2 of 5

To find the area between \( y = x^2 \) and \( y = 2x \), what are the correct limits of integration?

Set \( x^2 = 2x \), so \( x^2 - 2x = 0 \), giving \( x(x - 2) = 0 \). The intersection points are \( x = 0 \) and \( x = 2 \).

Question 3 of 5

If \( f(x) \geq g(x) \) on \([a, c]\) and \( g(x) \geq f(x) \) on \([c, b]\), which expression gives the total area?

When curves switch, you split at the crossing and add the two positive areas. On each sub-interval, subtract the bottom from the top.

Question 4 of 5

A student computes \( \displaystyle\int_0^{2\pi}\sin x\,dx = 0 \) and reports the area between \( \sin x \) and the \( x \)-axis is 0. What went wrong?

The integral \( \int_0^{2\pi}\sin x\,dx = 0 \) because the positive area on \([0,\pi]\) cancels the negative area on \([\pi, 2\pi]\). To get geometric area, split at \( \pi \) and take absolute values: \( A = 2 + 2 = 4 \).

Question 5 of 5

What is the area enclosed between \( y = x^2 \) and \( y = 2x \)?

Intersections at \( x = 0 \) and \( x = 2 \). On \([0,2]\), \( 2x \geq x^2 \). So \( A = \int_0^2(2x - x^2)\,dx = \left[x^2 - \frac{x^3}{3}\right]_0^2 = 4 - \frac{8}{3} = \frac{4}{3} \).

Area Between CurvesFunctions of \( x \)

What Does "Area Between Curves" Mean?

From One Curve to Two Curves

Why This Matters

The Core Formula

What If Curves Switch Position?

Signed Area vs. Geometric Area

How to Apply the Formula

Example 1: Parabola and Line

Example 1: Computing the Integral

Example 2: Curves That Cross

Example 2: Evaluating Both Integrals

Example 3: Given Limits

Given Limits vs. Finding Limits

Common Mistakes

Summary Checklist

Worked practice: Section 9, Practice Problem 4 — Applied

Question 1 of 5

Question 2 of 5

Question 3 of 5

Question 4 of 5

Question 5 of 5

Area Between Curves
Functions of \( x \)