Cambridge International AS & A Level Mathematics 9709 (2026-2027) · Paper 1: Pure Mathematics 1 · official syllabus
Study notes
Use these worked notes to check exact formulas, follow the reasoning and learn the exam wording.
Antiderivatives and the power rule 1.8
Integration reverses differentiation. An indefinite integral is a family of antiderivatives, so include an arbitrary constant C: ∫xⁿ dx = xⁿ⁺¹/(n + 1) + C for rational n ≠ -1, on an interval where the functions are defined.
For a nonzero constant a, ∫(ax + b)ⁿ dx = (ax + b)ⁿ⁺¹/[a(n + 1)] + C, with n ≠ -1. The denominator a accounts for the inner derivative. Check by differentiating the result.
Example: ∫(2x + 5)³ dx = (2x + 5)⁴/8 + C. Also ∫[3x² - 4/√x] dx = x³ - 8√x + C for x > 0.
The exponent -1 is excluded from this power formula. Do not divide by n + 1 when it is zero. Integrals involving logarithms belong to later Pure Mathematics content, not this section.
Finding the constant from a point 1.8
If a gradient function is supplied, integrate it to obtain the curve, then substitute a known point to find C. An indefinite integral without C can miss the required curve.
Given dy/dx = 2x + 1 and a curve through (1,-2), integration gives y = x² + x + C. Substitution gives -2 = 2 + C, so C = -4 and y = x² + x - 4.
If a second derivative is supplied, integrate twice and use two independent conditions. For y″ = 6x, y′(1) = 5 and y(1) = 4: y′ = 3x² + C₁ gives C₁ = 2; y = x³ + 2x + C₂ gives C₂ = 1.
Definite integrals and endpoint limits 1.8
If F′ = f, then ∫ from a to b of f(x) dx = F(b) - F(a), provided the integral exists. Parentheses matter when subtracting the lower-bound value. The constant cancels.
Example: ∫ from 1 to 3 of (2x + 1) dx = [x² + x] from 1 to 3 = 12 - 2 = 10. Reversing the limits changes the sign.
A function can be unbounded at an endpoint and still have a finite integral. Interpret ∫ from 0 to 1 of x^(-1/2) dx as the limit of [2√x] from ε to 1 as ε tends to 0 from above; it equals 2.
Do not treat every singular endpoint as harmless. For ∫ from 0 to 1 of x^(-2) dx, the same limiting method gives -1 + 1/ε, which grows without bound. The improper integral does not converge.
Areas between curves and lines 1.8
A definite integral is signed area: parts below the x-axis count negatively. Geometric area is nonnegative, so split at each crossing of the axis and add the absolute areas.
For y = x² - 1 from x = 0 to 2, split at x = 1. The area below the axis is 2/3 and above it 4/3, for total area 2. The signed integral alone is 2/3, not the total area.
Between two curves, find their intersections and integrate upper minus lower. For y = 2x and y = x², intersections are x = 0 and 2; on that interval 2x is higher, so area = ∫ from 0 to 2 of (2x - x²) dx = 4/3.
If integrating with respect to y is simpler, write x as a function of y and use right boundary minus left boundary. State the limits in the same variable as the differential.
Volumes of revolution 1.8
For a region between y = f(x) and the x-axis rotated about the x-axis, V = π∫[f(x)]² dx over the x-limits. The square is the radius squared, not the derivative or a second integral of f.
Rotating the region under y = x from x = 0 to 2 about the x-axis gives V = π∫ from 0 to 2 of x² dx = 8π/3. Volume has cubic units.
When a region does not touch the rotation axis, subtract the inner disc from the outer disc: V = π∫(R² - r²) dx. Do not square the difference of the radii.
The region between y = 9 - x² and y = 5 has intersections x = ±2. Rotating it about the x-axis gives V = π∫ from -2 to 2 of [(9 - x²)² - 25] dx = 704π/5.
About the y-axis, express the radius as x in terms of y and use V = π∫x² dy (or outer radius squared minus inner radius squared). All radii are distances from the stated axis.
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