Mathematics Advanced • Year 12 • Module 6 • Lesson 4
The Definite Integral
Apply definite integrals to distance/displacement, signed-area, and accumulation problems, including the displacement-vs-distance distinction.
Problem 1 — Odometer (the car analogy)
A car has velocity
v(t) = 3 t² metres per second, 0 ≤ t ≤ 4 s.
Set up: What are we solving for?
(i) Use a definite integral to find the distance travelled from t = 0 to t = 4. 2 marks
(ii) Find the distance travelled from t = 2 to t = 4 using the splitting property and your answer to (i). 2 marks
(iii) Compare your answer in (ii) with the distance travelled from t = 0 to t = 2. Explain in one sentence why the later 2 seconds covers more distance than the first 2 seconds. 2 marks
Stuck? Revisit lesson § Real-World Anchor — Velocity, Distance, and the Odometer.Problem 2 — Displacement vs distance (the crucial distinction)
A particle moves on a straight line with velocity
v(t) = 12 − 2 t m/s, for 0 ≤ t ≤ 8 s.
Set up: What are we solving for?
(i) Find the time at which v(t) = 0 and explain in one line what happens to the direction of motion before and after that time. 2 marks
(ii) Compute the displacement ∫08(12 − 2t) dt. 2 marks
(iii) Compute the total distance using |∫06| + |∫68| and explain in one sentence why this differs from your answer in (ii). 3 marks
Problem 3 — Tank draining (accumulation by integration)
A tank loses water at a rate (litres per minute)
L(t) = 4 + t, 0 ≤ t ≤ 10.
Set up: What are we solving for?
(i) Use a definite integral to find the total water lost between t = 0 and t = 5. 2 marks
(ii) Find the total water lost between t = 5 and t = 10. 2 marks
(iii) Use the splitting property to confirm that the total water lost between t = 0 and t = 10 equals the sum of your answers in (i) and (ii). 2 marks
Stuck? Splitting: ∫010 = ∫05 + ∫510.Problem 4 — Signed area between curve and axis
Consider f(x) = x² − 4 on the interval −3 ≤ x ≤ 3.
Set up: What are we solving for?
(i) Find the x-values at which f(x) = 0 in this interval. 1 mark
(ii) Evaluate the signed area ∫−33 (x² − 4) dx. 2 marks
(iii) Now find the total physical area enclosed between the curve and the x-axis on this interval (always positive). State your strategy in one sentence and show working. 3 marks
Problem 5 — Using properties to avoid recomputation
Given the following information about a function f:
∫02 f(x) dx = 5, ∫26 f(x) dx = 11, ∫06 g(x) dx = 3.
Set up: What are we solving for?
(i) Find ∫06 f(x) dx. State the property used. 1 mark
(ii) Find ∫60 f(x) dx. State the property used. 1 mark
(iii) Find ∫06 (2 f(x) − 3 g(x)) dx. State each property used. 3 marks
Stuck? Linearity (constant multiple + sum) and splitting cover everything here.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Odometer
Set up. Integrate v(t) on the given interval; v(t) ≥ 0 throughout (the car always moves forward), so distance = ∫v dt.
(i) ∫04 3t² dt = [t³]04 = 64 − 0 = 64 m.
(ii) ∫24 3t² dt = [t³]24 = 64 − 8 = 56 m. (Splitting: 64 = ∫02 + ∫24, and the first piece is 8.)
(iii) First two seconds: 8 m; last two seconds: 56 m. The car is accelerating: v(t) = 3t² is increasing, so each later second covers more ground than each earlier second.
Problem 2 — Displacement vs distance
Set up. Find where v changes sign (so we can split the integral into "forward" and "backward" stages), then compute displacement (signed integral) and distance (sum of absolute integrals).
(i) v(t) = 0 when 12 − 2t = 0 ⇒ t = 6 s. For 0 ≤ t < 6 we have v > 0 (forward motion); for 6 < t ≤ 8 we have v < 0 (backward motion).
(ii) ∫08(12 − 2t) dt = [12t − t²]08 = (96 − 64) − 0 = 32 m (this is the net displacement).
(iii) ∫06(12 − 2t) dt = [12t − t²]06 = 72 − 36 = 36 m (forward). ∫68(12 − 2t) dt = [12t − t²]68 = 32 − 36 = −4 m. Total distance = 36 + |−4| = 40 m. This is more than the displacement (32 m) because after t = 6 the particle moves backward, which the signed integral counts negatively but distance counts positively.
Problem 3 — Tank draining
Set up. Total water lost on [a, b] = ∫ab L(t) dt; use splitting to combine sub-intervals.
(i) ∫05(4 + t) dt = [4t + t²/2]05 = (20 + 12.5) − 0 = 32.5 L.
(ii) ∫510(4 + t) dt = [4t + t²/2]510 = (40 + 50) − (20 + 12.5) = 90 − 32.5 = 57.5 L.
(iii) ∫010(4 + t) dt = [4t + t²/2]010 = 40 + 50 = 90 L; and 32.5 + 57.5 = 90 ✓ — confirms the splitting property ∫010 = ∫05 + ∫510.
Problem 4 — Signed area for x² − 4
Set up. Find sign changes, compute signed integral, then split at zeros and use absolute values for total physical area.
(i) x² − 4 = 0 ⇒ x = ±2.
(ii) ∫−33(x² − 4) dx = [x³/3 − 4x]−33 = (9 − 12) − (−9 + 12) = −3 − 3 = −6.
(iii) Strategy: split at x = ±2 (where the curve crosses the axis) and take absolute values.
∫−3−2(x² − 4) dx = [x³/3 − 4x]−3−2 = (−8/3 + 8) − (−9 + 12) = (16/3) − 3 = 7/3 (positive — curve is above axis here).
∫−22(x² − 4) dx = [x³/3 − 4x]−22 = (8/3 − 8) − (−8/3 + 8) = (−16/3) − (16/3) = −32/3 (negative — curve below axis).
∫23(x² − 4) dx = [x³/3 − 4x]23 = (9 − 12) − (8/3 − 8) = −3 + 16/3 = 7/3 (positive).
Total physical area = 7/3 + 32/3 + 7/3 = 46/3.
Problem 5 — Properties toolkit
Set up. Each part is a one-step application of a single property (or two properties combined).
(i) ∫06 f = ∫02 f + ∫26 f = 5 + 11 = 16. Property used: splitting.
(ii) ∫60 f = −∫06 f = −16 = −16. Property used: reversed limits.
(iii) ∫06(2 f − 3 g) dx = 2 ∫06 f dx − 3 ∫06 g dx = 2 · 16 − 3 · 3 = 32 − 9 = 23. Properties used: sum rule + constant multiple (linearity), plus the splitting result from (i).