Edexcel A-Level Maths: Integration — Complete Revision Guide (9MA0)

Integration is one of the highest-yield topics on Edexcel A-Level Maths (9MA0). It runs through the back half of the pure specification and tends to dominate the longer questions on Paper 2. Unlike differentiation, where each rule largely stands alone, integration techniques compound: a single 8-12 mark question might ask you to use partial fractions to break apart a rational expression, integrate the resulting pieces, apply limits to evaluate a definite integral, and then interpret the result as an area or a volume of revolution. Get any one step wrong and the rest collapses. Get them all right and the marks come quickly.

This guide is a topic-by-topic walkthrough of the integration content in the 9MA0 specification. It covers everything Edexcel can examine in this section: indefinite integration of standard functions, definite integration and the fundamental theorem of calculus, area between curves, integration by substitution, integration by parts, integration using partial fractions, the trapezium rule, separable differential equations, volumes of revolution, and the wider applications of integration. For each topic you will see the core skills, the typical pitfalls, a short worked example, and a link to the full lesson on the LearningBro course.

The aim is not to replace problem practice. Integration becomes intuitive only after dozens of worked questions, with the patterns of substitution choices and parts orderings becoming second nature. The aim is to give you a clear map of what you need to know, in the order Edexcel teaches it, so your revision is targeted rather than scattered. Use this guide as a checklist, a refresher, and a launchpad into focused practice.

What the Edexcel 9MA0 Specification Covers

The Edexcel A-Level Maths qualification (9MA0) is assessed through three two-hour papers, each worth 100 marks. Papers 1 and 2 cover pure mathematics, and Paper 3 covers statistics and mechanics. Integration sits in Section 8 of the pure specification, immediately after differentiation, and is most heavily examined on Paper 2. There is no choice of questions and no coursework, so every integration mark must be earned in the exam.

Integration is one of the highest-frequency topic areas across the pure papers, and almost every Paper 2 sitting includes at least one extended integration question worth 8-12 marks. The table below shows the sub-topics of Section 8, the part of the specification they sit under, and a realistic estimate of how many marks across a Paper 1 / Paper 2 sitting come from each.

Topic	Spec Section	Typical Paper 2 marks weight
Indefinite integration of standard functions	8.1	4-6 marks
Definite integration	8.2	4-6 marks
Area between curves	8.3	5-8 marks
Integration by substitution	8.4	5-7 marks
Integration by parts	8.4	5-7 marks
Integration using partial fractions	8.4	5-7 marks
The trapezium rule	8.5	3-5 marks
Separable differential equations	8.6	6-10 marks
Volumes of revolution	8.7	6-9 marks
Applications of integration	8.8	4-6 marks

These weights are estimates based on the spread of typical 9MA0 papers — not guarantees for any single year. What is reliable, however, is that integration is consistently the most heavily examined area of Paper 2 pure, and that techniques compound across questions. Mastering this section is high-leverage revision.

Indefinite Integration of Standard Functions

Indefinite integration is the reverse of differentiation. If $\frac{d}{dx}[F(x)] = f(x)$dxd​[F(x)]=f(x), then $\int f(x)\,dx = F(x) + C$∫f(x)dx=F(x)+C, where $C$C is the constant of integration. The constant is non-negotiable on indefinite integrals — Edexcel mark schemes routinely deduct a method or accuracy mark for forgetting it. Train yourself to write $+ C$+C before you finish the rest of the line.

The core skills are integrating each of the standard functions on the formula sheet from memory, including powers, exponentials, trigonometric functions, and reciprocal functions. The fundamental power rule is

$\int x^n\,dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1$∫xndx=n+1xn+1​+C,n=−1

with the special case $\int x^{-1}\,dx = \ln|x| + C$∫x−1dx=ln∣x∣+C when $n = -1$n=−1. You should also know $\int e^x\,dx = e^x + C$∫exdx=ex+C, $\int \sin x\,dx = -\cos x + C$∫sinxdx=−cosx+C, $\int \cos x\,dx = \sin x + C$∫cosxdx=sinx+C, $\int \sec^2 x\,dx = \tan x + C$∫sec2xdx=tanx+C, and the small family of related results that drop out of the chain rule in reverse, such as $\int e^{ax}\,dx = \frac{1}{a}e^{ax} + C$∫eaxdx=a1​eax+C.

