Cambridge International AS & A Level Mathematics 9709: Integration (Pure Mathematics 1) – Revision Notes

These notes cover section 1.8 Integration of the Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2026 and 2027 (Version 4). The content is Pure Mathematics 1 and is examined on Paper 1, which every AS and A Level candidate sits; Papers 2 and 3 assume it. For full explanations and worked examples, use the Integration study guide.

Course hub: Cambridge A Level Mathematics. Printable list: 9709 checklist. Free check-up: AS 10-minute diagnostic. When you are ready, try the practice questions.

What 1.8 asks of you

• Understand integration as the reverse of differentiation.
• Integrate (ax + b)ⁿ for any rational n except −1, with constant multiples, sums and differences.
• Find a constant of integration from given information.
• Evaluate definite integrals, including simple improper ones.
• Find areas: a curve with lines parallel to the axes, a curve and a line, two curves.
• Find a volume of revolution about the x-axis or the y-axis, including a region not bounded by that axis.

Calculator: a scientific calculator is allowed on Paper 1, but no marks are given for unsupported calculator answers. Show every integration and every substitution of limits.

Key formulae

Result	In MF19?
∫ xⁿ dx = xⁿ⁺¹/(n + 1) + c, n ≠ −1	Yes
∫ (ax + b)ⁿ dx = (ax + b)ⁿ⁺¹/(a(n + 1)) + c, n ≠ −1	No
∫ₐᵇ f(x) dx = F(b) − F(a)	No
Area against the x-axis = ∫ y dx (curve above the axis)	No
Area against the y-axis = ∫ x dy (curve to the right of the axis)	No
Area between graphs = ∫ (upper − lower) dx	No
Volume about the x-axis = π ∫ y² dx	No
Volume about the y-axis = π ∫ x² dy	No
Volume with a hole = π ∫ (outer² − inner²) dx	No

Method in steps

Indefinite integral

1. Rewrite each term as a power: √x = x^(1/2), 5/x² = 5x⁻².
2. Expand brackets or divide through if the integrand is a product or a quotient.
3. Add 1 to each power and divide by the new power.
4. Add + c.
5. Check by differentiating.

Curve from its gradient

1. Integrate dy/dx, including + c.
2. Substitute the given point.
3. Solve for c and write y = … in full.

Definite integral

1. Integrate (no c).
2. Write [F(x)] with the limits.
3. Substitute the upper limit, then the lower, in separate brackets.
4. Subtract.

Area

1. Sketch. Mark roots and intersection points.
2. Split at every root where the curve crosses the axis you are integrating against.
3. Integrate each part; take the size of any negative part.
4. Add.

Volume

1. Decide which axis. About x: use y² and dx. About y: use x² and dy.
2. Make y (or x) the subject and square it before integrating.
3. If the region has a hole, find both radii and use outer² − inner².
4. Integrate between the correct limits, then multiply by π.

Small worked reminders

(ax + b)ⁿ: divide by a and by the new power.

∫ (5x + 2)³ dx = (5x + 2)⁴/20 + c
∫ 1/(1 − x)² dx = ∫ (1 − x)⁻² dx = (1 − x)⁻¹/((−1)(−1)) = 1/(1 − x) + c

Definite integral with a root:

∫₁⁴ 3√x dx = [2x^(3/2)]₁⁴ = 2(8) − 2(1) = 14

Improper integral: write the limit in words.

∫₁^∞ 2/x² dx = [−2/x]₁^∞.  As x → ∞, −2/x → 0.  Value = 0 − (−2) = 2

Two constants from a second derivative: d²y/dx² = 6x, and the curve has a stationary point at (1, 0).

dy/dx = 3x² + c₁;  dy/dx = 0 at x = 1, so c₁ = −3
y = x³ − 3x + c₂;  y = 0 at x = 1, so c₂ = 2
y = x³ − 3x + 2

Area between a curve and a line: y = x² and y = x + 2.

x² = x + 2 gives x = −1 and x = 2; the line is on top.
∫₋₁² (x + 2 − x²) dx = [x²/2 + 2x − x³/3]₋₁² = 10/3 − (−7/6) = 9/2

Volume about the y-axis: the region between y = x² (x ≥ 0), the y-axis and y = 4.

x² = y, so V = π ∫₀⁴ y dy = π[y²/2]₀⁴ = 8π

Spot the question type

Wording in the question	What to do
“Find ∫ … dx”	Indefinite integral: rewrite as powers, integrate, + c
“The curve passes through …” with dy/dx given	Integrate with + c, substitute the point, write y = …
“Find the exact value of ∫ₐᵇ …”	Definite integral; keep fractions, surds and π
An upper limit of ∞	Integrate, then say the term tends to 0 as x → ∞
“Area of the region enclosed by …”	Sketch, find intersections, integrate upper − lower
“By integrating with respect to y”	Make x the subject; use y-limits and dy
“Rotated through 360° about the x-axis”	π ∫ y² dx, or π ∫ (outer² − inner²) dx if there is a gap
“Rotated through 360° about the y-axis”	π ∫ x² dy with y-limits
“Show that …”	Every line of working written out; the given answer earns nothing on its own

