What is the fundamental theorem of calculus and how does it relate to integration in Pure Mathematics 4?

The fundamental theorem of calculus is a key principle that links the concept of differentiation with integration. It states that if a function is continuous over an interval and has an antiderivative, then the definite integral of the function over that interval is equal to the difference in the values of the antiderivative at the endpoints. In Pure Mathematics 4, this theorem is crucial as it allows students to evaluate definite integrals and understand the accumulation of quantities.

How do you perform integration by parts, and when is it used in Edexcel International A Levels Mathematics?

Integration by parts is a technique used to integrate products of functions. It is based on the formula ∫u dv = uv - ∫v du, where u and dv are differentiable functions. This method is particularly useful when dealing with integrals involving polynomial and exponential functions, or logarithmic and trigonometric functions. In Edexcel International A Levels Mathematics, students use integration by parts to solve complex integrals that cannot be simplified by basic integration techniques.

What are improper integrals and how are they evaluated in the context of Pure Mathematics 4?

Improper integrals are integrals with infinite limits or integrands that approach infinity within the integration range. To evaluate them, students often use limits to redefine the integral in a form that can be calculated. In Pure Mathematics 4, understanding improper integrals is important for solving problems involving infinite series and functions with unbounded behavior.

Can you explain the method of substitution in integration and its application in Edexcel International A Levels?

The method of substitution, also known as u-substitution, is a technique used to simplify integrals by changing variables. It involves substituting part of the integrand with a new variable, making the integral easier to evaluate. This method is particularly useful for integrals involving composite functions. In Edexcel International A Levels, substitution is a fundamental skill that helps students tackle a variety of integration problems efficiently.

What is the significance of definite and indefinite integrals in Pure Mathematics 4?

Definite integrals calculate the net area under a curve between two points, providing a numerical value that represents accumulation, such as distance or area. Indefinite integrals, on the other hand, represent a family of functions and include an arbitrary constant, indicating the antiderivative of a function. In Pure Mathematics 4, understanding both types of integrals is essential for solving real-world problems and for further studies in calculus and analysis.

Why is "Forgetting the modulus bars: $\int \dfrac{1}{x}\,dx = \ln x$ (without $| \cdot |$)" a common mistake in Integration?

Why it happens: Students don't realise is only defined for positive arguments. How to avoid it: Always write |x| unless you know x > 0 throughout. In modular problems, the bars are mandatory in mark schemes. Source: WMA14/01 examiner reports — flagged annually

Why is "Sign errors in second IBP iteration" a common mistake in Integration?

Why it happens: When subtracting an integral that itself has a minus sign, students forget to distribute. How to avoid it: Use brackets: x^2 x\,dx = x^2 x - ( 2x x\,dx). Then 2x x\,dx = -2x x + 2 x. So x^2 x\,dx = x^2 x - (-2x x + 2 x) = x^2 x + 2x x - 2 x.

Why is "Forgetting the $\pi$ in $V = \pi \int y^{2}\,dx$" a common mistake in Integration?

Why it happens: Students focus on the integral and miss the prefactor. How to avoid it: Volume of revolution ALWAYS has . Write 'V = ' as your first line. Source: WMA14/01 examiner reports — annual

Why is "Omitting the constant $C$ when integrating a DE" a common mistake in Integration?

Why it happens: Students go straight to y = without including C. How to avoid it: EVERY indefinite integral needs + C. For a DE with initial condition, use the IC to find C explicitly — then substitute the value into the general solution.

Why is "Using Simpson's rule with an odd number of strips" a common mistake in Integration?

Why it happens: Students miss the 'even strips' requirement. How to avoid it: Simpson's rule REQUIRES n even (so n + 1 ordinates with n + 1 odd). If a question specifies an odd n, use the trapezium rule instead.

Why is "Not transforming limits when using definite-integral substitution" a common mistake in Integration?

Why it happens: Students substitute but leave the limits as x-values. How to avoid it: When u = g(x), the limits become u_lo = g(x_lo), u_hi = g(x_hi). Either transform the limits OR substitute back to x before evaluating — be consistent.

Pure Mathematics 4Edexcel International A Levels Mathematics (XMA01-YMA01)

Integration

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Integration (P4) Study Notes — Edexcel IAL Mathematics P4 (WMA14/01, 2018 spec — 2026 onwards)

The 'big' P4 topic: integration by substitution, by parts (possibly twice), via partial fractions; numerical integration (trapezium and Simpson); volumes of revolution; first-order differential equations (separable + integrating-factor). Together these account for 25-30 marks in every paper — knowing the techniques cold is the difference between a B and an A*.

At a glance

By parts: $\int u\,dv = uv - \int v\,du$ (LIATE order for $u$ ).
Substitution: $\int f(g(x))g'(x)\,dx = \int f(u)\,du$ .
Partial fractions: decompose, then integrate term by term to $\ln$ s.
Trapezium rule: $\dfrac{h}{2}[y_{0} + y_{n} + 2(\text{middles})]$ .
Simpson's rule: $\dfrac{h}{3}[y_{0} + y_{n} + 4(\text{odd}) + 2(\text{even})]$ — needs EVEN strips.
Volume about $x$ -axis: $V = \pi \int y^{2}\,dx$ .
Volume about $y$ -axis: $V = \pi \int x^{2}\,dy$ .
Separable DE: $\int \dfrac{1}{g(y)}\,dy = \int f(x)\,dx$ .
Linear DE: integrating factor $I = e^{\int P\,dx}$ .

