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v0.1.1298
NotesMath AITopic 5.5
Unit 5 · Calculus · Topic 5.5

IB Math AI — 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. 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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