Integration (Edexcel IAL P4)

Edexcel International A-Level Mathematics· 2018 Specification Issue 3, first assessment 2019· 45 min read

1. Volume of Revolution Around the x-axis★★☆☆☆⏱ 8 min

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The volume of a solid formed by rotating the region under the curve $y = f (x)$ , between $x = a$ and $x = b$ , around the x-axis is given by the formula:

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

📘 Definition

Volume of Revolution for Parametric Curves

For a curve defined parametrically by $x = x (t)$ , $y = y (t)$ , where $x$ takes values $a$ when $t = t_{1}$ and $b$ when $t = t_{2}$ , the volume is calculated by substituting $y = y (t)$ and $d x = \frac{d x}{d t} d t$ into the standard formula.

Example:

$V = π \int_{t_{1}}^{t_{2}} y (t)^{2} \frac{d x}{d t} d t$

📐 Worked Example

Find the volume of revolution formed when the curve $y = e^{2 x}$ between $x = 0$ and $x = 1$ is rotated around the x-axis, giving your answer in terms of $π$ .

1
Step 1: Write down the volume formula, substitute $y = e^{2 x}$
$V = \pi \int_{0}^{1} (e^{2x})^2 dx = \pi \int_{0}^{1} e^{4x} dx$
2
Step 2: Integrate using P3 recognition rule for $e^{k x}$
$\int e^{4x} dx = \frac{1}{4}e^{4x} + C$
3
Step 3: Evaluate the definite integral between limits 0 and 1
$V = \pi \left[ \frac{1}{4}e^{4x} \right]_0^1 = \pi \left( \frac{1}{4}e^4 - \frac{1}{4}e^0 \right) = \frac{\pi}{4}(e^4 - 1)$

Exam tip:

Always square y before integrating, and remember to include the π factor outside the integral — this is one of the most common missed marks in volume questions.

2. Integration by Substitution and Integration by Parts★★★☆☆⏱ 12 min

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Integration by substitution reverses the chain rule, and integration by parts reverses the product rule. For P4, substitutions are provided for complex integrals, and you may need to apply integration by parts more than once for functions like $x^{2} e^{x}$ or $e^{x} sin x$ .

📘 Definition

Integration by Parts

Given in the formula booklet as $\int u \frac{d v}{d x} d x = uv - \int v \frac{d u}{d x} d x$ . Choose $u$ to be the function that simplifies when differentiated (e.g. polynomials, ln x).

Example:

For $\int x e^{x} d x$ , choose $u = x$ (simplifies to 1 when differentiated) and $\frac{d v}{d x} = e^{x}$ .

📐 Worked Example

Evaluate $\int x x - 2 d x$ using the substitution $u = x - 2$ .

1
Step 1: Rearrange substitution to find $x = u + 2$ , and differentiate to get $d x = d u$ .
2
Step 2: Substitute into the integral:
$\int (u + 2) \sqrt{u} du = \int (u^{3/2} + 2u^{1/2}) du$
3
Step 3: Integrate term by term:
$= \frac{2}{5}u^{5/2} + \frac{4}{3}u^{3/2} + C$
4
Step 4: Substitute back $u = x - 2$ :
$= \frac{2}{5}(x - 2)^{5/2} + \frac{4}{3}(x - 2)^{3/2} + C$

📐 Worked Example

Evaluate $\int e^{x} sin x d x$

1
Step 1: First application of by parts: let $u = sin x$ , $\frac{d v}{d x} = e^{x}$ , so $\frac{d u}{d x} = cos x$ , $v = e^{x}$ .
$I = \int e^x \sin x dx = e^x \sin x - \int e^x \cos x dx$
2
Step 2: Second application of by parts on the remaining integral: let $u = cos x$ , $\frac{d v}{d x} = e^{x}$ , so $\frac{d u}{d x} = - sin x$ , $v = e^{x}$ .
$I = e^x \sin x - \left( e^x \cos x - \int e^x (-\sin x) dx \right) = e^x \sin x - e^x \cos x - \int e^x \sin x dx$
3
Step 3: Notice the remaining integral is equal to the original integral I, so rearrange to solve for I:
$I + I = e^x(\sin x - \cos x) \implies 2I = e^x(\sin x - \cos x) \implies I = \frac{e^x}{2}(\sin x - \cos x) + C$

Exam tip:

For integrals of ln x, always set $u = ln x$ and $\frac{d v}{d x} = 1$ , since ln x is difficult to integrate directly but differentiates to $1/ x$ which simplifies the remaining integral.

