What does the limit notation mean informally in IB Maths AA HL calculus?

As x gets arbitrarily close to a value a, the function output f(x) gets arbitrarily close to L.

What is L'Hopital's rule?

If a limit gives the indeterminate form 0/0 or infinity over infinity, differentiate the numerator and denominator separately and take the limit of that ratio instead.

What does the Intermediate Value Theorem guarantee?

If a function is continuous on a closed interval and the endpoints have opposite signs, there is at least one value in that interval where the function equals zero.

IB HL

IB Math AA HL Calculus: Complete Guide 2026 (Limits, Derivatives, Integration)

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Syllabus Topics: SL 5.1–5.11 limits, derivatives, integrals · AHL 5.12–5.19 continuity, implicit differentiation, differential equations, Maclaurin series
Paper 1: 2 h, no calculator, show working (30%).
Paper 2: 2 h, GDC required, not every question (30%).
Paper 3: 1 h, two problem-solving questions (20%).

Topic 5 code details

SL 5.1 Limits; derivative as a gradient and a rate of change
SL 5.2 Increasing and decreasing functions
SL 5.3 Derivatives of integer powers
SL 5.4 Tangents and normals
SL 5.5 Anti-differentiation, definite integrals using technology, area under a curve
SL 5.6 Standard derivatives; chain, product and quotient rules
SL 5.7 Second derivative; graphs of f, f’ and f”
SL 5.8 Local maxima and minima, optimisation, points of inflexion
SL 5.9 Kinematics
SL 5.10 Standard integrals and the reverse chain rule
SL 5.11 Definite integrals and areas between curves
AHL 5.12 Continuity, differentiability, limits, first principles, higher derivatives
AHL 5.13 Limits by l’Hôpital’s rule or Maclaurin series
AHL 5.14 Implicit differentiation, related rates, optimisation
AHL 5.15 Derivatives of tan, sec, cosec, cot, exponential, log and inverse trig; partial fractions
AHL 5.16 Integration by substitution and by parts
AHL 5.17 Area with the y-axis and volumes of revolution
AHL 5.18 First-order differential equations
AHL 5.19 Maclaurin series

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IB Math AA HL — Calculus

Complete Study Guide

Topics Covered

Limits & Continuity
Differentiation — Rules & Techniques
Applications of Differentiation (optimisation, related rates, kinematics)
Integration — Techniques & Standard Integrals
Applications of Integration (areas, volumes of revolution)
Differential Equations
Maclaurin Series
Practice MCQs & Exam Alerts

Topic 5 of the IB Math AA HL syllabus — Papers 1, 2 and 3

Videos on this page: Differentiation · Integration

Watch: Differentiation — Chain, Product, Quotient Rules & Applications

Watch: Integration — Techniques, Areas, Volumes & Substitutions

How to approach calculus on exams: The IB rewards structured method. Even if your final answer is wrong, showing correct working earns method marks. Always write the rule or theorem you are applying before you apply it, and show every intermediate line. A one-line answer — even if numerically correct — can score zero if no method is visible.

Prerequisites: Calculus builds on functions, algebra, and trigonometry. If you’re struggling with function notation, exponents, or trig identities, review the Functions and Geometry & Trigonometry guides first.

What is and is not in the formula booklet: Derivatives of $\sin x$ , $\cos x$ , $e^x$ , $\ln x$ , and $x^n$ are given. The product rule, quotient rule, and chain rule are given. Standard integrals (including $\int \frac{1}{x}\,dx$ , $\int e^x\,dx$ , $\int \sin x\,dx$ , $\int \cos x\,dx$ ) are given. The integration by parts formula is given. NOT given: L’Hôpital’s rule, the formula for volumes of revolution, the Maclaurin series of $e^x$ , $\sin x$ , $\cos x$ , and $\ln(1+x)$ (you must know how to derive them), and all kinematics interpretations.

Section 1: Limits and Continuity

A limit describes the value a function approaches as the input approaches some value, without necessarily reaching it. The notation $\displaystyle\lim_{x \to a} f(x) = L$ means “as $x$ gets arbitrarily close to $a$ (from either side), $f(x)$ gets arbitrarily close to $L$ .” Limits are the rigorous foundation for both derivatives (limit of a difference quotient) and integrals (limit of a Riemann sum).

Formula booklet entries for this section: L’Hôpital’s rule is NOT in the booklet — you must state it explicitly when using it. The definition of the derivative as a limit IS given.

1.1 Evaluating Limits

Direct substitution works whenever $f$ is continuous at $a$ :

$\lim_{x \to a} f(x) = f(a) \qquad \text{(if } f \text{ is continuous at } a\text{)}$

When direct substitution gives $\frac{0}{0}$ or $\frac{\infty}{\infty}$ (an indeterminate form), use one of three strategies:

Strategy	When to use	Method
Factoring	Rational functions, $\frac{0}{0}$	Factor numerator and denominator, cancel common factor
Rationalising	Square roots, $\frac{0}{0}$	Multiply by conjugate over conjugate
L’Hôpital’s rule	$\frac{0}{0}$ or $\frac{\infty}{\infty}$	Differentiate top and bottom separately, then substitute

One-sided limits: $\displaystyle\lim_{x \to a^-} f(x)$ (approach from the left) and $\displaystyle\lim_{x \to a^+} f(x)$ (from the right). The two-sided limit exists if and only if both one-sided limits exist and are equal.

Infinite limits and limits at infinity:

$\lim_{x \to \infty} \frac{p(x)}{q(x)} = \begin{cases} \dfrac{\text{leading coeff of } p}{\text{leading coeff of } q} & \deg p = \deg q \\[6pt] 0 & \deg p < \deg q \\[6pt] \pm\infty & \deg p > \deg q \end{cases}$

Limits by factoring

Evaluate $\displaystyle\lim_{x \to 3} \dfrac{x^2 - 9}{x - 3}$ .

Step 1: Direct substitution gives $\frac{9-9}{3-3} = \frac{0}{0}$ — indeterminate form.

Step 2: Factor the numerator: $x^2 - 9 = (x-3)(x+3)$ .

Step 3: Cancel: $\dfrac{(x-3)(x+3)}{x-3} = x + 3$ for $x \neq 3$ .

Step 4: Substitute: $\displaystyle\lim_{x \to 3}(x+3) = 6$

$\lim_{x \to 3} \frac{x^2 - 9}{x - 3} = 6$

1.2 L’Hôpital’s Rule HL

Sometimes when you plug a value into a fraction, you get $\frac{0}{0}$ or $\frac{\infty}{\infty}$ — which tells you nothing about the limit. L’Hôpital’s Rule gives you a way out: instead of the original fraction, take the derivative of the top and bottom separately, then try the limit again.

If $\displaystyle\lim_{x \to a} f(x) = \lim_{x \to a} g(x) = 0$ or both equal $\pm\infty$ , then:

$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$

provided the right-hand limit exists. You may apply L’Hôpital’s rule repeatedly until the indeterminate form resolves.

L’Hôpital’s rule applies only when the limit is in the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ . If the form is $\frac{3}{0}$ , the limit is $\pm\infty$ (or does not exist) — L’Hôpital’s rule does NOT apply. Always verify the form before using it.

L’Hôpital’s Rule

Evaluate $\displaystyle\lim_{x \to 0} \dfrac{\sin x}{x}$ .

Step 1: Direct substitution: $\frac{\sin 0}{0} = \frac{0}{0}$ — L’Hôpital applies.

Step 2: Differentiate numerator and denominator separately:

$\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\cos x}{1} = \cos 0 = 1$

This fundamental limit appears inside proofs of derivative formulas, so knowing the result $\displaystyle\lim_{x \to 0}\frac{\sin x}{x} = 1$ by heart saves time.

Repeated L’Hôpital

Evaluate $\displaystyle\lim_{x \to 0} \dfrac{e^x - 1 - x}{x^2}$ .

Step 1: $\frac{e^0 - 1 - 0}{0} = \frac{0}{0}$ — apply L’Hôpital.

Step 2: $\displaystyle\lim_{x \to 0} \frac{e^x - 1}{2x}$ — still $\frac{0}{0}$ , apply again.

Step 3: $\displaystyle\lim_{x \to 0} \frac{e^x}{2} = \frac{1}{2}$

1.3 Continuity

Continuity captures the idea that you can draw a function’s graph without lifting your pen — no jumps, holes, or sudden breaks. It matters because many important theorems (like the Intermediate Value Theorem) only work when a function is continuous.

A function $f$ is continuous at $x = a$ if all three conditions hold:

$f(a)$ is defined
$\displaystyle\lim_{x \to a} f(x)$ exists (both one-sided limits are equal)
$\displaystyle\lim_{x \to a} f(x) = f(a)$

Types of discontinuity:

Type	Description	Example
Removable	Limit exists, but $f(a)$ is missing or wrong	$f(x) = \frac{x^2-1}{x-1}$ at $x=1$
Jump	One-sided limits exist but differ	Piecewise function with gap
Infinite	Limit is $\pm\infty$	$f(x) = \frac{1}{x}$ at $x=0$

Intermediate Value Theorem (IVT): If $f$ is continuous on $[a, b]$ and $k$ is any value strictly between $f(a)$ and $f(b)$ , then there exists at least one $c \in (a, b)$ with $f(c) = k$ . This is used to prove roots exist.

The IVT proves existence — it tells you a root exists, not where it is. To show a root of $f$ exists on $(a, b)$ , show $f(a)$ and $f(b)$ have opposite signs and state that $f$ is continuous on $[a, b]$ .

Quick Recall — Section 1

Try to answer without scrolling up:

What does $\lim_{x \to a} f(x) = L$ mean informally?
State L’Hopital’s rule.
What does the Intermediate Value Theorem guarantee?

Reveal answers

As $x$ gets arbitrarily close to $a$ , $f(x)$ gets arbitrarily close to $L$ .
If $\lim_{x \to a} f(x)/g(x)$ gives $0/0$ or $\infty/\infty$ , then $\lim_{x \to a} f(x)/g(x) = \lim_{x \to a} f'(x)/g'(x)$ (provided the latter limit exists).
If $f$ is continuous on $[a, b]$ and $f(a)$ and $f(b)$ have opposite signs, then there exists at least one $c$ in $(a, b)$ where $f(c) = 0$ .

Section 2: Differentiation

The derivative of $f$ at $x$ measures the instantaneous rate of change — equivalently, the gradient of the tangent to $y = f(x)$ at that point. It is defined as a limit:

$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

This is called differentiation from first principles.

2.1 First Principles

Derivative of $f(x) = x^2$ from first principles

Step 1: Write the difference quotient:

$\frac{f(x+h) - f(x)}{h} = \frac{(x+h)^2 - x^2}{h}$

Step 2: Expand: $= \dfrac{x^2 + 2xh + h^2 - x^2}{h} = \dfrac{2xh + h^2}{h} = 2x + h$

Step 3: Take the limit: $f'(x) = \displaystyle\lim_{h \to 0}(2x + h) = 2x$

In Paper 1 questions asking for first principles, you must write the limit definition with $\lim_{h \to 0}$ , expand fully, cancel $h$ from numerator and denominator, and then substitute $h = 0$ . Skipping any step loses marks. The expression $\frac{f(x+h)-f(x)}{h}$ must appear explicitly.

