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Lesson 88 — Area Between Curves

A = ∫ₐᵇ [f(x) − g(x)] dx, with f ≥ g on [a, b]. Finding intersection points, choosing the integration axis, curve crossings.

Used in: Calculus II (Brazil) · Equiv. Math III Japanese · Equiv. Analysis LK German · AP Calculus BC (USA)

A = \int_a^b [f(x) - g(x)]\, dx, \quad f(x) \geq g(x) \text{ on } [a, b]

Area between curves: integrate upper minus lower. If the curves cross, divide the interval at the crossing points. In some problems, integrating with respect to y simplifies the computation.

Choose your door

Rigorous notation, full derivation, hypotheses

Definition, justification, and procedure

Definition and justification via Riemann

Definition· Area between two curves

Let $f$ and $g$ be continuous on $[a, b]$ with $f(x) \geq g(x)$ for all $x \in [a, b]$ . The area of the region $R = \{(x, y) : a \leq x \leq b,\ g(x) \leq y \leq f(x)\}$ is

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

what this means · Fundamental formula for area between curves — integration of upper minus lower.

Justification: for each $x_i^*$ in a subinterval of width $\Delta x$ , a rectangle of height $[f(x_i^*) - g(x_i^*)]$ and width $\Delta x$ approximates the local area. The Riemann sum $\sum_i [f(x_i^*) - g(x_i^*)]\Delta x$ converges to the integral as $n \to \infty$ .

"The area of the region between the graphs of $f$ and $g$ is found by integrating the difference $f - g$ over the interval, provided $f \geq g$ throughout. If the graphs cross, break the interval at the crossing points." — Active Calculus §6.1

Integration with respect to $y$

Left: integration in x (vertical rectangles). Right: integration in y (horizontal rectangles).

General procedure

"Finding the area of a region between two curves requires careful attention to the sign of the integrand. Always determine which function is greater on the interval of integration." — APEX Calculus §7.1

Worked examples

Example— 1· Classic area x² and x (basic)

Problem. Compute the area between $y = x$ and $y = x^2$ on the interval $[0, 1]$ .

Strategy. Find intersection, check which is above, integrate the difference.

Solution.

Intersection: $x = x^2 \Rightarrow x(x-1) = 0 \Rightarrow x = 0$ or $x = 1$ .

On $(0, 1)$ : $x > x^2$ (check: $0.5 > 0.25$ ). So $f = x$ , $g = x^2$ .

$A = \int_0^1 (x - x^2)\, dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}.$

Verification. $1/6 \approx 0.167$ . The sketch shows the region covering roughly $17\%$ of the square $[0,1]^2$ . Plausible.

Source. Active Calculus §6.1, Activity 6.1.1 — CC-BY-NC-SA.

Example— 2· Curves with crossing — sin and cos (intermediate)

Problem. Compute the area between $y = \sin x$ and $y = \cos x$ on $[0, \pi/2]$ .

Strategy. The curves cross at $x = \pi/4$ . Divide and integrate each section separately.

Solution.

Crossing: $\sin x = \cos x \Rightarrow \tan x = 1 \Rightarrow x = \pi/4$ .

On $[0, \pi/4]$ : $\cos x \geq \sin x$ . On $[\pi/4, \pi/2]$ : $\sin x \geq \cos x$ .

$A = \int_0^{\pi/4}(\cos x - \sin x)\, dx + \int_{\pi/4}^{\pi/2}(\sin x - \cos x)\, dx$ .

$= [\sin x + \cos x]_0^{\pi/4} + [-\cos x - \sin x]_{\pi/4}^{\pi/2}$

$= (\sqrt{2} - 1) + ((-0 - 1) - (-\sqrt{2}/2 - \sqrt{2}/2))$

$= (\sqrt{2} - 1) + (-1 + \sqrt{2}) = 2\sqrt{2} - 2 \approx 0.83$ .

Verification. By symmetry of the problem, the two parts are equal: each is $\sqrt{2} - 1$ . Sum $= 2(\sqrt{2} - 1)$ . Consistent.

Source. APEX Calculus §7.1, Example 7.1.3 — CC-BY-NC.