3. Integration Using Partial Fractions and Recognition Integrals★★★☆☆⏱ 8 min

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You can integrate rational functions by first decomposing them into partial fractions, then integrating each term using standard P3 recognition rules. Common forms include linear denominators, repeated linear denominators, and functions of the form $\frac{a x + b}{c x ^{2} + d}$ or $\frac{k}{( m x + n ) ^{p}}$ .

📐 Worked Example

Evaluate $\int \frac{4 x + 5}{( x - 1 ) ( x + 2 ) ^{2}} d x$

1
Step 1: Decompose the fraction into partial fractions:
$\frac{4x + 5}{(x - 1)(x + 2)^2} = \frac{A}{x - 1} + \frac{B}{x + 2} + \frac{C}{(x + 2)^2}$
2
Step 2: Solve for constants: multiply both sides by denominator, substitute x = 1, x = -2, and compare coefficients to find A = 1, B = -1, C = 1.
3
Step 3: Integrate each term separately:
$\int \left( \frac{1}{x - 1} - \frac{1}{x + 2} + \frac{1}{(x + 2)^2} \right) dx$
4
Step 4: Apply recognition integrals:
$= \ln|x - 1| - \ln|x + 2| - \frac{1}{x + 2} + C = \ln \left| \frac{x - 1}{x + 2} \right| - \frac{1}{x + 2} + C$

Exam tip:

For integrals of the form $\frac{f ^{'} ( x )}{f ( x )}$ , the result is $ln ∣ f (x) ∣ + C$ — this is the most common recognition integral tested in P4, so always check if the numerator is a multiple of the derivative of the denominator first before attempting partial fractions or substitution.

4. First-Order Separable Differential Equations★★★★☆⏱ 9 min

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A first-order differential equation is separable if you can rearrange it to group all terms in y on one side and all terms in x on the other side. You will need to find both general solutions (with arbitrary constant C) and particular solutions (using given boundary conditions to find C).

📘 Definition

Separable Differential Equation Form

All separable ODEs can be written as $\frac{d y}{d x} = f (x) g (y)$ , which rearranges to $\int \frac{1}{g ( y )} d y = \int f (x) d x$ for integration.

Example:

For $\frac{d y}{d x} = x^{2} y$ , rearrange to $\int \frac{1}{y} d y = \int x^{2} d x$ .

📐 Worked Example

Solve the differential equation $\frac{d y}{d x} = 2 x cos^{2} y$ , given that $y = \frac{π}{4}$ when $x = 0$ . Give your answer in the form $y = f (x)$ .

1
Step 1: Rearrange to separate variables, dividing both sides by $cos^{2} y$ :
$\int \sec^2 y dy = \int 2x dx$
2
Step 2: Integrate both sides, adding the constant of integration to one side:
$\tan y = x^2 + C$
3
Step 3: Use the boundary condition $x = 0, y = π /4$ to find C:
$\tan(\pi/4) = 0 + C \implies C = 1$
4
Step 4: Rearrange to the required form:
$y = \arctan(x^2 + 1)$

Exam tip:

Always add the constant of integration immediately after integrating both sides, before applying boundary conditions. You will lose marks if you add C after substituting boundary values.

5. Area Under Parametric Curves★★★☆☆⏱ 8 min

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For a curve defined parametrically by $x = x (t)$ and $y = y (t)$ , the area under the curve between $x = a$ (when $t = t_{1}$ ) and $x = b$ (when $t = t_{2}$ ) is calculated by substituting $d x = \frac{d x}{d t} d t$ into the standard area formula $\int y d x$ . You do not need to sketch the curve for these questions.

📐 Worked Example

A curve has parametric equations $x = t^{2} + 1$ , $y = 3 t + 2$ , where $t \geq 0$ . Find the area under the curve between $x = 1$ and $x = 5$ .

1
Step 1: Find the values of t corresponding to the x limits: when x = 1, $t^{2} + 1 = 1 ⟹ t = 0$ ; when x=5, $t^{2} + 1 = 5 ⟹ t = 2$ .
2
Step 2: Differentiate x with respect to t:
$\frac{dx}{dt} = 2t \implies dx = 2t dt$
3
Step 3: Substitute into the area formula, replacing y, dx, and limits:
$A = \int_{t=0}^{t=2} (3t + 2) \cdot 2t dt = \int_{0}^{2} (6t^2 + 4t) dt$
4
Step 4: Integrate and evaluate:
$A = \left[ 2t^3 + 2t^2 \right]_0^2 = (2*8 + 2*4) - 0 = 16 + 8 = 24$

Exam tip:

Make sure your limits for t are in the correct order corresponding to increasing x values. If $\frac{d x}{d t}$ is negative, you may need to reverse the limits to get a positive area, but this will usually be handled automatically if you use the t values matching the lower and upper x limits.