2.2 Standard Derivatives

Standard Derivative Table

Function $f(x)$	Derivative $f'(x)$	Notes
$x^n$	$nx^{n-1}$	All real $n$ , including fractions and negatives
$e^x$	$e^x$	Unique self-derivative
$e^{kx}$	$ke^{kx}$	Chain rule applied
$a^x$	$a^x \ln a$	$a > 0$ , $a \neq 1$
$\ln x$	$\dfrac{1}{x}$	$x > 0$
$\ln\lvert x\rvert$	$\dfrac{1}{x}$	All $x \neq 0$
$\log_a x$	$\dfrac{1}{x \ln a}$	Change of base
$\sin x$	$\cos x$	Radians only
$\cos x$	$-\sin x$	Note the minus sign
$\tan x$	$\sec^2 x$	Memorise
$\sec x$	$\sec x \tan x$	Memorise
$\csc x$	$-\csc x \cot x$	Memorise
$\cot x$	$-\csc^2 x$	Memorise
$\arcsin x$	$\dfrac{1}{\sqrt{1-x^2}}$	$\lvert x\rvert < 1$
$\arccos x$	$-\dfrac{1}{\sqrt{1-x^2}}$	$\lvert x\rvert < 1$
$\arctan x$	$\dfrac{1}{1+x^2}$	All real $x$

$\frac{d}{dx}(\cos x) = -\sin x$ , NOT $+\sin x$ . This sign error is one of the most common on IB Paper 1. Write it out explicitly every time until it is automatic.

2.3 The Three Combination Rules

Most functions you encounter are combinations of simpler pieces — two things multiplied, one divided by another, or one function plugged inside another. The three rules below tell you how to differentiate each of these combinations without expanding everything out first.

Power Rule (already covered above): $\frac{d}{dx}[x^n] = nx^{n-1}$

Product Rule: For $y = u(x) \cdot v(x)$ :

$\frac{dy}{dx} = u'v + uv'$

Quotient Rule: For $y = \dfrac{u(x)}{v(x)}$ :

$\frac{dy}{dx} = \frac{u'v - uv'}{v^2}$

Chain Rule: For $y = f(g(x))$ :

$\frac{dy}{dx} = f'(g(x)) \cdot g'(x) \qquad \text{or equivalently} \qquad \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

Quotient rule: the numerator is $u'v - uv'$ (top derivative first, then bottom derivative). The order is not interchangeable — $uv' - u'v$ gives the wrong sign. A memory aid: “low d-high minus high d-low, square the bottom and away you go.”

Product Rule

Differentiate $y = x^3 e^{2x}$ .

Let $u = x^3$ , $v = e^{2x}$ , so $u' = 3x^2$ , $v' = 2e^{2x}$ .

$\frac{dy}{dx} = 3x^2 \cdot e^{2x} + x^3 \cdot 2e^{2x} = e^{2x}(3x^2 + 2x^3) = x^2 e^{2x}(3 + 2x)$

Always factorise the final answer — IB mark schemes expect a simplified form.

Quotient Rule

Differentiate $y = \dfrac{\sin x}{x^2 + 1}$ .

$u = \sin x$ , $v = x^2 + 1$ , $u' = \cos x$ , $v' = 2x$ .

$\frac{dy}{dx} = \frac{\cos x (x^2+1) - \sin x \cdot 2x}{(x^2+1)^2} = \frac{(x^2+1)\cos x - 2x\sin x}{(x^2+1)^2}$

Chain Rule

Differentiate $y = \sin(x^3 + 2x)$ .

Let $u = x^3 + 2x$ , so $y = \sin u$ .

$\frac{dy}{dx} = \cos u \cdot \frac{du}{dx} = \cos(x^3+2x) \cdot (3x^2 + 2)$

Nested Chain Rule

Differentiate $y = e^{\sin(2x)}$ .

Layer 1 (outermost): $\frac{d}{du}(e^u) = e^u$ where $u = \sin(2x)$

Layer 2: $\frac{d}{dv}(\sin v) = \cos v$ where $v = 2x$

Layer 3: $\frac{d}{dx}(2x) = 2$

$\frac{dy}{dx} = e^{\sin(2x)} \cdot \cos(2x) \cdot 2 = 2\cos(2x)\,e^{\sin(2x)}$

2.4 Implicit Differentiation

Sometimes a curve is defined by an equation like $x^2 + y^2 = 25$ where you can’t easily isolate $y$ — both variables are tangled together. Implicit differentiation lets you find the gradient anyway by differentiating both sides of the equation at once, without rearranging first.

When $y$ is defined implicitly by an equation in $x$ and $y$ , differentiate both sides with respect to $x$ , treating $y$ as a function of $x$ . Every time $y$ is differentiated, multiply by $\dfrac{dy}{dx}$ (chain rule).

$\frac{d}{dx}[y^n] = ny^{n-1}\frac{dy}{dx} \qquad \frac{d}{dx}[\sin y] = \cos y \cdot \frac{dy}{dx}$

Implicit Differentiation

Find $\dfrac{dy}{dx}$ for $x^2 + y^2 = 25$ .

Step 1: Differentiate both sides w.r.t. $x$ :

$2x + 2y\frac{dy}{dx} = 0$

Step 2: Solve for $\frac{dy}{dx}$ :

$\frac{dy}{dx} = -\frac{x}{y}$

This is the gradient at any point $(x, y)$ on the circle. At $(3, 4)$ : $\frac{dy}{dx} = -\frac{3}{4}$ .

Implicit Differentiation — Product Term

Find $\dfrac{dy}{dx}$ for $x^2 y + y^3 = 5$ .

Differentiate each term:

$\frac{d}{dx}(x^2 y) = 2xy + x^2\frac{dy}{dx}$ (product rule)
$\frac{d}{dx}(y^3) = 3y^2\frac{dy}{dx}$
$\frac{d}{dx}(5) = 0$

Collect $\frac{dy}{dx}$ terms:

$2xy + x^2\frac{dy}{dx} + 3y^2\frac{dy}{dx} = 0$

$\frac{dy}{dx}(x^2 + 3y^2) = -2xy$

$\frac{dy}{dx} = \frac{-2xy}{x^2 + 3y^2}$

When differentiating a product like $xy$ implicitly, the product rule gives TWO terms: $\frac{d}{dx}(xy) = y + x\frac{dy}{dx}$ . Students frequently write just $x\frac{dy}{dx}$ and lose the $y$ term.

2.5 Derivatives of Inverse Trigonometric Functions

Inverse trig functions like $\arctan x$ and $\arcsin x$ appear frequently in integration and in problems involving angles. Their derivatives look surprising at first, but each one can be derived using implicit differentiation — and they are worth knowing because they appear often as antiderivatives.

These arise from implicit differentiation of the definitions:

$\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1-x^2}} \qquad \frac{d}{dx}[\arctan x] = \frac{1}{1+x^2}$

With chain rule:

$\frac{d}{dx}[\arctan(g(x))] = \frac{g'(x)}{1 + [g(x)]^2}$

Inverse Trig with Chain Rule

Differentiate $y = \arctan(e^{2x})$ .

Outer function: $\arctan(u)$ where $u = e^{2x}$ (inner function).

Apply the derivative formula for $\arctan$ :

$\frac{dy}{dx} = \frac{1}{1 + (e^{2x})^2} \cdot \frac{d}{dx}[e^{2x}]$

Differentiate the inner function using chain rule on the exponential:

$\frac{d}{dx}[e^{2x}] = e^{2x} \cdot 2 = 2e^{2x}$

Combine:

$\frac{dy}{dx} = \frac{2e^{2x}}{1 + e^{4x}}$

This pattern appears frequently in Paper 2 when combining transcendental functions. A common error is forgetting to square the entire inner function in the denominator — it’s $1 + (e^{2x})^2 = 1 + e^{4x}$ , not $1 + e^{2x}$ .

Quick Recall — Section 2

Try to answer without scrolling up:

State the chain rule for $\frac{d}{dx}[f(g(x))]$ .
What is $\frac{d}{dx}[\ln x]$ ?
When do you use implicit differentiation?

Reveal answers

$f'(g(x)) \cdot g'(x)$ — differentiate the outer function, keep the inner, multiply by the derivative of the inner.
$\frac{1}{x}$ .
When $y$ is defined implicitly as a function of $x$ (not isolated), or when differentiating equations like $x^2 + y^2 = r^2$ .

Section 3: Applications of Differentiation

3.1 Tangent and Normal Lines

At the point $(a, f(a))$ on $y = f(x)$ :

Tangent gradient: $m_T = f'(a)$
Normal gradient: $m_N = -\dfrac{1}{f'(a)}$ (negative reciprocal, since tangent $\perp$ normal)
Tangent equation: $y - f(a) = f'(a)(x - a)$
Normal equation: $y - f(a) = -\dfrac{1}{f'(a)}(x - a)$

If $f'(a) = 0$ (horizontal tangent), the normal is a vertical line $x = a$ — it has no finite gradient. If $f'(a)$ is undefined (vertical tangent), the tangent is $x = a$ and the normal is horizontal with gradient 0. Never write the normal as $y = \frac{1}{0}x + c$ .

Tangent and Normal

Find the equations of the tangent and normal to $y = x^3 - 2x$ at the point where $x = 2$ .

Step 1: $y(2) = 8 - 4 = 4$ so the point is $(2, 4)$ .

Step 2: $y' = 3x^2 - 2$ , so $m_T = 3(4) - 2 = 10$ .

Step 3: Tangent: $y - 4 = 10(x - 2) \Rightarrow y = 10x - 16$

Step 4: Normal: $m_N = -\frac{1}{10}$ , so $y - 4 = -\frac{1}{10}(x-2) \Rightarrow y = -\frac{x}{10} + \frac{21}{5}$

3.2 Stationary Points and Curve Sketching

A stationary point occurs where $f'(x) = 0$ . The nature of a stationary point is determined by the second derivative test or by a sign chart on $f'$ :

Test	Local minimum	Local maximum	Inconclusive
Second derivative	$f''(a) > 0$	$f''(a) < 0$	$f''(a) = 0$
Sign chart on $f'$	$f'$ changes $- \to +$	$f'$ changes $+ \to -$	$f'$ doesn’t change sign

Increasing/decreasing: $f' > 0$ means increasing; $f' < 0$ means decreasing.

Concavity and inflection points:

Condition	Meaning
$f'' > 0$	Concave up (bowl shape)
$f'' < 0$	Concave down (cap shape)
$f''$ changes sign at $x = c$	Inflection point at $x = c$

$f''(a) = 0$ does NOT mean $x = a$ is an inflection point. You must also verify that $f''$ changes sign at $x = a$ . The function $f(x) = x^4$ has $f''(0) = 0$ but $x = 0$ is a minimum, not an inflection point.

Full curve-sketching checklist:

Domain and any restrictions
$x$ -intercepts (set $y = 0$ ) and $y$ -intercept (set $x = 0$ )
Vertical asymptotes (values where denominator $= 0$ )
Horizontal/oblique asymptotes (behaviour as $x \to \pm\infty$ )
Stationary points: solve $f'(x) = 0$ , classify each
Inflection points: solve $f''(x) = 0$ , verify sign change in $f''$
Sketch, labelling all key features

Curve Sketching

Sketch $f(x) = \dfrac{x^2}{x^2 - 4}$ , identifying all key features.

Domain: $x \neq \pm 2$

$y$ -intercept: $f(0) = 0$

$x$ -intercept: $x^2 = 0 \Rightarrow x = 0$

Vertical asymptotes: $x = 2$ and $x = -2$

Horizontal asymptote: As $x \to \pm\infty$ , $f(x) \to 1$ , so $y = 1$

Derivative: Using the quotient rule:

$f'(x) = \frac{2x(x^2-4) - x^2 \cdot 2x}{(x^2-4)^2} = \frac{-8x}{(x^2-4)^2}$

Stationary point at $x = 0$ : $f(0) = 0$ , $f''(0) > 0$ (local minimum at $(0,0)$ )

Behaviour: $f(x) > 1$ for $\lvert x\rvert > 2$ ; $f(x) \leq 0$ for $\lvert x\rvert < 2$ .

