Difficulty: Intermediate | Prerequisites: Definite integrals, antiderivative rules, basic graphing of polynomials and square root functions.
This topic extends definite integrals beyond "area under a curve" to "area between two curves." Up to now, you have been computing the area between a function and the x-axis. Here you learn to compute the area of a region trapped between two functions, which requires deciding which function sits on top, choosing sensible bounds, and sometimes deciding whether to integrate with respect to x or y. This is one of the first major applications of integration and it appears heavily on the AP exam.
To find the area between two curves, integrate the top function minus the bottom function over the interval. You can integrate with respect to x (vertical slices) or y (horizontal slices), and picking the right variable can save you a lot of work.
Upper function / lower function
On a given interval, the upper function is the one with the larger y-value. The area formula subtracts the lower from the upper so the result stays positive.
In simple terms, "upper minus lower" just means "the one on top minus the one on the bottom" at every x-value in the interval.
Integrating with respect to x
Slicing the region into thin vertical rectangles of width dx. All curves must be written as y = (something in terms of x).
Think of it as stacking vertical strips side by side, like a bar chart, to fill up the region.
Integrating with respect to y
Slicing the region into thin horizontal rectangles of width dy. All curves must be written as x = (something in terms of y).
Think of it as stacking horizontal strips from bottom to top instead. Sometimes this gives a single, simpler integral where the x approach would need two.
The area A of the region between f(x) and g(x) from x = a to x = b is:
A = ∫ from a to b of [f(x) – g(x)] dx, where f(x) ≥ g(x) on [a, b].
The requirement f(x) ≥ g(x) means f is the upper curve on that interval. If the curves swap (one goes above the other partway through), you must split the integral at the crossing point.
Find the area bounded by y = x² + 2, y = –x, x = 0, and x = 1.
On [0, 1], the parabola x² + 2 sits above the line –x, so the integrand is (x² + 2) – (–x) = x² + x + 2.
Integrate:
∫ from 0 to 1 of (x² + x + 2) dx = [x³/3 + x²/2 + 2x] from 0 to 1
= 1/3 + 1/2 + 2 = 17/6
When curves are given as x = (something in y), or when integrating with respect to x would force you to split into multiple integrals, switching to dy can simplify things considerably.
With respect to x: rectangles are vertical, expressions must be in terms of x.
With respect to y: rectangles are horizontal, expressions must be in terms of y.
Curves: x = 4 – y² and x = y – 2. Intersection points: solve 4 – y² = y – 2, which gives y = –3 and y = 2.
With respect to x (harder path): rewrite as y = ±√(4 – x) and y = x + 2. The square root splits into two branches, so you need two separate integrals:
∫ from –5 to 0 of [(x + 2) + √(4 – x)] dx + ∫ from 0 to 4 of 2√(4 – x) dx = 20.833
With respect to y (easier path): keep the original forms. On –3 ≤ y ≤ 2, the parabola x = 4 – y² is to the right of the line x = y – 2, so:
∫ from –3 to 2 of [(4 – y²) – (y – 2)] dy = 20.833
Both methods give the same answer, but the y approach uses a single integral with no square roots. Always check which variable gives a cleaner setup.
Area with respect to x:
A = ∫ from a to b of [f(x) – g(x)] dx, where f(x) ≥ g(x)
Area with respect to y:
A = ∫ from c to d of [f(y) – g(y)] dy, where f(y) ≥ g(y) (i.e. the rightmost curve minus the leftmost)
Finding the area between curves is the basis for computing quantities like the volume of a solid of revolution (coming in later units), the work done between two force curves, and consumer/producer surplus in economics. Any time you need the "difference" accumulated over an interval, this is the tool.
Students often set up f(x) – g(x) without checking which function is actually on top across the entire interval. If they swap, you get a negative contribution that shrinks your answer. Always verify or sketch.
Forgetting to rewrite both curves in terms of the same variable. If you integrate with respect to y, every expression must be x = (something in y), not a mix.
Assuming you must always integrate with respect to x. The exam regularly gives problems where integrating with respect to y is far simpler, and choosing x leads to messy split integrals.
Confusing the intersection points of the curves with the given boundary values. The bounds of integration come from where the region starts and ends, which may be intersections, may be given lines, or may be both.
⚠️ Area between curves is tested almost every year on the AP Calc AB exam, often as a free-response question that also asks about volume.
⚠️ You will frequently be expected to decide whether to integrate with respect to x or y. If you pick the wrong one, you can still get the answer, but you will spend much more time.
⚠️ Setting up the integral correctly (identifying bounds and which curve is on top) is worth significant partial credit even if you make an arithmetic error.
True or false: When computing area between two curves, a negative result is acceptable. (False. Area is always positive; if you get a negative number, you likely subtracted in the wrong order.)
Fill in the blank: When integrating with respect to y, the rectangles used to approximate the area are ________. (Horizontal.)
True or false: You must always integrate with respect to x when finding area between curves. (False. You can integrate with respect to y, and sometimes it is much easier.)
Fill in the blank: The area formula A = ∫[f(x) – g(x)] dx requires that ________ on the interval [a, b]. (f(x) ≥ g(x), i.e. f is the upper curve.)
Q: Given y = x² and y = x, find the area between the curves from x = 0 to x = 1.
A: On [0, 1], x ≥ x², so the area is ∫ from 0 to 1 of (x – x²) dx = [x²/2 – x³/3] from 0 to 1 = 1/2 – 1/3 = 1/6.
Q: Why might you choose to integrate with respect to y instead of x?
A: When the curves are more naturally expressed as functions of y, or when integrating with respect to x would require splitting the region into two or more sub-integrals due to branch changes (e.g. ±√), switching to y often gives a single, cleaner integral.
Q: Two curves intersect at x = –1 and x = 3. On [–1, 1] the parabola is on top; on [1, 3] the line is on top. How do you set up the total area?
A: Split into two integrals: ∫ from –1 to 1 of [parabola – line] dx + ∫ from 1 to 3 of [line – parabola] dx. You cannot use a single integral without absolute value because the "top" function changes.
Q: Set up (but do not evaluate) the integral for the area between x = 4 – y² and x = y – 2, integrating with respect to y.
A: The curves intersect at y = –3 and y = 2, and 4 – y² ≥ y – 2 on that interval. Area = ∫ from –3 to 2 of [(4 – y²) – (y – 2)] dy.
This connects directly to volumes of solids of revolution (disc, washer, and shell methods), which use the same "set up an integral of a geometric cross-section" logic. It also links back to Riemann sums from earlier units, since the area formula is really the limit of a sum of thin rectangles. In economics courses, the same technique computes consumer and producer surplus.
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