Unit 8 · Topic 8.4

8.4 Finding the Area Between Curves Expressed as Functions of x

To find the area between two curves y = f(x) and y = g(x), integrate the top curve minus the bottom curve with respect to x. The limits of integration are where the region starts and ends, which is often where the curves cross.

Key terms

area between curves
points of intersection
top minus bottom
vertical slices (dx)

Thin vertical rectangles

Slice the region into thin vertical strips. Each strip has width dx and height (top curve) − (bottom curve). Its area is [f(x) − g(x)] dx. Adding infinitely many strips gives

Area = ∫ₐᵇ [f(x) − g(x)] dx, where f(x) ≥ g(x) on [a, b].

This works even if one or both curves are below the x-axis. The difference top − bottom is always the strip's height, so the area comes out positive.

The steps

Sketch the curves, even roughly, to see the region.
Find the intersection points by setting f(x) = g(x). These are often the limits a and b. If the problem gives a vertical boundary like x = 4, use that.
Decide which curve is on top. Test any x between the limits: the bigger value is the top.
Set up ∫ₐᵇ (top − bottom) dx and evaluate.
Check: the area must be positive.

With a calculator

When the intersection points can't be found by algebra, like for cos x = x², use the calculator to solve. Store the x-values in memory rather than rounding them, and use the stored values as limits. Rounding a limit to two decimals can change the third decimal of the area.

Write the setup on paper (the integral with the limits) before writing the number. Scorers grade the setup.

When the top or bottom boundary changes

Sometimes a region has one top curve but a bottom boundary that switches partway across, or the reverse. For example, the region bounded by y = √x, y = 2 − x and the x-axis has the x-axis on the bottom throughout, but its top is y = √x for 0 ≤ x ≤ 1 and y = 2 − x for 1 ≤ x ≤ 2. Split at x = 1 and add two integrals: ∫₀¹ √x dx + ∫₁² (2 − x) dx = 2/3 + 1/2 = 7/6. Or slice horizontally instead (8.5), which may need only one integral.

Why it works

If f(x) ≥ 0 and g(x) ≥ 0, the area under f minus the area under g leaves exactly the region between them. When the curves dip below the axis, shift both up by the same constant: the difference f − g doesn't change, so the formula still holds. This is why top minus bottom always works, regardless of where the x-axis is.

Worked examples

Try each one yourself first, then open the solution.

Example 1
Curves that cross twice
Find the area of the region bounded by y = 4 − x² and y = x + 2.
Show the solutionHide the solution
1. Step 1: Intersections: 4 − x² = x + 2 → x² + x − 2 = 0 → (x + 2)(x − 1) = 0, so x = −2 and x = 1.
2. Step 2: Which is on top? At x = 0: 4 − 0 = 4 and 0 + 2 = 2, so the parabola is on top.
3. Step 3: Area = ∫ from −2 to 1 of [(4 − x²) − (x + 2)] dx = ∫ from −2 to 1 of (2 − x − x²) dx.
4. Step 4: Antiderivative: 2x − x²/2 − x³/3.
5. Step 5: At 1: 2 − 1/2 − 1/3 = 7/6. At −2: −4 − 2 + 8/3 = −10/3.
6. Step 6: Area = 7/6 − (−10/3) = 7/6 + 20/6 = 27/6 = 9/2.
Answer: 9/2
Example 2Calculator allowed
Calculator: intersections you can't solve by hand
Find the area of the region between y = cos x and y = x².
Show the solutionHide the solution
1. Step 1: Solve cos x = x² with the calculator: x ≈ −0.824 and x ≈ 0.824. Store them as A and B.
2. Step 2: At x = 0, cos 0 = 1 > 0, so cos x is on top.
3. Step 3: Area = ∫ from A to B of (cos x − x²) dx ≈ 1.095.
Answer: About 1.095
Example 3
Trap: bottom minus top
A student finds the area between y = x and y = x² on [0, 1] as ∫₀¹ (x² − x) dx = −1/6. What went wrong?
Show the solutionHide the solution
1. Step 1: On (0, 1), x > x² (for example, 0.5 > 0.25), so y = x is on top.
2. Step 2: The student subtracted top from bottom, so the answer came out negative.
3. Step 3: Correct setup: ∫₀¹ (x − x²) dx = 1/2 − 1/3 = 1/6.
Answer: The area is 1/6. The integrand must be top minus bottom.

Common mistakes

Using the x-intercepts of one curve as limits when the region is bounded by where the curves meet.
Assuming the curve with the bigger formula is on top. Test a point.
Rounding intersection points before integrating with a calculator.
Reporting a negative area. Area between curves is always positive.

On the exam

Area questions are a free-response favorite, often followed by a volume question about the same region. Get the intersection points right first, because every later part uses them.
On calculator questions, the setup with limits earns its own point. Write it out.

Connected topics

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