Mathematics Advanced • Year 11 • Module 3 • Lesson 10
Integration
Apply integration to reconstruct functions from their derivatives in kinematics, finance, and curve-of-best-fit contexts.
Problem 1 — Velocity to position (kinematics)
A drone moves vertically with velocity (in m/s)
v(t) = 6t² − 12t + 4, t ≥ 0
and starts at height s(0) = 3 m above the launch pad.
Set up: What are we solving for?
(i) Integrate v(t) to obtain a general expression for s(t) in the form s(t) = ... + C. 2 marks
(ii) Use the initial condition s(0) = 3 to find C, hence write the specific s(t). 2 marks
(iii) Find the drone's height at t = 2 s. 1 mark
Stuck? The initial condition lets you solve for C exactly.Problem 2 — Rainfall accumulation (rate to total)
During a Sydney summer storm, rain falls at a rate
r(t) = 2t − 0.4t² mm per hour, 0 ≤ t ≤ 5
where t is hours since the storm started. Let R(t) be the cumulative rainfall (mm) collected in a gauge, with R(0) = 0.
Set up: What are we solving for?
(i) Integrate r(t) to obtain R(t) (use R(0) = 0 to fix the constant). 3 marks
(ii) How much rain has fallen after 3 hours? 2 marks
(iii) At t = 5 hours the rain stops; how much in total fell during the storm? Comment on the reasonableness of this figure for a Sydney summer storm. 2 marks
Problem 3 — Marginal cost to total cost (finance)
The marginal cost (the cost of producing one extra item when x items have already been made) for a Sydney workshop is
C′(x) = 6x + 20 dollars per item, x ≥ 0
The fixed cost (when x = 0) is $100.
Set up: What are we solving for?
(i) Integrate C′(x) and use C(0) = 100 to find the specific total cost function C(x). 3 marks
(ii) Find the total cost of producing 10 items. 2 marks
(iii) What is the additional cost of producing the 11th item (i.e. C(11) − C(10))? Compare this to C′(10), the predicted marginal cost. Are they close? Why? 2 marks
Stuck on (iii)? Compute both numbers separately, then compare; "marginal" approximates "incremental" because the cost function is differentiable.Problem 4 — Recover a curve from gradient + point (geometric)
A curve y = f(x) satisfies dy/dx = 3x² − 6x + 1 and passes through the point (1, 0).
Set up: What are we solving for?
(i) Find the general form y = f(x) + C by integration. 2 marks
(ii) Use the point (1, 0) to find C, and write the specific function. 2 marks
(iii) Find the y-intercept of the curve (i.e. the value of f at x = 0). 1 mark
Problem 5 — When simplifying makes a hard integral easy (technique)
Consider four integrals:
A. ∫ (x³ + 2x² − x)/x dx
B. ∫ (2x − 1)² dx
C. ∫ x · (x + 4) dx
D. ∫ (x⁴ − 3x³ + x²)/x² dx
Set up: What are we solving for?
(i) For each, write the integrand in simplified form (as a sum of power terms xn) before integrating. 2 marks
(ii) Integrate each. Remember +C in every answer. 4 marks
(iii) Verify one of your answers (your choice) by differentiating, and show the steps. 1 mark
Stuck? Always check if the integrand can be rewritten using basic algebra before reaching for a rule.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Drone height from velocity
Set up. We are recovering position from velocity by anti-differentiation, then pinning down the constant using an initial condition.
(i) s(t) = ∫ v(t) dt = ∫ (6t² − 12t + 4) dt = 2t³ − 6t² + 4t + C.
(ii) s(0) = 0 − 0 + 0 + C = 3 ⇒ C = 3. So s(t) = 2t³ − 6t² + 4t + 3.
(iii) s(2) = 2(8) − 6(4) + 4(2) + 3 = 16 − 24 + 8 + 3 = 3 m above the launch pad.
Problem 2 — Rainfall
Set up. We are turning a rate (mm/hour) into a cumulative total (mm) by integration.
(i) R(t) = ∫ (2t − 0.4t²) dt = t² − 0.4t³/3 + C = t² − (2/15)t³ + C. (Equivalently: t² − 0.1333t³ + C.) Use R(0) = 0 ⇒ C = 0. So R(t) = t² − (2/15)t³.
(ii) R(3) = 9 − (2/15)(27) = 9 − 54/15 = 9 − 3.6 = 5.4 mm.
(iii) R(5) = 25 − (2/15)(125) = 25 − 250/15 = 25 − 16.67 ≈ 8.33 mm. That total over five hours is on the light side for a Sydney summer thunderstorm (intense storms can deliver 20-50 mm/hour in short bursts), but plausible for a steady drizzly afternoon shower. The model also has r(5) = 10 − 10 = 0, consistent with the rain stopping at t = 5.
Problem 3 — Marginal to total cost
Set up. We are integrating a marginal cost to get a total cost, with the fixed cost as the initial condition.
(i) C(x) = ∫ (6x + 20) dx = 3x² + 20x + K. Use C(0) = 100 ⇒ K = 100. So C(x) = 3x² + 20x + 100.
(ii) C(10) = 3(100) + 200 + 100 = 300 + 200 + 100 = $600.
(iii) C(11) = 3(121) + 220 + 100 = 363 + 220 + 100 = $683. Incremental cost C(11) − C(10) = 683 − 600 = $83. Marginal prediction C′(10) = 6(10) + 20 = $80. They differ by $3 because the marginal cost is the instantaneous derivative at x = 10, but the actual cost of the 11th item averages the derivative over the interval [10, 11]; for a quadratic cost the discrepancy is exactly (1/2) · C″ · 1² = (1/2)(6)(1) = 3, which matches.
Problem 4 — Curve from gradient + point
Set up. We are reconstructing a function from its derivative plus one known value.
(i) y = ∫ (3x² − 6x + 1) dx = x³ − 3x² + x + C.
(ii) Through (1, 0): 1 − 3 + 1 + C = 0 ⇒ −1 + C = 0 ⇒ C = 1. So y = x³ − 3x² + x + 1.
(iii) f(0) = 0 − 0 + 0 + 1 = 1. The y-intercept is at (0, 1).
Problem 5 — Simplify first
Set up. We are exploiting algebraic simplification so the power rule applies directly to each term.
(i) Simplified integrands:
A. (x³ + 2x² − x)/x = x² + 2x − 1.
B. (2x − 1)² = 4x² − 4x + 1.
C. x(x + 4) = x² + 4x.
D. (x⁴ − 3x³ + x²)/x² = x² − 3x + 1.
(ii) Integrals:
A. ∫ (x² + 2x − 1) dx = x³/3 + x² − x + C.
B. ∫ (4x² − 4x + 1) dx = 4x³/3 − 2x² + x + C.
C. ∫ (x² + 4x) dx = x³/3 + 2x² + C.
D. ∫ (x² − 3x + 1) dx = x³/3 − 3x²/2 + x + C.
(iii) Verification (e.g. for A): d/dx (x³/3 + x² − x + C) = x² + 2x − 1, which matches the simplified integrand from (i)A. ✓