A common pitfall is sign errors on $\int \sin x\,dx = -\cos x + C$∫sinxdx=−cosx+C. Many candidates write $+\cos x$+cosx on autopilot. Another is forgetting that the power rule fails at $n = -1$n=−1 and trying to apply $\frac{x^0}{0}$0x0​, which is undefined. A third is the missing constant of integration, which is the cheapest mark in the exam to lose.

A short worked example. Find $\int (4x^3 - 6x^{-2} + 2e^x)\,dx$∫(4x3−6x−2+2ex)dx. Integrate term by term: $\int 4x^3\,dx = x^4$∫4x3dx=x4, $\int -6x^{-2}\,dx = \frac{-6 x^{-1}}{-1} = 6 x^{-1} = \frac{6}{x}$∫−6x−2dx=−1−6x−1​=6x−1=x6​, and $\int 2e^x\,dx = 2e^x$∫2exdx=2ex. So the answer is $x^4 + \frac{6}{x} + 2e^x + C$x4+x6​+2ex+C. Notice the sign flip on the negative power — this is where most arithmetic errors creep in.

For full coverage with practice questions and worked solutions, see the Indefinite Integration lesson.

Definite Integration

A definite integral has limits and evaluates to a number rather than a function. The fundamental theorem of calculus tells you how to compute it:

$\int_a^b f(x)\,dx = F(b) - F(a)$∫ab​f(x)dx=F(b)−F(a)

where $F$F is any antiderivative of $f$f. The constant of integration cancels in the subtraction, so you do not need to write $+ C$+C on definite integrals. This is the only place the constant disappears, so be careful not to forget it elsewhere.

The core skills are evaluating definite integrals of standard functions, applying the linearity rules $\int_a^b [f(x) + g(x)]\,dx = \int_a^b f(x)\,dx + \int_a^b g(x)\,dx$∫ab​[f(x)+g(x)]dx=∫ab​f(x)dx+∫ab​g(x)dx and $\int_a^b k\,f(x)\,dx = k\int_a^b f(x)\,dx$∫ab​kf(x)dx=k∫ab​f(x)dx, and using the limit-swap rule $\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx$∫ab​f(x)dx=−∫ba​f(x)dx. You should also know that a definite integral evaluates to zero if the limits are equal, and that integrals can be split: $\int_a^c f(x)\,dx = \int_a^b f(x)\,dx + \int_b^c f(x)\,dx$∫ac​f(x)dx=∫ab​f(x)dx+∫bc​f(x)dx.

A common pitfall is substituting the lower limit before the upper limit and dropping a sign. Always write $F(b) - F(a)$F(b)−F(a) in that order, and use brackets around each term so the subtraction is unambiguous. Another is mishandling negative values inside logarithms when the antiderivative is $\ln|x|$ln∣x∣ — for the limits to be valid, the integrand must not blow up between them.

A short worked example. Evaluate $\int_1^3 (2x + \frac{1}{x})\,dx$∫13​(2x+x1​)dx. The antiderivative is $x^2 + \ln|x|$x2+ln∣x∣. Substituting: $[(9 + \ln 3) - (1 + \ln 1)] = 8 + \ln 3$[(9+ln3)−(1+ln1)]=8+ln3. The answer is exact and contains a logarithm — a typical 9MA0 question format will expect the exact form rather than a decimal.

For practice with the fundamental theorem and limit-handling, see the Definite Integration lesson.

Area Between Curves

The geometric meaning of a definite integral is the signed area between a curve and the x-axis. For a curve $y = f(x)$y=f(x) that lies entirely above the x-axis between $x = a$x=a and $x = b$x=b, the area is simply $\int_a^b f(x)\,dx$∫ab​f(x)dx. When the curve dips below the x-axis, the integral counts that piece negatively, which is not what you usually want geometrically.

For an area between two curves $y = f(x)$y=f(x) (upper) and $y = g(x)$y=g(x) (lower) between $x = a$x=a and $x = b$x=b, the formula is

$A = \int_a^b [f(x) - g(x)]\,dx$A=∫ab​[f(x)−g(x)]dx

The trick is identifying which curve is on top across the relevant interval. If the curves cross between $a$a and $b$b, you need to split the integral at the crossing point and take the appropriate difference in each region.

