Mathematics Advanced • Year 12 • Module 6 • Lesson 4

The Definite Integral

Practise HSC-style writing on definite integrals — evaluation, exact form, and the distinction between displacement and distance.

Master · Past-Paper Style

1. Short-answer questions

1.1 Evaluate ∫₀²(3x² + 4x − 1) dx. Show all working. 2 marks Band 3

1.2 Evaluate ∫₁³(e^x + 1/x) dx. Give your answer in exact form. 3 marks Band 3-4

1.3 Given that ∫₀⁴ f(x) dx = 10 and ∫₂⁴ f(x) dx = 7, find:
(a) ∫₀² f(x) dx, (b) ∫₄⁰ f(x) dx, (c) ∫₀⁴(2 f(x) + 3) dx. 4 marks Band 4

Stuck on 1.3(c)? Use linearity: ∫(2 f + 3) = 2 ∫f + ∫3, and ∫₀⁴ 3 dx = 12.

2. Extended response

2.1 A car's velocity is v(t) = 12 − 2t m/s for 0 ≤ t ≤ 8.
(a) Identify the time at which v(t) = 0 and describe what happens to the direction of motion.
(b) Compute the displacement ∫₀⁸(12 − 2t) dt.
(c) Compute the total distance travelled in the first 8 seconds.
(d) Explain in two sentences why the definite integral ∫₀⁸(12 − 2t) dt does not equal the total distance, and what it does represent. 7 marks Band 5-6

Explicit marking criteria

Part (a) — 1 mark

• 1 mark — sets 12 − 2t = 0, finds t = 6, and states that the car moves forward for 0 ≤ t < 6 then backward for 6 < t ≤ 8.

Part (b) — 2 marks

• 1 mark — correct antiderivative [12t − t²]₀⁸.

• 1 mark — value 96 − 64 = 32 m, identified as displacement (not distance).

Part (c) — 3 marks

• 1 mark — splits at t = 6: total distance = |∫₀⁶| + |∫₆⁸|.

• 1 mark — evaluates ∫₀⁶(12 − 2t) dt = 36, ∫₆⁸(12 − 2t) dt = −4.

• 1 mark — total distance = 36 + 4 = 40 m.

Part (d) — 1 mark

• 1 mark — clearly explains that the signed integral counts backward motion negatively, so it gives the net displacement (32 m) rather than the total distance (40 m).

Your response:

Stuck on (c)? After t = 6 the car moves backward; absolute value the second piece before adding.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — sample responses + marking notes

1.1 — ∫₀²(3x² + 4x − 1) dx (2 marks)

Sample response. F(x) = x³ + 2x² − x. [F(x)]₀² = (8 + 8 − 2) − 0 = 14.

Marking notes. 1 mark — correct antiderivative. 1 mark — correct substitution and arithmetic. Writing "+ C" in a definite-integral answer should not be penalised if the numerical answer is correct, but is poor form.

1.2 — ∫₁³(e^x + 1/x) dx (3 marks)

Sample response. F(x) = e^x + ln|x|. [F(x)]₁³ = (e³ + ln 3) − (e¹ + ln 1) = e³ − e + ln 3.

Marking notes. 1 mark — correct antiderivative (with ln|x|). 1 mark — correct substitution and exact form (ln 1 = 0). 1 mark — final exact form without unnecessary decimal approximation. Forgetting "exact" and giving a decimal: deduct 1 mark.

1.3 — Properties (4 marks)

Sample response.

(a) Splitting: ∫₀⁴ = ∫₀² + ∫₂⁴, so ∫₀² f = 10 − 7 = 3.
(b) Reversed limits: ∫₄⁰ f = −∫₀⁴ f = −10.
(c) Linearity: ∫₀⁴(2 f + 3) dx = 2 · 10 + 3 · (4 − 0) = 20 + 12 = 32.

Marking notes. 1 mark each for (a) and (b). 2 marks for (c): 1 for separating into 2 ∫f + ∫3, 1 for the final value. Forgetting to multiply 3 by the interval length (giving 20 + 3 = 23) is a common slip.

2.1 — Extended response (7 marks): sample Band-6 response with annotations

Sample Band-6 response.

(a) v(t) = 0 when 12 − 2t = 0, i.e. t = 6 s. For 0 ≤ t < 6, v > 0 (forward motion); for 6 < t ≤ 8, v < 0 (backward motion). [1 mark.]

(b) Displacement = ∫₀⁸(12 − 2t) dt = [12t − t²]₀⁸. [1 mark — antiderivative.]

= (96 − 64) − 0 = 32 m (net displacement, not total distance). [1 mark — value identified as displacement.]

(c) Since v changes sign at t = 6, split the integral:

Total distance = |∫₀⁶(12 − 2t) dt| + |∫₆⁸(12 − 2t) dt|. [1 mark — splits at t = 6 and uses absolute values.]

∫₀⁶(12 − 2t) dt = [12t − t²]₀⁶ = 72 − 36 = 36.
∫₆⁸(12 − 2t) dt = [12t − t²]₆⁸ = (96 − 64) − (72 − 36) = 32 − 36 = −4. [1 mark — evaluates both pieces.]

Total distance = 36 + |−4| = 36 + 4 = 40 m. [1 mark — final value.]

(d) The definite integral ∫₀⁸(12 − 2t) dt counts the backward motion (after t = 6) negatively, so it gives the net change in position — the displacement (32 m). The total distance (40 m) is larger because it counts every metre the particle moves, regardless of direction. [1 mark.]

Total: 7/7.

Band descriptors for marker.

Band 3: Computes (b) correctly but treats it as "distance"; cannot split the integral in (c). ≈ 2-3 marks.

Band 4: Identifies t = 6 in (a); computes (b); attempts (c) by splitting but with arithmetic errors; (d) is loosely worded. ≈ 4-5 marks.

Band 5: Full (a)(b)(c) with correct values; (d) distinguishes displacement vs distance but does not explain the sign-counting mechanism. ≈ 5-6 marks.

Band 6: Full working in all four parts, explicit absolute values in (c), and (d) gives both the mechanistic reason ("counts backward motion negatively") and the conceptual identification ("displacement" vs "distance"). 7/7.