Mathematics Advanced • Year 12 • Module 6 • Lesson 6
Areas Between Curves
Apply area-between-curves techniques to financial, geometric, biological and physical contexts.
Problem 1 — Profit between break-even points (financial)
A small Australian manufacturer models its weekly revenue and cost (in dollars) by
R(x) = 100x and C(x) = x² + 20x + 400
where x is the number of units produced per week.
Set up: What are we solving for?
(i) Find the break-even points by solving R(x) = C(x). Give exact values then 1 d.p. approximations. 2 marks
(ii) Set up (but do not yet evaluate) the integral that gives total profit over production levels between the break-even points. State which function is on top and how you tested it. 2 marks
(iii) Evaluate the profit integral and interpret the value in dollars per week of production span. 3 marks
Stuck? Revisit lesson § Real-World Anchor — Profit Region in Economics.Problem 2 — Sail-cloth between two parabolas (geometric)
A boat designer cuts a sail panel out of cloth whose top edge follows y = 8 − x² and whose bottom edge follows y = x² (both in metres above the boom).
Set up: What are we solving for?
(i) Find the x-coordinates of the corners of the panel (the intersection points). 2 marks
(ii) Find the area of the sail panel in square metres. 3 marks
(iii) Cloth costs $42/m². The designer says "the panel area is symmetric — I'll just compute ∫02 (8 − x² − x²) dx and double it." Is this valid? Justify, and state the cloth cost. 2 marks
Problem 3 — Two competing populations (biological)
An ecologist models two competing kangaroo populations on neighbouring reserves. The instantaneous growth rates (in animals per year) at time t (years) are
RA(t) = 60 − 4t, RB(t) = 20 + 4t, for 0 ≤ t ≤ 10
Set up: What are we solving for?
(i) Find the time t at which the two growth rates are equal. 1 mark
(ii) Find the area between the two growth curves on [0, 10] (you may need to split). Interpret what each piece represents biologically: who out-grew whom and by how many animals. 4 marks
(iii) Without recomputing, explain how you know the two pieces of the split must be equal in this case. 1 mark
Stuck? Revisit lesson § More Examples — When Curves Cross Multiple Times.Problem 4 — Dam cross-section (physical)
An engineer sketches a cross-section of a small concrete dam. The outer (upstream) face follows y = (1/4)x² from the base at y = 0 up to y = 4, and the inner (downstream) face follows y = x on the same height interval (x and y in metres, x measured from the dam's leading edge).
Set up: What are we solving for?
(i) Find the heights at which the two faces meet (the intersection points of the two curves). 2 marks
(ii) Find the cross-sectional area of concrete (in m²) — the region between the two curves. 3 marks
(iii) If the dam is 50 m long and concrete masses 2400 kg/m³, estimate the total mass (in tonnes) of concrete in the dam. 2 marks
Problem 5 — Net displacement vs total distance (kinematics)
A particle moves with velocity v(t) = 3t² − 12 m/s for 0 ≤ t ≤ 3 s. A second particle has velocity u(t) = 0 m/s (stationary). The "area between v(t) and u(t)" interpretation matters because v(t) changes sign.
Set up: What are we solving for?
(i) Find the time at which v(t) = 0. State on which subinterval v(t) is negative. 1 mark
(ii) Compute the signed area ∫03 v(t) dt. This is the net displacement of the moving particle. 2 marks
(iii) Compute the total area between v(t) and u(t) on [0, 3] by splitting at the sign change. Why does this equal the total distance travelled, and not equal the answer in (ii)? 3 marks
Stuck? Net displacement = signed integral; total distance = unsigned area. Revisit § Setting Up.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Profit between break-evens
Set up. We are finding production levels at which revenue equals cost (break-even), then summing R − C between those levels to get total profit.
(i) 100x = x² + 20x + 400 ⇒ x² − 80x + 400 = 0 ⇒ x = (80 ± √(6400 − 1600))/2 = 40 ± √1200 = 40 ± 20√3 ≈ 5.4 and 74.6 units.
(ii) Test x = 40: R(40) = 4000, C(40) = 1600 + 800 + 400 = 2800, so R > C. The profit integral is P = ∫40−20√340+20√3 (100x − x² − 20x − 400) dx = ∫40−20√340+20√3 (−x² + 80x − 400) dx.
