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Learning objectives
Core idea
Key formulae
Worked examples
Explore the graph
Common mistakes
Exam tip
Guided practice
Practice questions
Challenge
Summary
Quick quiz
Same topic, different board
Past paper questions

Learn9FM0Option · Further Pure Mathematics 2Further calculusReduction formulae, arc length and surface area

Further calculus · Subtopic 1 of 1

Reduction Formulae, Arc Length and Surface Area

A reduction formula links $I_n=\int f_n$ to $I_{n-1}$ or $I_{n-2}$ (usually by parts), so you climb from an easy $I_0$ to any $I_n$ . The length of a curve is $\int\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx$ and the surface it sweeps out is $2\pi\int y\,ds$ .

Cambridge 9231 Paper 2Edexcel YFM01 WFM03Edexcel 9FM0 Option≈ 50 minRevision 1 · updated 2026-10-27

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Section 1 of 14

Learning objectives

Derive reduction formulae using integration by parts.
Use a reduction formula to find a specific integral.
Find arc lengths (Cartesian and parametric).
Find surface areas of revolution.

Pause and think

How long is the curve $y=x^2$ from $0$ to $1$ ? Is it more or less than $\sqrt2$ ?

Section 2 of 14

Core idea

Start hereA ladder of integrals

$\int x^8e^x\,dx$ needs integration by parts eight times. A reduction formula does one integration by parts in general, linking $I_n$ to $I_{n-1}$ . Then you just climb the ladder from $I_0$ .

The big idea

Reduction formula: $I_n$ in terms of $I_{n-1}$ or $I_{n-2}$ , usually by parts. Arc length $s=\int\sqrt{1+y'^2}\,dx$ . Surface area $S=2\pi\int y\sqrt{1+y'^2}\,dx$ .

1. Reduction formulae

I_n=\int_0^1x^ne^xdx

$u=x^n,\ v'=e^x$
Parts: differentiate the power.
$I_n=\left[x^ne^x\right]_0^1-\int_0^1nx^{n-1}e^xdx$
$\int uv'=uv-\int u'v$ .
$I_n=e-nI_{n-1}$
Boundary term $e-0$ .

Wallis:

I_n=\int_0^{\pi/2}\sin^nx\,dx

$I_n=\int\sin^{n-1}x\cdot\sin x\,dx$
Split off one sine.
$=\left[-\sin^{n-1}x\cos x\right]_0^{\pi/2}+(n-1)\int\sin^{n-2}x\cos^2x\,dx$
By parts.
$=(n-1)\int\sin^{n-2}x(1-\sin^2x)\,dx=(n-1)(I_{n-2}-I_n)$
Boundary term zero; $\cos^2=1-\sin^2$ .
$I_n=\frac{n-1}nI_{n-2}$
Collect $I_n$ .

Worked example $I_4$

Find $\int_0^{\pi/2}\sin^4x\,dx$ .

Goal: $I_4$ .
Strategy: $I_0=\frac\pi2$ ; climb.

$I_2=\tfrac12I_0=\tfrac\pi4$
$n=2$ .
$I_4=\tfrac34I_2=\tfrac{3\pi}{16}$
$n=4$ .

Answer

\frac{3\pi}{16}

✓ Check

\sin^4x=\frac38-\frac12\cos2x+\frac18\cos4x

integrates to

\frac38\cdot\frac\pi2

✓.

2. Arc length and surface area

Arc length

$\delta s^2\approx\delta x^2+\delta y^2$
Pythagoras on a tiny piece of curve.
$\delta s\approx\sqrt{1+\left(\frac{\delta y}{\delta x}\right)^2}\delta x$
Factor out $\delta x$ .
$s=\int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx$
Limit of the sum.

Parametric:

s=\int\sqrt{\dot x^2+\dot y^2}\,dt

. Surface area: each piece

\delta s

sweeps a band of area

2\pi y\,\delta s

Worked example $y=\frac23x^{3/2}$ , $0\le x\le3$

Arc length.

Goal: $s$ .
Strategy: Simplify $1+y'^2$ .

$y'=x^{1/2},\ 1+y'^2=1+x$
Differentiate.
$s=\int_0^3(1+x)^{1/2}dx=\left[\tfrac23(1+x)^{3/2}\right]_0^3$
Integrate.
$=\tfrac23(8-1)=\tfrac{14}3$
Evaluate.

Answer

\frac{14}3

✓ CheckChord length

\sqrt{9+(2\sqrt3)^2}=\sqrt{21}\approx4.58<4.67

✓.

Common misconceptionForgetting the square root

s=\int\sqrt{1+y'^2}\,dx

, not

\int(1+y')\,dx

Summary

Parts once, then iterate.
Arc length by Pythagoras.
Surface $2\pi\int y\,ds$ .

Reduction formula: Relation between $I_n$ and lower $I$ 's.
Arc length: Length measured along a curve.

HistoryRectifying curves. Finding the length of a curve ("rectification") was thought impossible until William Neile rectified the semicubical parabola

y^2=x^3

in 1657.