A common pitfall is forgetting to find the intersection points before integrating. The limits of an area-between-curves integral are usually not given directly — you have to set $f(x) = g(x)$f(x)=g(x), solve, and use the resulting x-values as your limits. Another is integrating $f(x) - g(x)$f(x)−g(x) when $g$g is actually above $f$f on part of the interval, producing a negative answer. If your area comes out negative, check the order of subtraction.

A short worked example. Find the area enclosed between $y = x^2$y=x2 and $y = 4 - x^2$y=4−x2. Setting them equal: $x^2 = 4 - x^2$x2=4−x2, so $2x^2 = 4$2x2=4 and $x = \pm\sqrt{2}$x=±2​. Between these limits the parabola $y = 4 - x^2$y=4−x2 is on top. The area is

$A = \int_{-\sqrt{2}}^{\sqrt{2}} [(4 - x^2) - x^2]\,dx = \int_{-\sqrt{2}}^{\sqrt{2}} (4 - 2x^2)\,dx$A=∫−2​2​​[(4−x2)−x2]dx=∫−2​2​​(4−2x2)dx

Evaluating: $[4x - \frac{2x^3}{3}]_{-\sqrt{2}}^{\sqrt{2}} = (4\sqrt{2} - \frac{4\sqrt{2}}{3}) - (-4\sqrt{2} + \frac{4\sqrt{2}}{3}) = \frac{16\sqrt{2}}{3}$[4x−32x3​]−2​2​​=(42​−342​​)−(−42​+342​​)=3162​​.

For sketching practice and worked examples on enclosed regions, see the Area Between Curves lesson.

Integration by Substitution

Integration by substitution is the reverse of the chain rule. When you cannot integrate $\int f(x)\,dx$∫f(x)dx directly, you make a change of variable $u = g(x)$u=g(x) that turns it into a simpler integral $\int h(u)\,du$∫h(u)du. The substitution will often be suggested by the structure of the integrand: the presence of a function and its derivative, or an expression nested inside another function.

The mechanical steps are: choose $u = g(x)$u=g(x), compute $\frac{du}{dx}$dxdu​, rearrange to get $dx$dx in terms of $du$du, substitute everything in terms of $u$u, integrate, and then substitute back to express the answer in terms of $x$x. For a definite integral, the cleaner approach is to also change the limits — if $u = g(x)$u=g(x) then the new limits are $g(a)$g(a) and $g(b)$g(b), and you do not need to substitute back at the end.

The core insight is to look for a function and (a constant multiple of) its derivative inside the integrand. For $\int 2x \cos(x^2)\,dx$∫2xcos(x2)dx, the derivative of $x^2$x2 is $2x$2x, which is sitting right there. So $u = x^2$u=x2 gives $du = 2x\,dx$du=2xdx and the integral becomes $\int \cos u\,du = \sin u + C = \sin(x^2) + C$∫cosudu=sinu+C=sin(x2)+C.

A common pitfall is choosing a substitution that does not actually simplify the integral. If after substituting you still have $x$x terms hanging around, the substitution was wrong or incomplete. Another is forgetting to change the limits on a definite integral and then accidentally evaluating at the original $x$x-limits instead of the new $u$u-limits. A third is sign errors when rearranging $du/dx$du/dx to express $dx$dx in terms of $du$du.

A short worked example. Evaluate $\int_0^1 x(1 + x^2)^3\,dx$∫01​x(1+x2)3dx by substitution. Let $u = 1 + x^2$u=1+x2, so $du = 2x\,dx$du=2xdx and $x\,dx = \frac{1}{2}du$xdx=21​du. The new limits: when $x = 0$x=0, $u = 1$u=1; when $x = 1$x=1, $u = 2$u=2. The integral becomes $\frac{1}{2}\int_1^2 u^3\,du = \frac{1}{2} \cdot \left[\frac{u^4}{4}\right]_1^2 = \frac{1}{8}(16 - 1) = \frac{15}{8}$21​∫12​u3du=21​⋅[4u4​]12​=81​(16−1)=815​.

