Unit 8: Applications of Integration

Area Between Curves

Finding the area between two curves using a definite integral.

Advanced25 min lesson3 min readUpdated August 12, 2026Author not yet attributed

Prerequisites

Average Value of a Function

The Gap Between Two Curves

A single definite integral measures area between one curve and the x-axis. Before reading on: if you subtracted one function from another before integrating, would the result measure the area between the two curves themselves, instead of between a curve and the axis?

Definition — Area Between Two Curves

The area enclosed between two curves y = f(x) and y = g(x), where f(x) ≥ g(x) on [a, b], is ∫ₐᵇ [f(x) − g(x)] dx — the integral of the vertical gap between them, top curve minus bottom curve. The bounds a and b are usually the curves' intersection points, found by solving f(x) = g(x).

A = \int_a^b \big[f(x) - g(x)\big]\,dx

Worked Example — Finding the Area Between Two Curves

Find the area enclosed between y = x + 2 and y = x². First find the intersection points: x + 2 = x², so x² − x − 2 = 0, which factors as (x − 2)(x + 1) = 0, giving x = −1 and x = 2. Test which curve is on top between them, at x = 0: the line gives 2, the parabola gives 0 — the line is on top. Area = ∫₋₁² [(x + 2) − x²] dx. An antiderivative is x²/2 + 2x − x³/3. At x = 2: 2 + 4 − 8/3 = 6 − 8/3 = 10/3. At x = −1: 1/2 − 2 + 1/3 = −7/6. Area = 10/3 − (−7/6) = 20/6 + 7/6 = 27/6 = 4.5.

This same integral can be computed as the exact area under a single new function — the difference h(x) = (x + 2) − x² = −x² + x + 2 — rather than thinking of it as 'between two curves' at all. Graphing h(x) and finding the exact area under it from x = −1 to x = 2 gives the identical integral, and therefore the identical answer, 4.5.

Integral Visualizer

f(x) =

x^2

Integration bounds

a — lower bound0

b — upper bound2

Animate b (the play button above) to watch the shaded area — and the accumulation curve below — grow.

Approximation method

Method

Subdivisions (n)10

Show approximationShow exact area

Trace mode

f(x), with the region between a and b shaded

Show accumulation function A(x)

Accumulation function A(x) = ∫ₐˣ f(t) dt — linked to the graph above; panning or zooming either moves both

What the Fundamental Theorem of Calculus tells us

A(x) = ∫ from a to x of f(t) dt is the accumulation function. At x = b ≈ 2, A(b) ≈ 2.67 — the signed area under f from a ≈ 0 to b.
The Fundamental Theorem of Calculus says A'(x) = f(x): differentiating the accumulation function gives the original function back.
Confirmed numerically here: differentiating the accumulation curve gives A'(b) ≈ 4, matching f(b) ≈ 4.

Numerical approximation comparison

Every method's estimate of ∫ₐᵇ f(x) dx at n = 10 subdivisions, against a high-resolution reference — all should converge toward the same value.
Method	Estimate
Left Riemann sum	2.28
Right Riemann sum	3.08
Midpoint Riemann sum	2.66
Trapezoidal Rule	2.68
Simpson's Rule	2.66667
Exact (high-resolution reference)	2.66667

Analysis of f(x)

y-intercept: (0, 0)
x-intercepts: (0, 0)
Extrema: local min at (0, 0)
Inflection points: none found in view
Vertical asymptotes: none found in view
Horizontal asymptotes: none found
Domain: all real numbers in view
Range (estimated): approximately [0, 100]

Tip

Always confirm which curve is actually on top with a single test point inside the interval — a sign error here (integrating bottom-minus-top instead of top-minus-bottom) produces the exact negative of the correct area.

Common Mistakes

Using the wrong order of subtraction, computing g(x) − f(x) when f is actually the top curve.
Subtracting in the wrong order integrates a negative quantity across the whole interval, producing the negative of the true area — always confirm which function is larger with a test point before setting up the integral.
Using the original functions' domain endpoints as the bounds instead of their actual intersection points.
The area between two curves is only well-defined between the x-values where they actually cross — using arbitrary endpoints instead of the true intersection points integrates over a region where the 'between' relationship no longer holds.

Key Takeaways

Area between two curves is ∫ₐᵇ[top(x) − bottom(x)]dx, with the bounds usually found by solving for the curves' intersection points.
A test point between the intersection points confirms which curve is actually on top before setting up the integral.
The area between two curves equals the area under their difference function — the same integral, viewed as a single-function area problem.

Summary

This same 'integrate a cross-sectional quantity' idea extends beyond flat area. The final lesson uses it to find the volume of a three-dimensional solid.

Track your progress

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