Calculus — IB Math AA HL AA HL Study Guide

For: IB Math AA HL candidates sitting IB Math: Analysis & Approaches HL.

Covers: Limits and continuity, first principles derivatives, differentiation rules, HL integration techniques, volumes of revolution, separable differential equations, and HL Maclaurin series, aligned to the 2020-2026 IB Math AA HL syllabus.

You should already know: IGCSE / pre-DP math, comfort with proof and algebra.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the IB Math AA HL style for educational use. They are not reproductions of past IBO papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official IBO mark schemes for grading conventions.

1. What Is Calculus?

Calculus is the branch of mathematics focused on quantifying change and motion, split into two core subfields: differential calculus (which measures instantaneous rates of change) and integral calculus (which measures accumulated change over an interval). It makes up ~35% of the IB Math AA HL assessment, appearing in both calculator and non-calculator papers, as well as the Paper 3 investigative task. It is also the most widely applied math topic in university STEM, economics, and social science courses.

2. Limits, continuity, and derivatives from first principles

A limit describes the value a function approaches as its input gets arbitrarily close to a given point, written as ${\lim }_{x\to a}f(x)=L$, where $L$ is the limit value, independent of the actual value of $f(a)$. A function is continuous at $x=a$ if three conditions hold: 1) $f(a)$ is defined, 2) ${\lim }_{x\to a}f(x)$ exists, and 3) ${\lim }_{x\to a}f(x)=f(a)$.

The derivative of a function $f^{\prime }(x)$ is the gradient of the tangent to $f(x)$ at any point $x$, derived from first principles by taking the limit of the gradient of a secant line connecting two points on the curve as the distance between the points approaches 0: $$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$$

Worked example (4 marks, Paper 1): Find the derivative of $f(x)=2x^2+3x$ from first principles.

1. Compute $f(x+h)=2(x+h{)}^2+3(x+h)=2x^2+4xh+2h^2+3x+3h$
2. Subtract $f(x)$ to get $f(x+h)-f(x)=4xh+2h^2+3h$
3. Divide by $h$ to get $\frac{f(x+h)-f(x)}{h}=4x+2h+3$
4. Take the limit as $h\to 0$ to eliminate the $h$ term: $f^{\prime }(x)=4x+3$

Exam tip: Examiners award a separate mark for correct limit notation, so never skip writing ${\lim }_{h\to 0}$ before the final substitution step.

3. Differentiation rules — product, quotient, chain

These shortcut rules eliminate the need for first principles for all standard functions, and are often used in combination for complex expressions:

1. Product rule: For $y=u(x)v(x)$, $\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$. Example: Differentiate $y=x^2\sin x$: $u=x^2,v=\sin x$, so $\frac{dy}{dx}=x^2\cos x+2x\sin x$
2. Quotient rule: For $y=\frac{u(x)}{v(x)}$, $\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$. Use the mnemonic "low d high minus high d low, over the square of what's below" to avoid mixing up numerator order. Example: Differentiate $y=\frac{e^x}{2x+1}$: $\frac{dy}{dx}=\frac{(2x+1)e^x-2e^x}{(2x+1{)}^2}=\frac{(2x-1)e^x}{(2x+1{)}^2}$
3. Chain rule: For composite functions $y=f(g(x))$, $\frac{dy}{dx}=f^{\prime }(g(x))\times g^{\prime }(x)$. Differentiate the outer function first (keeping the inner function unchanged), then multiply by the derivative of the inner function. Example: Differentiate $y=\ln (\cos x)$: $\frac{dy}{dx}=\frac{1}{\cos x}\times (-\sin x)=-\tan x$

4. Integration — substitution, by parts (HL)

Integration is the reverse process of differentiation, and all indefinite integrals include an arbitrary constant of integration $C$ to account for the family of antiderivatives.

