Lesson 17: Area Between CurvesAP

The definite integral's headline application (CED topics 8.4–8.6) is the area between two curves: integrate the *difference* of the functions — top minus bottom with respect to x, or right minus left with respect to y — over the interval where the two curves enclose a region. The setup discipline matters more than the integration: find where the curves cross, decide which is on top on each subinterval, and only then integrate.

The principle

If f(x) ≥ g(x) on [a, b], the area between them is A = ∫ from a to b of [f(x) − g(x)] dx. Because a length (f − g) is integrated along x, the result is genuinely an area — always nonnegative, unlike a bare definite integral. If the curves cross inside the interval, split at the crossing: "top minus bottom" must be re-decided on each piece. When the region is described more naturally sideways (or a single x-slice would need splitting), integrate with respect to y: A = ∫ from c to d of [right(y) − left(y)] dy.

Worked Example 1: the starter region

Problem. Find the area enclosed by y = x and y = x^2.

Intersections: x = x^2 → x(1 − x) = 0 → x = 0 and x = 1.
On (0, 1): x > x^2 (test x = 1/2: 1/2 > 1/4), so top is y = x.
A = ∫ from 0 to 1 of (x − x^2) dx = [x^2/2 − x^3/3] from 0 to 1 = 1/2 − 1/3 = 1/6

Answer: 1/6. Two habits on display: the integration limits came from *solving the intersection equation* (they are rarely given), and the top/bottom decision was made with an explicit test value rather than a guess.

Worked Example 2: parabola versus line

Problem. Find the area enclosed by y = 4 − x^2 and y = x + 2.

Intersections: 4 − x^2 = x + 2 → x^2 + x − 2 = 0 → (x + 2)(x − 1) = 0 → x = −2, 1.
On (−2, 1): test x = 0: parabola 4 vs line 2 → parabola on top.
Integrand: (4 − x^2) − (x + 2) = 2 − x − x^2.
A = [2x − x^2/2 − x^3/3] from −2 to 1
  At 1: 2 − 1/2 − 1/3 = 7/6.   At −2: −4 − 2 + 8/3 = −10/3.
  A = 7/6 − (−10/3) = 7/6 + 20/6 = 27/6 = 9/2

Answer: 9/2. Note that the integrand is the *difference function*; the most common structural error is integrating each curve separately against the x-axis and subtracting areas that were never the region asked for.

Worked Example 3: slicing with respect to y

Problem. Find the area enclosed by x = y^2 and x = y + 2.

Intersections: y^2 = y + 2 → y^2 − y − 2 = 0 → y = −1 and y = 2.
On (−1, 2): test y = 0: right curve x = 2, left curve x = 0 → right is x = y + 2.
A = ∫ from −1 to 2 of [(y + 2) − y^2] dy = [y^2/2 + 2y − y^3/3] from −1 to 2
  At 2: 2 + 4 − 8/3 = 10/3.   At −1: 1/2 − 2 + 1/3 = −7/6.
  A = 10/3 + 7/6 = 27/6 = 9/2

Answer: 9/2 — the same value as Example 2, and not by accident: the two regions are congruent, viewed from perpendicular directions. (The parabola x = y^2 facing right, cut by the line, encloses the mirror of the downward parabola cut by its line.) The lesson: when y-slices make the boundary description simpler — one right function, one left function, no splitting — switch variables without hesitation.

Worked Example 4: estimating the area from a table (numerical)

Problem. Two functions f and g are tabulated at x = 0, 1, 2, 3, 4: f takes values 5, 6, 4, 7, 5 and g takes values 1, 2, 3, 2, 4, with f ≥ g throughout. Estimate the area between the curves on [0, 4].

Differences d = f − g at the nodes: 4, 4, 1, 5, 1.
Trapezoidal rule with Δx = 1 on d:
  A ≈ 1 × (4/2 + 4 + 1 + 5 + 1/2) = 12.5

Answer: ≈ 12.5. Area between curves from data is just Lesson 14 applied to the difference — one more reminder that the whole of Unit 8 is "integrate the right quantity".

Real-world application: the production gap

Two factories' output rates (units per day) over a quarter are modelled by curves p(t) (the established plant) and q(t) (a retooled plant that starts slower but improves). The area between the curves from the day q overtakes p to the quarter's end is the *total extra production* the retooling ultimately delivered — an accumulated difference, in units, not a mere comparison of final rates. Land surveyors use the identical mathematics when a parcel is bounded by a river (one curve) and a straight road (another): the integral of the gap is the acreage on the deed. "Integrate the difference" is the mathematics of every accumulated gap.