3.3 Optimisation

Optimisation is how you use calculus to answer “what is the best possible outcome?” questions — the largest area you can enclose with a fixed fence, the box with maximum volume from a sheet of card, the speed that minimises fuel use. The key insight is that at a maximum or minimum, the rate of change equals zero, so you find it by solving $f'(x) = 0$ .

Optimisation problems ask for the maximum or minimum value of some quantity. The standard approach:

Define the variable(s) and write the objective function (the quantity to optimise)
Use any constraint to write the objective function in terms of a single variable
Differentiate and solve $f'(x) = 0$
Use the second derivative or domain analysis to confirm it is a maximum/minimum
Answer the question — often the maximum/minimum value is required, not just $x$

Always check the domain for the optimisation problem. A critical point at $x = 5$ is useless if the physical constraint restricts $x$ to $[0, 4]$ . In that case, check the endpoints as well.

Optimisation — Closed Box

A closed rectangular box has a square base of side $x$ cm and height $h$ cm. Its surface area is $600 \text{ cm}^2$ . Find the dimensions that maximise the volume.

Objective function: $V = x^2 h$

Constraint: $2x^2 + 4xh = 600 \Rightarrow h = \dfrac{600 - 2x^2}{4x} = \dfrac{300 - x^2}{2x}$

Single-variable form:

$V(x) = x^2 \cdot \frac{300 - x^2}{2x} = \frac{x(300-x^2)}{2} = 150x - \frac{x^3}{2}$

Differentiate: $V'(x) = 150 - \frac{3x^2}{2}$

Set to zero: $150 = \frac{3x^2}{2} \Rightarrow x^2 = 100 \Rightarrow x = 10$ (positive dimension)

Confirm maximum: $V''(x) = -3x < 0$ for $x > 0$ — concave down, so this is a maximum.

Dimensions: $x = 10$ cm, $h = \frac{300-100}{20} = 10$ cm. The optimal box is a cube.

$V_{\max} = 1000 \text{ cm}^3$

When two quantities are connected by a formula, changing one forces the other to change too — and you can find exactly how fast. For example, if a balloon’s radius is growing, how fast is its volume growing? Related rates use the chain rule to link these speeds together.

In related-rates problems, two or more quantities change with time. The chain rule links their rates of change.

Strategy:

Identify what is changing, assign variables and their rates (typically as $\frac{dy}{dt}$ , $\frac{dx}{dt}$ , etc.)
Write a geometric or physical relationship between the variables
Differentiate implicitly with respect to $t$
Substitute known values and solve

Related Rates — Expanding Circle

The radius of a circle is increasing at $3\text{ cm s}^{-1}$ . Find the rate of increase of the area when the radius is $5$ cm.

Let $r$ = radius, $A = \pi r^2$ .

$\frac{dA}{dt} = 2\pi r \cdot \frac{dr}{dt}$

When $r = 5$ and $\frac{dr}{dt} = 3$ :

$\frac{dA}{dt} = 2\pi(5)(3) = 30\pi \approx 94.2 \text{ cm}^2\text{s}^{-1}$

Related Rates — Conical Water Tank (8 marks)

A water tank has the shape of an inverted circular cone with base radius $6$ m and height $8$ m. Water is being pumped into the tank at a constant rate of $2\text{ m}^3\text{min}^{-1}$ .

(a) [2 marks] Express the radius $r$ of the water surface in terms of the depth $h$ of the water.

(b) [3 marks] Find an expression for the volume $V$ of water in the tank in terms of $h$ only.

(c) [3 marks] Find the rate at which the water level is rising when the depth is $4$ m.

Solution:

(a) By similar triangles, the ratio of radius to height is constant throughout the cone.

For the full tank: $\dfrac{6}{8} = \dfrac{3}{4}$

For the water at depth $h$ :

$\frac{r}{h} = \frac{3}{4}$

Therefore:

$r = \frac{3h}{4}$

(b) The volume of a cone is given by:

$V = \frac{1}{3}\pi r^2 h$

Substituting $r = \dfrac{3h}{4}$ :

$V = \frac{1}{3}\pi \left(\frac{3h}{4}\right)^2 h$

$V = \frac{1}{3}\pi \cdot \frac{9h^2}{16} \cdot h$

$V = \frac{3\pi h^3}{16} \text{ m}^3$

(c) Differentiate $V = \dfrac{3\pi h^3}{16}$ with respect to time $t$ :

$\frac{dV}{dt} = \frac{3\pi}{16} \cdot 3h^2 \cdot \frac{dh}{dt}$

$\frac{dV}{dt} = \frac{9\pi h^2}{16} \cdot \frac{dh}{dt}$

We are given that $\dfrac{dV}{dt} = 2\text{ m}^3\text{min}^{-1}$ and we want to find $\dfrac{dh}{dt}$ when $h = 4$ m.

Substitute known values:

$2 = \frac{9\pi (4)^2}{16} \cdot \frac{dh}{dt}$

$2 = \frac{9\pi \cdot 16}{16} \cdot \frac{dh}{dt}$

$2 = 9\pi \cdot \frac{dh}{dt}$

$\frac{dh}{dt} = \frac{2}{9\pi} \approx 0.0707 \text{ m min}^{-1}$

The water level is rising at approximately $0.071\text{ m min}^{-1}$ (or $7.1\text{ cm min}^{-1}$ ) when the depth is $4$ m.

Related rates questions require units in the answer. If $r$ is in cm and $t$ is in seconds, then $\frac{dA}{dt}$ is in $\text{cm}^2\text{s}^{-1}$ . Missing units on a rates answer will lose the final mark.

3.5 Kinematics

For a particle moving in a straight line with displacement $s(t)$ :

$v(t) = s'(t) = \frac{ds}{dt} \qquad a(t) = v'(t) = s''(t) = \frac{d^2s}{dt^2}$

Key interpretations:

Quantity	Sign	Meaning
$v > 0$	Positive	Moving in positive direction
$v < 0$	Negative	Moving in negative direction
$v = 0$	Zero	Particle is at rest (momentarily)
$a > 0$	Positive	Velocity increasing
$a < 0$	Negative	Velocity decreasing (decelerating if $v > 0$ )

Speed is $\lvert v \rvert$ . A particle decelerates when $v$ and $a$ have opposite signs.

Total distance (not displacement): integrate $\lvert v(t) \rvert$ , or split the integral at points where $v = 0$ and add absolute values.

Distance $\neq$ displacement. Displacement is $\int_a^b v(t)\,dt$ (signed). Distance is $\int_a^b \lvert v(t)\rvert\,dt$ (unsigned). If the particle reverses, the displacement integral underestimates the total path length. Always find where $v = 0$ to check for reversal.

Kinematics

A particle has displacement $s(t) = t^3 - 6t^2 + 9t$ metres, $t \geq 0$ seconds. Find: (a) when it is at rest, (b) its acceleration when $t = 3$ , (c) the total distance in $0 \leq t \leq 4$ .

Part (a): $v = 3t^2 - 12t + 9 = 3(t-1)(t-3) = 0 \Rightarrow t = 1$ or $t = 3$

Part (b): $a = v' = 6t - 12$ . At $t = 3$ : $a = 18 - 12 = 6 \text{ m s}^{-2}$

Part (c): Position values: $s(0) = 0$ , $s(1) = 4$ , $s(3) = 0$ , $s(4) = 4$

Distances: $\lvert 4-0\rvert + \lvert 0-4\rvert + \lvert 4-0\rvert = 4 + 4 + 4 = 12$ m

Quick Recall — Section 3

Try to answer without scrolling up:

How do you find stationary points of $f(x)$ ?
How do you determine if a stationary point is a maximum or minimum?
In kinematics, what is velocity in terms of displacement $s(t)$ ?

Reveal answers

Solve $f'(x) = 0$ .
Use the second derivative test: if $f''(x) > 0$ it is a minimum; if $f''(x) < 0$ it is a maximum. Or use a sign diagram of $f'$ .
$v(t) = \frac{ds}{dt} = s'(t)$ .

Section 4: Integration

Integration is the reverse of differentiation. The indefinite integral $\int f(x)\,dx$ gives a family of antiderivatives $F(x) + C$ . The definite integral $\int_a^b f(x)\,dx = F(b) - F(a)$ gives a specific number (the net signed area under the curve).

Fundamental Theorem of Calculus:

$\frac{d}{dx}\int_a^x f(t)\,dt = f(x) \qquad \text{and} \qquad \int_a^b f'(x)\,dx = f(b) - f(a)$

4.1 Standard Integrals

Standard Integral Table

Integrand $f(x)$	$\int f(x)\,dx$	Condition
$x^n$	$\dfrac{x^{n+1}}{n+1} + C$	$n \neq -1$
$\dfrac{1}{x}$	$\ln\lvert x\rvert + C$	$x \neq 0$
$e^x$	$e^x + C$
$e^{kx}$	$\dfrac{1}{k}e^{kx} + C$	$k \neq 0$
$\sin x$	$-\cos x + C$
$\cos x$	$\sin x + C$
$\tan x$	$\ln\lvert\sec x\rvert + C$
$\sec^2 x$	$\tan x + C$
$\dfrac{1}{\sqrt{1-x^2}}$	$\arcsin x + C$	$\lvert x\rvert < 1$
$\dfrac{1}{1+x^2}$	$\arctan x + C$
$\dfrac{1}{x^2 + a^2}$	$\dfrac{1}{a}\arctan\!\left(\dfrac{x}{a}\right) + C$
$\dfrac{1}{\sqrt{a^2 - x^2}}$	$\arcsin\!\left(\dfrac{x}{a}\right) + C$

$\int \frac{1}{x}\,dx = \ln\lvert x\rvert + C$ , NOT $\ln(x) + C$ . The absolute value is essential when $x$ can be negative. On IB exams, omitting the absolute value bars inside a logarithm loses a mark.

4.2 Integration by Substitution

Substitution (reverse chain rule): choose $u = g(x)$ , then $du = g'(x)\,dx$ .

$\int f(g(x)) \cdot g'(x)\,dx = \int f(u)\,du$

Key step: Always express $dx$ in terms of $du$ by computing $\frac{du}{dx}$ , then substitute. For definite integrals, either change the limits (using $u = g(x)$ ) or convert back to $x$ at the end.

Substitution — Indefinite

Find $\displaystyle\int 2x\cos(x^2)\,dx$ .

Let $u = x^2$ , so $du = 2x\,dx$ .

$\int 2x\cos(x^2)\,dx = \int \cos u\,du = \sin u + C = \sin(x^2) + C$

Substitution — Definite

Evaluate $\displaystyle\int_0^1 \frac{x}{(x^2+1)^3}\,dx$ .

Let $u = x^2 + 1$ , $du = 2x\,dx \Rightarrow x\,dx = \frac{1}{2}du$ .

Change limits: $x = 0 \Rightarrow u = 1$ ; $x = 1 \Rightarrow u = 2$ .