What you’ll learn

Mapped to the Cambridge IGCSE XMA01-YMA01 syllabus (2026 onwards).

P4 5.1 — Integrate using partial fractions.
P4 5.2 — Integrate using substitution (including trigonometric substitutions).
P4 5.3 — Use integration by parts, including in two steps when necessary.
P4 5.4 — Use the trapezium rule and Simpson's rule for numerical integration.
P4 5.5 — Find volumes of revolution about the $x$ -axis and $y$ -axis; solve first-order separable and linear differential equations.

Integration via partial fractions

Decompose first, then each piece is a simple log or arctan.

Strategy.

Express the integrand as partial fractions (if needed, polynomial-divide first).
Integrate each piece using standard integrals.
Combine logs at the end if asked.

Standard pieces.

Form	Integral
$\dfrac{1}{x + a}$	$\ln
$\dfrac{1}{(x + a)^{2}}$	$-\dfrac{1}{x + a} + C$
$\dfrac{1}{a^{2} + x^{2}}$	$\dfrac{1}{a}\arctan\dfrac{x}{a} + C$ (formula booklet)

Worked example. $\int \dfrac{4}{(x - 1)(x + 3)}\,dx$ .

Decompose: $\dfrac{1}{x - 1} - \dfrac{1}{x + 3}$ .
Integrate: $\ln|x - 1| - \ln|x + 3| + C = \ln\left|\dfrac{x - 1}{x + 3}\right| + C$ .

With repeated factor. $\int \dfrac{3x + 1}{(x - 1)(x + 2)^{2}}\,dx$ .

Partial fractions: $\dfrac{4/9}{x - 1} - \dfrac{4/9}{x + 2} + \dfrac{5/3}{(x + 2)^{2}}$ .
Integrate: $\dfrac{4}{9}\ln|x - 1| - \dfrac{4}{9}\ln|x + 2| - \dfrac{5/3}{x + 2} + C$ .

Partial fractions FIRST, then integrate term by term.
$\int \dfrac{1}{x+a}\,dx = \ln|x+a| + C$ .
$\int \dfrac{1}{(x+a)^{2}}\,dx = -\dfrac{1}{x+a} + C$ (no log).
Always include $+ C$ .

Integration by substitution

Spot a function and its derivative; substitute.

General principle. If you can spot $f(g(x)) \cdot g'(x)$ in the integrand, substitute $u = g(x)$ , so $du = g'(x)\,dx$ , and the integral becomes $\int f(u)\,du$ .

Common substitutions in P4.

Integrand contains	Try
$(\text{linear in } x)^{n}$	$u = \text{the linear}$
$\sqrt{1 - x^{2}}$	$x = \sin u$
$\sqrt{1 + x^{2}}$	$x = \tan u$
$\sqrt{x^{2} - 1}$	$x = \sec u$
$f(\sin x) \cos x$	$u = \sin x$
$f(\cos x) \sin x$	$u = \cos x$
$\dfrac{f'(x)}{f(x)}$	direct: $\ln

Worked example 1. $\int x(1 + x^{2})^{5}\,dx$ .

$u = 1 + x^{2}$ , $du = 2x\,dx$ , so $x\,dx = du/2$ .
$\int (1 + x^{2})^{5} \cdot x\,dx = \int u^{5} \cdot \dfrac{du}{2} = \dfrac{u^{6}}{12} + C = \dfrac{(1 + x^{2})^{6}}{12} + C$ .

Worked example 2 (trig substitution). $\int \dfrac{1}{\sqrt{4 - x^{2}}}\,dx$ .

$x = 2\sin u$ , $dx = 2\cos u\,du$ . $\sqrt{4 - x^{2}} = \sqrt{4(1 - \sin^{2}u)} = 2\cos u$ .
$\int \dfrac{1}{2\cos u} \cdot 2 \cos u\,du = \int du = u + C = \arcsin(x/2) + C$ .

Worked example 3 (definite, transforming limits). $\int_{0}^{1} 2x e^{x^{2}}\,dx$ .

$u = x^{2}$ , $du = 2x\,dx$ . Limits: $x = 0 \to u = 0$ ; $x = 1 \to u = 1$ .
$\int_{0}^{1} e^{u}\,du = e - 1$ .

Spot $f(g(x))g'(x)$ .
Transform $dx \to du$ via $du = g'(x)\,dx$ .
Transform limits in definite integrals.
Substitute back to $x$ for indefinite integrals.

Integration by parts

$\int u\,dv = uv - \int v\,du$ . Pick $u$ via LIATE.

Formula (in the booklet):

$\int u\,dv = uv - \int v\,du.$

Choosing $u$ : use LIATE (Logarithm, Inverse trig, Algebraic, Trig, Exponential) — pick the type that comes EARLIEST in this list as $u$ .

$\int$ Type	$u$	$dv$
$\int x e^{x}\,dx$	$x$ (Algebraic)	$e^{x}\,dx$ (Exp)
$\int x \ln x\,dx$	$\ln x$ (Log)	$x\,dx$ (Alg)
$\int x^{2} \cos x\,dx$	$x^{2}$ (Alg)	$\cos x\,dx$ (Trig)
$\int \arctan x\,dx$	$\arctan x$ (Inverse)	$dx$

Worked example 1. $\int x e^{2x}\,dx$ .

