Difficulty: Intermediate | Prerequisites: Calculus I (basic integration, finding antiderivatives, working with polynomial and root functions).
TL;DR: To find the area between two curves, integrate the difference of the "top" minus the "bottom" function (or "right" minus "left" when integrating with respect to y). The main work is choosing the right variable, finding intersection points, and keeping track of which curve is on which side across the interval.
Finding the area between curves is one of the first "applications of integration" topics in Calculus II, and it builds directly on definite integrals from Calculus I. Where Calc I asked you to find the area under a single curve, Calc II asks: what is the area trapped between two curves? This idea underpins later topics like volumes of revolution (where you rotate that enclosed region around an axis). If you are comfortable setting up and evaluating definite integrals, you have the prerequisite skills. The new challenge is geometric: deciding which function is "on top," finding where the curves cross, and choosing whether to integrate with respect to x or y.
Enclosed region
The finite area bounded on all sides by two or more curves. You need to identify this region before setting up any integral.
Top curve / bottom curve (integrating with respect to x)
When you slice vertically (dx), the "top" curve has the larger y-value on that interval. The integrand is always (top) minus (bottom). In simple terms, you are stacking thin vertical rectangles and measuring their heights.
Right curve / left curve (integrating with respect to y)
When you slice horizontally (dy), the "right" curve has the larger x-value. The integrand is (right) minus (left). Think of it as laying thin horizontal rectangles side by side.
Intersection points
The x- or y-values where the two curves meet. These typically form the limits of integration. You find them by setting the two functions equal and solving.
Integrating with respect to y
Rewriting or treating the curves as functions of y (i.e., x = g(y)) and integrating over a y-interval. This is sometimes far simpler than integrating with respect to x, especially when the curves are more naturally expressed as x = f(y).
Method 1: Integrating with respect to x
Sketch the curves and identify the enclosed region.
Find intersection points by setting f(x) = g(x) and solving for x.
Determine which curve is on top across the interval. If the "top" function changes partway through, split the integral at that crossing point.
Set up the integral:
A = \int_a^b \bigl[ f(x) - g(x) \bigr]\, dxwhere f(x) is the top curve and g(x) is the bottom curve on [a, b].
Example approach (Problem 1): Area between y = -x³ + 6x and y = -x²
Set -x³ + 6x = -x² to find intersections. This gives x³ - x² - 6x = 0, so x(x² - x - 6) = 0, yielding x = 0, x = 3, x = -2.
On [-2, 0], check which curve is on top (plug in a test value like x = -1).
On [0, 3], check again.
Set up two integrals (or use absolute value / careful sign work) and evaluate.
Method 2: Integrating with respect to y
Sometimes the region is more naturally bounded by curves written as x = f(y).
Find intersection points by setting f(y) = g(y) and solving for y.
Determine which curve is to the right on the interval.
Set up the integral:
A = \int_c^d \bigl[ x_{\text{right}}(y) - x_{\text{left}}(y) \bigr]\, dyExample approach (Problem 2): Area between x = -y² + 10 and x = (y - 2)²
Both curves are already expressed as functions of y, so integrate with respect to y.
Set -y² + 10 = (y - 2)² and solve: -y² + 10 = y² - 4y + 4, giving 2y² - 4y - 6 = 0, so y = 3, y = -1.
On [-1, 3], the parabola x = -y² + 10 is to the right.
Integrate:
A = \int_{-1}^{3} \bigl[(-y^2 + 10) - (y - 2)^2\bigr]\, dy = \frac{64}{3}Choosing your variable
If both curves are given as y = f(x), integrating in x is the natural first choice.
If both curves are given as x = g(y), or if integrating in x would require splitting into multiple integrals while integrating in y would not, switch to y.
The goal is always: one integral with a clean integrand, rather than two or three patched together.
Forgetting to check which curve is on top across the whole interval. The top and bottom functions can switch at an intersection point. If you blindly integrate f(x) - g(x) without checking the sign, you will subtract area instead of adding it.
Using the wrong variable. When curves are given as x = g(y), students sometimes try to solve for y and integrate in x, creating unnecessary complexity or algebraic errors. Look at the form of the equations first.
Confusing intersection points with limits of integration. The intersection points give you the limits only when those are the points where the enclosed region begins and ends. Occasionally a boundary is a vertical or horizontal line, not an intersection.
Forgetting absolute value or sign. The area is always positive. If your integrand comes out negative on a sub-interval, it means you have the curves in the wrong order for that piece.
⚠️ Expect at least one problem where integrating with respect to y is clearly simpler. The exam tests whether you can recognise when to switch variables.
⚠️ Intersection-point algebra is where most errors happen. Double-check by plugging your solutions back into both equations.
⚠️ If a problem gives you curves as x = f(y), that is a strong hint to integrate in y. Do not waste time solving for y.
⚠️ Sketching the region, even roughly, prevents sign errors and helps confirm limits of integration. The exam review includes graphs for a reason.
True or false: When integrating with respect to x, the integrand is always (upper curve) - (lower curve). True.
Fill in the blank: To find where two curves intersect, set ______ and solve. The two functions equal to each other.
True or false: If both curves are given as x = g(y), you must solve for y before integrating. False. Integrate with respect to y directly.
Fill in the blank: If the "top" function switches partway through the interval, you need to ______ the integral. Split.
True or false: The area between two curves can be negative. False. Area is always non-negative; a negative result means the curves were subtracted in the wrong order.
Q: Find the area of the region enclosed by y = -x³ + 6x and y = -x².
A: Set -x³ + 6x = -x² and rearrange to x³ - x² - 6x = 0. Factor: x(x - 3)(x + 2) = 0, giving x = -2, 0, 3. Check which curve is on top on each sub-interval. On [-2, 0], y = -x² is above y = -x³ + 6x; on [0, 3], y = -x³ + 6x is above y = -x². Set up two integrals and evaluate. The total area is the sum of the two pieces.
Q: Determine the area of the region bounded by x = -y² + 10 and x = (y - 2)². (Answer: 64/3)
A: Both curves are functions of y, so integrate with respect to y. Set -y² + 10 = (y - 2)² to find intersections: -y² + 10 = y² - 4y + 4, which simplifies to 2y² - 4y - 6 = 0, giving y = -1 and y = 3. On [-1, 3], x = -y² + 10 is the right curve. Compute A = ∫ from -1 to 3 of [(-y² + 10) - (y² - 4y + 4)] dy = ∫ from -1 to 3 of (-2y² + 4y + 6) dy = 64/3.
Q: Suppose two curves y = f(x) and y = g(x) intersect at x = 1 and x = 5, and f(x) ≥ g(x) on that interval. Write the integral for the enclosed area.
A: A = ∫ from 1 to 5 of [f(x) - g(x)] dx.
This connects directly to volumes of revolution (Sections 6.1-6.2), because the region whose area you find here is the same region you will later rotate around an axis to compute a volume using disks, washers, or shells. The skill of identifying "top vs bottom" and "right vs left" carries over exactly.
Arc length (Section 8.2) also builds on the idea of slicing a curve into small pieces and summing, though the integrand changes from a height to a length element.
Tags: area between curves, enclosed region, integrating with respect to y, integrating with respect to x, intersection points, top minus bottom, right minus left, Calculus II, Calc 2, NYU, sections 6.1 6.2, area under a curve, area between two functions