For substitution-spotting practice and worked examples, see the Integration by Substitution lesson.

Integration by Parts

Integration by parts is the reverse of the product rule. The formula is

$\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx$∫udxdv​dx=uv−∫vdxdu​dx

You choose one factor of the integrand to be $u$u (which you will differentiate) and the other to be $\frac{dv}{dx}$dxdv​ (which you will integrate). The choice matters: get it wrong and the new integral is harder than the original.

A useful mnemonic for the choice of $u$u is LIATE — Logarithm, Inverse trig, Algebraic, Trigonometric, Exponential. Pick whichever factor comes first in this list to be $u$u. So in $\int x \ln x\,dx$∫xlnxdx, $\ln x$lnx comes before $x$x (logarithm before algebraic), so $u = \ln x$u=lnx and $\frac{dv}{dx} = x$dxdv​=x. In $\int x e^x\,dx$∫xexdx, $x$x comes before $e^x$ex (algebraic before exponential), so $u = x$u=x and $\frac{dv}{dx} = e^x$dxdv​=ex.

Some integrals require parts twice. $\int x^2 e^x\,dx$∫x2exdx becomes a problem in $\int 2x e^x\,dx$∫2xexdx after one application, which itself requires parts. Some require a clever trick: $\int e^x \sin x\,dx$∫exsinxdx requires parts twice and then solving for the original integral as if it were an algebraic unknown. A 9MA0 question can include any of these patterns.

A common pitfall is choosing $u$u and $\frac{dv}{dx}$dxdv​ the wrong way round. If after one application of the formula the new integral looks worse than the original, swap your choices. Another is sign errors in the formula — note the minus sign in front of the second integral, which is a frequent place to drop a mark. A third is forgetting that $\int \ln x\,dx$∫lnxdx requires parts with $\frac{dv}{dx} = 1$dxdv​=1, which is non-obvious.

A short worked example. Evaluate $\int x \cos x\,dx$∫xcosxdx. By LIATE, $u = x$u=x (algebraic) and $\frac{dv}{dx} = \cos x$dxdv​=cosx. So $\frac{du}{dx} = 1$dxdu​=1 and $v = \sin x$v=sinx. Applying the formula: $\int x\cos x\,dx = x\sin x - \int \sin x\,dx = x\sin x + \cos x + C$∫xcosxdx=xsinx−∫sinxdx=xsinx+cosx+C.

For LIATE practice and twice-applied parts examples, see the Integration by Parts lesson.

Integration Using Partial Fractions

When the integrand is a rational function — a polynomial divided by another polynomial — and the denominator factorises, the right move is usually partial fractions. You decompose the rational function into a sum of simpler fractions, each of which you can integrate using a standard rule, and then integrate term by term.

The three cases are exactly the same as in the algebra section of the spec: distinct linear factors, repeated linear factors, and irreducible quadratic factors. The partial-fraction templates are identical. The only new element is what each piece integrates to.

Partial fraction form	Integrates to
$\frac{A}{ax + b}$ax+bA​	$\frac{A}{a}\ln\|ax + b\| + C$aA​ln∥ax+b∥+C
$\frac{A}{(ax + b)^2}$(ax+b)2A​	$-\frac{A}{a(ax + b)} + C$−a(ax+b)A​+C
$\frac{Ax + B}{x^2 + p^2}$x2+p2Ax+B​	(Combination of $\ln$ln and $\arctan$arctan pieces)

The most common Edexcel pattern is two distinct linear factors in the denominator, decomposed into two logarithm-yielding pieces. Repeated factors and irreducible quadratics appear less often but are within the spec.

A common pitfall is forgetting to factorise the denominator before attempting partial fractions. If you write $\frac{1}{x^2 - 1}$x2−11​ as $\frac{A}{x^2 - 1}$x2−1A​, you have made no progress — you need $\frac{1}{(x - 1)(x + 1)}$(x−1)(x+1)1​ first. Another is using the wrong template for repeated factors, as in the algebra section. A third is dropping the modulus signs inside logarithms; for definite integrals where the integrand is positive throughout, the modulus often does not change the answer, but it is required notationally.

