Difficulty: Intermediate | Prerequisites: Calc 1 (antiderivatives, definite integrals, Fundamental Theorem of Calculus)
This topic sits right at the start of Calc 2, building directly on definite integrals from Calc 1. You are applying the Fundamental Theorem of Calculus to geometric problems: instead of finding the area under a single curve, you find the area trapped between two curves. It matters because it is the foundation for later topics like volumes of revolution, arc length, and applications in physics and probability. You should be comfortable evaluating definite integrals and finding intersection points of functions before working through this material.
To find the area between two curves, integrate the difference of the "top" function minus the "bottom" function across the interval where they overlap. You can slice vertically (integrate with respect to x) or horizontally (integrate with respect to y), depending on whichever setup is simpler. The key steps are always: find the intersection points, determine which function is on top, set up the integral, and evaluate.
Definite integral
The limit of Riemann sums; it computes a net "accumulated quantity" between two bounds. Formally: the integral from a to b of f(x) dx = F(b) - F(a), where F is any antiderivative of f.
In simple terms, it gives you the total area (signed) under a curve between two x-values.
Riemann sum
An approximation of a definite integral formed by slicing the region into thin rectangles and adding their areas. The definite integral is the limit of this sum as the number of rectangles approaches infinity.
Think of it as chopping up a shape into thin strips, measuring each one, and adding them up.
Area between curves
If f(x) >= g(x) on [a, b], the area of the region trapped between the two curves is the integral from a to b of [f(x) - g(x)] dx.
In simple terms, you subtract the lower curve from the upper curve and integrate.
Vertical integration (slicing with respect to x)
Setting up the integral as dx slices, where you integrate along the x-axis. Each slice is a thin vertical rectangle whose height is [top function] minus [bottom function].
Horizontal integration (slicing with respect to y)
Setting up the integral as dy slices, where you integrate along the y-axis. Each slice is a thin horizontal rectangle whose width is [right function] minus [left function]. You express both curves as x = f(y) and x = g(y).
Think of it as turning the problem on its side.
Intersection points
The x-values (or y-values) where two curves meet. You find them by setting f(x) = g(x) and solving. These become the limits of integration.
Fundamental Theorem of Calculus (FTC)
The theorem linking differentiation and integration: if F'(x) = f(x), then the integral from a to b of f(x) dx = F(b) - F(a). This is the engine that lets you evaluate definite integrals by finding antiderivatives.
A definite integral is defined as the limit of a Riemann sum: integral from a to b of f(x) dx = lim (n to infinity) of the sum of f(x_i*) times delta x.
By the Fundamental Theorem of Calculus: integral from a to b of f(x) dx = F(b) - F(a), where F'(x) = f(x).
Definite integrals compute bulk quantities: area, volume, mass, population, probability, and work.
When f(x) >= g(x) on [a, b], the area between the curves is:
Area = integral from a to b of [f(x) - g(x)] dx
The steps:
Identify the top function f(x) and the bottom function g(x) on the interval.
Find the intersection points by solving f(x) = g(x). These become your limits a and b.
Integrate [f(x) - g(x)] from a to b.
Example 1: f(x) = 6x - x², g(x) = 0.
Intersection: 6x - x² = 0, so x(6 - x) = 0, giving x = 0 and x = 6.
Area = integral from 0 to 6 of (6x - x²) dx = [3x² - x³/3] from 0 to 6 = 108 - 72 = 36.
Example 2: f(x) = -x² + 4x + 5, g(x) = 2x + 5.
Intersection: 2x + 5 = -x² + 4x + 5, so 0 = -x² + 2x, giving x(x - 2) = 0, so x = 0 and x = 2.
Integrand: (-x² + 4x + 5) - (2x + 5) = -x² + 2x.
Area = integral from 0 to 2 of (-x² + 2x) dx = [-x³/3 + x²] from 0 to 2 = -8/3 + 4 = 4/3.
When curves are easier to express as x = f(y) and x = g(y), integrate with respect to y:
Area = integral from c to d of [f(y) - g(y)] dy
Here f(y) is the rightmost curve and g(y) is the leftmost.
Example: x = 0 and x = 12y - 2y².
Intersection (set equal): 0 = 12y - 2y², so 0 = 2y(6 - y), giving y = 0 and y = 6.
Area = integral from 0 to 6 of (12y - 2y²) dy = [6y² - 2y³/3] from 0 to 6 = 216 - 144 = 72.
Use whichever orientation lets you write the area as a single integral. If the "top" and "bottom" curves swap partway through the x-interval, it may be simpler to slice horizontally (or vice versa). The enclosed region is the same either way; only the setup changes.