Common mistakes

• Assuming the same curve stays on top. If the curves cross inside the interval and you do not split, the negative piece silently cancels part of the positive piece and the "area" comes out wrong — sometimes even negative, which should always trigger a re-check.
• Using the given interval without finding intersections. Enclosed-region problems set their own limits; [a, b] endpoints in the statement are often just where the picture sits, not where the region is.
• Mixing variables in y-slices: writing ∫ [right − left] dx with y-limits. The variable of integration, the limits, and the boundary functions must all speak the same variable.
• Reporting a signed integral as an area. Area is nonnegative; a negative result is a diagnostic, not an answer.

AP Exam Experience

1. (No calculator.) The area enclosed by y = x and y = x^2 is

• (A) 1/6
• (B) 1/3
• (C) 1/2
• (D) 2/3

Answer: (A). Intersections at 0 and 1; on (0, 1) the line is above the parabola: ∫ from 0 to 1 of (x − x^2) dx = 1/2 − 1/3 = 1/6.

2. (No calculator.) The area enclosed by y = 4 − x^2 and the x-axis is

• (A) 16/3
• (B) 8
• (C) 64/3
• (D) 32/3

Answer: (D). The parabola meets the axis at ±2 and is above it between: ∫ from −2 to 2 of (4 − x^2) dx = [4x − x^3/3] = (8 − 8/3) − (−8 + 8/3) = 16 − 16/3 = 32/3.

3. The region enclosed by y = x^2 − 2x and y = x has area

• (A) 3/2
• (B) 9/2
• (C) 9
• (D) 27/2

Answer: (B). Intersections: x^2 − 2x = x gives x = 0 and x = 3; the line is on top between them: ∫ from 0 to 3 of (3x − x^2) dx = [3x^2/2 − x^3/3] from 0 to 3 = 27/2 − 9 = 9/2.

Free-response question. Let R be the region enclosed by the graphs of y = sqrt(x) and y = x/4.

(a) Find all points of intersection of the two graphs. (b) Write an integral expression for the area of R. Do not evaluate. (c) Evaluate the area of R. (d) Set up an integral with respect to y for the area of R. Do not evaluate.

Scoring guideline.

• (a) 1 point. sqrt(x) = x/4 gives x = x^2/16, so x^2 − 16x = 0 and x = 0 and x = 16 (points (0, 0) and (16, 4)).
• (b) 1 point. On (0, 16), sqrt(x) ≥ x/4 (test x = 4: 2 ≥ 1), so the setup is ∫ from 0 to 16 of (sqrt(x) − x/4) dx. Reversed order earns no point — the setup must encode top minus bottom.
• (c) 2 points. 1 point for a correct antiderivative (2/3) x^(3/2) − x^2/8; 1 point for the value (2/3)(64) − 256/8 = 128/3 − 32 = 32/3.
• (d) 1 point. With x = y^2 (from y = sqrt(x)) as the left boundary and x = 4y (from y = x/4) as the right, y runs 0 to 4: ∫ from 0 to 4 of (4y − y^2) dy. Both boundary conversions and the y-limits must be present.

Extension: crossings inside the interval — the split that saves the problem

When two curves cross *inside* the stated interval, the area formula must be applied piecewise. Consider the region between y = sin x and the line y = 1/2 on [0, π] (the area enclosed between them across the whole arch). The curves meet where sin x = 1/2: at x = π/6 and x = 5π/6. The line is on top on [0, π/6] and [5π/6, π]; the sine is on top between. So:

A = ∫ from 0 to π/6 of (1/2 − sin x) dx + ∫ from π/6 to 5π/6 of (sin x − 1/2) dx + ∫ from 5π/6 to π of (1/2 − sin x) dx.

The middle integral evaluates to [−cos x − x/2] from π/6 to 5π/6 = sqrt(3) − π/3 ≈ 0.685, and by symmetry the two side pieces are equal, each [x/2 + cos x] from 0 to π/6 = π/12 + sqrt(3)/2 − 1 ≈ 0.128; the total is ≈ 0.94. A student who integrates (sin x − 1/2) over the whole [0, π] without splitting gets [−cos x − x/2] from 0 to π = (1 − π/2) − (−1) = 2 − π/2 ≈ 0.43, a number with no geometric meaning: the negative lobes cancelled against the positive one. Whenever the two curves' graphs cross inside the interval, split first. A ten-second sketch (or a sign test at one interior point per subinterval) is the cheapest insurance in Unit 8.