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NotesMath AI HLTopic 5.5
Unit 5 · Calculus · Topic 5.5

IB Math AI HL — Introduction to integration

IB Mathematics AI SL topic covering core concepts and exam-style applications.

Exam technique guidePractice questions

Key concepts in Introduction to integration

Key Idea: Integration is the reverse of differentiation. Topic 5.5 covers three key skills: indefinite integration (finding the family of antiderivatives, always + C), definite integration (calculating the exact signed area under a curve between two limits), and using an initial condition to find the specific constant C. These appear on both papers.

✅ Integration rules

Example: Indefinite integral: ∫(3x² + 4x − 1) dx = x³ + 2x² − x + C Definite integral: ∫[1 to 4] (2x + 3) dx = [x² + 3x]₁⁴ = (16+12) − (1+3) = 28 − 4 = 24 Area between curves: ∫[0 to 3] (upper − lower) dx where upper = x + 4 and lower = x² = ∫[0 to 3] (x + 4 − x²) dx = [x²/2 + 4x − x³/3]₀³ = (4.5 + 12 − 9) − 0 = 7.5 Initial condition: f'(x) = 6x + 2 and f(1) = 5. f(x) = 3x² + 2x + C. Substitute: 3 + 2 + C = 5 → C = 0. So f(x) = 3x² + 2x.
Always include + C for indefinite integrals. Omitting it loses marks. For area between curves: sketch first to identify which function is 'upper' (higher on the y-axis) in the interval. The integral always goes (upper) − (lower) to stay positive.
Paper 1 (GDC allowed): Show each integration step (add 1 to power, divide). Show the substitution F(b) − F(a) in full for definite integrals. Paper 2 (GDC allowed): Use the GDC's definite integral function for complex functions. For area between curves, verify the intersection points first using the GDC, then integrate over the correct interval.

IB-style question [6 marks]

Find the area of the region enclosed between the curve y = x² and the line y = 2x.

Step by step:

  1. Find the x-coordinates where the curve and line meet.

    x2=2x⇒x(x−2)=0⇒x=0, 2x^2 = 2x \Rightarrow x(x-2) = 0 \Rightarrow x = 0,\ 2x2=2x⇒x(x−2)=0⇒x=0, 2
  2. Between these, the line is above the curve, so integrate (line − curve).

    A=∫02(2x−x2) dxA = \int_{0}^{2} (2x - x^2)\,dxA=∫02​(2x−x2)dx
  3. Integrate and evaluate.

    =[ x2−x33 ]02=4−83=43= \left[\,x^2 - \tfrac{x^3}{3}\,\right]_{0}^{2} = 4 - \tfrac{8}{3} = \tfrac{4}{3}=[x2−3x3​]02​=4−38​=34​
Final answer:

Area = 4/3 ≈ 1.33 square units.

What you'll learn in Topic 5.5

  • 5.5.1 Indefinite Integration — The Power Rule
  • 5.5.2 Definite Integration and Area Under a Curve
  • 5.5.3 Integration with Initial Conditions
Suggested study order: Read the notes for each sub-topic below → test yourself with flashcards → attempt practice questions → review exam technique.

Study resources — 5.5 Introduction to integration

5.5.1

Indefinite Integration — The Power Rule

Notes
5.5.2

Definite Integration and Area Under a Curve

Notes
5.5.3

Integration with Initial Conditions

Notes

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Topic 5.5 Introduction to integration forms a core part of Unit 5: Calculus in IB Math AI HL. Mastering these concepts will strengthen your understanding of connected topics across the syllabus and prepare you for exam questions that require analysis, evaluation, and real-world application.

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