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Area Between Curves
Functions of \( x \)

Master the technique of computing enclosed areas between two curves using definite integrals of functions of \( x \).

AP Calculus BC

What Does "Area Between Curves" Mean?

When two continuous functions \( f(x) \) and \( g(x) \) are defined on the same interval, they may enclose a bounded region. The "area between curves" is the total positive area of that region, regardless of whether the curves are above or below the \( x \)-axis.

x y f(x) g(x) f(x) - g(x) a b

Signed Area vs. Actual Area

A definite integral \( \int_a^b f(x)\,dx \) gives signed area: regions below the \( x \)-axis count as negative. But the area between two curves is always non-negative. We achieve this by always subtracting the lower curve from the upper curve.

Signed Area
\( \int_a^b f(x)\,dx \) can be positive, negative, or zero depending on where \( f \) lies relative to the \( x \)-axis.
Area Between Curves
\( \int_a^b [f(x) - g(x)]\,dx \) where \( f(x) \geq g(x) \) on \([a, b]\) is always \( \geq 0 \).

Why Does This Matter?

Computing the area between curves is one of the most fundamental applications of integration. It appears throughout physics, economics, and engineering, and is a core skill on the AP Calculus exam.

Physics
Work done by a variable force, net displacement from velocity curves
Economics
Consumer/producer surplus between supply and demand curves
Engineering
Cross-sectional area for volumes of revolution

The Core Formula

x y dx f(x) - g(x) f(x) = upper g(x) = lower a b

Think of it as summing infinitely many thin rectangular strips of height \( f(x) - g(x) \) and width \( dx \).

When Curves Cross

If the curves switch which is on top, you must split the integral at each intersection point. On each sub-interval, identify which function is greater and integrate accordingly.

f(x) g(x) f > g g > f f > g cross cross

The General Formula

In practice, you don't integrate the absolute value directly. Instead, you find all intersection points, then on each sub-interval you determine which function is larger and write the integrand as (upper) - (lower).

Area Under a Curve vs. Area Between Curves

Area Under a Curve Area Between Curves
Formula \( \int_a^b f(x)\,dx \) \( \int_a^b [f(x)-g(x)]\,dx \)
Reference The \( x \)-axis (\( g(x)=0 \)) Another function \( g(x) \)
Sign Can be negative (signed area) Always non-negative (use \( |f-g| \))
Special case When \( g(x) = 0 \) General: any two functions

How to Apply the Formula

1
Sketch the curves. Graph \( f(x) \) and \( g(x) \) to visualize the enclosed region.
2
Find intersection points. Solve \( f(x) = g(x) \) to determine the bounds \( a \) and \( b \).
3
Identify upper and lower. On each sub-interval, determine which function has larger values.
4
Set up the integral. Write \( \int_a^b [\text{upper} - \text{lower}]\,dx \).
5
Evaluate. Compute the antiderivative and apply the Fundamental Theorem of Calculus.

Example 1: Parabola and Line

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

x y -1 1 2 3 f(x) = x² g(x) = x+2 (-1, 1) (2, 4)

Example 1: Solution

1
Intersections: Solve \( x^2 = x + 2 \). We get \( x^2 - x - 2 = 0 \), so \( (x-2)(x+1) = 0 \), giving \( x = -1 \) and \( x = 2 \).
2
Which is upper? Test \( x = 0 \): \( g(0) = 2 \) and \( f(0) = 0 \). So \( g(x) = x+2 \) is on top.
3
Set up: \( A = \int_{-1}^{2} [(x+2) - x^2]\,dx \).
4
Antiderivative: \( \frac{x^2}{2} + 2x - \frac{x^3}{3} \).
5
Evaluate: \( \Bigl[\frac{x^2}{2}+2x-\frac{x^3}{3}\Bigr]_{-1}^{2} = \Bigl(2+4-\frac{8}{3}\Bigr) - \Bigl(\frac{1}{2}-2+\frac{1}{3}\Bigr) = \frac{10}{3} - \bigl(-\frac{7}{6}\bigr) = \frac{9}{2} \).

