Mathematics Advanced • Year 12 • Module 6 • Lesson 1
Introduction to Integration
Apply antiderivatives to physical and geometric contexts where a rate of change must be turned into a total amount.
Problem 1, Free fall (kinematics)
A small steel ball is released from rest at the top of a stairwell. Its downward velocity is modelled by
v(t) = 9.8 t metres per second, t in seconds.
Set up: What are we solving for?
(i) Find the displacement function s(t), measuring s = 0 from the release point with downward positive. 2 marks
(ii) How far has the ball fallen after 3 seconds? 1 mark
(iii) Now suppose instead the ball is thrown upward from a height of 10 m above the release point with initial speed 5 m/s, using the same gravity. State the new v(t) and s(t), and explain in one sentence how the constant of integration encodes the initial conditions. 3 marks
Stuck? Revisit lesson § Real-World Anchor, Velocity and Displacement.Problem 2, Reconstructing a curve from its gradient
A curve has gradient function
dy/dx = 6x² − 4x + 1
and is known to pass through the point (1, 4).
Set up: What are we solving for?
(i) Find the general antiderivative y in terms of x, including + C. 2 marks
(ii) Use the point (1, 4) to determine C, and state the equation of the curve. 2 marks
(iii) Explain in one sentence why a second curve passing through (0, 0) with the same gradient function would have a different value of C. 1 mark
Problem 3, Revenue accumulation (business)
A small online business records that on day t after launch, sales revenue accrues at a rate of
R'(t) = 120 − 6t dollars per day, 0 ≤ t ≤ 20.
Set up: What are we solving for?
(i) Find R(t), the total revenue function, given R(0) = 0. 2 marks
(ii) Use R(t) to find the total revenue earned over the first 10 days. 2 marks
(iii) The owner thinks revenue is "growing all the time" because R(t) is positive. Identify the day on which R'(t) = 0, state what happens to the rate after that, and explain in one sentence what R(t) does after that day. 2 marks
Stuck? Set R'(t) = 0 and solve; remember R'(t) is the rate, not the total.Problem 4, Water tank with changing inflow
Water flows into a 1000 L holding tank at a rate that decreases over time:
dV/dt = 50 − 2t litres per minute, for 0 ≤ t ≤ 25.
The tank is empty at t = 0.
Set up: What are we solving for?
(i) Find V(t), the volume of water in the tank at time t. 2 marks
(ii) How much water is in the tank after 10 minutes? 1 mark
(iii) The model is only valid while dV/dt ≥ 0. Find the time at which inflow stops, and use V(t) to find the maximum volume reached. Comment in one sentence on whether this exceeds the tank's 1000 L capacity. 3 marks
Problem 5, Auditing four classmates' answers
Four classmates each present an answer to ∫ f'(x) dx. For each, check by differentiating and write ✓ if correct, or ✗ with a one-line correction.
Set up: What are we solving for?
(i) Audit the four answers. 4 marks (1 each)
| Q | Claim | ✓ / ✗ and one-line note |
|---|---|---|
| A | ∫ 5x⁴ dx = x⁵ + C | |
| B | ∫ (6x² − 4) dx = 2x³ − 4x | |
| C | ∫ 7 dx = 7 | |
| D | ∫ (3x² + 2x) dx = x³ + x² + 5 |
(ii) Identify the single most common error type appearing in the wrong answers above and describe in one sentence how to avoid it. 2 marks
Stuck? Remember every indefinite integral needs + C, but a specific value of C (like + 5) needs justification from a condition.How did this worksheet feel?
What I'll revisit before next class:
Problem 1, Free fall
Set up. Integrate the velocity v(t) to obtain s(t), then use initial conditions to determine the constant of integration, and finally evaluate at t = 3.
(i) s(t) = ∫9.8t dt = 4.9t² + C. At t = 0, s = 0, so C = 0. s(t) = 4.9t² m.
(ii) s(3) = 4.9 × 9 = 44.1 m.
(iii) With upward positive: v(t) = 5 − 9.8t (initial 5 m/s up, gravity 9.8 m/s² down). s(t) = 5t − 4.9t² + C; s(0) = 10 ⇒ C = 10, so s(t) = 5t − 4.9t² + 10. The constant of integration captures the initial position; the antiderivative alone determines displacement only up to this constant, which the initial condition fixes.
Problem 2, Curve from gradient
Set up. Antidifferentiate dy/dx to get y in terms of x with an unknown C, then use the given point to solve for C.
(i) y = ∫(6x² − 4x + 1) dx = 2x³ − 2x² + x + C.
(ii) At (1, 4): 4 = 2 − 2 + 1 + C, so C = 3. y = 2x³ − 2x² + x + 3.
(iii) A curve through (0, 0) with the same gradient has the same general form, but the point forces C = 0, every different boundary point shifts the curve vertically and so changes C.
Problem 3, Revenue accumulation
Set up. Antidifferentiate R'(t) to obtain R(t) with C, fix C from R(0) = 0, then evaluate R(10) and locate the day R' = 0.
(i) R(t) = ∫(120 − 6t) dt = 120t − 3t² + C. R(0) = 0 ⇒ C = 0. R(t) = 120t − 3t².
(ii) R(10) = 1200 − 300 = $900.
(iii) R'(t) = 0 when 120 − 6t = 0, i.e. t = 20. For t > 20, R'(t) < 0, meaning revenue would decrease day-on-day; correspondingly R(t) has a maximum at t = 20 and falls thereafter (so the model is only credible up to day 20, consistent with the stated domain).
Problem 4, Tank inflow
Set up. Antidifferentiate dV/dt to obtain V(t), fix C from V(0) = 0, evaluate V(10), then find the time at which inflow stops (dV/dt = 0) and compute V at that time.
(i) V(t) = 50t − t² + C; V(0) = 0 ⇒ C = 0. V(t) = 50t − t².
(ii) V(10) = 500 − 100 = 400 L.
(iii) Inflow stops when dV/dt = 50 − 2t = 0 ⇒ t = 25 min. V(25) = 1250 − 625 = 625 L, which is below the 1000 L capacity, the tank never overflows under this model.
Problem 5, Audit
Set up. For each claim, differentiate the proposed antiderivative; if it matches the integrand it is correct (the constant of integration is the only remaining question).
(i)
A. d/dx(x⁵ + C) = 5x⁴ ✓. Correct.
B. d/dx(2x³ − 4x) = 6x² − 4 ✓ for the derivative, but the answer is ✗ missing + C. Correct: 2x³ − 4x + C.
C. d/dx(7) = 0 ≠ 7. ✗ Correct: ∫7 dx = 7x + C (constant integrand integrates to a linear function).
D. d/dx(x³ + x² + 5) = 3x² + 2x ✓, but the "+ 5" is a specific constant not justified by any boundary condition. ✗, should be + C, not + 5.
(ii) The most common error is handling the constant of integration either omitting + C (B), or replacing it with an arbitrary specific value (D), or treating a constant integrand as if integration leaves it unchanged (C). Avoid this by always writing + C on every indefinite integral and only fixing its value when a condition is given.