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.)