1. Substitution: Used for integrands that are a product of a composite function and the derivative of its inner function. Set $u=g(x)$, rewrite the integral in terms of $u$, integrate, then substitute back. For definite integrals, you can adjust the bounds to match $u$ values to avoid re-substituting later. Worked example: Evaluate $\int 2x{e}^{x^2}dx$. Let $u=x^2$, so $du=2xdx$. The integral becomes $\int e^{u}du=e^u+C=e^{x^2}+C$.
2. Integration by parts (HL only): Derived from the product rule, used for products of functions that cannot be solved with substitution. The formula is: $$\int u\frac{dv}{dx}dx=uv-\int v\frac{du}{dx}dx$$ Choose $u$ using the LIATE priority rule: Logarithmic > Inverse trigonometric > Algebraic > Trigonometric > Exponential. Worked example: Evaluate $\int x\ln xdx$. Let $u=\ln x$ (logarithmic, higher priority than algebraic $x$), $\frac{dv}{dx}=x$. So $\frac{du}{dx}=\frac{1}{x}$, $v=\frac{x^2}{2}$. Substitute into the formula: $\frac{x^2\ln x}{2}-\int \frac{x}{2}dx=\frac{x^2\ln x}{2}-\frac{x^2}{4}+C$.

5. Volumes of revolution

Volumes of revolution are 3D shapes formed when a 2D curve is rotated 360° around the x-axis or y-axis, with two core formulas:

1. Rotation around the x-axis, between $x=a$ and $x=b$: $V=\pi {\int }_a^b[f(x){]}^{2}dx$
2. Rotation around the y-axis, between $y=c$ and $y=d$: $V=\pi {\int }_c^d[f^{-1}(y){]}^{2}dx$, where you rearrange $y=f(x)$ to get $x$ as a function of $y$.

Worked example (5 marks, Paper 2): Find the volume when $y=\sqrt{x}$ between $x=0$ and $x=4$ is rotated around the x-axis.

1. Square the function: $[f(x){]}^2=x$
2. Integrate between 0 and 4: ${\int }_0^{4}xdx={\left[\frac{x^2}{2}\right]}_0^4=8-0=8$
3. Multiply by $\pi$: $V=8\pi$ cubic units.

Exam tip: 60% of students lose marks by forgetting to square the function before integrating, so always write the full formula with the square explicitly before substituting values.

6. Differential equations — separable (HL extra)

A differential equation relates a function to its derivative, e.g. $\frac{dy}{dx}=2xy$. Separable differential equations can be rearranged to isolate all $y$ terms on one side and all $x$ terms on the other, then integrated to solve for $y$. The steps are:

1. Separate variables: Move all $y$ terms to the left with $dy$, all $x$ terms to the right with $dx$
2. Integrate both sides, adding a single constant of integration $C$ on one side
3. Rearrange to solve for $y$ if required, using any given initial conditions to find the value of $C$

Worked example (6 marks, Paper 1): Solve $\frac{dy}{dx}=y^2\cos x$, given $y(0)=1$.

1. Separate: $\frac{1}{y^2}dy=\cos xdx$
2. Integrate: $-\frac{1}{y}=\sin x+C$
3. Apply initial condition $x=0,y=1$: $-1=0+C$, so $C=-1$
4. Rearrange: $y=\frac{1}{1-\sin x}$

7. Maclaurin series (HL extra)

Maclaurin series are infinite polynomial approximations of functions centered at $x=0$, derived from the values of the function and its derivatives at $x=0$. The general formula is: $$f(x)=f(0)+f^{\prime }(0)x+\frac{f^{\prime \prime }(0)}{2!}{x}^2+\frac{f^{\prime \prime \prime }(0)}{3!}{x}^3+...+\frac{f^{(n)}(0)}{n!}{x}^n+...$$

Standard Maclaurin series to memorize (given in the formula booklet, but memorization saves time):

• $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$ for all $x$
• $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...$ for all $x$
• $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-...$ for all $x$
• $\ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-...$ for $|x|<1$

Worked example: Find the first 3 non-zero terms of the Maclaurin series for $e^{\sin x}$. Instead of differentiating repeatedly, substitute the series for $\sin x$ into the series for $e^x$: $e^u=1+u+\frac{u^2}{2!}+...$ where $u=\sin x=x-\frac{x^3}{6}+...$. Substitute to get $e^{\sin x}=1+x+\frac{x^2}{2}+...$.