A short worked example. Evaluate $\int \frac{3x + 5}{(x - 1)(x + 2)}\,dx$∫(x−1)(x+2)3x+5​dx. Decompose: $\frac{3x + 5}{(x - 1)(x + 2)} = \frac{A}{x - 1} + \frac{B}{x + 2}$(x−1)(x+2)3x+5​=x−1A​+x+2B​. Multiplying through: $3x + 5 = A(x + 2) + B(x - 1)$3x+5=A(x+2)+B(x−1). Setting $x = 1$x=1: $8 = 3A$8=3A, so $A = 8/3$A=8/3. Setting $x = -2$x=−2: $-1 = -3B$−1=−3B, so $B = 1/3$B=1/3. The integral is $\frac{8}{3}\ln|x - 1| + \frac{1}{3}\ln|x + 2| + C$38​ln∣x−1∣+31​ln∣x+2∣+C.

For decomposition practice and worked integrations, see the Partial Fractions Integration lesson.

The Trapezium Rule

When an integrand cannot be integrated in closed form, or when the question explicitly asks for a numerical estimate, the standard A-Level method is the trapezium rule. You divide the interval $[a, b]$[a,b] into $n$n equal-width strips of width $h = \frac{b - a}{n}$h=nb−a​, evaluate the integrand at each strip boundary, and approximate the area as a sum of trapezium areas.

The formula is

$\int_a^b f(x)\,dx \approx \frac{h}{2}\left[y_0 + y_n + 2(y_1 + y_2 + \ldots + y_{n-1})\right]$∫ab​f(x)dx≈2h​[y0​+yn​+2(y1​+y2​+…+yn−1​)]

where $y_i = f(a + ih)$yi​=f(a+ih). The end-point values appear with a coefficient of 1; the interior values appear with a coefficient of 2. Most arithmetic errors on trapezium-rule questions come from miscounting which $y$y-values are end-points and which are interior.

You also need to be able to comment on whether the trapezium rule overestimates or underestimates the true value of the integral. The rule joins consecutive points by straight lines; when the curve is concave up (curving like a smile), the chord lies above the curve and the rule overestimates. When the curve is concave down (curving like a frown), the chord lies below the curve and the rule underestimates. Examiners sometimes ask you to sketch the curve and explain which case applies.

A common pitfall is using $n + 1$n+1 strips instead of $n$n when $n + 1$n+1 ordinates are listed. With $n$n strips you have $n + 1$n+1 ordinates — count them carefully. Another is rounding intermediate $y$y-values too aggressively, especially when the question asks for an answer to 4 significant figures. Carry full precision through the sum and round only at the end.

A short worked example. Estimate $\int_0^1 \sqrt{1 + x^3}\,dx$∫01​1+x3​dx using the trapezium rule with 4 strips. So $h = 0.25$h=0.25 and the ordinates are at $x = 0, 0.25, 0.5, 0.75, 1$x=0,0.25,0.5,0.75,1. The $y$y-values are $\sqrt{1}, \sqrt{1.015625}, \sqrt{1.125}, \sqrt{1.421875}, \sqrt{2}$1​,1.015625​,1.125​,1.421875​,2​, which are approximately $1.0000, 1.0078, 1.0607, 1.1924, 1.4142$1.0000,1.0078,1.0607,1.1924,1.4142. Applying the rule: $\frac{0.25}{2}[1.0000 + 1.4142 + 2(1.0078 + 1.0607 + 1.1924)] \approx 0.125 \cdot [2.4142 + 6.5218] \approx 1.117$20.25​[1.0000+1.4142+2(1.0078+1.0607+1.1924)]≈0.125⋅[2.4142+6.5218]≈1.117.

For trapezium-rule practice and over-/underestimate justifications, see the Trapezium Rule lesson.

Separable Differential Equations

A first-order separable differential equation is one of the form $\frac{dy}{dx} = f(x)g(y)$dxdy​=f(x)g(y), where the right-hand side splits as a product of a function of $x$x alone and a function of $y$y alone. The solution method is to separate the variables and integrate both sides:

$\int \frac{1}{g(y)}\,dy = \int f(x)\,dx$∫g(y)1​dy=∫f(x)dx

The result is an implicit relation between $x$x and $y$y, which you may or may not be able to rearrange to give $y$y explicitly as a function of $x$x.

