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.