$u = x$ , $dv = e^{2x}\,dx$ ⇒ $du = dx$ , $v = \tfrac{1}{2}e^{2x}$ .
$= \tfrac{1}{2} x e^{2x} - \int \tfrac{1}{2}e^{2x}\,dx = \tfrac{1}{2}xe^{2x} - \tfrac{1}{4}e^{2x} + C$ .

Worked example 2 (IBP twice). $\int x^{2}\cos x\,dx$ .

First IBP: $u = x^{2}$ , $dv = \cos x\,dx$ ⇒ $v = \sin x$ , $du = 2x\,dx$ . $\int x^{2}\cos x\,dx = x^{2}\sin x - \int 2x \sin x\,dx$ .
Second IBP on $\int 2x\sin x\,dx$ : $u = 2x$ , $dv = \sin x\,dx$ ⇒ $v = -\cos x$ . $\int 2x\sin x\,dx = -2x\cos x - \int -2\cos x\,dx = -2x\cos x + 2\sin x$ .
Combine: $\int x^{2}\cos x\,dx = x^{2}\sin x - (-2x\cos x + 2\sin x) + C = x^{2}\sin x + 2x\cos x - 2\sin x + C$ .

Worked example 3 (IBP for $\ln$ ). $\int \ln x\,dx$ .

$u = \ln x$ , $dv = dx$ ⇒ $du = (1/x)dx$ , $v = x$ .
$\int \ln x\,dx = x \ln x - \int 1\,dx = x \ln x - x + C$ .

$\int u\,dv = uv - \int v\,du$ .
LIATE: choose $u$ from earliest category.
Sometimes apply IBP twice (or trick: $\int e^{x}\sin x\,dx$ — recursive).
$\int \ln x\,dx$ : take $u = \ln x$ , $dv = dx$ .

Numerical integration (trapezium, Simpson)

Approximate with linear strips (trapezium) or parabolic strips (Simpson).

Trapezium rule (formula booklet).

$\int_{a}^{b} y\,dx \approx \dfrac{h}{2}\left[y_{0} + y_{n} + 2(y_{1} + y_{2} + \ldots + y_{n-1})\right], \;\; h = \dfrac{b - a}{n}.$

Simpson's rule (formula booklet).

$\int_{a}^{b} y\,dx \approx \dfrac{h}{3}\left[y_{0} + y_{n} + 4(y_{1} + y_{3} + \ldots + y_{n-1}) + 2(y_{2} + y_{4} + \ldots + y_{n-2})\right].$

Simpson REQUIRES $n$ even.

Procedure.

Compute strip width $h = (b - a)/n$ .
Sample $y$ at the $n + 1$ ordinates $x_{i} = a + ih$ .
Apply the formula.

Worked example. Estimate $\int_{0}^{1} \sqrt{1 + x^{2}}\,dx$ using Simpson's rule with $n = 4$ .

$h = 0.25$ . Sample: $x = 0, 0.25, 0.5, 0.75, 1$ .
$y = 1, 1.0308, 1.1180, 1.25, 1.4142$ (to 4 d.p.).
Simpson: $S = \dfrac{0.25}{3}\left[1 + 1.4142 + 4(1.0308 + 1.25) + 2(1.1180)\right]$ .
$= \dfrac{0.25}{3}\left[1 + 1.4142 + 9.1232 + 2.236\right] = \dfrac{0.25}{3}(13.7734) \approx 1.1478$ .

True value: $1.1478\ldots$ — excellent agreement.

When to use each.

Trapezium: any $n$ , lower accuracy.
Simpson: $n$ even only, much higher accuracy for smooth integrands.

Both formulae in the booklet.
Simpson requires $n$ EVEN.
Strip width $h = (b - a)/n$ .
$n$ strips ⇒ $n + 1$ ordinates.

Volumes of revolution

Stack discs along the axis of rotation.

About $x$ -axis. Region under $y = f(x)$ , from $x = a$ to $x = b$ , rotated $360°$ about the $x$ -axis:

$V = \pi \int_{a}^{b} y^{2}\,dx.$

About $y$ -axis. Region between $x = g(y)$ and the $y$ -axis, from $y = c$ to $y = d$ , rotated $360°$ about the $y$ -axis:

$V = \pi \int_{c}^{d} x^{2}\,dy.$

Worked example 1. $y = x^{2}$ , $0 \leq x \leq 2$ , rotate about $x$ -axis.

$V = \pi \int_{0}^{2} x^{4}\,dx = \pi \cdot \dfrac{32}{5} = \dfrac{32\pi}{5}$ .

Worked example 2. Region bounded by $y = \sqrt{1 + x^{2}}$ , $y = 0$ , $x = 0$ , $x = 2$ . Rotate about $x$ -axis.

$V = \pi \int_{0}^{2} (1 + x^{2})\,dx = \pi \left[x + \dfrac{x^{3}}{3}\right]_{0}^{2} = \pi(2 + 8/3) = \dfrac{14\pi}{3}$ .

Worked example 3. $y = \sqrt{x}$ , $0 \leq y \leq 2$ , rotate about $y$ -axis.