Example 2: Crossing Curves

Find the area between \( f(x) = x^3 \) and \( g(x) = x \) on \([-1, 1]\).

x y -1 1 x³ x f > g g > f

Example 2: Solution

The curves cross at \( x = -1, 0, 1 \). We must split the integral:

1
First piece: \( \int_{-1}^{0}(x^3-x)\,dx = \Bigl[\frac{x^4}{4}-\frac{x^2}{2}\Bigr]_{-1}^{0} = 0 - \bigl(\frac{1}{4}-\frac{1}{2}\bigr) = \frac{1}{4} \).
2
Second piece: \( \int_{0}^{1}(x-x^3)\,dx = \Bigl[\frac{x^2}{2}-\frac{x^4}{4}\Bigr]_{0}^{1} = \frac{1}{2}-\frac{1}{4} = \frac{1}{4} \).
3
Total: \( A = \frac{1}{4} + \frac{1}{4} = \) \( \frac{1}{2} \).

Example 3: Two Parabolas

Find the area enclosed between \( f(x) = 2x - x^2 \) and \( g(x) = x^2 - 4x + 4 \).

Common Mistakes to Avoid

✕ Subtracting in the Wrong Order
Always subtract lower from upper. If you get a negative answer, you likely swapped the functions.
✕ Ignoring Crossings
If you integrate straight through a crossing point, positive and negative areas cancel, giving a wrong (smaller) answer.
✕ Using the Wrong Bounds
The bounds come from solving \( f(x) = g(x) \), not from where the functions cross the \( x \)-axis.

Summary Cheat Sheet

Scenario Setup Key Action
\( f \geq g \) everywhere \( \int_a^b [f-g]\,dx \) Single integral
Curves cross once at \( c \) Split at \( x = c \) Two integrals
Curves cross multiple times Split at each crossing Sum all pieces
Bounds not given Solve \( f = g \) Find intersections first

Worked practice: Practice Problems, Problem 4 — applied

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

A civil engineer models the cross-section of a river channel. The river surface is at height y = 6 and the riverbed follows y = x² − 1 for −√7 ≤ x ≤ √7 (all measurements in meters). Set up and evaluate a definite integral to find the cross-sectional area of the water in the channel. Show all work, including finding the limits of integration, setting up the integral, and evaluating it. Then interpret the result in context.

Show answer and explanation

Point 1 (Limits of Integration): Set x² − 1 = 6 → x² = 7 → x = ±√7. The water fills the cross-section from x = −√7 to x = √7. Point 2 (Integral Setup): The top curve is y = 6 (the water surface) and the bottom curve is y = x² − 1 (the riverbed). The cross-sectional area is A = ∫₋√₇^√7 [6 − (x² − 1)] dx = ∫₋√₇^√7 (7 − x²) dx. Point 3 (Antidifferentiation and Evaluation): By the even symmetry of the integrand, A = 2 ∫₀^√7 (7 − x²) dx = 2 [7x − x³/3] from 0 to √7 = 2 [7√7 − (√7)³/3] = 2 [7√7 − 7√7/3] = 2 · 7√7 · (1 − 1/3) = 2 · 7√7 · (2/3) = 28√7/3 ≈ 24.69 square meters. Point 4 (Interpretation): The cross-sectional area of approximately 24.69 m² represents the area of water visible in a vertical slice perpendicular to the river's flow. This value can be multiplied by the river's flow speed to estimate volumetric discharge rate (m³/s), a key quantity in hydrology and flood management.

Parabolic bed y=x²−1 under waterline y=6, between x=±√7. The strip at x=1 has height 6; area is 28√7/3 m².
The channel’s level water surface and parabolic bed enclose area 28√7/3 m² in the stated cross-section. The water face is an explanatory cutaway, not an unsupported real water wall.

Question 1 of 5

What is the area between \( f(x) = x^2 \) and \( g(x) = x \) on \([0, 1]\)?

Question 2 of 5

To find the area between two curves that cross within the interval, you must:

Question 3 of 5

Find the area enclosed by \( f(x) = 4 - x^2 \) and \( g(x) = 0 \) (the \( x \)-axis).

Question 4 of 5

If \( \int_1^4 [f(x) - g(x)]\,dx = -5 \), which of the following must be true on \([1, 4]\)?

Question 5 of 5

Which integral represents the area between \( y = \sin x \) and \( y = \cos x \) on \(\bigl[\frac{\pi}{4}, \frac{5\pi}{4}\bigr]\)?