Mathematics Advanced • Year 12 • Module 6 • Lesson 1

Introduction to Integration

Build procedural fluency in finding antiderivatives of polynomials using the power rule and the constant of integration.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Complete the power rule for integration:

∫ xⁿ dx = ____________________ + C, for n ≠ ______.

Q1.2 Explain in one short sentence why the "+ C" appears in every indefinite integral.

__________________________________________________________________________________

Q1.3 The best way to check an indefinite integral is to ____________________ your answer and confirm you recover the original integrand.

Stuck? Revisit lesson § Formula Reference and § Checking Your Answers.

2. Worked example — ∫(3x² + 4x − 5) dx

Follow every line. Each step has a reason on the right.

Problem. Find ∫ (3x² + 4x − 5) dx.

Step 1 — Integrate term by term using the sum rule.

∫(3x² + 4x − 5) dx = ∫3x² dx + ∫4x dx + ∫(−5) dx

Reason: the integral of a sum is the sum of the integrals.

Step 2 — Pull each constant out, then apply the power rule.

∫3x² dx = 3 · x³/3 = x³

∫4x dx = 4 · x²/2 = 2x²

∫(−5) dx = −5x

Reason: ∫kxⁿ dx = k · x^(n+1)/(n+1); for n = 0 this gives ∫k dx = kx.

Step 3 — Combine and add a single + C at the end.

x³ + 2x² − 5x + C

Step 4 — Check by differentiating.

d/dx (x³ + 2x² − 5x + C) = 3x² + 4x − 5 ✓

Answer. ∫(3x² + 4x − 5) dx = x³ + 2x² − 5x + C.

3. Faded example — fill in the missing steps

Find ∫ (4x³ − 6x² + 2x − 7) dx. Fill in each blank line. 4 marks

Step 1 — Sum rule, term by term:

∫4x³ dx = 4 · x⁴/4 = ____________

∫(−6x²) dx = −6 · x³/3 = ____________

∫2x dx = 2 · x²/2 = ____________

∫(−7) dx = ____________

Step 2 — Combine with a single + C:

∫(4x³ − 6x² + 2x − 7) dx = ______________________________________ + C

Step 3 — Check by differentiating your answer. You should recover the original integrand:

d/dx(your answer) = ______________________________________

Stuck? Revisit lesson § Worked Example for the model layout.

4. Graduated practice — find each indefinite integral

Show at least one line of working for the Standard and Extension questions. Always include + C.

Foundation — single-term power rule (4 questions)

Q	Integral	Answer (with + C)
4.1 1	∫ x⁴ dx
4.2 1	∫ 6x² dx
4.3 1	∫ 8 dx
4.4 1	∫ (−5x) dx

Standard — typical HSC difficulty (6 questions)

Show the term-by-term integration on the line below each part.

4.5 ∫ (2x³ − 5x + 3) dx 2 marks

4.6 ∫ (5x⁴ + 4x³ − 3x²) dx 2 marks

4.7 ∫ (10x − 7) dx 2 marks

4.8 ∫ (3x² − 4x + 6) dx, then evaluate the antiderivative at x = 2. 2 marks

4.9 Find f(x) given f'(x) = 4x − 3 and f(1) = 5. 2 marks

4.10 A particle has velocity v(t) = 2t + 3 m/s. Find the displacement function s(t) given s(0) = 0. 2 marks

Extension — combine concepts (2 questions)

4.11 The gradient of a curve is dy/dx = 3x² − 4x + 2 and the curve passes through (1, 5). Find the equation of the curve. 3 marks

4.12 A student writes "∫(2x + 1)² dx = (2x + 1)³ / 3 + C". Check by differentiating, identify the error, then find the correct antiderivative by expanding first. 3 marks

Stuck on 4.12? Differentiating (2x+1)³/3 needs the chain rule — it gives an extra factor of 2.

5. Self-check the easy 3

Tick the first three once you've checked your method works.

For 4.1 (∫x⁴ dx) I obtained x⁵/5 + C and confirmed d/dx(x⁵/5) = x⁴.

For 4.2 (∫6x² dx) I pulled the 6 out and used x³/3 (not x²/2) — final answer 2x³ + C.

For 4.3 (∫8 dx) I integrated the constant correctly as 8x + C (not just 8 or 8 + C).

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Power rule

∫ xⁿ dx = xⁿ⁺¹/(n+1) + C, for n ≠ −1.

Q1.2 — Why + C

Differentiating any constant gives 0, so two antiderivatives can differ by any constant; the "+ C" represents this unknown constant that is lost in differentiation.

Q1.3 — How to check

Differentiate your answer and confirm you get back the original integrand. This catches sign, coefficient, and power errors instantly.

Q3 — Faded example ∫(4x³ − 6x² + 2x − 7) dx

Step 1: ∫4x³ dx = x⁴; ∫(−6x²) dx = −2x³; ∫2x dx = x²; ∫(−7) dx = −7x.
Step 2: ∫(4x³ − 6x² + 2x − 7) dx = x⁴ − 2x³ + x² − 7x + C.
Step 3 check: d/dx(x⁴ − 2x³ + x² − 7x + C) = 4x³ − 6x² + 2x − 7 ✓.

Q4.1 — ∫ x⁴ dx

x⁵/5 + C. Check: d/dx(x⁵/5) = 5x⁴/5 = x⁴ ✓.

Q4.2 — ∫ 6x² dx

6 · x³/3 + C = 2x³ + C.

Q4.3 — ∫ 8 dx

8x + C. The integrand 8 = 8x⁰, so the antiderivative is 8 · x¹/1 = 8x.

Q4.4 — ∫ (−5x) dx

−5 · x²/2 + C = −5x²/2 + C.

Q4.5 — ∫(2x³ − 5x + 3) dx

2 · x⁴/4 − 5 · x²/2 + 3x + C = x⁴/2 − 5x²/2 + 3x + C.

Q4.6 — ∫(5x⁴ + 4x³ − 3x²) dx

x⁵ + x⁴ − x³ + C. (5 · x⁵/5 = x⁵; 4 · x⁴/4 = x⁴; 3 · x³/3 = x³.)

Q4.7 — ∫(10x − 7) dx

5x² − 7x + C.

Q4.8 — ∫(3x² − 4x + 6) dx; value at x = 2

F(x) = x³ − 2x² + 6x + C. F(2) = 8 − 8 + 12 + C = 12 + C. (The value depends on C unless a boundary condition is given.)

Q4.9 — f'(x) = 4x − 3, f(1) = 5

f(x) = 2x² − 3x + C. At x = 1: 5 = 2 − 3 + C ⇒ C = 6. f(x) = 2x² − 3x + 6.

Q4.10 — v(t) = 2t + 3, s(0) = 0

s(t) = t² + 3t + C. s(0) = 0 ⇒ C = 0. s(t) = t² + 3t.

Q4.11 — Curve with dy/dx = 3x² − 4x + 2 through (1, 5)

y = ∫(3x² − 4x + 2) dx = x³ − 2x² + 2x + C. At (1, 5): 5 = 1 − 2 + 2 + C ⇒ C = 4. y = x³ − 2x² + 2x + 4.

Q4.12 — Error in ∫(2x + 1)² dx

Check: d/dx[(2x+1)³/3] = (1/3) · 3(2x+1)² · 2 = 2(2x+1)², which is twice the integrand. So the student's answer is wrong by a factor of 2.
Correct (expand first): (2x+1)² = 4x² + 4x + 1, so ∫(4x² + 4x + 1) dx = 4x³/3 + 2x² + x + C. (Equivalent form: (2x+1)³/6 + C.)