Difficulty: Intermediate | Prerequisites: Sections 13.1 through 13.4 (indefinite integrals, substitution, definite integrals).
This section extends the area-under-a-curve idea to the area trapped between two curves. Instead of measuring the region between a function and the x-axis, you measure the region between two functions. The key skill is setting up the integral correctly: identifying which function is on top, finding where they cross, and splitting the integral when the curves switch positions. This is the capstone application of integration in MAT 210.
To find the area between two curves, set the functions equal to find intersection points, determine which is on top in each interval, and integrate (top minus bottom). If the curves cross within the interval, split into separate integrals at each crossing point.
Area between curves
The total non-negative area of the region enclosed between two functions f(x) and g(x) over an interval. Think of it as the space sandwiched between the two graphs.
Upper function / lower function
On a given interval, the upper function is the one with the larger y-values (the one on top when you look at the graph). The integral is always set up as ∫ (upper − lower) dx so the result is positive.
Intersection points
The x-values where the two curves meet, found by setting f(x) = g(x) and solving. These determine the bounds of integration when the problem asks for the "enclosed" region, and they mark where the upper and lower functions may swap.
Splitting the integral
When curves cross inside the interval, you must break the integral at the crossing point and integrate each piece separately, swapping which function is on top as needed. Failing to split gives the net signed area (which can be smaller than the true area or even zero).
Set f(x) = g(x) and solve to find intersection points (these become your limits of integration if the problem says "enclosed region")
Determine which function is on top in each subinterval (plug in a test point between the intersection values)
Set up the integral as ∫ (top − bottom) dx
If the curves swap positions, split into separate integrals at the crossing point
Evaluate each integral and add the results
Set equal: −x² + 6x + 2 = 2x² + 9x − 4, which simplifies to −3x² − 3x + 6 = 0, or x² + x − 2 = 0. Factor: (x + 2)(x − 1) = 0, so x = −2 and x = 1.
Test x = 0: first curve gives 2, second gives −4. So the first curve is on top.
Area = ∫ from −2 to 1 of [(−x² + 6x + 2) − (2x² + 9x − 4)] dx = ∫ from −2 to 1 of (−3x² − 3x + 6) dx.
Antiderivative: −x³ − (3/2)x² + 6x. Evaluate at 1: −1 − 3/2 + 6 = 7/2. At −2: 8 − 6 − 12 = −10. Answer: 7/2 − (−10) = 27/2.
Set equal: x² − x + 5 = x + 8 gives x² − 2x − 3 = 0, so (x − 3)(x + 1) = 0. Intersection at x = −1 and x = 3.
Test x = 0: f(0) = 5, g(0) = 8. So g is on top.
Area = ∫ from −1 to 3 of [(x + 8) − (x² − x + 5)] dx = ∫ from −1 to 3 of (−x² + 2x + 3) dx.
Antiderivative: −(1/3)x³ + x² + 3x. At x = 3: −9 + 9 + 9 = 9. At x = −1: 1/3 + 1 − 3 = −5/3. Answer: 9 − (−5/3) = 32/3.
For the area between y = x² and y = 1 from x = −1 to x = 2:
The curves cross at x = −1 and x = 1. On [−1, 1], the line y = 1 is above y = x². On [1, 2], y = x² is above y = 1.
The correct setup is: ∫ from −1 to 1 of (1 − x²) dx + ∫ from 1 to 2 of (x² − 1) dx. This is answer C from the multiple-choice problem.
For y = x² and y = −1 from x = −1 to x = 1: since x² is always above −1 on this interval, the area is ∫ from −1 to 1 of (x² − (−1)) dx = ∫ from −1 to 1 of (x² + 1) dx = [(1/3)x³ + x] from −1 to 1 = (1/3 + 1) − (−1/3 − 1) = 8/3.
\text{Area} = \int_a^b [f(x) - g(x)]\,dx \quad \text{where } f(x) \geq g(x) \text{ on } [a,b]When curves swap, split at crossing point c:
\text{Area} = \int_a^c [f(x) - g(x)]\,dx + \int_c^b [g(x) - f(x)]\,dxStudents integrate (f − g) over the whole interval without checking whether the curves cross. If they do cross, you must split the integral, or you get the net signed area instead of the total area.
Students set up the integral as (lower − upper), which gives a negative result. Area is always positive; always put the larger function first.
Students forget to find the intersection points and use the wrong bounds. When the problem says "enclosed region," the bounds come from solving f(x) = g(x), not from numbers given in the problem.
Students confuse this with definite integral area problems from Section 13.4. In Section 13.4, the area is between one curve and the x-axis. Here, it is between two curves.
⚠️ The multiple-choice format ("which integral represents the area?") tests whether you know when to split. Sketch the curves or test points to see which is on top in each subinterval.
⚠️ Always show the step where you find intersection points by setting the functions equal.
⚠️ "Give the exact value" means fractions and radicals, not decimals.
Fill in the blank: to find where two curves intersect, set ____ = ____. (f(x) = g(x))
True or false: area between curves can be negative. (False. Area is always non-negative; if your integral gives a negative number, you subtracted in the wrong order.)
True or false: if the curves cross at x = 1 inside the interval [−1, 2], you need two separate integrals. (True.)
Fill in the blank: the area between y = x² and y = x from x = 0 to x = 1 is ∫ from 0 to 1 of ____ dx. (x − x², because x ≥ x² on [0, 1])
Q: Find the area enclosed by y = −x² + 6x + 2 and y = 2x² + 9x − 4.
A: Intersection at x = −2 and x = 1 (from x² + x − 2 = 0). First curve is on top. Area = ∫ from −2 to 1 of (−3x² − 3x + 6) dx = [−x³ − (3/2)x² + 6x] from −2 to 1 = 7/2 − (−10) = 27/2.
Q: Find the area enclosed by f(x) = x² − x + 5 and g(x) = x + 8.
A: Intersection at x = −1 and x = 3. g is on top. Area = ∫ from −1 to 3 of (−x² + 2x + 3) dx = [−(1/3)x³ + x² + 3x] from −1 to 3 = 9 − (−5/3) = 32/3.
Q: Find the area between y = x² and y = −1 from x = −1 to x = 1.
A: x² is always above −1 on this interval. Area = ∫ from −1 to 1 of (x² + 1) dx = [(1/3)x³ + x] from −1 to 1 = (4/3) − (−4/3) = 8/3.
Q: Which integral gives the area between y = x² and y = 1 from x = −1 to x = 2?
A: Answer C. The curves cross at x = −1 and x = 1. On [−1, 1], y = 1 is on top, so integrate (1 − x²). On [1, 2], y = x² is on top, so integrate (x² − 1). The correct expression is ∫ from −1 to 1 of (1 − x²) dx + ∫ from 1 to 2 of (x² − 1) dx.
Area between curves is the geometric payoff for everything in Chapters 13 and 14. It relies on finding antiderivatives (Section 13.1), sometimes requires substitution (Section 13.2), and always uses the FTC to evaluate the definite integral (Section 13.3). In more advanced courses, this extends to volumes of revolution and probability density functions.
area between curves, enclosed region, intersection of curves, upper minus lower function, splitting integrals, signed area vs total area, MAT 210, brief calculus, Section 14.2, ASU calculus, area bounded by curves