Mathematics Advanced • Year 11 • Module 3 • Lesson 12
Areas
Apply definite integrals to rainfall, oil spills, garden beds, energy bills, and competitive runner pacing — including regions partly below the axis and regions between two curves.
Problem 1 — Total rainfall from a rate model
During a storm, the rate of rainfall in millimetres per hour at time t hours is modelled by
r(t) = 6t − t², for 0 ≤ t ≤ 6.
Set up: What are we solving for?
(i) Total rainfall is the area under r(t) on [0, 6]. Evaluate ∫₀⁶ r(t) dt. 2 marks
(ii) Find the total rainfall during the first 3 hours, and during the last 3 hours. 2 marks
(iii) One forecaster claims "the rain falls evenly throughout the storm". Use parts (i) and (ii) to evaluate that claim in one sentence. 2 marks
Stuck? The total rain over each subinterval is ∫ r(t) dt over that subinterval.Problem 2 — Net displacement during a leak-and-repair
An oil spill rate (positive = oil escaping, negative = oil being pumped back), in litres per hour, is modelled by
f(t) = t² − 4t + 3, for 0 ≤ t ≤ 4.
Set up: What are we solving for?
(i) Compute the signed integral ∫₀⁴ f(t) dt. What does the sign of your answer mean physically? 2 marks
(ii) Find the times in [0, 4] when f(t) = 0. Split the integral at these times and compute the total volume of oil moved (the total area between f and the t-axis). 3 marks
(iii) A clean-up crew is paid per litre of oil they handle (in either direction). Which is the relevant figure — the signed integral or the total area — and why? 2 marks
Problem 3 — Curved garden bed
A landscape architect designs a garden bed bounded by the curves y = x² and y = 2x (all measurements in metres) for 0 ≤ x ≤ 2.
Set up: What are we solving for?
(i) Find the x-coordinates of the curves' intersections. 1 mark
(ii) Determine which curve lies above the other on the enclosed region, and write the area as a single definite integral. 2 marks
(iii) Evaluate the integral to find the area of the garden bed. Then, if topsoil costs $42 per square metre, find the total topsoil cost. 3 marks
Stuck? Revisit lesson § Worked Example 3 — area between two curves uses ∫ (top − bottom) dx.Problem 4 — Total energy from a power curve
The power consumption of a small workshop, in kilowatts at time t hours after 9 am, is modelled by
P(t) = 10 + 6t − t², for 0 ≤ t ≤ 8.
(Energy used = ∫ P dt, measured in kilowatt-hours.)
Set up: What are we solving for?
(i) Find the total energy consumed during the 8-hour workday. 2 marks
(ii) The retailer charges $0.32 per kWh from 9 am to 1 pm (peak) and $0.18 per kWh from 1 pm to 5 pm (off-peak). Calculate the total bill. 3 marks
(iii) The owner is offered a flat $0.27/kWh deal. State whether they should accept, justifying with the totals above. 2 marks
Problem 5 — Lead between two runners
Two runners A and B set off from the same point. Their velocities (m/s) at time t seconds are
v_A(t) = 6t, v_B(t) = 3t² − 6t + 9, for 0 ≤ t ≤ 3.
(Position is the integral of velocity.)
Set up: What are we solving for?
(i) Find the position of each runner at t = 3 s by computing ∫₀³ v_A(t) dt and ∫₀³ v_B(t) dt. 2 marks
(ii) The gap (distance B is ahead of A) at time T is ∫₀ᵀ (v_B − v_A) dt. Write this integrand as a simple quadratic in t and find the times in [0, 3] at which the gap is momentarily not changing. 2 marks
(iii) Sketch a graph of v_A and v_B on the same axes for 0 ≤ t ≤ 3, then describe what the area between the two curves represents physically. 3 marks
Stuck on (ii)? Set v_B(t) − v_A(t) = 0; the t-values where this is zero are when the two runners have equal speed (the gap stops changing instantaneously).How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Rainfall r(t) = 6t − t²
Set up. Total rainfall is the area under the rate curve, i.e. the definite integral.
(i) ∫₀⁶ (6t − t²) dt = [3t² − t³/3]₀⁶ = (108 − 72) − 0 = 36 mm.
(ii) ∫₀³ (6t − t²) dt = [3t² − t³/3]₀³ = (27 − 9) = 18 mm (first 3 hours). ∫₃⁶ (6t − t²) dt = (108 − 72) − (27 − 9) = 36 − 18 = 18 mm (last 3 hours).