(iii) Antiderivative: −x³/3 + 40x² − 400x. Evaluating between the symmetric limits (or using the fact that the integrand factors as −(x − (40−20√3))(x − (40+20√3)) so the integral equals (1/6)(b − a)³ with b − a = 40√3): P = (1/6)(40√3)³ = (1/6)(64000 · 3√3) = 32000√3 ≈ $55 425. The area literally represents the total accumulated profit over the production range between break-evens.
Problem 2 — Sail panel
Set up. We are finding the area of the region enclosed by the two parabolic edges, then converting to cloth cost.
(i) 8 − x² = x² ⇒ 2x² = 8 ⇒ x = ±2. The corners are at x = −2 and x = 2.
(ii) On (−2, 2), 8 − x² > x² (test x = 0: 8 > 0). A = ∫−22 ((8 − x²) − x²) dx = ∫−22 (8 − 2x²) dx = [8x − 2x³/3]−22 = (16 − 16/3) − (−16 + 16/3) = 64/3 m² ≈ 21.33 m².
(iii) Yes — the integrand 8 − 2x² is even and the interval is symmetric about 0, so ∫−22 = 2∫02. Cloth cost = (64/3) × $42 = $896.
Problem 3 — Competing populations
Set up. We are finding when the two growth-rate curves cross, then quantifying the cumulative "lead" each population had over the 10-year window.
(i) 60 − 4t = 20 + 4t ⇒ 40 = 8t ⇒ t = 5 years.
(ii) Split at t = 5. On [0, 5], RA > RB; on [5, 10], RB > RA.
Area on [0, 5] = ∫05 ((60 − 4t) − (20 + 4t)) dt = ∫05 (40 − 8t) dt = [40t − 4t²]05 = 200 − 100 = 100 animals.
Area on [5, 10] = ∫510 ((20 + 4t) − (60 − 4t)) dt = ∫510 (8t − 40) dt = [4t² − 40t]510 = (400 − 400) − (100 − 200) = 100 animals.
Total area between the curves = 200 animals. Biologically: over the first 5 years population A added 100 more animals than B; over the next 5 years population B added 100 more than A. They finish at the same net.
(iii) The two growth-rate lines are reflections of each other about the horizontal line y = 40 (60 − 4t and 20 + 4t average to 40). By that mirror symmetry, the "lead" triangle on [0, 5] is congruent to the "deficit" triangle on [5, 10].
Problem 4 — Dam cross-section
Set up. We are finding the cross-sectional area between two curves describing the dam's two faces, then scaling to mass using length and density.
(i) (1/4)x² = x ⇒ x² − 4x = 0 ⇒ x = 0 or x = 4 (which corresponds to y = 4). Faces meet at the base (0, 0) and the top (4, 4).
(ii) On (0, 4), x > (1/4)x² (test x = 2: 2 > 1). A = ∫04 (x − x²/4) dx = [x²/2 − x³/12]04 = 8 − 64/12 = 8 − 16/3 = 8/3 m² ≈ 2.667 m².
(iii) Volume = (8/3) × 50 = 400/3 m³. Mass = 400/3 × 2400 = 320 000 kg = 320 tonnes.
Problem 5 — Displacement vs distance
Set up. We are distinguishing the signed integral of velocity (net displacement) from the unsigned area between v(t) and 0 (total distance).
(i) 3t² − 12 = 0 ⇒ t = 2 (positive root). On [0, 2), v(t) < 0 (test t = 1: 3 − 12 = −9). On (2, 3], v(t) > 0.
(ii) Net displacement = ∫03 (3t² − 12) dt = [t³ − 12t]03 = 27 − 36 = −9 m. (Particle ends 9 m behind the start.)
(iii) Split at t = 2: Total distance = ∫02 −(3t² − 12) dt + ∫23 (3t² − 12) dt = [−t³ + 12t]02 + [t³ − 12t]23 = (−8 + 24) + ((27 − 36) − (8 − 24)) = 16 + (−9 − (−16)) = 16 + 7 = 23 m. Unsigned area between v and 0 ignores direction, so it adds the backwards travel (16 m) to the forwards travel (7 m). The net answer in (ii) instead subtracts them because the curve dips below the axis.