$\int_1^2 \frac{1}{u^3} \cdot \frac{1}{2}\,du = \frac{1}{2}\int_1^2 u^{-3}\,du = \frac{1}{2}\left[\frac{u^{-2}}{-2}\right]_1^2 = -\frac{1}{4}\left[\frac{1}{4} - 1\right] = -\frac{1}{4} \cdot \left(-\frac{3}{4}\right) = \frac{3}{16}$

4.3 Integration by Parts

When you have an integral that is a product of two different types of functions — like $x \cdot e^x$ or $x \cdot \ln x$ — substitution doesn’t work. Integration by parts is the technique for these cases: it splits the integral into a simpler one by trading complexity between the two factors.

The integration-by-parts formula follows from reversing the product rule:

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

Choosing $u$ — LIATE order (highest priority to lowest):

Priority	Type	Example
1st	Logarithms	$\ln x$
2nd	Inverse trig	$\arctan x$
3rd	Algebraic	$x^n$
4th	Trigonometric	$\sin x$ , $\cos x$
5th	Exponential	$e^x$

Make $u$ the term with the higher-priority type; let $dv$ be the rest.

LIATE is a guide, not a law. For $\int x e^x\,dx$ , take $u = x$ (algebraic, priority 3) and $dv = e^x\,dx$ (exponential, priority 5) — then $du = dx$ and $v = e^x$ . If you take $u = e^x$ and $dv = x\,dx$ , the integral becomes more complex, not simpler.

Integration by Parts — Standard

Find $\displaystyle\int x e^x\,dx$ .

$u = x$ , $dv = e^x\,dx$ $\Rightarrow$ $du = dx$ , $v = e^x$

$\int x e^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C = e^x(x - 1) + C$

Integration by Parts — Twice (IBP $^2$ )

Find $\displaystyle\int x^2 e^x\,dx$ .

First IBP: $u = x^2$ , $dv = e^x\,dx \Rightarrow du = 2x\,dx$ , $v = e^x$

$\int x^2 e^x\,dx = x^2 e^x - 2\int x e^x\,dx$

Second IBP: From above, $\int x e^x\,dx = e^x(x-1) + C$

$= x^2 e^x - 2e^x(x-1) + C = e^x(x^2 - 2x + 2) + C$

Integration by Parts — Cyclic

Find $\displaystyle\int e^x \cos x\,dx$ .

First IBP: $u = e^x$ , $dv = \cos x\,dx \Rightarrow v = \sin x$

$I = e^x\sin x - \int e^x\sin x\,dx$

Second IBP: $u = e^x$ , $dv = \sin x\,dx \Rightarrow v = -\cos x$

$I = e^x\sin x - \left(-e^x\cos x + \int e^x\cos x\,dx\right) = e^x\sin x + e^x\cos x - I$

Solve for $I$ : $2I = e^x(\sin x + \cos x) \Rightarrow I = \dfrac{e^x(\sin x + \cos x)}{2} + C$

Practice: Integration by Parts — Fading Sequence

The worked examples above showed the full method. Now try these progressively: the first shows all steps, the second hides the final steps for you to attempt, and the third gives only the setup.

WORKED EXAMPLEFull worked example — all steps shown

Find $\displaystyle\int x \sin x\,dx$ .

Step 1

Identify $u$ and $dv$ : Using LIATE, $u = x$ (algebraic) and $dv = \sin x\,dx$ .

Step 2

Compute $du$ and $v$ : $du = dx$ , $v = \int \sin x\,dx = -\cos x$

Step 3

Apply the formula: $\int x \sin x\,dx = x(-\cos x) - \int (-\cos x)\,dx = -x\cos x + \int \cos x\,dx$

Step 4

Evaluate the remaining integral: $= -x\cos x + \sin x + C$

YOUR TURN (PARTIAL)Partial example — try the last steps yourself

Find $\displaystyle\int x^2 \ln x\,dx$ .

Steps 1—2 are shown. Try steps 3—4 before revealing.

Step 1

Identify $u$ and $dv$ : Using LIATE, $u = \ln x$ (log, highest priority) and $dv = x^2\,dx$ .

Step 2

Compute $du$ and $v$ : $du = \frac{1}{x}\,dx$ , $v = \frac{x^3}{3}$

Step 3 — Apply the IBP formula and simplify the remaining integral

Apply the formula:

$\int x^2 \ln x\,dx = \frac{x^3}{3}\ln x - \int \frac{x^3}{3} \cdot \frac{1}{x}\,dx = \frac{x^3}{3}\ln x - \frac{1}{3}\int x^2\,dx$

Step 4 — Evaluate and write the final answer

Evaluate: $= \dfrac{x^3}{3}\ln x - \dfrac{1}{3} \cdot \dfrac{x^3}{3} + C = \dfrac{x^3}{3}\ln x - \dfrac{x^3}{9} + C$

YOUR TURN (SCAFFOLDED)Scaffolded — only the setup is given

Find $\displaystyle\int \ln x\,dx$ .

Work through the full solution, then reveal each step to check.

Step 1 — Choose u and dv (hint: write ln x as ln x times 1)

The trick: Write $\int \ln x\,dx = \int 1 \cdot \ln x\,dx$ . Let $u = \ln x$ , $dv = 1\,dx$ , so $du = \frac{1}{x}\,dx$ , $v = x$ .

Step 2 — Apply the IBP formula

Apply: $\int \ln x\,dx = x\ln x - \int x \cdot \frac{1}{x}\,dx = x\ln x - \int 1\,dx$

Step 3 — Write the final answer

Result: $= x\ln x - x + C = x(\ln x - 1) + C$

4.4 Partial Fractions HL

A fraction like $\frac{3x+1}{(x+1)(x-2)}$ is hard to integrate directly, but if you can split it into two simpler fractions each with a single linear denominator, each piece integrates to a natural log. Partial fractions is the technique that does this splitting.

Partial fractions decompose a rational function into simpler fractions before integrating. The method depends on the nature of the denominator’s factors.

Case 1 — Distinct linear factors: $\dfrac{f(x)}{(x-a)(x-b)} = \dfrac{A}{x-a} + \dfrac{B}{x-b}$

Case 2 — Repeated linear factor: $\dfrac{f(x)}{(x-a)^2} = \dfrac{A}{x-a} + \dfrac{B}{(x-a)^2}$

Case 3 — Irreducible quadratic factor: $\dfrac{f(x)}{(x-a)(x^2+bx+c)} = \dfrac{A}{x-a} + \dfrac{Bx+C}{x^2+bx+c}$

Important: If the degree of the numerator $\geq$ degree of the denominator, perform polynomial long division first to obtain a proper fraction.

Partial Fractions Integration

Find $\displaystyle\int \frac{3x+1}{(x+1)(x-2)}\,dx$ .

Step 1: Decompose: $\dfrac{3x+1}{(x+1)(x-2)} = \dfrac{A}{x+1} + \dfrac{B}{x-2}$

Step 2: Multiply through: $3x+1 = A(x-2) + B(x+1)$

Step 3: Substitute $x = 2$ : $7 = 3B \Rightarrow B = \frac{7}{3}$

Step 4: Substitute $x = -1$ : $-2 = -3A \Rightarrow A = \frac{2}{3}$

Step 5: Integrate:

$\int \frac{3x+1}{(x+1)(x-2)}\,dx = \frac{2}{3}\ln\lvert x+1\rvert + \frac{7}{3}\ln\lvert x-2\rvert + C$

Before partial fractions, always check whether the rational function is proper (degree of numerator < degree of denominator). If it is improper, divide first. Attempting partial fractions on an improper fraction without dividing first gives a wrong decomposition.

Section 5: Applications of Integration

5.1 Area Under a Curve

The area between $y = f(x)$ and the $x$ -axis from $x = a$ to $x = b$ is:

$A = \int_a^b \lvert f(x)\rvert\,dx$

If $f(x) \geq 0$ on $[a, b]$ , this simplifies to $\int_a^b f(x)\,dx$ . If $f$ changes sign, split the interval at the zeros and add absolute values.

$\int_a^b f(x)\,dx$ gives the net signed area (regions below the axis subtract). For a total area, split at roots and sum the absolute values. IB questions usually ask explicitly for “area” (unsigned) or “value of the integral” (signed) — do not confuse them.

5.2 Area Between Two Curves

For $f(x) \geq g(x)$ on $[a, b]$ :

$A = \int_a^b \bigl[f(x) - g(x)\bigr]\,dx$

If the curves cross within $[a, b]$ , find the crossing points (intersections), then split and add.

Area Between Two Curves

Find the area enclosed by $y = x^2$ and $y = x + 2$ .

Step 1: Find intersections: $x^2 = x+2 \Rightarrow x^2 - x - 2 = 0 \Rightarrow (x-2)(x+1) = 0 \Rightarrow x = -1, 2$

Step 2: On $[-1, 2]$ : $x+2 \geq x^2$ (check at $x = 0$ : $2 > 0$ ). So $f = x+2$ , $g = x^2$ .

Step 3:

$A = \int_{-1}^{2}(x+2-x^2)\,dx = \left[\frac{x^2}{2} + 2x - \frac{x^3}{3}\right]_{-1}^{2}$

$= \left(2 + 4 - \frac{8}{3}\right) - \left(\frac{1}{2} - 2 + \frac{1}{3}\right) = \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{10}{3} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}$

5.3 Volume of Revolution HL

If you take a curve and spin it around an axis, you carve out a 3D solid — like a vase or a bullet shape. The volume of revolution formula calculates that solid’s volume by treating it as a stack of thin circular discs, each with radius equal to the function’s value at that point.

Rotation about the $x$ -axis: The volume generated by rotating $y = f(x)$ about the $x$ -axis from $x = a$ to $x = b$ :

$V = \pi\int_a^b [f(x)]^2\,dx$

Rotation about the $y$ -axis: Express $x$ as a function of $y$ :

$V = \pi\int_c^d [g(y)]^2\,dy$

Volume between two curves (shell method or washer method):

$V = \pi\int_a^b \bigl([f(x)]^2 - [g(x)]^2\bigr)\,dx \qquad \text{(washer method, outer}^2 - \text{inner}^2\text{)}$

The volume formula is $V = \pi\int y^2\,dx$ — it is $y^2$ , not $y$ . Students commonly write $V = \pi \int y\,dx$ . The factor of $\pi$ comes from the cross-sectional area $\pi r^2 = \pi y^2$ . This formula is NOT in the IB formula booklet for AA HL — you must recall it.

Volume of Revolution

Find the volume generated when $y = \sqrt{x}$ , $0 \leq x \leq 4$ , is rotated $360°$ about the $x$ -axis.

$V = \pi\int_0^4 (\sqrt{x})^2\,dx = \pi\int_0^4 x\,dx = \pi\left[\frac{x^2}{2}\right]_0^4 = \pi \cdot 8 = 8\pi$

5.4 Kinematics — Integration

Given acceleration $a(t)$ or velocity $v(t)$ :

$v(t) = \int a(t)\,dt + C_1 \qquad s(t) = \int v(t)\,dt + C_2$

The constants are found from initial conditions: $v(0) = v_0$ and $s(0) = s_0$ .

Displacement over $[t_1, t_2]$ : $\displaystyle\int_{t_1}^{t_2} v(t)\,dt = s(t_2) - s(t_1)$

Total distance over $[t_1, t_2]$ : $\displaystyle\int_{t_1}^{t_2} \lvert v(t)\rvert\,dt$

Kinematics — Finding Position from Acceleration

A particle starts at rest at the origin. Its acceleration is $a(t) = 6t - 4$ . Find its position at $t = 3$ .

Step 1: $v(t) = \int(6t-4)\,dt = 3t^2 - 4t + C_1$ . Initial rest: $v(0) = 0 \Rightarrow C_1 = 0$ .