(iii) The two halves are equal (18 mm each), but the rate r(t) is not constant — it peaks at t = 3 and is small near the start and end. So while equal totals make it sound even, the storm is actually heaviest mid-storm; "evenly" depends on how you measure it (totals vs intensity).
Problem 2 — Oil rate f(t) = t² − 4t + 3
Set up. f = (t − 1)(t − 3): positive on [0, 1] and [3, 4], negative on [1, 3].
(i) ∫₀⁴ (t² − 4t + 3) dt = [t³/3 − 2t² + 3t]₀⁴ = (64/3 − 32 + 12) − 0 = 64/3 − 20 = 4/3 L. Positive sign means the net effect over the 4 hours is that 4/3 L of oil escaped (more escape than was pumped back).
(ii) Zeros at t = 1 and t = 3. F(t) = t³/3 − 2t² + 3t. F(0) = 0; F(1) = 1/3 − 2 + 3 = 4/3; F(3) = 9 − 18 + 9 = 0; F(4) = 64/3 − 32 + 12 = 4/3. Pieces: ∫₀¹ = 4/3 (positive, oil escaping); ∫₁³ = 0 − 4/3 = −4/3 (negative, oil pumped back); ∫₃⁴ = 4/3 − 0 = 4/3 (positive). Total area moved = 4/3 + 4/3 + 4/3 = 4 L of oil moved in total.
(iii) Pay per litre handled = total area (4 L), not the signed integral (4/3 L). The signed integral only tells net displacement; the crew physically moved 4 litres back and forth, and they should be paid for handling all of it.
Problem 3 — Garden bed between y = x² and y = 2x
Set up. Find intersections, decide top curve, integrate the difference.
(i) x² = 2x ⇒ x(x − 2) = 0 ⇒ x = 0 and x = 2.
(ii) Test x = 1: y = 2x gives 2; y = x² gives 1. So 2x is on top. Area = ∫₀² (2x − x²) dx.
(iii) = [x² − x³/3]₀² = 4 − 8/3 = 4/3 m². Topsoil cost = 4/3 × $42 = $56.
Problem 4 — Workshop power P(t) = 10 + 6t − t²
Set up. Energy = ∫ P dt. Bill = energy × rate per kWh, split by time-of-day.
(i) ∫₀⁸ (10 + 6t − t²) dt = [10t + 3t² − t³/3]₀⁸ = 80 + 192 − 512/3 = 272 − 170.67 ≈ 101.33 kWh (exactly 304/3 kWh).
(ii) Peak (0 ≤ t ≤ 4): ∫₀⁴ P dt = [10t + 3t² − t³/3]₀⁴ = 40 + 48 − 64/3 = 88 − 64/3 = 200/3 ≈ 66.67 kWh. Off-peak (4 ≤ t ≤ 8): total − peak = 304/3 − 200/3 = 104/3 ≈ 34.67 kWh. Bill = 200/3 × $0.32 + 104/3 × $0.18 = $21.33 + $6.24 = $27.57 (to nearest cent).
(iii) Flat-rate bill = 304/3 × $0.27 ≈ $27.36. The flat-rate deal saves about 21 c per day on this usage profile — yes, accept the flat deal (and the saving grows if peak usage rises in future).
Problem 5 — Runners v_A = 6t, v_B = 3t² − 6t + 9
Set up. Position is ∫ v dt; the gap (B ahead of A) is ∫₀ᵀ (v_B − v_A) dt.
(i) ∫₀³ 6t dt = [3t²]₀³ = 27 m (runner A). ∫₀³ (3t² − 6t + 9) dt = [t³ − 3t² + 9t]₀³ = 27 − 27 + 27 = 27 m (runner B). Both end at 27 m — a tie after 3 s.
(ii) v_B − v_A = 3t² − 6t + 9 − 6t = 3t² − 12t + 9 = 3(t² − 4t + 3) = 3(t − 1)(t − 3). Set equal to zero: t = 1 and t = 3. At these times the runners have the same velocity, so the gap is instantaneously not changing.
(iii) v_A is a straight line from (0, 0) through (3, 18); v_B is an upward-opening parabola through (0, 9), (1, 6) (minimum at t = 1), passing through (3, 18). For 0 ≤ t ≤ 1, v_B > v_A so the gap (B − A) is growing; for 1 ≤ t ≤ 3, v_A > v_B so the gap is shrinking. The area between v_B and v_A from 0 to 1 equals how far B pulled ahead of A in that first second; the same-sized area between v_A and v_B from 1 to 3 represents how much A made it back to draw level at t = 3 s.