Fun factWallis's formula. The reduction formula for

\int_0^{\pi/2}\sin^nx\,dx

leads to Wallis's product

\frac\pi2=\frac21\cdot\frac23\cdot\frac43\cdot\frac45\cdot\frac65\cdots

Section 3 of 14

Key formulae

Arc length

s=\int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx=\int\sqrt{\dot x^2+\dot y^2}\,dt

Surface area (about x-axis)

S=2\pi\int_a^by\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx

Wallis

\int_0^{\pi/2}\sin^nx\,dx=I_n,\quad I_n=\frac{n-1}nI_{n-2}

TrickPerfect squares. Arc-length questions are designed so

1+y'^2

becomes a perfect square. Expand it and look for

(\ldots)^2

Section 4 of 14

Worked examples

Example 1 · A reduction formula0 / 3 steps

I_n=\int_0^1x^ne^x\,dx

. Show

I_n=e-nI_{n-1}

and find

I_2

Example 2 · Arc length0 / 3 steps

Find the length of

y=\frac23x^{3/2}

from

x=0

3

Section 5 of 14

Explore the graph

Arc length as many tiny hypotenuses

Loading the interactive graph…

Try this

Length from $0$ to $b$ is $\frac23\left((1+b)^{3/2}-1\right)$ .
Compare with the straight chord to the end point: the curve is always longer.

$\delta s\approx\sqrt{\delta x^2+\delta y^2}$ .

In the real worldCables and pipes. Engineers find the length of a hanging cable (a catenary) by arc length:

\int\sqrt{1+\sinh^2x}\,dx=\int\cosh x\,dx

Section 6 of 14

Common mistakes

Incorrect

s=\int\left(1+y'\right)dx

Correct

\int\sqrt{1+y'^2}\,dx

Pythagoras on a tiny segment.

IncorrectLosing the boundary term in parts

CorrectEvaluate

[uv]

at both limits

It's often the

e

1

in the formula.

Incorrect

S=\pi\int y^2dx

Correct

2\pi\int y\,ds

That's volume, not surface area.

Integrate x⁸ eˣ by parts eight times

Reduction formula: or just climb the ladder.

Section 7 of 14

Exam tip

Exam tipStart from the bottom. Compute

I_0

(or

I_1

) directly, then apply the formula step by step. Show each rung.

Section 8 of 14

Guided practice

SA-FP-RA-G012 marksL2 StandardNon-calculatorReduction formulae

I_n=\int_0^1x^ne^xdx

with

I_n=e-nI_{n-1}

and

I_0=e-1

. Find

I_1

Salam Academy Original Practice · Proposed mark scheme

Section 9 of 14

Practice questions

SA-FP-RA-0012 marksL2 StandardCalculatorReduction formulae

Using the same formula, find

I_2

(to 4 s.f.).

Salam Academy Original Practice · Proposed mark scheme

SA-FP-RA-0023 marksL3 ExamCalculatorReduction formulae

I_n=\int_0^{\pi/2}\sin^nx\,dx

satisfies

I_n=\frac{n-1}nI_{n-2}

I_0=\frac\pi2

. Find

I_4

(to 4 s.f.).

Salam Academy Original Practice · Proposed mark scheme

SA-FP-RA-0033 marksL3 ExamNon-calculatorArc length

Find the length of

y=\frac23x^{3/2}

from

x=0

x=3

Salam Academy Original Practice · Proposed mark scheme

SA-FP-RA-0042 marksL1 FoundationCalculatorArc length

Use the arc length formula to find the length of

y=2x

from

x=0

x=1

(to 4 s.f.).

Salam Academy Original Practice · Proposed mark scheme

More questions on every skill in this subtopic are in the question bank under Practise.

Section 10 of 14

Challenge

SA-FP-RA-0053 marksL4 ChallengingCalculatorSurface area

The line

y=x

0\le x\le1

, is rotated about the

x

-axis. Find the curved surface area (to 3 s.f.).

Salam Academy Original Practice · Proposed mark scheme

Section 11 of 14

Summary

Reduction: by parts, then iterate.
$s=\int\sqrt{1+y'^2}dx$ .
$S=2\pi\int y\,ds$ .

Section 12 of 14

Quick quiz

Q1 · SA-FP-RA-Q011 markL1 FoundationNon-calculatorArc length

The arc length of

y=f(x)

from

a

b

Salam Academy Original Practice · Proposed mark scheme

Section 13 of 14

Same topic, different board

Specification: IAL Further Mathematics (YFM01) and A level Further Mathematics (9FM0)
Where: Further Pure 2 · Further calculus (IAL: WFM03 Integration)
Expected: Reduction formulae, arc length and surface area of revolution, including parametric and polar forms.
Calculator: Permitted; show enough working to earn the method marks.
Typical style: Edexcel often opens with a routine part and builds to a "show that" or "prove" part.
Open in: YFM01 · WFM03 9FM0 · Option

Section 14 of 14

Past paper questions

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Salam Academy original content · Content status: reviewed and published. Syllabus mapping: Pearson Edexcel 9FM0, First teaching 2017 (official specification).