Difficulty: Intermediate | Prerequisites: Basic integration (antiderivatives, definite integrals, Fundamental Theorem of Calculus)
Area between curves is one of the core applications of definite integration in AP Calculus AB. It takes the idea of "area under a curve" and extends it to the area trapped between two (or more) functions. This topic sits squarely in the integration unit, right after you learn the Fundamental Theorem of Calculus, and it forms the foundation for the volumes and cross-section problems that follow. If you can set up an area integral correctly, you already have the hardest part of most FRQ volume questions handled.
You should already be comfortable evaluating definite integrals and finding intersection points of functions (by setting them equal and solving). If that feels shaky, revisit those topics first.
To find the area between two curves, integrate the difference of the "outer" and "inner" functions across the interval where one sits above the other. You can integrate with respect to x (top minus bottom) or with respect to y (right minus left), and the right choice depends on which setup avoids splitting the integral. Always find the intersection points first.
Bounded region
The closed area enclosed by two or more curves (and possibly the axes). On the AP exam, this is usually labelled R or S in the problem diagram.
In simple terms, it is the "trapped" shape between the graphs.
Integrating with respect to x (dx)
Slicing the region into thin vertical rectangles. The height of each rectangle is (top function) minus (bottom function).
Think of it as stacking vertical strips side by side from left to right.
Integrating with respect to y (dy)
Slicing the region into thin horizontal rectangles. The width of each rectangle is (right function) minus (left function).
Think of it as stacking horizontal strips from bottom to top.
Intersection point
A point where two curves meet, found by setting their equations equal and solving. Intersection points give you the limits of integration.
Limits of integration
The start and end values (of x or y) that bound the region. These come from the intersection points or from boundaries stated in the problem (such as an axis).
Every area-between-curves problem starts the same way:
Identify the curves that bound the region.
Find where they intersect (set the equations equal, solve for x or y).
Sketch the region, even roughly. This tells you which function is on top/right and which is on the bottom/left.
Integrate with respect to x when the top and bottom functions are consistent across the interval. The integral looks like:
\int_a^b \bigl[f(x) - g(x)\bigr]\,dxwhere f(x) is the top curve and g(x) is the bottom curve, from x = a to x = b.
Integrate with respect to y when the right and left functions are consistent across the interval, or when integrating with respect to x would force you to split into two (or more) separate integrals. The integral looks like:
\int_c^d \bigl[r(y) - l(y)\bigr]\,dywhere r(y) is the rightmost curve and l(y) is the leftmost curve, from y = c to y = d.
The choice of variable is about convenience. Both give the same answer if set up correctly.
The region R is in the first quadrant, bounded by the x-axis and the graphs of y = ln x and y = 5 − x.
Step 1: Find the intersection. Set ln x = 5 − x. Solving (by calculator or numerical method) gives x ≈ 3.69344, y ≈ 1.30656. Call this point (A, B).
Step 2: Choose your variable.
Option 1 (dy): Integrate with respect to y from 0 to B. Rewrite the curves in terms of y: x = eʸ (from y = ln x) and x = 5 − y (from y = 5 − x). For every horizontal strip, the right boundary is x = 5 − y and the left boundary is x = eʸ.
\text{Area} = \int_0^B \bigl[(5 - y) - e^y\bigr]\,dy = 2.986This is a single integral, clean and direct.
Option 2 (dx): Integrate with respect to x, but the top function changes at x = A. From x = 1 to x = A the top curve is y = ln x. From x = A to x = 5 the top curve is y = 5 − x. You need two integrals:
\text{Area} = \int_1^A \ln x\,dx + \int_A^5 (5 - x)\,dx = 2.986Both approaches yield 2.986 (or 2.985, depending on rounding). The dy approach is simpler here because it avoids the split.
Area using vertical slices (dx):
A = \int_a^b \bigl[f(x) - g(x)\bigr]\,dxwhere f(x) ≥ g(x) on [a, b].
Area using horizontal slices (dy):
A = \int_c^d \bigl[r(y) - l(y)\bigr]\,dywhere r(y) ≥ l(y) on [c, d].