6. Common Pitfalls

Wrong move:

Forgetting to include the π factor when calculating volume of revolution.

Why:

The volume formula requires multiplying the integral of y² by π, which is often missed in rushed calculations.

Correct move:

Write the π outside the integral immediately when starting a volume question, before doing any integration steps.

Wrong move:

Choosing the wrong u term for integration by parts, e.g. setting u = eˣ instead of u = x when integrating ∫x eˣ dx.

Why:

This leads to a more complex integral that cannot be solved easily, as differentiating eˣ leaves it unchanged.

Correct move:

Use the mnemonic LIPET (Logarithms, Inverse trig, Polynomials, Exponentials, Trig) to choose u: pick the function that comes first in this list as your u term.

Wrong move:

Forgetting to change limits when integrating parametric equations or using substitution.

Why:

If you leave x limits in an integral written in terms of t or u, you will get an incorrect numerical result.

Correct move:

As soon as you change variable in an integral, immediately convert the original limits to match the new variable before evaluating.

Wrong move:

Adding the constant of integration after applying boundary conditions for differential equations.

Why:

The constant of integration is part of the general solution, so omitting it before substituting boundaries leads to the wrong value of C.

Correct move:

Add the constant C immediately after integrating both sides of the differential equation, before substituting any given values of x and y.

Wrong move:

Squaring only the x term instead of the entire y expression for volume of revolution, e.g. writing (2x + 3)² as 2x² + 9 instead of 4x² + 12x +9.

Why:

Incorrect expansion of y² changes the integral completely, leading to lost method and accuracy marks.

Correct move:

Write out the full expansion of y² explicitly before integrating, even if it seems simple, to avoid algebraic errors.

7. Quick Reference Cheatsheet

Concept	Formula / Rule	Key Tip
Volume of Revolution (x-axis)	$V = π \int_{a}^{b} y^{2} d x$ , parametric: $V = π \int_{t 1}^{t 2} y (t)^{2} \frac{d x}{d t} d t$	Never forget the π factor outside the integral
Integration by Parts	$\int u \frac{d v}{d x} d x = uv - \int v \frac{d u}{d x} d x$ (given)	Choose u = ln x for ∫ln x dx, set dv/dx = 1
Partial Fraction Integration	Integrate each partial fraction term separately, use $ln ∣ f (x) ∣$ for 1/(ax+b) terms	Check if numerator is derivative of denominator first to avoid unnecessary partial fraction decomposition
Separable Differential Equations	Rearrange to $\int g (y) d y = \int f (x) d x$ , add C then apply boundaries	Always add C before substituting boundary values
Parametric Area	$A = \int_{t 1}^{t 2} y (t) \frac{d x}{d t} d t$	Convert x limits to t limits before integrating

8. Frequently Asked

Do I need to memorise the integration by parts formula for P4?

No, the integration by parts formula is provided in your exam formula booklet, along with integrals of sec x and cosec x. You only need to memorise P3 recognition integrals such as $\int \frac{x}{x ^{2} + c} d x = \frac{1}{2} ln ∣ x^{2} + c ∣ + C$ .

Do I need to calculate volumes of revolution around the y-axis for P4?

No, Edexcel explicitly states that only volume of revolution around the x-axis (using the formula $π \int y^{2} d x$ ) is required for P4; volumes around the y-axis are out of scope.

Can I use a graphical calculator for integration questions in P4?

Yes, calculators are permitted for all Edexcel IAL Maths papers, but CAS (symbolic algebra) calculators are forbidden. You must show all working to gain full marks, even if you can check your answer on a calculator.

Going deeper

formula_bookEdexcel IAL Mathematics Formula BookletOfficial formula booklet for all P1-P4 units

What's Next

Now that you have mastered all P4 integration content, you are ready to practice full exam-style questions that combine multiple integration techniques, which are frequently tested as the final long question in P4 papers. Make sure you also practice connecting integration to other P4 topics like parametric equations and differential equations, as cross-topic questions are common. Be sure to show all working even if you check your answer with a calculator, as method marks make up the majority of marks for integration questions. Finally, practice applying your integration skills to real-world modelling questions, which often accompany differential equation problems in exams.