The single constant of integration appears on one side only — combine the two constants into one and place it on the $x$x-side by convention. If the question gives an initial condition like $y(0) = 2$y(0)=2, substitute this in immediately to find the constant. The full solution then follows.

A common pitfall is treating $\frac{dy}{dx}$dxdy​ as a fraction and getting the algebra wrong. Strictly speaking, you are using the chain rule in reverse, but the manipulation $\frac{1}{g(y)}\,dy = f(x)\,dx$g(y)1​dy=f(x)dx is the standard A-Level shorthand and Edexcel accepts it. Another is forgetting the constant of integration entirely or putting one on each side; you only need one. A third is failing to rearrange to the form the question asks for — an implicit answer is sometimes acceptable, but if the question says "express $y$y in terms of $x$x", you must rearrange.

A short worked example. Solve $\frac{dy}{dx} = xy$dxdy​=xy with $y(0) = 1$y(0)=1. Separating: $\int \frac{1}{y}\,dy = \int x\,dx$∫y1​dy=∫xdx. Integrating: $\ln|y| = \frac{x^2}{2} + C$ln∣y∣=2x2​+C. Applying the initial condition $y(0) = 1$y(0)=1: $\ln 1 = 0 + C$ln1=0+C, so $C = 0$C=0. Therefore $\ln|y| = \frac{x^2}{2}$ln∣y∣=2x2​, or $y = e^{x^2/2}$y=ex2/2.

Differential-equation questions often appear in modelling contexts — Newton's law of cooling, exponential growth and decay, mixing problems. The setup is half the question; once the equation is written down, the integration is usually routine.

For full coverage with modelling examples, see the Differential Equations lesson.

Volumes of Revolution

When you rotate a curve $y = f(x)$y=f(x) between $x = a$x=a and $x = b$x=b through $360°$360° (i.e. $2\pi$2π radians) about the x-axis, you generate a 3D solid called a solid of revolution. The volume of this solid is given by

$V = \pi \int_a^b y^2\,dx = \pi \int_a^b [f(x)]^2\,dx$V=π∫ab​y2dx=π∫ab​[f(x)]2dx

The intuition is to slice the solid into thin disks perpendicular to the axis of rotation, each of thickness $dx$dx and radius $f(x)$f(x). The volume of each disk is $\pi [f(x)]^2\,dx$π[f(x)]2dx, and integrating sums the disk volumes.

For rotation about the y-axis, the formula switches: $V = \pi \int_c^d x^2\,dy$V=π∫cd​x2dy, with the curve now expressed as $x = g(y)$x=g(y) and limits in terms of $y$y. Edexcel usually examines rotation about the x-axis but can include either.

The hardest part of these questions is usually not the integration but the algebra of squaring the integrand. If $y = 2x + 3$y=2x+3, then $y^2 = 4x^2 + 12x + 9$y2=4x2+12x+9, and you must expand carefully. Sign errors and missed cross-terms are the standard failure mode. Another pitfall is forgetting the factor of $\pi$π outside the integral — Edexcel mark schemes will dock for an otherwise correct answer that omits it.

A short worked example. Find the volume generated when the curve $y = \sqrt{x}$y=x​ is rotated through $2\pi$2π about the x-axis between $x = 0$x=0 and $x = 4$x=4. Here $y^2 = x$y2=x, so $V = \pi \int_0^4 x\,dx = \pi \left[\frac{x^2}{2}\right]_0^4 = \pi \cdot 8 = 8\pi$V=π∫04​xdx=π[2x2​]04​=π⋅8=8π. Leave the answer in exact form unless the question specifies a decimal.

For rotation-about-y-axis examples and harder integrands, see the Volumes of Revolution lesson.

Applications of Integration

The integration toolkit feeds into a wider set of applications across the 9MA0 specification. The most important of these are computing distance from velocity, computing displacement and total distance, computing work done, and computing the mean value of a function. These appear both in the pure papers as direct applications and in the mechanics paper as part of kinematics.