Step 2: $s(t) = \int(3t^2 - 4t)\,dt = t^3 - 2t^2 + C_2$ . Starts at origin: $s(0) = 0 \Rightarrow C_2 = 0$ .

Step 3: $s(3) = 27 - 18 = 9$ m

Section 6: Differential Equations

A differential equation is an equation that contains a function and its derivative at the same time — it describes how something changes rather than what it equals. They are the language of real-world modelling: population growth, radioactive decay, temperature cooling, and spread of disease are all described by differential equations.

A differential equation (DE) relates a function to its derivatives. The order is the highest derivative appearing. The general solution contains arbitrary constants; an initial condition (or boundary condition) pins down a particular solution.

6.1 Separable Differential Equations

The simplest differential equations to solve are ones where you can get all the $y$ terms on one side and all the $x$ terms on the other — then integrate both sides separately. This is called “separating variables.”

A first-order DE is separable if it can be written as $\dfrac{dy}{dx} = f(x) g(y)$ . Separate variables and integrate both sides:

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

Separable DE

Solve $\dfrac{dy}{dx} = xy$ , given $y(0) = 2$ .

Step 1: Separate: $\dfrac{dy}{y} = x\,dx$

Step 2: Integrate: $\ln\lvert y\rvert = \dfrac{x^2}{2} + C$

Step 3: Exponentiate: $\lvert y\rvert = e^C \cdot e^{x^2/2}$ , so $y = A e^{x^2/2}$ where $A = \pm e^C$

Step 4: Apply $y(0) = 2$ : $2 = A e^0 = A$

Particular solution: $y = 2e^{x^2/2}$

After separating and integrating, you get $\ln\lvert y\rvert = \ldots + C$ . Exponentiating gives $\lvert y\rvert = e^C e^{\ldots}$ , and the $\pm$ from the absolute value is absorbed into a new constant $A$ , so $y = Ae^{\ldots}$ where $A \neq 0$ . Do not lose the $\pm$ or the constant — they combine to give $A$ which is determined by the initial condition.

6.2 Initial Value Problems

A differential equation’s general solution contains an unknown constant, giving a whole family of curves. An initial value problem pins down which specific curve you want by giving one known point on it — for example, the population at time zero, or the temperature at the start of an experiment.

An initial value problem (IVP) specifies the DE and a condition such as $y(x_0) = y_0$ . The general solution’s constant is uniquely determined by substituting $x_0$ and $y_0$ .

Initial Value Problem

Solve $\dfrac{dy}{dx} = \dfrac{x}{y}$ , $y(0) = 3$ .

Separate: $y\,dy = x\,dx$

Integrate: $\dfrac{y^2}{2} = \dfrac{x^2}{2} + C$

Apply IC: $\dfrac{9}{2} = 0 + C \Rightarrow C = \dfrac{9}{2}$

Particular solution: $y^2 = x^2 + 9$ , or $y = \sqrt{x^2 + 9}$ (positive since $y(0) = 3 > 0$ )

6.3 Modelling with Differential Equations

Many natural phenomena are modelled by DEs. Key models:

Model	DE	Solution
Exponential growth	$\dfrac{dN}{dt} = kN$ , $k > 0$	$N = N_0 e^{kt}$
Exponential decay	$\dfrac{dN}{dt} = -kN$ , $k > 0$	$N = N_0 e^{-kt}$
Newton’s Law of Cooling	$\dfrac{dT}{dt} = -k(T - T_{\text{env}})$	$T = T_{\text{env}} + (T_0 - T_{\text{env}})e^{-kt}$
Logistic growth	$\dfrac{dP}{dt} = rP\!\left(1 - \dfrac{P}{K}\right)$	S-shaped curve, $P \to K$

For Newton’s Law of Cooling, let $\theta = T - T_{\text{env}}$ (excess temperature). Then $\frac{d\theta}{dt} = -k\theta$ , which is a standard exponential decay. Always substitute the environmental temperature first before integrating.

Modelling — Radioactive Decay

A radioactive substance has half-life 8 years. Find: (a) the decay constant $k$ , (b) the fraction remaining after 20 years.

Part (a): At $t = 8$ , $N = \frac{1}{2}N_0$ :

$\frac{N_0}{2} = N_0 e^{-8k} \Rightarrow \frac{1}{2} = e^{-8k} \Rightarrow -8k = \ln\frac{1}{2} = -\ln 2 \Rightarrow k = \frac{\ln 2}{8}$

Part (b):

$\frac{N}{N_0} = e^{-20k} = e^{-20\ln 2/8} = e^{-5\ln 2/2} = 2^{-5/2} = \frac{1}{4\sqrt{2}} \approx 0.177$

About 17.7% remains after 20 years.

Section 7: Maclaurin Series HL

Polynomials are easy to work with — you can add, multiply, differentiate, and integrate them by hand. A Maclaurin series takes a complicated function like $e^x$ or $\sin x$ and rewrites it as an infinite polynomial. This makes it possible to approximate function values, evaluate limits that look like $\frac{0}{0}$ , and integrate functions that have no closed-form antiderivative.

A Maclaurin series expresses a function as an infinite power series centred at $x = 0$ :

$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n$

The series converges to $f(x)$ within a radius of convergence $R$ . Outside this radius the series diverges.

7.1 Deriving the Four Standard Series

The Four Essential Maclaurin Series (must be able to derive)

$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots = \sum_{n=0}^{\infty}\frac{x^n}{n!} \qquad \text{(all } x\text{)}$

$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!} \qquad \text{(all } x\text{)}$

$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{(2n)!} \qquad \text{(all } x\text{)}$

$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots = \sum_{n=1}^{\infty}\frac{(-1)^{n+1} x^n}{n} \qquad (-1 < x \leq 1)$

Deriving the Maclaurin Series for $e^x$

Let $f(x) = e^x$ . Since $\frac{d}{dx}(e^x) = e^x$ , all derivatives equal $e^x$ .

At $x = 0$ : $f^{(n)}(0) = e^0 = 1$ for all $n$ .

Substituting into the Maclaurin formula:

$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$

Deriving the Maclaurin Series for $\sin x$

$f(x) = \sin x$ . Compute derivatives at $x = 0$ :

$n$	$f^{(n)}(x)$	$f^{(n)}(0)$
0	$\sin x$	0
1	$\cos x$	1
2	$-\sin x$	0
3	$-\cos x$	$-1$
4	$\sin x$	0

The pattern is $0, 1, 0, -1, 0, 1, \ldots$ Only odd powers survive:

$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$

Deriving the Maclaurin Series for $\cos x$

$f(x) = \cos x$ . Compute derivatives at $x = 0$ :

$n$	$f^{(n)}(x)$	$f^{(n)}(0)$
0	$\cos x$	1
1	$-\sin x$	0
2	$-\cos x$	$-1$
3	$\sin x$	0
4	$\cos x$	1

The pattern is $1, 0, -1, 0, 1, \ldots$ Only even powers survive, alternating in sign:

$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$

Common Mistakes — Maclaurin Series

Applying series outside the radius of convergence: Each series has a specific domain of validity. For example, $\ln(1+x)$ only converges for $-1 < x \leq 1$ . Using it for $x = 2$ produces nonsense.
Confusing which series applies where: Students often mix up $e^x$ , $e^{-x}$ , and $e^{x^2}$ . Always substitute carefully and check signs.
Forgetting factorials: The Maclaurin formula has $\frac{f^{(n)}(0)}{n!}$ , not just $f^{(n)}(0)$ . Forgetting the factorial gives wildly incorrect coefficients.
Truncating too early: Using only the first two terms of a series can produce large errors. Check the remainder term or compare with the exact value to verify accuracy.

7.2 Applications of Maclaurin Series

Substitution: Replace $x$ with a multiple or power to get new series.

$e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \cdots$

Evaluating limits: Series give an exact expansion near $x = 0$ , resolving indeterminate forms.

Approximation: The first few terms give accurate approximations for small $x$ .

Multiplying series: Multiply term by term, collecting powers.

Limit Using Maclaurin Series

Evaluate $\displaystyle\lim_{x \to 0} \dfrac{e^x - 1 - x - \frac{x^2}{2}}{x^3}$ .

Substitute the series for $e^x$ :

$e^x - 1 - x - \frac{x^2}{2} = \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots\right) - 1 - x - \frac{x^2}{2} = \frac{x^3}{6} + \cdots$

$\lim_{x \to 0} \frac{x^3/6 + \cdots}{x^3} = \frac{1}{6}$

Integration Using Maclaurin Series

Find $\displaystyle\int_0^{0.1} e^{-x^2}\,dx$ to 5 decimal places.

Using $e^{-x^2} = 1 - x^2 + \dfrac{x^4}{2!} - \cdots$ :

$\int_0^{0.1} e^{-x^2}\,dx \approx \int_0^{0.1}\left(1 - x^2 + \frac{x^4}{2}\right)dx = \left[x - \frac{x^3}{3} + \frac{x^5}{10}\right]_0^{0.1}$

$= 0.1 - \frac{0.001}{3} + \frac{0.00001}{10} \approx 0.1 - 0.000333 + 0.000001 = 0.099668$

Maclaurin series for $\ln(1+x)$ converges only for $-1 < x \leq 1$ . At $x = 1$ : the alternating harmonic series converges conditionally to $\ln 2$ . At $x = -1$ : it diverges. Always state the interval of validity when writing the series. Applying the series outside its radius gives nonsensical results.

Maclaurin Series Multiplication

Find the Maclaurin series for $\sin x \cdot e^x$ up to and including the $x^3$ term.

$\sin x \approx x - \frac{x^3}{6} + \cdots \qquad e^x \approx 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots$

Multiply term by term, collecting up to $x^3$ :

$x^1: \; x \cdot 1 = x$ $x^2: \; x \cdot x = x^2$ $x^3: \; x \cdot \frac{x^2}{2} + \left(-\frac{x^3}{6}\right) \cdot 1 = \frac{x^3}{2} - \frac{x^3}{6} = \frac{x^3}{3}$

$\sin x \cdot e^x \approx x + x^2 + \frac{x^3}{3} + \cdots$

Section 8: MCQ Practice

Q1. The derivative of $f(x) = \ln(\sin x)$ is:

(A) $\cos x$ $\quad$ (B) $\cot x$ $\quad$ (C) $\tan x$ $\quad$ (D) $\dfrac{1}{\sin x}$

Answer: (B) Chain rule: $f'(x) = \dfrac{1}{\sin x} \cdot \cos x = \dfrac{\cos x}{\sin x} = \cot x$ .

Q2. $\displaystyle\int_0^{\pi/2} \sin^2 x\,dx =$

(A) $0$ $\quad$ (B) $\dfrac{\pi}{4}$ $\quad$ (C) $\dfrac{\pi}{2}$ $\quad$ (D) $1$

Answer: (B) Use $\sin^2 x = \dfrac{1 - \cos 2x}{2}$ :

$\int_0^{\pi/2} \frac{1-\cos 2x}{2}\,dx = \left[\frac{x}{2} - \frac{\sin 2x}{4}\right]_0^{\pi/2} = \frac{\pi}{4} - 0 = \frac{\pi}{4}$

Q3. If $y = x^x$ , then $\dfrac{dy}{dx} =$

(A) $x \cdot x^{x-1}$ $\quad$ (B) $x^x \ln x$ $\quad$ (C) $x^x(1 + \ln x)$ $\quad$ (D) $x^x \cdot x$

Answer: (C) Use logarithmic differentiation: $\ln y = x \ln x$ .