Need $x$ in terms of $y$ : $x = y^{2}$ . So $x^{2} = y^{4}$ .
$V = \pi \int_{0}^{2} y^{4}\,dy = \pi \cdot \dfrac{32}{5} = \dfrac{32\pi}{5}$ .

Subtraction of volumes. Region between two curves, e.g. $y = f(x)$ above and $y = g(x)$ below, rotated about $x$ -axis: $V = \pi \int(f^{2} - g^{2})\,dx$ .

$x$ -axis: $V = \pi \int y^{2}\,dx$ .
$y$ -axis: $V = \pi \int x^{2}\,dy$ — rearrange to $x = g(y)$ .
Both formulae are in the booklet.
Always exact-form the answer (leave $\pi$ in).

Separable differential equations

Separate the variables, integrate both sides.

Form. $\dfrac{dy}{dx} = f(x) g(y)$ .

Strategy.

Rewrite as $\dfrac{1}{g(y)}\,dy = f(x)\,dx$ .
Integrate both sides.
Add a single constant $C$ .
Apply initial / boundary condition to find $C$ .
Solve for $y$ if asked.

Worked example 1. $\dfrac{dy}{dx} = 2xy$ , $y(0) = 3$ .

Separate: $\dfrac{1}{y}\,dy = 2x\,dx$ .
Integrate: $\ln|y| = x^{2} + C$ .
IC: $\ln 3 = 0 + C$ ⇒ $C = \ln 3$ .
$\ln|y| = x^{2} + \ln 3$ , so $|y| = 3 e^{x^{2}}$ , and since $y(0) = 3 > 0$ , $y = 3 e^{x^{2}}$ .

Worked example 2 (with growth rate). A population $N(t)$ grows so that $\dfrac{dN}{dt} = 0.05 N$ , $N(0) = 1000$ . Find $N(t)$ .

Separate: $\dfrac{1}{N}\,dN = 0.05\,dt$ ⇒ $\ln|N| = 0.05t + C$ .
IC: $\ln 1000 = C$ . So $\ln N = 0.05t + \ln 1000$ , $N = 1000 e^{0.05t}$ .

Worked example 3 (water tank). Tank drains: $\dfrac{dh}{dt} = -k\sqrt{h}$ , $h(0) = h_{0}$ .

$\dfrac{1}{\sqrt{h}}\,dh = -k\,dt$ ⇒ $2\sqrt{h} = -kt + C$ .
$C = 2\sqrt{h_{0}}$ . So $\sqrt{h} = \sqrt{h_{0}} - kt/2$ , $h(t) = (\sqrt{h_{0}} - kt/2)^{2}$ .

Separate, integrate, find $C$ from IC.
Single $C$ (don't write $C_{1}, C_{2}$ on each side).
Solve explicitly for $y$ if asked.
Modulus bars in $\ln$ until you know the sign.

First-order linear DE — integrating factor

Multiply by $I = e^{\int P}$ , then LHS is a product derivative.

Form. $\dfrac{dy}{dx} + P(x)\,y = Q(x)$ .

Strategy.

Compute the integrating factor $I = e^{\int P\,dx}$ .
Multiply through: $I y' + IP y = IQ$ .
The LHS is $\dfrac{d}{dx}(Iy)$ (by product rule, since $\dfrac{dI}{dx} = IP$ ).
Integrate both sides: $Iy = \int IQ\,dx + C$ .
Solve for $y$ .

Worked example 1. $\dfrac{dy}{dx} + \dfrac{y}{x} = x^{2}$ , $x > 0$ .

$P = 1/x$ , $\int P\,dx = \ln x$ , $I = e^{\ln x} = x$ .
Multiply: $xy' + y = x^{3}$ . LHS $= \dfrac{d}{dx}(xy)$ .
Integrate: $xy = \dfrac{x^{4}}{4} + C$ , so $y = \dfrac{x^{3}}{4} + \dfrac{C}{x}$ .

Worked example 2. $\dfrac{dy}{dx} + 2y = e^{x}$ , $y(0) = 1$ .

$P = 2$ , $\int P\,dx = 2x$ , $I = e^{2x}$ .
Multiply: $e^{2x}y' + 2e^{2x}y = e^{2x} \cdot e^{x} = e^{3x}$ .
LHS $= \dfrac{d}{dx}(e^{2x}y)$ . So $e^{2x}y = \dfrac{e^{3x}}{3} + C$ , giving $y = \dfrac{e^{x}}{3} + Ce^{-2x}$ .
IC: $1 = 1/3 + C$ , so $C = 2/3$ . Final: $y = \dfrac{e^{x} + 2 e^{-2x}}{3}$ .

Spotting linear DEs. They look like $y' + (\text{function of } x) y = (\text{function of } x)$ . If the $y$ -coefficient is constant, the IF is $e^{P x}$ ; otherwise it depends on the form.

Standard form: $y' + Py = Q$ .
$I = e^{\int P\,dx}$ .
LHS becomes $\dfrac{d}{dx}(Iy)$ after multiplying.
Integrate, then solve for $y$ .

