Mathematics Advanced • Year 12 • Module 6 • Lesson 15

Module Synthesis — Further Calculus

Build fluency in selecting the right tool from the Module 6 toolbox: power rule, FTC, substitution, by parts, separable DE, motion.

Build · Skill Drill

1. Quick recall — pick the right tool

Match each integral to the best technique. 1 mark each

Q1.1 ∫ x · e^x² dx → Best technique: ____________________

Q1.2 ∫ x cos(x) dx → Best technique: ____________________

Q1.3 State the two parts of the Fundamental Theorem of Calculus in one phrase each:

Part 1: ________________________________; Part 2: ____________________________________________.

Stuck? Revisit lesson § Formula Reference and § HSC Tips — "Know when each technique applies".

2. Worked example — ∫ x e^x² dx via substitution

Problem. Evaluate ∫ x e^x² dx using the substitution u = x².

Step 1 — Substitute.

u = x², du/dx = 2x ⇒ du = 2x dx ⇒ x dx = du/2

Step 2 — Rewrite the integral in u.

∫ x e^x² dx = ∫ e^u · (du/2) = (1/2) ∫ e^u du

Step 3 — Integrate, then back-substitute.

= (1/2) e^u + C = (1/2) e^x² + C

Check. d/dx [(1/2) e^x²] = (1/2) · 2x · e^x² = x e^x² ✓.

3. Faded example — ∫ x sin(x) dx via by parts

Use the LIATE rule to choose u. 4 marks

Step 1 — Choose u and dv.

u = ________ (Algebraic), dv = ________ dx (Trig)

du = ________ dx, v = ________

Step 2 — Apply ∫ u dv = u v − ∫ v du.

∫ x sin(x) dx = ________ − ∫ ________ dx

Step 3 — Evaluate the remaining integral.

= ________ + C

Stuck? L-I-A-T-E ordering: Log, Inverse, Algebraic, Trig, Exp. Choose u from the earlier category.

4. Graduated practice — choose your tool

Foundation — direct antiderivatives (4 questions)

Q	Integral	Antiderivative + C
4.1 1	∫ 3x² dx
4.2 1	∫ (1/x) dx
4.3 1	∫ e^2x dx
4.4 1	∫ sin(x) dx

Standard — apply a technique (6 questions)

4.5 Find ∫ (3x² + 1/x + e^2x) dx. 2 marks

4.6 Evaluate ∫₀² x √(x² + 1) dx using u = x² + 1. 3 marks

4.7 Find the area between y = x² and y = 2 − x (in the region where the line is above the parabola). 3 marks

4.8 Solve dy/dx = 3x² y with y(0) = 2. 3 marks

4.9 A particle has a(t) = 2t + 1, v(0) = 2, x(0) = 1. Find x(3). 3 marks

4.10 Find the volume when y = x³ from x = 0 to x = 1 is rotated about the x-axis. 2 marks

Extension (2 questions)

4.11 Evaluate ∫₀^π/2 x sin(x) dx using by parts. 3 marks

4.12 A common error: a student writes ∫ 1/x dx = x⁰/0 + C using the power rule. Identify the error in one sentence, give the correct antiderivative, and explain why the power rule fails at n = −1. 3 marks

Stuck on 4.12? The power rule formula has a denominator (n + 1) that vanishes when n = −1.

5. Self-check the easy 3

Tick once verified by differentiation (the gold-standard check).

For 4.1 I checked d/dx [x³ + C] = 3x² ✓.

For 4.3 I checked d/dx [(1/2) e^2x + C] = e^2x ✓ — the 1/2 cancels the inside-derivative 2.

For 4.4 I checked d/dx [−cos(x) + C] = sin(x) ✓ (sign trap!).

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Q1.2 Pick the technique

Q1.1: substitution (u = x², because du = 2x dx — the integrand contains the inside function and its derivative). Q1.2: integration by parts (Algebraic × Trig — choose u = x by LIATE).

Q1.3 — FTC

Part 1: differentiation undoes integration (d/dx ∫_a^x f(t) dt = f(x)). Part 2: definite integrals are evaluated using antiderivatives (∫_a^b f(x) dx = F(b) − F(a)).

Q3 — Faded by-parts example

u = x, dv = sin(x) dx. du = 1 dx, v = −cos(x).
∫ x sin(x) dx = −x cos(x) − ∫ −cos(x) dx = −x cos(x) + sin(x) + C.

Q4.1 — 4.4 Foundation

4.1 x³ + C. 4.2 ln|x| + C. 4.3 (1/2) e^2x + C. 4.4 −cos(x) + C.

Q4.5 — Sum of antiderivatives

∫ 3x² dx + ∫ 1/x dx + ∫ e^2x dx = x³ + ln|x| + (1/2) e^2x + C.

Q4.6 — Definite by substitution

u = x² + 1, du = 2x dx ⇒ x dx = du/2. Bounds: x = 0 → u = 1; x = 2 → u = 5.
∫₀² x √(x² + 1) dx = (1/2) ∫₁⁵ u^1/2 du = (1/2) · [(2/3) u^3/2]₁⁵ = (1/3)(5^3/2 − 1) = (5√5 − 1)/3 ≈ 3.39.

Q4.7 — Area between curves

Intersections: x² = 2 − x ⇒ x² + x − 2 = 0 ⇒ x = 1 or x = −2. On [−2, 1] the line lies above the parabola.
A = ∫₋₂¹ ((2 − x) − x²) dx = [2x − x²/2 − x³/3]₋₂¹ = (2 − 1/2 − 1/3) − (−4 − 2 + 8/3) = 7/6 − (−10/3) = 7/6 + 20/6 = 9/2.

Q4.8 — Separable DE

dy/y = 3x² dx ⇒ ln|y| = x³ + C ⇒ y = A e^x³. y(0) = 2 ⇒ A = 2. y = 2 e^x³.

Q4.9 — Motion

v(t) = t² + t + 2 (using v(0) = 2). x(t) = t³/3 + t²/2 + 2t + 1 (using x(0) = 1). x(3) = 9 + 4.5 + 6 + 1 = 20.5 m.

Q4.10 — Volume of revolution (disk)

V = π ∫₀¹ (x³)² dx = π ∫₀¹ x⁶ dx = π [x⁷/7]₀¹ = π/7.

Q4.11 — ∫₀^π/2 x sin(x) dx by parts

u = x, dv = sin(x) dx ⇒ du = dx, v = −cos(x).
I = [−x cos(x)]₀^π/2 + ∫₀^π/2 cos(x) dx = [0 − 0] + [sin(x)]₀^π/2 = 1.

Q4.12 — Why the power rule fails at n = −1

Error: the student applied ∫ xⁿ dx = x^{n + 1}/(n + 1) + C with n = −1, but then the denominator n + 1 = 0, which is undefined. Correct answer: ∫ (1/x) dx = ln|x| + C. The power rule fails because dividing by zero is undefined; this is exactly the exceptional case that motivates introducing ln as a separate antiderivative.