Must-know distinctions

• Indefinite vs definite. An indefinite integral is a function and needs + c. A definite integral is a number and has no c.
• Value of an integral vs area. ∫ y dx counts regions below the x-axis as negative. An area is never negative. Split at the roots.
• ∫ y dx vs ∫ x dy. Use dx for a region against the x-axis (limits are x-values). Use dy for a region against the y-axis (limits are y-values).
• Area between graphs vs volume between graphs. For area, subtract first: ∫ (y₁ − y₂) dx. For volume, square first: π ∫ (y₁² − y₂²) dx. Never π ∫ (y₁ − y₂)² dx.
• Linear bracket vs non-linear bracket. (3x − 1)⁴ can be integrated directly with the (ax + b)ⁿ rule. (x² − 1)⁴ cannot; expand it first.
• n = −1. ∫ x⁻¹ dx is not xᵒ/0. It gives ln x, which is Paper 2 and Paper 3 work (Pure Mathematics 2 revision notes).
• Exact vs 3 s.f. “Exact” means keep π, surds and fractions. Otherwise give 3 significant figures, and don’t round until the end.

Quick self-test

1. Find ∫ (x² + 1/x²) dx.
2. Find ∫ (1 − 3x)⁴ dx.
3. Find ∫ 10/(2x + 1)⁶ dx.
4. Evaluate ∫₀¹ (3x² − 2x) dx. What does your answer tell you about the area between y = 3x² − 2x and the x-axis for 0 ≤ x ≤ 1?
5. Evaluate ∫₁⁸ x^(−1/3) dx.
6. A curve has dy/dx = 4x − 3 and passes through (2, 1). Find its equation.
7. Evaluate ∫₃^∞ 18/x³ dx.
8. Evaluate ∫₀⁴ 1/√x dx.
9. Find the area enclosed by y = 4 − x² and the x-axis.
10. The region under y = √x from x = 0 to x = 4 is rotated about the x-axis. Find the exact volume.
11. The region between y = x (x ≥ 0), the y-axis and the line y = 3 is rotated about the y-axis. Find the exact volume.

Answers

1. x³/3 − 1/x + c (write 1/x² as x⁻², which integrates to −x⁻¹).
2. −(1 − 3x)⁵/15 + c (here a = −3, so divide by −3 × 5).
3. −(2x + 1)⁻⁵ + c, or −1/(2x + 1)⁵ + c (10 ÷ (2 × (−5)) = −1).
4. [x³ − x²]₀¹ = 0. The curve crosses the axis at x = 2/3, so equal areas lie below and above the axis. The area is not zero.
5. [3x^(2/3)/2]₁⁸ = (3/2)(4 − 1) = 9/2.
6. y = 2x² − 3x + c; 1 = 8 − 6 + c, so c = −1 and y = 2x² − 3x − 1.
7. [−9/x²]₃^∞; as x → ∞, −9/x² → 0, so the value is 0 − (−1) = 1.
8. [2√x]₀⁴ = 4 − 0 = 4 (the integrand is undefined at 0, but the integral has a value).
9. Roots x = ±2. ∫₋₂² (4 − x²) dx = 32/3.
10. π ∫₀⁴ x dx = π[x²/2]₀⁴ = 8π.
11. x = y, so π ∫₀³ y² dy = π[y³/3]₀³ = 9π. (Check: this is a cone of radius 3 and height 3, and ⅓π(3²)(3) = 9π.)

Where marks are usually lost

• Integrating before rewriting: 3/√x must become 3x^(−1/2) before you add 1 to the power.
• Integrating a product such as (2x + 1)(x − 3) as a product of two separate integrals.
• In (ax + b)ⁿ, dividing only by the new power and forgetting a, which loses the accuracy mark.
• No “+ c” on an indefinite integral, or c found but the final equation of the curve never written out.
• Substituting only the upper limit, or not showing the substitution at all, when the working is needed for the method mark.
• Writing ∞ into an expression as if it were a number, instead of stating that the term tends to 0.
• Reporting a negative area, or integrating in one go across a root of the curve.
• Mixing up dx and dy: using x-limits in a ∫ x dy integral.
• Squaring the difference of radii for a solid with a hole, instead of taking the difference of the squared radii.
• Dropping π in a volume, or giving a decimal when the question says “exact”.

Official syllabus

Cambridge International AS & A Level Mathematics 9709 syllabus for 2026 and 2027 (Version 4, published December 2025), Cambridge Assessment International Education. Section 1.8 Integration, Pure Mathematics 1 (Paper 1).