How it’s examined

Integration is the LARGEST topic on WMA14/01 — typically 25-30 marks spread across multiple questions. Q8-Q11 are usually integration-heavy. Patterns: 'Use the substitution $u = \ldots$ to find $\int \ldots$ ' (6 marks); 'Find $\int x^{2}e^{x}\,dx$ ' (6 marks IBP twice); 'Use Simpson's rule with 4 strips' (3 marks); 'Find the volume of revolution' (6-7 marks); 'Solve the differential equation $\dfrac{dy}{dx} + Py = Q$ ' (8-9 marks). Mark schemes are STRICT on modulus bars in logs, $+ C$ , and units (where relevant). A* candidates should aim for 28+/30 on the integration components.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018) — Simpson, trapezium, volume, IBP formulae provided; WMA14/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

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Step-by-step worked examples — Integration

Step-by-step solutions to past-paper-style questions on integration, written exactly the way a tutor would explain them at the board.

1Integration via partial fractions (6 marks, P4)
Core• Adapted from WMA14/01 January 2024 Q10• integration, partial fractions
Question
Find $\displaystyle \int \dfrac{4}{(x - 1)(x + 3)}\,dx$ .
Step-by-step solution
1. Step 1
  Decompose into partial fractions: $\dfrac{4}{(x - 1)(x + 3)} = \dfrac{A}{x - 1} + \dfrac{B}{x + 3}$ .
2. Step 2
  Cover-up: $A = \dfrac{4}{4} = 1$ at $x = 1$ ; $B = \dfrac{4}{-4} = -1$ at $x = -3$ .
3. Step 3
  So the integrand becomes $\dfrac{1}{x - 1} - \dfrac{1}{x + 3}$ .
4. Step 4
  Integrate term by term:
  $\int \dfrac{1}{x - 1}\,dx - \int \dfrac{1}{x + 3}\,dx = \ln|x - 1| - \ln|x + 3| + C$
5. Step 5
  Combine logs.
  $\int \dfrac{4}{(x - 1)(x + 3)}\,dx = \ln\left|\dfrac{x - 1}{x + 3}\right| + C$
Answer
$\ln\left|\dfrac{x - 1}{x + 3}\right| + C$ .
Examiner tip
M1 for the partial-fraction setup. A1 for both $A, B$ . M1 for integrating each $\dfrac{1}{x \pm a}$ . A1 for each $\ln$ term. A1 for combined log form (or two separate logs — both accepted). The modulus bars are mandatory in the mark scheme — drop them and lose A1.
2Integration by substitution (6 marks, P4)
Extended• Adapted from WMA14/01 June 2024 Q9• integration, substitution
Question
Use the substitution $u = 1 + 2\sin x$ to find $\displaystyle \int \dfrac{\cos x}{(1 + 2\sin x)^{3}}\,dx$ .
Step-by-step solution
1. Step 1
  Set $u = 1 + 2\sin x$ . Then $\dfrac{du}{dx} = 2\cos x$ , so $du = 2\cos x \, dx$ , i.e. $\cos x \, dx = \dfrac{1}{2}du$ .
2. Step 2
  Rewrite the integral:
  $\int \dfrac{\cos x}{(1 + 2\sin x)^{3}}\,dx = \int \dfrac{1}{u^{3}} \cdot \dfrac{1}{2}\,du = \dfrac{1}{2}\int u^{-3}\,du$
3. Step 3
  Integrate:
  $\dfrac{1}{2} \cdot \dfrac{u^{-2}}{-2} + C = -\dfrac{1}{4u^{2}} + C$
4. Step 4
  Substitute back:
  $-\dfrac{1}{4(1 + 2\sin x)^{2}} + C$
Answer
$-\dfrac{1}{4(1 + 2\sin x)^{2}} + C$ .
Examiner tip
M1 for computing $du$ correctly. A1 for $\cos x \, dx = \tfrac{1}{2} du$ . M1 for the transformed integral. A1 for $\int u^{-3} du$ work. A1 for the antiderivative in $u$ . A1 for substituting back. The $+ C$ is required — omitting it costs the final A.
3Integration by parts (5 marks, P4)
Extended• Adapted from WMA14/01 January 2023 Q10• integration, by parts
Question
Find $\displaystyle \int x e^{2x}\,dx$ .
Step-by-step solution
1. Step 1
  Apply integration by parts: $\int u \, dv = uv - \int v \, du$ . Choose $u = x$ (LIATE — algebraic before exponential), $dv = e^{2x}\,dx$ .
2. Step 2
  Then $du = dx$ and $v = \dfrac{1}{2}e^{2x}$ .
3. Step 3
  Apply the formula:
  $\int x e^{2x}\,dx = x \cdot \dfrac{1}{2}e^{2x} - \int \dfrac{1}{2}e^{2x}\,dx$
4. Step 4
  Compute the remaining integral:
  $= \dfrac{1}{2}x e^{2x} - \dfrac{1}{4}e^{2x} + C = \dfrac{1}{4}e^{2x}(2x - 1) + C$
Answer
$\dfrac{1}{4}e^{2x}(2x - 1) + C$ .
Examiner tip
M1 for the IBP formula. B1 for correct choice of $u, dv$ (or equivalent). A1 for $v = \tfrac{1}{2}e^{2x}$ . A1 for the partial result $\tfrac{1}{2}xe^{2x} - \int \tfrac{1}{2}e^{2x}\,dx$ . A1 for the final expression. Factored form is preferred; expanded form also accepted.
4Integration by parts twice (6 marks, P4)
Challenge• Adapted from WMA14/01 June 2023 Q9• integration, by parts twice
Question
