Mathematics Advanced • Year 12 • Module 6 • Lesson 4
The Definite Integral
Build fluency in evaluating definite integrals using F(b) − F(a), and applying the standard properties (same limits, reversed limits, splitting).
1. Quick recall
Answer each question in the space provided. 1 mark each
Q1.1 Complete the FTC evaluation rule:
∫ab f(x) dx = [F(x)]ab = ____________ − ____________ where F'(x) = f(x).
Q1.2 Why do definite integrals never include + C? (One sentence.) ___________________________________
Q1.3 Complete the two properties: ∫aa f(x) dx = ______ ∫ba f(x) dx = ____________ ∫ab f(x) dx.
2. Worked example — ∫02 (ex + x) dx
Follow every line. Each step has a reason on the right.
Problem. Evaluate ∫02 (ex + x) dx.
Step 1 — Find an antiderivative F(x).
F(x) = e^x + x²/2
Reason: integrate term by term; no + C is needed for definite integrals.
Step 2 — Evaluate at the upper and lower limits.
F(2) = e² + 2²/2 = e² + 2
F(0) = e⁰ + 0 = 1
Step 3 — Subtract: F(b) − F(a).
∫_0^2 (e^x + x) dx = (e² + 2) − 1 = e² + 1 ≈ 8.389
Answer. e² + 1 ≈ 8.39 (no + C — definite integrals give a number).
3. Faded example — fill in the missing steps
Evaluate ∫13 (2x² − 4x + 5) dx. Fill in each blank line. 4 marks
Step 1 — Antiderivative:
F(x) = ______________________________________
Step 2 — Evaluate at the limits:
F(3) = ____________ + ____________ + ____________ = ____________
F(1) = ____________ + ____________ + ____________ = ____________
Step 3 — Subtract:
∫13 (2x² − 4x + 5) dx = F(3) − F(1) = ____________
4. Graduated practice — evaluate each definite integral
Use [F(x)]ab = F(b) − F(a). Give exact answers unless told otherwise.
Foundation — single-term integrand (4 questions)
| Q | Integral | Value |
|---|---|---|
| 4.1 1 | ∫03 x² dx | |
| 4.2 1 | ∫12 ex dx | |
| 4.3 1 | ∫14 (1/x) dx | |
| 4.4 1 | ∫01 6x dx |
Standard — typical HSC difficulty (6 questions)
Show the antiderivative and the substitution into [ ]ab.
4.5 ∫01 (3x² + 2x + 1) dx 2 marks
4.6 ∫02 (3x² + 4x − 1) dx 2 marks
4.7 ∫1e (1/x) dx 2 marks
4.8 ∫01 e2x dx (exact) 2 marks
4.9 ∫−11 x³ dx 2 marks
4.10 ∫13 (ex + 1/x) dx (exact) 2 marks
Extension — apply properties (2 questions)
4.11 Given ∫05 f(x) dx = 12 and ∫35 f(x) dx = 7, use the splitting property to find ∫03 f(x) dx. State the property used. 3 marks
4.12 Evaluate ∫02π sin x dx and explain in one sentence why the answer is 0 from the geometric (signed-area) interpretation. 3 marks
5. Self-check the easy 3
Tick the first three once you've checked your method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — FTC rule
∫ab f(x) dx = [F(x)]ab = F(b) − F(a).
Q1.2 — Why no + C
The constants cancel: [F(x) + C]ab = (F(b) + C) − (F(a) + C) = F(b) − F(a), so any antiderivative gives the same number.
Q1.3 — Two properties
∫aa f(x) dx = 0; ∫ba f(x) dx = − ∫ab f(x) dx.
Q3 — Faded example ∫13(2x² − 4x + 5) dx
F(x) = (2/3) x³ − 2x² + 5x.
F(3) = 18 − 18 + 15 = 15.
F(1) = 2/3 − 2 + 5 = 2/3 + 3 = 11/3.
Subtract: 15 − 11/3 = 45/3 − 11/3 = 34/3.
Q4.1 — ∫03 x² dx
[x³/3]03 = 9 − 0 = 9.
Q4.2 — ∫12 ex dx
[ex]12 = e² − e.
Q4.3 — ∫14 (1/x) dx
[ln|x|]14 = ln 4 − ln 1 = ln 4 = 2 ln 2.
Q4.4 — ∫01 6x dx
[3x²]01 = 3 − 0 = 3.
Q4.5 — ∫01 (3x² + 2x + 1) dx
[x³ + x² + x]01 = (1 + 1 + 1) − 0 = 3.
Q4.6 — ∫02 (3x² + 4x − 1) dx
[x³ + 2x² − x]02 = (8 + 8 − 2) − 0 = 14.
Q4.7 — ∫1e (1/x) dx
[ln|x|]1e = ln e − ln 1 = 1.
Q4.8 — ∫01 e2x dx (exact)
[(1/2) e2x]01 = (1/2) e² − (1/2) = (e² − 1)/2.
Q4.9 — ∫−11 x³ dx
[x⁴/4]−11 = 1/4 − 1/4 = 0. Geometric reason: x³ is odd, so the area below the axis on (−1, 0) cancels the area above on (0, 1).
Q4.10 — ∫13 (ex + 1/x) dx (exact)
[ex + ln|x|]13 = (e³ + ln 3) − (e + 0) = e³ − e + ln 3.
Q4.11 — Splitting property
Splitting: ∫05 f = ∫03 f + ∫35 f. So ∫03 f = ∫05 f − ∫35 f = 12 − 7 = 5.
Q4.12 — ∫02π sin x dx
[−cos x]02π = −cos(2π) − (−cos 0) = −1 + 1 = 0. Geometrically, the positive lobe on (0, π) and the negative lobe on (π, 2π) have equal magnitude and opposite sign, so their signed areas cancel exactly.