$\frac{1}{y}\frac{dy}{dx} = \ln x + x \cdot \frac{1}{x} = \ln x + 1$

$\frac{dy}{dx} = y(\ln x + 1) = x^x(1 + \ln x)$

Q4. The value of $\displaystyle\lim_{x \to 0} \dfrac{1 - \cos x}{x^2}$ is:

(A) $0$ $\quad$ (B) $\dfrac{1}{4}$ $\quad$ (C) $\dfrac{1}{2}$ $\quad$ (D) $1$

Answer: (C) Using the Maclaurin series $\cos x = 1 - \dfrac{x^2}{2} + \cdots$ :

$\frac{1 - \cos x}{x^2} = \frac{x^2/2 - \cdots}{x^2} \to \frac{1}{2}$

Alternatively, apply L’Hôpital’s rule twice: $\dfrac{\sin x}{2x} \to \dfrac{\cos x}{2} \to \dfrac{1}{2}$ .

Q5. The function $f(x) = x^3 - 6x^2 + 9x$ has an inflection point at:

(A) $x = 1$ $\quad$ (B) $x = 2$ $\quad$ (C) $x = 3$ $\quad$ (D) $x = 0$

Answer: (B) $f'(x) = 3x^2 - 12x + 9$ , $f''(x) = 6x - 12 = 6(x-2)$ .

$f''(x) = 0$ at $x = 2$ . Sign change: $f''(1) = -6 < 0$ and $f''(3) = 6 > 0$ — confirmed sign change from negative to positive. Therefore $x = 2$ is the only inflection point.

Q6. $\displaystyle\int x \ln x\,dx =$

(A) $\dfrac{x^2}{2}\ln x - \dfrac{x^2}{4} + C$ $\quad$ (B) $\dfrac{x^2 \ln x}{2} + C$ $\quad$ (C) $x\ln x - x + C$ $\quad$ (D) $\dfrac{x^2}{4}(\ln x - 1) + C$

Answer: (A) IBP with $u = \ln x$ , $dv = x\,dx$ :

$\int x\ln x\,dx = \frac{x^2}{2}\ln x - \int\frac{x^2}{2} \cdot \frac{1}{x}\,dx = \frac{x^2}{2}\ln x - \frac{x^2}{4} + C$

Q7. The general solution of $\dfrac{dy}{dx} = 2y$ is:

(A) $y = Ce^x$ $\quad$ (B) $y = Ce^{2x}$ $\quad$ (C) $y = 2Ce^x$ $\quad$ (D) $y = x^2 + C$

Answer: (B) Separate: $\dfrac{dy}{y} = 2\,dx \Rightarrow \ln\lvert y\rvert = 2x + k \Rightarrow y = Ce^{2x}$ .

Q8. Which of the following is the correct Maclaurin expansion of $\cos(2x)$ up to the $x^4$ term?

(A) $1 - 2x^2 + \dfrac{2x^4}{3}$ $\quad$ (B) $1 - 2x^2 + \dfrac{x^4}{3}$ $\quad$ (C) $1 - x^2 + \dfrac{x^4}{6}$ $\quad$ (D) $1 - 4x^2 + \dfrac{8x^4}{3}$

Answer: (A) Substitute $2x$ into the standard $\cos x$ series:

$\cos(2x) = 1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} = 1 - \frac{4x^2}{2} + \frac{16x^4}{24} = 1 - 2x^2 + \frac{2x^4}{3}$

Common error: substituting $x \to 2x$ but forgetting to apply the exponents — writing $\frac{2x^2}{2}$ instead of $\frac{(2x)^2}{2} = \frac{4x^2}{2}$ .

Q9. If $f''(x) > 0$ on $(a, b)$ then on $(a, b)$ , $f$ is:

(A) decreasing $\quad$ (B) increasing $\quad$ (C) concave down $\quad$ (D) concave up

Answer: (D) $f'' > 0$ means the gradient $f'$ is increasing, which is the definition of concave up. The sign of $f''$ says nothing directly about whether $f$ is increasing or decreasing (that depends on $f'$ ).

Q10. A particle moves with $v(t) = t^2 - 3t$ . The particle first comes to rest at $t =$

(A) $0$ $\quad$ (B) $1$ $\quad$ (C) $3$ $\quad$ (D) $1.5$

Answer: (C) $v(t) = 0 \Rightarrow t(t-3) = 0 \Rightarrow t = 0$ or $t = 3$ . At $t = 0$ the particle starts from rest, so the first time it comes to rest again is at $t = 3$ .

Q11. The area enclosed by $y = e^x$ and the lines $x = 0$ , $x = 1$ , $y = 0$ is:

(A) $e$ $\quad$ (B) $e - 1$ $\quad$ (C) $e + 1$ $\quad$ (D) $1$

Answer: (B) $\displaystyle\int_0^1 e^x\,dx = [e^x]_0^1 = e^1 - e^0 = e - 1$ .

Q12. The volume generated by rotating $y = 2x$ , $0 \leq x \leq 3$ , about the $x$ -axis is:

(A) $12\pi$ $\quad$ (B) $24\pi$ $\quad$ (C) $36\pi$ $\quad$ (D) $72\pi$

Answer: (C) $V = \pi\displaystyle\int_0^3 (2x)^2\,dx = \pi\int_0^3 4x^2\,dx = 4\pi\left[\dfrac{x^3}{3}\right]_0^3 = 4\pi \cdot 9 = 36\pi$ .

Q13. The partial fraction decomposition of $\dfrac{5}{(x-1)(x+4)}$ is:

(A) $\dfrac{1}{x-1} - \dfrac{1}{x+4}$ $\quad$ (B) $\dfrac{1}{x-1} + \dfrac{1}{x+4}$ $\quad$ (C) $\dfrac{5}{x-1} - \dfrac{5}{x+4}$ $\quad$ (D) $\dfrac{1}{x-1} + \dfrac{4}{x+4}$

Answer: (A) Let $\dfrac{5}{(x-1)(x+4)} = \dfrac{A}{x-1} + \dfrac{B}{x+4}$ .

$5 = A(x+4) + B(x-1)$ . Set $x=1$ : $5 = 5A \Rightarrow A = 1$ . Set $x=-4$ : $5 = -5B \Rightarrow B = -1$ .

Q14. If $\displaystyle\int_0^a x^2\,dx = 9$ , then $a =$

(A) $3$ $\quad$ (B) $\sqrt{3}$ $\quad$ (C) $\sqrt[3]{27}$ $\quad$ (D) $9$

Answer: (A) $\left[\dfrac{x^3}{3}\right]_0^a = \dfrac{a^3}{3} = 9 \Rightarrow a^3 = 27 \Rightarrow a = 3$ .

Note: (C) is also equal to 3, since $\sqrt[3]{27} = 3$ — but (A) is the standard form.

Q15. Using the Maclaurin series for $\ln(1+x)$ , the approximate value of $\ln(1.1)$ to 4 decimal places (using up to the $x^4$ term) is:

(A) $0.0953$ $\quad$ (B) $0.1000$ $\quad$ (C) $0.0900$ $\quad$ (D) $0.0909$

Answer: (A) With $x = 0.1$ :

$\ln(1.1) \approx 0.1 - \frac{(0.1)^2}{2} + \frac{(0.1)^3}{3} - \frac{(0.1)^4}{4} = 0.1 - 0.005 + 0.000333 - 0.000025 = 0.095308$

Rounded to 4 d.p.: $0.0953$ .

Q16. For $f(x) = x^3 - 3x$ , the local minimum value is:

(A) $-2$ $\quad$ (B) $0$ $\quad$ (C) $2$ $\quad$ (D) $3$

Answer: (A) $f'(x) = 3x^2 - 3 = 3(x-1)(x+1) = 0 \Rightarrow x = \pm 1$ .

$f''(x) = 6x$ . At $x = 1$ : $f'' = 6 > 0$ (local min). $f(1) = 1 - 3 = -2$ .

Q17. Given the implicit equation $x^2y + y^3 = 2x$ , find $\dfrac{dy}{dx}$ at the point $(1, 1)$ .

(A) $\dfrac{1}{4}$ $\quad$ (B) $0$ $\quad$ (C) $-\dfrac{1}{4}$ $\quad$ (D) $-1$

Answer: (B) Differentiate both sides with respect to $x$ :

$\frac{d}{dx}(x^2y + y^3) = \frac{d}{dx}(2x)$

Using product rule on $x^2y$ and chain rule on $y^3$ :

$2xy + x^2\frac{dy}{dx} + 3y^2\frac{dy}{dx} = 2$

Collect terms with $\dfrac{dy}{dx}$ :

$\frac{dy}{dx}(x^2 + 3y^2) = 2 - 2xy$

$\frac{dy}{dx} = \frac{2 - 2xy}{x^2 + 3y^2}$

At $(1, 1)$ : $\dfrac{dy}{dx} = \dfrac{2 - 2(1)(1)}{1^2 + 3(1)^2} = \dfrac{0}{4} = 0$

Common error: Forgetting to apply the product rule to $x^2y$ , leading to option (C).

Q18. A ladder 5 m long leans against a vertical wall. The bottom of the ladder slides away from the wall at 1 m/s. How fast is the top of the ladder sliding down the wall when the bottom is 3 m from the wall?

(A) $-\dfrac{3}{8}$ m/s $\quad$ (B) $-\dfrac{1}{2}$ m/s $\quad$ (C) $-\dfrac{2}{3}$ m/s $\quad$ (D) $-\dfrac{3}{4}$ m/s

Answer: (D) Let $x$ = distance from wall to bottom of ladder, $y$ = height of top of ladder.

By Pythagoras: $x^2 + y^2 = 25$

Given: $\dfrac{dx}{dt} = 1$ m/s. Find: $\dfrac{dy}{dt}$ when $x = 3$ .

When $x = 3$ : $9 + y^2 = 25 \Rightarrow y = 4$ m.

Differentiate the constraint with respect to time:

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$

$\frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt}$

Substitute $x = 3$ , $y = 4$ , and $\dfrac{dx}{dt} = 1$ :

$\frac{dy}{dt} = -\frac{3}{4}(1) = -\frac{3}{4} \text{ m/s}$

The negative sign indicates the ladder is sliding down.

Common error: Forgetting the negative sign from differentiation, leading to option (C) with positive value, or incorrectly computing $\dfrac{x}{y}$ as $\dfrac{4}{3}$ , leading to option (B).