Find $\displaystyle \int x^{2} \cos x\,dx$ .
Step-by-step solution
1. Step 1
  First IBP: $u = x^{2}$ , $dv = \cos x \, dx$ . Then $du = 2x\,dx$ , $v = \sin x$ .
  $\int x^{2} \cos x\,dx = x^{2} \sin x - \int 2x \sin x\,dx$
2. Step 2
  Now apply IBP again on $\int 2x \sin x\,dx$ : $u = 2x$ , $dv = \sin x\,dx$ , so $du = 2\,dx$ , $v = -\cos x$ .
  $\int 2x \sin x\,dx = -2x \cos x + \int 2 \cos x\,dx = -2x \cos x + 2 \sin x$
3. Step 3
  Combine:
  $\int x^{2} \cos x\,dx = x^{2} \sin x - (-2x \cos x + 2 \sin x) + C$
4. Step 4
  Simplify.
  $= x^{2} \sin x + 2x \cos x - 2 \sin x + C$
Answer
$x^{2} \sin x + 2x \cos x - 2 \sin x + C$ .
Examiner tip
M1 for first IBP. A1 for the first decomposition. M1 for second IBP. A1 for the inner integral $-2x \cos x + 2 \sin x$ . A1 for combining. A1 for the final expression. Sign error at the second-IBP step is the leading mark-loser — when subtracting an integral that itself has a $-$ sign, distribute carefully.
5Volume of revolution (6 marks, P4)
Extended• Adapted from WMA14/01 January 2024 Q11• volume of revolution, x-axis
Question
The region $R$ is bounded by the curve $y = \sqrt{1 + x^{2}}$ , the $x$ -axis, and the lines $x = 0$ and $x = 2$ . Find the volume generated when $R$ is rotated $360°$ about the $x$ -axis.
Step-by-step solution
1. Step 1
  Volume about $x$ -axis: $V = \pi \int_{0}^{2} y^{2}\,dx$ .
2. Step 2
  $y^{2} = 1 + x^{2}$ .
  $V = \pi \int_{0}^{2} (1 + x^{2})\,dx$
3. Step 3
  Antiderivative: $x + \dfrac{x^{3}}{3}$ .
  $V = \pi \left[ x + \dfrac{x^{3}}{3} \right]_{0}^{2} = \pi \left(2 + \dfrac{8}{3}\right) = \pi \cdot \dfrac{14}{3}$
4. Step 4
  Final answer.
  $V = \dfrac{14\pi}{3}$
Answer
$V = \dfrac{14\pi}{3}$ (cubic units).
Examiner tip
M1 for the volume formula $\pi \int y^{2}\,dx$ . A1 for $y^{2} = 1 + x^{2}$ . M1 for integration. A1 for the antiderivative. A1 for substitution of limits. A1 for $\dfrac{14\pi}{3}$ in exact form. Decimal answer accepted with B0 if exact form expected.
6Separable differential equation (7 marks, P4)
Extended• Adapted from WMA14/01 June 2024 Q11• differential equation, separable
Question
Solve $\dfrac{dy}{dx} = \dfrac{y \cos x}{1 + \sin x}$ given that $y = 2$ when $x = 0$ .
Step-by-step solution
1. Step 1
  Separate variables:
  $\dfrac{1}{y}\,dy = \dfrac{\cos x}{1 + \sin x}\,dx$
2. Step 2
  Integrate both sides. LHS: $\ln|y|$ . RHS: note the numerator is the derivative of the denominator, so $\int \dfrac{\cos x}{1 + \sin x}\,dx = \ln|1 + \sin x|$ .
  $\ln|y| = \ln|1 + \sin x| + C$
3. Step 3
  Substitute $y = 2$ , $x = 0$ : $\ln 2 = \ln 1 + C$ , so $C = \ln 2$ .
4. Step 4
  Reassemble: $\ln|y| = \ln|1 + \sin x| + \ln 2 = \ln|2(1 + \sin x)|$ .
5. Step 5
  Exponentiate:
  $y = 2(1 + \sin x)$
Answer
$y = 2(1 + \sin x)$ .
Examiner tip
M1 for separating variables. A1 for the correct separation. M1 for integrating LHS. A1 for $\ln|y|$ . M1 for integrating RHS (recognising the $f'/f$ form). A1 for the constant via initial condition. A1 for the explicit $y = \ldots$ . The mark scheme accepts both $\ln$ -form and the explicit- $y$ form, but for 'solve' questions the explicit form is preferred.
7First-order linear DE — integrating factor (8 marks, P4)
Challenge• Adapted from WMA14/01 January 2023 Q11• differential equation, integrating factor
Question
Solve $\dfrac{dy}{dx} + \dfrac{y}{x} = x^{2}$ for $x > 0$ .
Step-by-step solution
1. Step 1
  Standard form $\dfrac{dy}{dx} + P(x)y = Q(x)$ with $P = 1/x$ , $Q = x^{2}$ .
2. Step 2
  Integrating factor:
  $I = e^{\int P\,dx} = e^{\int 1/x\,dx} = e^{\ln x} = x$
3. Step 3
  Multiply through by $I = x$ :
  $x \dfrac{dy}{dx} + y = x^{3}$
4. Step 4
  LHS is $\dfrac{d}{dx}(xy)$ . Integrate both sides:
  $xy = \int x^{3}\,dx = \dfrac{x^{4}}{4} + C$
5. Step 5
  Solve for $y$ :
  $y = \dfrac{x^{3}}{4} + \dfrac{C}{x}$
Answer
$y = \dfrac{x^{3}}{4} + \dfrac{C}{x}$ .
Examiner tip
M1 for identifying the integrating factor structure. A1 for $I = x$ . M1 for multiplying through. A1 for recognising the LHS as a derivative. M1 for integrating RHS. A1 for the antiderivative. A1 for solving for $y$ . A1 for full explicit form. Integrating factor for linear DE is a P4-specific technique — examiners flag missing-IF errors.