The absolute value of the difference is implied: always subtract the smaller from the larger for your chosen direction, so the integrand stays non-negative.
Engineers use area-between-curves calculations to find the work done by a varying force, the net displacement between velocity curves, or the material between two profiles in manufacturing. In economics, the area between a supply curve and a demand curve (up to the equilibrium point) gives consumer and producer surplus.
Students often subtract the functions in the wrong order and get a negative area. Area is always positive; if your integral comes out negative, you have the top/bottom (or right/left) functions reversed.
Forgetting to split the integral when the "top" and "bottom" functions swap partway through the interval. If one curve is on top for part of the interval and the other is on top for the rest, you need separate integrals for each sub-interval (or switch to dy if that avoids the split).
Using the wrong limits of integration. A common slip is integrating from 0 to the intersection x-value when the region does not actually start at x = 0 (for instance, in the worked example, the region starts at x = 1 because ln x = 0 at x = 1).
Mixing up which variable you are integrating with respect to. If you integrate with respect to y, the curves must be rewritten as functions of y, and the limits must be y-values, not x-values.
⚠️ Area between curves appears on the AP Calculus AB exam almost every year, typically as part (a) of an FRQ. Getting the setup right here earns you the integrand point and the limits point before you even evaluate.
⚠️ The College Board awards separate points for the integrand, the limits, and the numerical answer. Even if you cannot evaluate the integral, writing the correct expression with correct limits earns 2 out of 3 points.
⚠️ On calculator-active FRQs (like the 2012 Q2 shown here), you are expected to use your calculator to evaluate the integral and find intersection points numerically. Practice this workflow before exam day.
⚠️ Choosing dy over dx (or vice versa) is never penalised, but it can save time and reduce errors. The exam does not require one approach over the other.
True or False: When integrating with respect to y, the limits of integration should be x-values. Answer: False. The limits must be y-values.
Fill in the blank: To find the area between f(x) and g(x) where f(x) ≥ g(x), integrate ______ from a to b. Answer: f(x) − g(x)
True or False: If your area integral evaluates to a negative number, the area of the region is negative. Answer: False. Area is always non-negative. A negative result means the subtraction order is wrong.
Fill in the blank: The intersection points of the bounding curves give you the ______ of integration. Answer: limits
Q: The region S is bounded by y = x² and y = x + 2. Write an integral expression for the area of S using dx.
A: Find the intersections: x² = x + 2 gives x = −1 and x = 2. Since y = x + 2 is on top, Area = ∫ from −1 to 2 of [(x + 2) − x²] dx.
Q: For the 2012 FRQ region R (bounded by the x-axis, y = ln x, and y = 5 − x), why does the dy approach use a single integral while the dx approach requires two?
A: With respect to y, the right boundary is always x = 5 − y and the left boundary is always x = eʸ for the entire interval 0 ≤ y ≤ B. With respect to x, the top function changes from y = ln x (for 1 ≤ x ≤ A) to y = 5 − x (for A ≤ x ≤ 5), so you must split at x = A.
Q: A region is bounded by y = sin x, y = cos x, x = 0, and x = π/2. Set up an integral (or integrals) for the area.
A: The curves cross at x = π/4. From 0 to π/4, cos x ≥ sin x. From π/4 to π/2, sin x ≥ cos x. Area = ∫ from 0 to π/4 of (cos x − sin x) dx + ∫ from π/4 to π/2 of (sin x − cos x) dx.
Q: If you set up an area integral and your calculator returns a negative value, what went wrong?
A: The order of subtraction is reversed. You are subtracting the larger function from the smaller. Swap the functions in the integrand (or take the absolute value of the result, though fixing the setup is the proper approach).
This connects directly to volumes of solids with known cross sections and volumes of revolution (disc/washer method), because both begin by finding the area of a cross-sectional slice, which is an area-between-curves calculation. It also connects to net displacement vs total distance in kinematics: the area between a velocity curve and the t-axis gives displacement, and you split the integral wherever velocity changes sign, just as you split an area integral wherever the "top" function changes.
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