Key Formulas

Complete Calculus Formula Reference

Limits

Formula	Notes
$\displaystyle\lim_{x \to 0}\dfrac{\sin x}{x} = 1$	Fundamental limit
$\displaystyle\lim_{x \to 0}\dfrac{e^x - 1}{x} = 1$	Derived from series
L’Hôpital: $\displaystyle\lim_{x \to a}\dfrac{f(x)}{g(x)} = \lim_{x \to a}\dfrac{f'(x)}{g'(x)}$	Only for $\frac{0}{0}$ or $\frac{\infty}{\infty}$

Differentiation Rules

Rule	Formula
Power	$\frac{d}{dx}[x^n] = nx^{n-1}$
Product	$\frac{d}{dx}[uv] = u'v + uv'$
Quotient	$\frac{d}{dx}\!\left[\frac{u}{v}\right] = \frac{u'v - uv'}{v^2}$
Chain	$\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$
Implicit: $y^n$	$ny^{n-1}\frac{dy}{dx}$

Key Derivatives

$f(x)$	$f'(x)$
$\sin x$	$\cos x$
$\cos x$	$-\sin x$
$\tan x$	$\sec^2 x$
$e^x$	$e^x$
$\ln x$	$1/x$
$\arcsin x$	$1/\sqrt{1-x^2}$
$\arctan x$	$1/(1+x^2)$

Integration

Rule / Formula	Notes
$\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$	$n \neq -1$
$\int \frac{1}{x}\,dx = \ln\lvert x\rvert + C$	Absolute value required
$\int e^x\,dx = e^x + C$
$\int \sin x\,dx = -\cos x + C$	Note sign
$\int \cos x\,dx = \sin x + C$
$\int \sec^2 x\,dx = \tan x + C$
$\int u\,dv = uv - \int v\,du$	Integration by parts
Substitution: $\int f(g(x))g'(x)\,dx = \int f(u)\,du$	$u = g(x)$
$\int \frac{1}{x^2+a^2}\,dx = \frac{1}{a}\arctan\!\left(\frac{x}{a}\right) + C$
$\int \frac{1}{\sqrt{a^2-x^2}}\,dx = \arcsin\!\left(\frac{x}{a}\right) + C$

Applications

Quantity	Formula
Tangent gradient at $x=a$	$m = f'(a)$
Normal gradient at $x=a$	$m = -1/f'(a)$
Area under curve	$A = \int_a^b \lvert f(x)\rvert\,dx$
Area between curves	$A = \int_a^b [f(x)-g(x)]\,dx$
Volume of revolution ( $x$ -axis)	$V = \pi\int_a^b [f(x)]^2\,dx$
Displacement from $v$	$s = \int v\,dt$
Distance from $v$	$d = \int \lvert v\rvert\,dt$

Differential Equations

Type	Method
Separable: $\frac{dy}{dx} = f(x)g(y)$	$\int\frac{dy}{g(y)} = \int f(x)\,dx$
Exponential growth/decay	$N = N_0 e^{\pm kt}$
Newton’s cooling: $\frac{d\theta}{dt} = -k\theta$	$\theta = \theta_0 e^{-kt}$ , $\theta = T - T_{\text{env}}$

Maclaurin Series

Function	Series	Interval
$e^x$	$\displaystyle\sum_{n=0}^{\infty}\frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \cdots$	All $x$
$\sin x$	$\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots$	All $x$
$\cos x$	$\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots$	All $x$
$\ln(1+x)$	$\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$	$-1 < x \leq 1$

Mixed Practice — Exam Style

How to use this section: Unlike topic-specific practice, these questions are interleaved — they mix all topics from this guide in random order. Before answering, identify which concept or topic area the question is testing. This is exactly the skill you need on Paper 2 and Paper 3, where you don’t know in advance which topic each question covers.

[Integration] Evaluate $\displaystyle\int_0^1 x e^{x^2}\,dx$ .

A. $\dfrac{e-1}{2}$

B. $e - 1$

C. $\dfrac{1}{2}$

D. $e^2 - 1$
[Limits] Evaluate $\displaystyle\lim_{x \to 0} \dfrac{\sin 3x}{5x}$ .

A. $0$

B. $\dfrac{3}{5}$

C. $1$

D. $\dfrac{5}{3}$
[Differential Equations] The rate of change of a population $P$ satisfies $\dfrac{dP}{dt} = 0.04P$ . If $P(0) = 500$ , find $P(10)$ .

A. $500 + 0.04(10)$

B. $500 e^{0.4}$

C. $500 e^{4}$

D. $500(1.04)^{10}$
[Applications of Derivatives] A function $f$ has $f'(x) = (x-2)^2(x+1)$ . Which statement is correct?

A. $f$ has a local minimum at $x = 2$ and a local maximum at $x = -1$

B. $f$ has a local minimum at $x = -1$ only; $x = 2$ is a stationary point of inflection

C. $f$ has local minima at both $x = -1$ and $x = 2$

D. $f$ has a local maximum at $x = -1$ and no stationary point at $x = 2$
[Maclaurin Series] The first three non-zero terms of the Maclaurin series of $\cos(2x)$ are:

A. $1 - 2x^2 + \dfrac{2x^4}{3}$

B. $1 - 2x + 2x^2$

C. $2x - \dfrac{8x^3}{6} + \dfrac{32x^5}{120}$

D. $1 + 2x^2 - \dfrac{2x^4}{3}$
[Differentiation Rules] Find $\dfrac{dy}{dx}$ if $y = \ln(\sin x)$ .

A. $\dfrac{1}{\sin x}$

B. $\cos x$

C. $\cot x$

D. $-\cot x$
[Volumes of Revolution] The region bounded by $y = \sqrt{x}$ , the $x$ -axis, and $x = 4$ is rotated $2\pi$ radians about the $x$ -axis. The exact volume is:

A. $4\pi$

B. $8\pi$

C. $16\pi$

D. $32\pi$
[Limits — L’Hôpital] Use L’Hôpital’s rule to evaluate $\displaystyle\lim_{x \to 0} \dfrac{e^x - 1 - x}{x^2}$ .

A. $0$

B. $\dfrac{1}{4}$

C. $\dfrac{1}{2}$

D. $1$
[Differential Equations] The general solution of $\dfrac{dy}{dx} = \dfrac{x}{y}$ is:

A. $y = x + C$

B. $y^2 = x^2 + C$

C. $y = \dfrac{x^2}{2} + C$

D. $\ln y = \ln x + C$
[Integration by Parts] Evaluate $\displaystyle\int x \cos x\,dx$ .

A. $x \sin x + \cos x + C$

B. $x \sin x - \cos x + C$

C. $-x \sin x + \cos x + C$

D. $\sin x - x\cos x + C$

Show Answers

A — $\dfrac{e-1}{2}$ . Substitute $u = x^2$ , $du = 2x\,dx$ , so the integral becomes $\dfrac{1}{2}\int_0^1 e^u\,du = \dfrac{1}{2}[e^u]_0^1 = \dfrac{e-1}{2}$ . B is a common error from forgetting the factor of $\frac{1}{2}$ .
B — $\dfrac{3}{5}$ . Use the standard limit $\lim_{u\to 0}\frac{\sin u}{u}=1$ : rewrite as $\frac{3}{5}\cdot\frac{\sin 3x}{3x}\to\frac{3}{5}$ . C (answer of 1) is the trap when students forget to account for the coefficient 3 in the numerator.
B — $500e^{0.4}$ . Exponential growth $P = P_0 e^{kt}$ , so $P(10) = 500e^{0.04 \times 10} = 500e^{0.4}$ . D uses compound interest formula — a distractor for students who confuse continuous and discrete growth.
B — Local minimum at $x = -1$ only; $x = 2$ is a stationary point of inflection. At $x=-1$ , $f'$ changes from negative to positive (local min). At $x=2$ , $f' = 0$ but $(x-2)^2 \geq 0$ on both sides, so $f'$ does not change sign — it is a stationary point of inflection.
A — $1 - 2x^2 + \dfrac{2x^4}{3}$ . Substitute $2x$ into $\cos u = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \cdots$ : $1 - \frac{(2x)^2}{2} + \frac{(2x)^4}{24} = 1 - 2x^2 + \frac{2x^4}{3}$ .
C — $\cot x$ . Chain rule: $\frac{d}{dx}[\ln(\sin x)] = \frac{\cos x}{\sin x} = \cot x$ . D is a trap — the derivative of $\ln(\cos x)$ is $-\tan x$ , not $-\cot x$ here.
B — $8\pi$ . $V = \pi\int_0^4 (\sqrt{x})^2\,dx = \pi\int_0^4 x\,dx = \pi\left[\frac{x^2}{2}\right]_0^4 = \pi \cdot 8 = 8\pi$ .
C — $\dfrac{1}{2}$ . $\lim_{x\to 0}\frac{e^x - 1 - x}{x^2}$ : first application of L’Hôpital gives $\frac{e^x - 1}{2x}$ ( $\frac{0}{0}$ still), apply again: $\frac{e^x}{2}\to\frac{1}{2}$ . Students who apply L’Hôpital only once get $\frac{0}{0}$ and may incorrectly conclude the limit is 0.
B — $y^2 = x^2 + C$ . Separate: $y\,dy = x\,dx$ , integrate both sides: $\frac{y^2}{2} = \frac{x^2}{2} + C_1$ , multiply by 2: $y^2 = x^2 + C$ . Note the constant absorbs the factor of 2.
A — $x\sin x + \cos x + C$ . Let $u = x$ , $dv = \cos x\,dx$ ; then $du = dx$ , $v = \sin x$ . IBP: $x\sin x - \int\sin x\,dx = x\sin x + \cos x + C$ . B has the wrong sign on $\cos x$ .

IB Math IA Ideas — Calculus

Exploration topics from this chapter:

Modelling drug concentration — Use a first-order differential equation $\frac{dC}{dt} = -kC$ to model how a drug is eliminated from the bloodstream. Investigate how the elimination constant $k$ varies across drugs, optimise dosing intervals to keep concentration within a therapeutic window, and compare your model against real pharmacokinetic data.
Lorenz curves and the Gini coefficient — A Lorenz curve $L(x)$ describes income distribution; the Gini coefficient $G = 1 - 2\int_0^1 L(x)\,dx$ measures inequality. Download World Bank income-share data for two countries and use numerical integration to calculate and compare their Gini coefficients. Extend by fitting a functional form to the curve.
The Brachistochrone problem — Find the curve of fastest descent between two points under gravity. Derive the cycloid parametrically and verify it is faster than a straight line by comparing definite integrals of travel time. This connects differentiation, integration, and parametric equations in one elegant problem.
Population growth: logistic vs exponential — The logistic model $\frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right)$ accounts for carrying capacity. Source census data for a real population, solve both ODEs analytically, and use least-squares or residual analysis to determine which model fits better.
Optimising packaging — Minimise the surface area of a cylindrical tin for a fixed volume using $\frac{dS}{dr} = 0$ . Extend to elliptical cross-sections, conical lids, or prism shapes, and compare your theoretical optimum against the dimensions of real commercial products.
Volumes of revolution in architecture — Model a dome, arch, or vase using a curve $y = f(x)$ and compute the volume using $V = \pi\int_a^b [f(x)]^2\,dx$ . Photograph a real object, digitise its profile, fit a function, and compare the calculated volume to the manufacturer’s stated capacity.
The Mean Value Theorem and speed cameras — Prove rigorously that if a car’s average speed between two cameras exceeds the limit, then by the Mean Value Theorem it must have instantaneously exceeded the limit at some point. Extend to discuss the mathematics behind average-speed enforcement and how it differs from point-speed checks.

Tip: A strong IA has a clear personal engagement angle. Pick a topic that connects to something you genuinely find interesting — sports, medicine, economics, or architecture — and let the mathematics serve your question, not the other way around.

November 2026 Prediction Questions

These are NOT official IB questions. These are trend-based practice questions written to reflect the topic areas and question styles most likely to appear on the November 2026 IB Math AA HL Paper 2. Based on recent exam patterns (2022–2025), expect heavy weighting on: integration techniques, differential equations, optimization, and Maclaurin series.

Question 1 [Integration by Parts] [~7 marks]

Evaluate the definite integral

$\int_0^{\pi} x \sin x \, dx$

showing all working. State the integration technique used.

Show Solution

Technique: Integration by parts — $\int u \, dv = uv - \int v \, du$

Step 1: Choose $u$ and $dv$ .

Let $u = x$ and $dv = \sin x \, dx$ .

Then $du = dx$ and $v = -\cos x$ .

Step 2: Apply the integration by parts formula.