Definite integral (FTC)
\int_a^b f(x)\,dx = F(b) - F(a), \quad F'(x) = f(x)Area between curves, vertical slices
A = \int_a^b \bigl[f(x) - g(x)\bigr]\,dx, \quad f(x) \ge g(x) \text{ on } [a,b]Area between curves, horizontal slices
A = \int_c^d \bigl[f(y) - g(y)\bigr]\,dy, \quad f(y) \ge g(y) \text{ on } [c,d]In the horizontal version, f(y) is the right curve and g(y) is the left curve.
Engineers use area-between-curves calculations to find the net displacement between velocity profiles, which matters when designing anything from car suspensions to roller coasters. Economists use the same idea to compute consumer and producer surplus: the area between a supply curve and a demand curve represents the total economic benefit from trade at a given price.
Students often forget to check which function is on top across the entire interval. If the curves cross within [a, b], you must split the integral at the crossing point, or you will get cancellation and the wrong answer.
Confusing the area under a single curve with the area between two curves. The area "under" f(x) from a to b already assumes g(x) = 0 (the x-axis). If there is a second curve, you subtract it explicitly.
Assuming you must always integrate with respect to x. Some regions are far simpler when expressed as functions of y. If the curves are given as x = something in terms of y, integrate horizontally.
Dropping the absolute value. The formula f(x) - g(x) works only when f(x) >= g(x). If you are unsure which is on top, use |f(x) - g(x)|, or (better) sketch the curves and split the interval at crossings.
This topic appears on virtually every Calc 2 exam. Expect to be asked to:
Set up and evaluate an area integral given two functions (vertical slicing).
Recognise when horizontal slicing is the better approach and rewrite accordingly.
Find intersection points as part of the problem (they will not always be given).
Handle regions described by inequalities (e.g. y <= 1, y >= x², x >= 0) where you must sketch the region and decide on bounds yourself.
Prove a known geometric result via integration (the classic: prove the area of a unit circle is pi).
True or False: The area between f(x) and g(x) on [a, b] is always equal to the integral from a to b of f(x) - g(x) dx, regardless of which function is larger. (Answer: False. You need f(x) >= g(x), or you must use absolute value / split at crossings.)
Fill in the blank: To find the limits of integration, you set ______ and solve. (Answer: f(x) = g(x), i.e. the two functions equal to each other.)
True or False: Horizontal integration means integrating with respect to y, where each slice is a horizontal rectangle. (Answer: True.)
Fill in the blank: When integrating horizontally, the integrand is [______ function] minus [______ function]. (Answer: right, left.)
True or False: If two curves do not cross on [a, b], you can always write a single integral for the area between them on that interval. (Answer: True.)
Q: Find the area between f(x) = 6x - x² and the x-axis.
A: Intersection points: x = 0 and x = 6. Area = integral from 0 to 6 of (6x - x²) dx = [3x² - x³/3] from 0 to 6 = 108 - 72 = 36.
Q: Find the area between f(x) = -x² + 4x + 5 and g(x) = 2x + 5.
A: Intersection points: x = 0 and x = 2. Integrand: -x² + 2x. Area = [-x³/3 + x²] from 0 to 2 = -8/3 + 4 = 4/3.
Q: Find the area between x = 0 and x = 12y - 2y², integrating with respect to y.
A: Intersection points: y = 0 and y = 6. Area = integral from 0 to 6 of (12y - 2y²) dy = [6y² - 2y³/3] from 0 to 6 = 216 - 144 = 72.
Q: Find the area of the region where y >= x², y <= 1, and x >= 0.
A: Bounds: x = 0 to x = 1 (since 1 = x² gives x = 1). Top function: y = 1. Bottom function: y = x². Area = integral from 0 to 1 of (1 - x²) dx = [x - x³/3] from 0 to 1 = 1 - 1/3 = 2/3.
Q: Find the area between sqrt(x) and x² on [0, 1].
A: Intersection points: sqrt(x) = x² implies x = x⁴, so x(1 - x³) = 0, giving x = 0 and x = 1. Area = integral from 0 to 1 of (sqrt(x) - x²) dx = [2x^(3/2)/3 - x³/3] from 0 to 1 = 2/3 - 1/3 = 1/3.
This connects directly to volumes of revolution (disk, washer, and shell methods), where you rotate an area-between-curves region around an axis to produce a 3D solid. The same "top minus bottom" logic extends to computing arc length and surface area of revolution later in Calc 2. In probability and statistics, the area between a probability density function and the x-axis gives cumulative probability, and in economics, consumer/producer surplus is literally the area between supply and demand curves.
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