Example— 3· Integration in y — horizontal parabola (intermediate)

Problem. Compute the area of the region bounded by $x = y^2$ and $x = y + 2$ .

Strategy. The curves are more natural in $y$ . Find intersection in $y$ , integrate $h(y) - k(y)$ .

Solution.

Intersection: $y^2 = y + 2 \Rightarrow y^2 - y - 2 = 0 \Rightarrow (y-2)(y+1) = 0 \Rightarrow y = -1$ or $y = 2$ .

On $[-1, 2]$ : $y + 2 \geq y^2$ (check at $y = 0$ : $2 \geq 0$ ). So $h = y + 2$ , $k = y^2$ .

$A = \int_{-1}^2 [(y + 2) - y^2]\, dy = \left[\frac{y^2}{2} + 2y - \frac{y^3}{3}\right]_{-1}^2.$

At $y = 2$ : $2 + 4 - 8/3 = 6 - 8/3 = 10/3$ .

At $y = -1$ : $1/2 - 2 + 1/3 = -7/6$ .

$A = 10/3 - (-7/6) = 20/6 + 7/6 = 27/6 = 9/2$ .

Verification. Integration in $x$ would require two parts (the parabola $x = y^2$ has two "branches" in $x$ ) — more tedious. Integration in $y$ was more efficient.

Source. OpenStax Calculus Vol. 2, §2.1, Example 2.5 — CC-BY-NC-SA.

Example— 4· Consumer surplus (modeling)

Problem. Demand curve $D(q) = 100 - q$ , equilibrium price $p^* = 60$ , quantity $Q^* = 40$ . Compute consumer surplus.

Strategy. $CS = \int_0^{Q^*} [D(q) - p^*]\, dq$ — area between demand curve and horizontal price line.

Solution.

$CS = \int_0^{40} [(100 - q) - 60]\, dq = \int_0^{40}(40 - q)\, dq = \left[40q - \frac{q^2}{2}\right]_0^{40} = 1600 - 800 = 800.$

Verification. Geometrically, the region is a triangle of base $40$ and height $D(0) - p^* = 40$ . Area = $(40 \cdot 40)/2 = 800$ . Confirms.

Source. Active Calculus §6.1, Exercise 6.1.5 — CC-BY-NC-SA.

Example— 5· Area with subdivision — cubic polynomial (challenge)

Problem. Compute the total area of the region between $y = x^3 - 4x$ and the $x$ -axis on $[-2, 2]$ .

Strategy. Identify where $x^3 - 4x$ changes sign; divide the interval; use $|f|$ in each part.

Solution.

Zeros: $x^3 - 4x = x(x^2 - 4) = 0 \Rightarrow x = -2, 0, 2$ .

On $[-2, 0]$ : $f(x) = x^3 - 4x \geq 0$ (check $f(-1) = -1 + 4 = 3 > 0$ ).

On $[0, 2]$ : $f(x) \leq 0$ (check $f(1) = 1 - 4 = -3 < 0$ ).

$A = \int_{-2}^0 (x^3 - 4x)\, dx + \int_0^2 [0 - (x^3 - 4x)]\, dx$ .

For the 1st part: $\left[\frac{x^4}{4} - 2x^2\right]_{-2}^0 = 0 - (4 - 8) = 4$ .

For the 2nd part: $\left[-\frac{x^4}{4} + 2x^2\right]_0^2 = (-4 + 8) - 0 = 4$ .

$A = 4 + 4 = 8$ .

Verification. By odd symmetry of $f$ , the two areas are equal. Total $= 8$ .

Source. APEX Calculus §7.1, Example 7.1.4 — CC-BY-NC.

Exercise list

45 exercises · 11 with worked solution (25%)

Application 31Understanding 3Modeling 4Challenge 6Proof 1

Sources

Active Calculus 2.0 — Boelkins · 2024 · CC-BY-NC-SA · §6.1. Primary source.
APEX Calculus v5 — Hartman et al. · 2024 · CC-BY-NC · §7.1.
Calculus Volume 2 (OpenStax) — OpenStax · 2016 · CC-BY-NC-SA · §2.1.

Definition, justification, and procedure

Definition and justification via Riemann

Integration with respect to yyy

General procedure

Worked examples

Sources

Integration with respect to $y$