$\int_0^{\pi} x \sin x \, dx = \Bigl[-x\cos x\Bigr]_0^{\pi} - \int_0^{\pi}(-\cos x)\,dx$

$= \Bigl[-x\cos x\Bigr]_0^{\pi} + \int_0^{\pi}\cos x\,dx$

Step 3: Evaluate the remaining integral.

$\int_0^{\pi}\cos x\,dx = \Bigl[\sin x\Bigr]_0^{\pi} = \sin\pi - \sin 0 = 0 - 0 = 0$

Step 4: Evaluate the boundary term.

$\Bigl[-x\cos x\Bigr]_0^{\pi} = (-\pi\cos\pi) - (-0\cdot\cos 0) = (-\pi)(-1) - 0 = \pi$

Step 5: Combine.

$\int_0^{\pi} x \sin x \, dx = \pi + 0 = \pi$

Answer: $\displaystyle\int_0^{\pi} x \sin x \, dx = \pi$

Question 2 [Optimization] [~8 marks]

A closed cylindrical tin has a fixed volume of $V = 250\pi \text{ cm}^3$ . The total surface area of the tin is $S$ .

(a) Show that $S = 2\pi r^2 + \dfrac{500\pi}{r}$ , where $r$ is the radius of the base in cm.

(b) Find the value of $r$ that minimizes $S$ , and verify it is a minimum.

(c) Find the minimum surface area, giving your answer in exact form.

Show Solution

Part (a) — Setting up the expression for $S$

A closed cylinder has surface area $S = 2\pi r^2 + 2\pi r h$ .

Volume constraint: $\pi r^2 h = 250\pi$ , so $h = \dfrac{250}{r^2}$ .

Substitute:

$S = 2\pi r^2 + 2\pi r \cdot \frac{250}{r^2} = 2\pi r^2 + \frac{500\pi}{r} \qquad \checkmark$

Part (b) — Minimizing $S$

Step 1: Differentiate with respect to $r$ .

$\frac{dS}{dr} = 4\pi r - \frac{500\pi}{r^2}$

Step 2: Set $\dfrac{dS}{dr} = 0$ .

$4\pi r = \frac{500\pi}{r^2}$

$4r^3 = 500$

$r^3 = 125 \implies r = 5 \text{ cm}$

Step 3: Verify it is a minimum using the second derivative.

$\frac{d^2S}{dr^2} = 4\pi + \frac{1000\pi}{r^3}$

At $r = 5$ :

$\frac{d^2S}{dr^2} = 4\pi + \frac{1000\pi}{125} = 4\pi + 8\pi = 12\pi > 0$

Since $\dfrac{d^2S}{dr^2} > 0$ , this is a local (and global) minimum.

Part (c) — Minimum surface area

$S_{\min} = 2\pi(5)^2 + \frac{500\pi}{5} = 50\pi + 100\pi = 150\pi \text{ cm}^2$

Answer: $r = 5$ cm gives minimum surface area $S_{\min} = 150\pi \text{ cm}^2 \approx 471 \text{ cm}^2$ .

Question 3 [Separable Differential Equation] [~7 marks]

Consider the differential equation

$\frac{dy}{dx} = \frac{2x}{y+1}, \quad y > -1$

(a) Find the general solution, expressing $y$ explicitly in terms of $x$ .

(b) Find the particular solution satisfying $y(0) = 2$ .

Show Solution

Part (a) — General solution

Step 1: Separate variables.

$(y + 1)\,dy = 2x\,dx$

Step 2: Integrate both sides.

$\int (y+1)\,dy = \int 2x\,dx$

$\frac{(y+1)^2}{2} = x^2 + C_1$

$(y+1)^2 = 2x^2 + C \quad \text{where } C = 2C_1$

Step 3: Solve for $y$ (using $y > -1$ ).

$y + 1 = \sqrt{2x^2 + C}$

$y = \sqrt{2x^2 + C} - 1$

Part (b) — Particular solution

Apply initial condition $y(0) = 2$ :

$2 = \sqrt{2(0)^2 + C} - 1$

$3 = \sqrt{C} \implies C = 9$

Answer: The particular solution is $y = \sqrt{2x^2 + 9} - 1$ .

Verification: At $x = 0$ : $y = \sqrt{9} - 1 = 2$ ✓

Question 4 [Maclaurin Series] [~7 marks]

(a) Write down the Maclaurin series for $\sin u$ up to and including the term in $u^5$ .

(b) Hence find the first three non-zero terms of the Maclaurin series for $\sin(x^2)$ .

(c) State the interval of validity of the series in part (b).

(d) Use the series from part (b) to find an approximation for $\displaystyle\int_0^{0.5} \sin(x^2)\,dx$ , giving your answer to 4 significant figures.

Show Solution

Part (a) — Maclaurin series for $\sin u$

$\sin u = u - \frac{u^3}{3!} + \frac{u^5}{5!} - \cdots = u - \frac{u^3}{6} + \frac{u^5}{120} - \cdots$

Part (b) — Series for $\sin(x^2)$

Substitute $u = x^2$ :

$\sin(x^2) = x^2 - \frac{(x^2)^3}{6} + \frac{(x^2)^5}{120} - \cdots$

$\sin(x^2) = x^2 - \frac{x^6}{6} + \frac{x^{10}}{120} - \cdots$

Part (c) — Interval of validity

The Maclaurin series for $\sin u$ converges for all $u \in \mathbb{R}$ . Since $u = x^2$ is defined for all $x \in \mathbb{R}$ , the series for $\sin(x^2)$ is valid for all $x \in \mathbb{R}$ .

Part (d) — Approximation of the integral

$\int_0^{0.5}\sin(x^2)\,dx \approx \int_0^{0.5}\left(x^2 - \frac{x^6}{6} + \frac{x^{10}}{120}\right)dx$

$= \left[\frac{x^3}{3} - \frac{x^7}{42} + \frac{x^{11}}{1320}\right]_0^{0.5}$

Evaluate at $x = 0.5$ :

$= \frac{(0.5)^3}{3} - \frac{(0.5)^7}{42} + \frac{(0.5)^{11}}{1320}$

$= \frac{0.125}{3} - \frac{0.0078125}{42} + \frac{0.000488\ldots}{1320}$

$\approx 0.041667 - 0.000186 + 0.0000004$

$\approx 0.04148$

Answer: $\displaystyle\int_0^{0.5}\sin(x^2)\,dx \approx 0.04148$ (to 4 s.f.)

Question 5 [Area Between Curves] [~8 marks]

The curves $f(x) = x^2 - 2x$ and $g(x) = 4 - x^2$ intersect at two points.

(a) Find the $x$ -coordinates of the two intersection points.

(b) Find the exact area of the region enclosed between the two curves.

Show Solution

Part (a) — Intersection points

Set $f(x) = g(x)$ :

$x^2 - 2x = 4 - x^2$

$2x^2 - 2x - 4 = 0$

$x^2 - x - 2 = 0$

$(x-2)(x+1) = 0$

$x = 2 \quad \text{or} \quad x = -1$

The curves intersect at $x = -1$ and $x = 2$ .

Part (b) — Area between the curves

Step 1: Determine which curve is on top on $[-1, 2]$ .

Test $x = 0$ : $f(0) = 0$ , $g(0) = 4$ . Since $g(0) > f(0)$ , we have $g(x) \geq f(x)$ on $[-1, 2]$ .

Step 2: Set up the integral.

$A = \int_{-1}^{2}\bigl[g(x) - f(x)\bigr]\,dx = \int_{-1}^{2}\bigl[(4 - x^2) - (x^2 - 2x)\bigr]\,dx$

$= \int_{-1}^{2}(4 - 2x^2 + 2x)\,dx$

Step 3: Integrate.

$= \left[4x - \frac{2x^3}{3} + x^2\right]_{-1}^{2}$

Step 4: Evaluate at the bounds.

At $x = 2$ :

$4(2) - \frac{2(8)}{3} + 4 = 8 - \frac{16}{3} + 4 = 12 - \frac{16}{3} = \frac{36 - 16}{3} = \frac{20}{3}$

At $x = -1$ :

$4(-1) - \frac{2(-1)}{3} + 1 = -4 + \frac{2}{3} + 1 = -3 + \frac{2}{3} = \frac{-9 + 2}{3} = -\frac{7}{3}$

Step 5: Subtract.

$A = \frac{20}{3} - \left(-\frac{7}{3}\right) = \frac{20}{3} + \frac{7}{3} = \frac{27}{3} = 9$

Answer: The exact area enclosed between the two curves is $9$ square units.

GIVEN	z = r(cosθ + i sinθ) = r cis θ
GIVEN	z = re^iθ (Euler form)
MEMORISE	\|z\| = √(a² + b²)
MEMORISE	arg(z) — sketch point, use quadrant formula

GIVEN	z&sub1;z&sub2; = r&sub1;r&sub2; cis(θ&sub1; + θ&sub2;)
GIVEN	z&sub1;/z&sub2; = (r&sub1;/r&sub2;) cis(θ&sub1; − θ&sub2;)

GIVEN	(r cis θ)ⁿ = rⁿ cis(nθ)
MEMORISE	z + 1/z = 2cosθ (when \|z\|=1)
MEMORISE	z − 1/z = 2i sinθ (when \|z\|=1)

GIVEN	w^1/n = r^1/n cis((θ + 2πk)/n), k=0..n-1
MEMORISE	Sum of nth roots of unity = 0
MEMORISE	1 + ω + ω² = 0 (cube roots)

MEMORISE	z* = a − bi
MEMORISE	z · z* = \|z\|² (always real)
MEMORISE	z + z* = 2Re(z)
MEMORISE	z − z* = 2i Im(z)

MEMORISE	\|z − a\| = r → Circle, centre a, radius r
MEMORISE	\|z − a\| = \|z − b\| → Perpendicular bisector
MEMORISE	arg(z − a) = θ → Ray from a

MEMORISE	z² + az + b = 0: sum = −a, product = b
MEMORISE	Conjugate root theorem: real coeff → roots come in conjugate pairs

IB Math AA HL — Calculus

Section 1: Limits and Continuity

1.1 Evaluating Limits

1.2 L’Hôpital’s Rule HL

1.3 Continuity

Section 2: Differentiation

2.1 First Principles

2.2 Standard Derivatives

2.3 The Three Combination Rules

2.4 Implicit Differentiation

2.5 Derivatives of Inverse Trigonometric Functions

Section 3: Applications of Differentiation

3.1 Tangent and Normal Lines

3.2 Stationary Points and Curve Sketching

3.3 Optimisation

3.4 Related Rates

3.5 Kinematics

Section 4: Integration

4.1 Standard Integrals

4.2 Integration by Substitution

4.3 Integration by Parts

Practice: Integration by Parts — Fading Sequence

4.4 Partial Fractions HL

Section 5: Applications of Integration

5.1 Area Under a Curve

5.2 Area Between Two Curves

5.3 Volume of Revolution HL

5.4 Kinematics — Integration

Section 6: Differential Equations

6.1 Separable Differential Equations

6.2 Initial Value Problems

6.3 Modelling with Differential Equations

Section 7: Maclaurin Series HL

7.1 Deriving the Four Standard Series

7.2 Applications of Maclaurin Series

Section 8: MCQ Practice

Key Formulas

Limits

Differentiation Rules

Key Derivatives

Integration

Applications

Differential Equations

Maclaurin Series

Mixed Practice — Exam Style

IB Math IA Ideas — Calculus

November 2026 Prediction Questions

IB Formula Booklet — Complex Numbers

Modulus & Polar Form

Polar Multiplication & Division

De Moivre's Theorem

nth Roots

Conjugate & Arithmetic

Loci

Vieta's Formulas

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