The mean value of a function $f(x)$f(x) over the interval $[a, b]$[a,b] is

$\bar{f} = \frac{1}{b - a}\int_a^b f(x)\,dx$fˉ​=b−a1​∫ab​f(x)dx

This is the average height of the curve across the interval. Geometrically, it is the height of a rectangle on the same base whose area equals the area under the curve. Mean-value questions are usually short — 3-4 marks — and reduce to a definite integral followed by a division.

For kinematics, the relationships are: position is the integral of velocity, and velocity is the integral of acceleration. So if a particle has velocity $v(t) = 3t^2 - 6t$v(t)=3t2−6t, its displacement from $t = 0$t=0 to $t = 4$t=4 is $\int_0^4 (3t^2 - 6t)\,dt = [t^3 - 3t^2]_0^4 = 64 - 48 = 16$∫04​(3t2−6t)dt=[t3−3t2]04​=64−48=16. Total distance is different from displacement when the velocity changes sign; you need to integrate $|v(t)|$∣v(t)∣ piecewise.

A common pitfall is confusing displacement and total distance. If a particle moves forward and then back, the displacement may be small or zero, but the total distance travelled is the sum of the forward and backward movements. Always check whether $v(t)$v(t) changes sign in the interval, and split the integral at the sign change if so.

For mean-value problems and kinematics applications, see the Applications of Integration lesson.

Exam Strategy and High-Yield Topics for May 2026

Integration questions on Paper 2 reward a particular kind of preparation: pattern-recognition speed and arithmetic care. The techniques themselves are not deep — substitution, parts, partial fractions and separable equations are all routine in isolation. What makes the questions hard is that they combine, and the candidates who score full marks are those who recognise the right technique within seconds and execute it cleanly.

Three patterns are worth flagging specifically for May 2026.

The first is synoptic technique combination. A typical 9MA0 question format will set up an integral that requires partial fractions to decompose and then integration of each piece. Or an integral that benefits from a substitution and then parts on the result. These 8-12 mark questions reward candidates who can stand back, recognise the structure, and pick the right opening move. Drilling individual techniques in isolation is necessary but not sufficient — practise mixed problem sets where you have to choose the technique yourself.

The second is differential equations in modelling contexts. Setting up the equation from a worded scenario — exponential growth, cooling, mixing — is half the question, and many candidates lose marks here before they even begin to integrate. Read the question slowly, identify the rate, and translate it into $\frac{dy}{dx} = \ldots$dxdy​=… before reaching for the integration. Once the equation is written, separable techniques handle it routinely.

The third is volumes of revolution with messier integrands. The standard $V = \pi\int y^2\,dx$V=π∫y2dx is straightforward; what catches candidates is the algebra of squaring. A curve like $y = x + \frac{1}{x}$y=x+x1​ has $y^2 = x^2 + 2 + \frac{1}{x^2}$y2=x2+2+x21​, and missing the cross-term loses the question. Slow down on the squaring step and write each term explicitly.

Across all integration questions, three habits make the difference between a mid-range and a top-end performance. Show every step of the working — Edexcel mark schemes award method marks generously for clear setup, even when arithmetic goes astray at the end. Check the constant of integration on every indefinite integral — it is the single cheapest mark to lose. Carry exact forms through to the final line unless the question explicitly asks for a decimal — most 9MA0 integration answers are intended to be exact.

Where to Go from Here

LearningBro's Edexcel A-Level Maths: Integration course is built around the structure of this guide. Each of the ten lessons covers one section of the 9MA0 specification, in the order Edexcel teaches it, with worked examples, practice questions and full mark-scheme-style solutions. Lessons end with a short review and quick-recall questions designed for spaced revisits.

The course is designed to be used in two ways. As a first pass, you can work through the lessons in order, building each technique on the last. As a revision tool, you can drop into any lesson and work the practice independently — for example, drilling integration by parts for a week before mocks. The AI tutor is available throughout to give targeted hints when you get stuck, without giving away full solutions, and to mark your written working with structured feedback.

If you want one place to revise this section of the spec well, with realistic practice and clean explanations of every topic, the full course is the right next step. Start with the Edexcel A-Level Maths: Integration course.

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• Edexcel A-Level Maths: Algebra and Functions Guide
• Edexcel A-Level Maths: Trigonometry Guide
• A-Level Final Fortnight Revision Plan, May 2026