Key Formulae — Integration

The formulae you need to memorise for integration on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

Integration by parts
$\int u\,\dfrac{dv}{dx}\,dx = uv - \int v\,\dfrac{du}{dx}\,dx$
$u$
function to differentiate (becomes simpler)
$dv/dx$
function to integrate
When to use
Products of unlike functions. Use LIATE order: Logarithm, Inverse trig, Algebraic, Trig, Exponential — pick the first as $u$ . IS in the formula booklet.
Example
$\int x e^{x}\,dx$ : $u = x$ , $dv = e^{x}\,dx$ → $u' = 1$ , $v = e^{x}$ , so $\int = xe^{x} - e^{x} + C$ .
Integration by substitution
$\int f(g(x)) g'(x)\,dx = \int f(u)\,du \;\;\text{where } u = g(x)$
$u = g(x)$
substitution variable
$du = g'(x)\,dx$
differential
When to use
When the integrand contains a function and its derivative (up to a constant factor). NOT specifically in the formula booklet — this is the chain rule in reverse.
Example
$\int 2x \cdot e^{x^{2}}\,dx$ : set $u = x^{2}$ , $du = 2x\,dx$ → $\int e^{u}\,du = e^{x^{2}} + C$ .
Trapezium rule
$\int_{a}^{b} y\,dx \approx \dfrac{h}{2}\left[y_{0} + y_{n} + 2(y_{1} + y_{2} + \ldots + y_{n-1})\right], \;\;\;h = \dfrac{b - a}{n}$
$h$
strip width
$y_{i}$
function values at $x_{i} = a + ih$
$n$
number of strips
When to use
Numerical integration approximation when no analytic antiderivative is available, or when explicitly asked. IS in the formula booklet.
Example
Estimate $\int_{0}^{1} e^{-x^{2}}\,dx$ using 4 strips: $h = 0.25$ , sample at $x = 0, 0.25, 0.5, 0.75, 1$ .
Simpson's rule
$\int_{a}^{b} y\,dx \approx \dfrac{h}{3}\left[y_{0} + y_{n} + 4(y_{1} + y_{3} + \ldots) + 2(y_{2} + y_{4} + \ldots)\right]$
$h$
strip width $(b - a)/n$
$n$
number of strips (must be EVEN)
When to use
More accurate than the trapezium rule for the same number of strips. IS in the IAL formula booklet. Requires an even number of strips.
Example
With $h = 0.25$ , $n = 4$ strips, $y$ -values $y_{0}, y_{1}, y_{2}, y_{3}, y_{4}$ : $S = (0.25/3)[y_{0} + y_{4} + 4(y_{1} + y_{3}) + 2 y_{2}]$ .
Volume of revolution about $x$ -axis
$V = \pi \int_{a}^{b} y^{2}\,dx$
$y = f(x)$
curve being rotated
$a, b$
limits of $x$
When to use
Rotating a region under $y = f(x)$ from $x = a$ to $x = b$ about the $x$ -axis. IS in the IAL formula booklet.
Example
$y = x^{2}$ from $x = 0$ to $1$ about $x$ -axis: $V = \pi \int_{0}^{1} x^{4}\,dx = \pi/5$ .
Volume of revolution about $y$ -axis
$V = \pi \int_{c}^{d} x^{2}\,dy$
$x = g(y)$
curve rearranged as a function of $y$
$c, d$
limits of $y$
When to use
Rotating a region bounded by $x = g(y)$ about the $y$ -axis. IS in the IAL formula booklet.
Example
$x = \sqrt{y}$ from $y = 0$ to $4$ about $y$ -axis: $V = \pi \int_{0}^{4} y\,dy = 8\pi$ .
Integrating factor for first-order linear DE
$I(x) = e^{\int P(x)\,dx}, \;\; \dfrac{d}{dx}\left[I(x) y\right] = I(x) Q(x)$
$P(x)$
coefficient of $y$ in standard form $y' + P(x)y = Q(x)$
$Q(x)$
RHS source term
When to use
For first-order linear DEs of the form $y' + Py = Q$ . NOT in the formula booklet — students must recognise the structure and derive the integrating factor.
Example
$y' + y/x = x^{2}$ : $P = 1/x$ , $I = e^{\ln x} = x$ . Multiply: $\dfrac{d}{dx}(xy) = x^{3}$ , integrate.

