IB Math AA HL — Calculus

Complete Study Guide

Topics Covered

1. Limits & Continuity
2. Differentiation — Rules & Techniques
3. Applications of Differentiation (optimisation, related rates, kinematics)
4. Integration — Techniques & Standard Integrals
5. Applications of Integration (areas, volumes of revolution)
6. Differential Equations
7. Maclaurin Series
8. Practice MCQs & Exam Alerts

Topic 5 of the IB Math AA HL syllabus — Paper 2 and Paper 3

Videos on this page: Differentiation · Integration

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

IB Maths Resources · Differentiation rules (chain, product, quotient), optimisation, implicit differentiation, rates of change, and equation of tangents

Watch: Integration — Techniques, Areas, Volumes & Substitutions

IB Maths Resources · Integration rules, definite integrals, areas between curves, volumes of revolution, trig substitutions, integration by parts, and u-substitutions

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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).

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:

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

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.

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:

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

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.

Quick Recall — Section 1

Try to answer without scrolling up:

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

Reveal answers

1. As $x$ gets arbitrarily close to $a$, $f(x)$ gets arbitrarily close to $L$.
2. 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).
3. 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 $f(c) = 0$.

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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

2.2 Standard Derivatives

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

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

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).

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

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:

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

Quick Recall — Section 2

Try to answer without scrolling up:

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

Reveal answers

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

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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: $y - f(a) = -\dfrac{1}{f'(a)}(x - a)$

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 $x = c$

Full curve-sketching checklist:

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

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:

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

3.4 Related Rates

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:

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

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.

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Quick Recall — Section 3

Try to answer without scrolling up:

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

Reveal answers

1. Solve $f'(x) = 0$.
2. 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 $f'$.
3. $v(t) = \frac{ds}{dt} = s'(t)$.

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

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.

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.

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.

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.

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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.

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.

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):

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

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

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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:

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

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 $y_0$.

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, $P \to K$

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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

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.

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Section 8: MCQ Practice

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Key Formulas

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Mixed Practice — Exam Style

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

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

   B. $e - 1$

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

   D. $e^2 - 1$

2. [Limits] Evaluate $\displaystyle\lim_{x \to 0} \dfrac{\sin 3x}{5x}$.

   A. $0$

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

   C. $1$

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

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. $500(1.04)^{10}$

4. [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 $x = 2$

5. [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. $1 + 2x^2 - \dfrac{2x^4}{3}$

6. [Differentiation Rules] Find $\dfrac{dy}{dx}$ if $y = \ln(\sin x)$.

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

   B. $\cos x$

   C. $\cot x$

   D. $-\cot x$

7. [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. $32\pi$

8. [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. $1$

9. [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. $\ln y = \ln x + C$

10. [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

1. 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 $\frac{1}{2}$.

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.

3. 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.

4. 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.

5. A — $1 - 2x^2 + \dfrac{2x^4}{3}$. Substitute $2x$ into $\cos u = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \cdots$: $1 - \frac{(2x)^2}{2} + \frac{(2x)^4}{24} = 1 - 2x^2 + \frac{2x^4}{3}$.

6. 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.

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

8. 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.

9. 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.

10. 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 $\cos x$.

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IB Math IA Ideas — Calculus

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May 2026 Prediction Questions

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