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:
Find the x-coordinates where the curve and line meet.
Between these, the line is above the curve, so integrate (line − curve).
Integrate and evaluate.
Area = 4/3 ≈ 1.33 square units.