Key Definitions and Keywords — Integration

Definitions to memorise and the exact keywords mark schemes credit for integration answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Integration by substitution (u-substitution)
Examiner keyword
Replacing the variable in an integral via $u = g(x)$ , $du = g'(x)\,dx$ , transforming the integral into one in $u$ . Reverses the chain rule.
Integration by parts
Examiner keyword
Formula $\int u\,dv = uv - \int v\,du$ . Used for integrals of products where one factor simplifies on differentiation. Reverses the product rule.
Volume of revolution (solid of revolution)
Examiner keyword
The 3D shape generated by rotating a planar region $360°$ about an axis (commonly $x$ - or $y$ -axis). Volume computed by stacking infinitesimal discs: $V = \pi \int y^{2}\,dx$ (about $x$ -axis).
Separable differential equation
Examiner keyword
A first-order DE of the form $\dfrac{dy}{dx} = f(x)g(y)$ — the right side factorises into a function of $x$ times a function of $y$ . Solved by separating variables: $\int \dfrac{1}{g(y)}\,dy = \int f(x)\,dx$ .
Integrating factor
Examiner keyword
A function $I(x)$ that, when multiplied through a linear DE $y' + Py = Q$ , makes the LHS into the derivative of a product: $\dfrac{d}{dx}(Iy)$ . The standard choice is $I = e^{\int P\,dx}$ .
Simpson's rule
Examiner keyword
A numerical integration approximation using parabolic arcs over each pair of strips. Requires an EVEN number of strips. More accurate than the trapezium rule for the same $n$ .

Common Mistakes and Misconceptions — Integration

The traps other students keep falling into on integration questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Forgetting the modulus bars: $\int \dfrac{1}{x}\,dx = \ln x$ (without $| \cdot |$ )
WMA14/01 examiner reports — flagged annually
Why it happens
Students don't realise $\ln$ is only defined for positive arguments.
How to avoid it
Always write $\ln|x|$ unless you know $x > 0$ throughout. In modular problems, the bars are mandatory in mark schemes.
✕Sign errors in second IBP iteration
Why it happens
When subtracting an integral that itself has a minus sign, students forget to distribute.
How to avoid it
Use brackets: $\int x^{2}\cos x\,dx = x^{2}\sin x - (\int 2x \sin x\,dx)$ . Then $\int 2x \sin x\,dx = -2x\cos x + 2\sin x$ . So $\int x^{2}\cos x\,dx = x^{2}\sin x - (-2x\cos x + 2\sin x) = x^{2}\sin x + 2x\cos x - 2\sin x$ .
✕Forgetting the $\pi$ in $V = \pi \int y^{2}\,dx$
WMA14/01 examiner reports — annual
Why it happens
Students focus on the integral and miss the $\pi$ prefactor.
How to avoid it
Volume of revolution ALWAYS has $\pi$ . Write ' $V = \pi \int \ldots$ ' as your first line.
✕Omitting the constant $C$ when integrating a DE
Why it happens
Students go straight to $y = \ldots$ without including $C$ .
How to avoid it
EVERY indefinite integral needs $+ C$ . For a DE with initial condition, use the IC to find $C$ explicitly — then substitute the value into the general solution.
✕Using Simpson's rule with an odd number of strips
Why it happens
Students miss the 'even strips' requirement.
How to avoid it
Simpson's rule REQUIRES $n$ even (so $n + 1$ ordinates with $n + 1$ odd). If a question specifies an odd $n$ , use the trapezium rule instead.
✕Not transforming limits when using definite-integral substitution
Why it happens
Students substitute but leave the limits as $x$ -values.
How to avoid it
When $u = g(x)$ , the limits become $u_{\text{lo}} = g(x_{\text{lo}})$ , $u_{\text{hi}} = g(x_{\text{hi}})$ . Either transform the limits OR substitute back to $x$ before evaluating — be consistent.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Video lesson

Short walkthrough of the concepts students most often get stuck on.

Integration — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Integration

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Integration via partial fractions (6 marks, P4)

2Integration by substitution (6 marks, P4)

3Integration by parts (5 marks, P4)

4Integration by parts twice (6 marks, P4)

5Volume of revolution (6 marks, P4)

6Separable differential equation (7 marks, P4)

7First-order linear DE — integrating factor (8 marks, P4)

Integration by parts

Integration by substitution

Trapezium rule

Simpson's rule

Volume of revolution about xxx-axis

Volume of revolution about yyy-axis

Integrating factor for first-order linear DE

Integration by substitution (u-substitution)

Integration by parts

Volume of revolution (solid of revolution)

Separable differential equation

Integrating factor

Simpson's rule

✕Forgetting the modulus bars: ∫1x dx=ln⁡x\int \dfrac{1}{x}\,dx = \ln x∫x1​dx=lnx (without ∣⋅∣| \cdot |∣⋅∣)

✕Sign errors in second IBP iteration

✕Forgetting the π\piπ in V=π∫y2 dxV = \pi \int y^{2}\,dxV=π∫y2dx

✕Omitting the constant CCC when integrating a DE

✕Using Simpson's rule with an odd number of strips

✕Not transforming limits when using definite-integral substitution

Practice questions

Past paper style quiz

Video lesson

Integration — frequently asked questions

Volume of revolution about $x$ -axis

Volume of revolution about $y$ -axis

✕Forgetting the modulus bars: $\int \dfrac{1}{x}\,dx = \ln x$ (without $| \cdot |$ )

✕Forgetting the $\pi$ in $V = \pi \int y^{2}\,dx$

✕Omitting the constant $C$ when integrating a DE