8. Common Pitfalls (and how to avoid them)

• Wrong move: Forgetting the constant of integration $C$ for indefinite integrals. Why it happens: Rushing, confusing definite and indefinite integrals. Correct move: Write $C$ immediately after every indefinite integral, even if you will solve for it later using boundary conditions.
• Wrong move: Mixing up the numerator order in the quotient rule, writing $udv-vdu$ instead of $vdu-udv$. Why it happens: Poor memorization. Correct move: Test the rule on a simple function like $y=\frac{x}{\sin x}$ to confirm you get the right derivative, or use the "low d high" mnemonic.
• Wrong move: Leaving out the limit statement in first principles derivative proofs. Why it happens: Assuming the algebra alone is enough. Correct move: Explicitly write ${\lim }_{h\to 0}$ at every step until you substitute $h=0$, as examiners award a separate mark for correct limit notation.
• Wrong move: Forgetting to rearrange the function to $x$ in terms of $y$ for y-axis volume of revolution calculations. Why it happens: Mindlessly applying the x-axis formula. Correct move: Circle the axis of rotation in every question before starting working.
• Wrong move: Omitting the absolute value when integrating $\frac{1}{y}$ to get $\ln |y|$ for differential equations. Why it happens: Assuming $y$ is always positive. Correct move: Keep the absolute value until you rearrange to an explicit form for $y$, where the arbitrary constant absorbs the sign.

9. Practice Questions (IB Math AA HL Style)

Question 1 (Paper 1, 6 marks)

a) Find the derivative of $f(x)=3x^2-5x+2$ from first principles. (4 marks) b) Find the gradient of the tangent to $f(x)$ at $x=2$. (2 marks)

Solution: a) Step 1: $f(x+h)=3(x+h{)}^2-5(x+h)+2=3x^2+6xh+3h^2-5x-5h+2$ Step 2: $f(x+h)-f(x)=6xh+3h^2-5h$ Step3: $\frac{f(x+h)-f(x)}{h}=6x+3h-5$ Step4: ${\lim }_{h\to 0}(6x+3h-5)=6x-5$, so $f^{\prime }(x)=6x-5$ (1 mark per step, 1 mark for correct limit notation) b) Substitute $x=2$ into $f^{\prime }(x)$: $6(2)-5=7$, so the gradient is 7. (2 marks)

Question 2 (Paper 1, HL, 7 marks)

Evaluate the indefinite integral $\int x^2{e}^{x}dx$.

Solution: Use integration by parts twice:

1. First iteration: Let $u=x^2$, $\frac{dv}{dx}=e^x$. So $\frac{du}{dx}=2x$, $v=e^x$. Integral becomes $x^2{e}^x-\int 2x{e}^{x}dx$ (2 marks for correct u/dv choice)
2. Second iteration on $\int 2x{e}^{x}dx$: Let $u=2x$, $\frac{dv}{dx}=e^x$. So $\frac{du}{dx}=2$, $v=e^x$. The integral becomes $2x{e}^x-\int 2e^{x}dx=2x{e}^x-2e^x+C$ (2 marks for correct second iteration)
3. Combine terms: $\int x^2{e}^{x}dx=x^2{e}^x-2x{e}^x+2e^x+C=e^x(x^2-2x+2)+C$ (2 marks for simplification, 1 mark for +C)

Question 3 (Paper 2, HL, 6 marks)

Solve the separable differential equation $\frac{dy}{dx}=3x^{2}y$, given $y(1)=2e$. Give your answer in the form $y=f(x)$.

Solution:

1. Separate variables: $\frac{1}{y}dy=3x^{2}dx$ (2 marks for correct separation)
2. Integrate both sides: $\ln |y|=x^3+C$ (2 marks for correct integration)
3. Apply initial condition: $\ln (2e)=1^3+C$, so $\ln 2+1=1+C$, so $C=\ln 2$ (1 mark for correct C value)
4. Rearrange: $y=e^{x^3+\ln 2}=2e^{x^3}$ (1 mark for final simplified form)

10. Quick Reference Cheatsheet

11. What's Next

Calculus is a foundational topic that connects to almost every other part of the IB Math AA HL syllabus, including kinematics (using derivatives to find acceleration and integrals to find displacement), optimization problems (using derivatives to find maximum and minimum values), and probability distributions (integrating probability density functions to find cumulative probabilities). You will also use calculus extensively in Paper 3 investigation tasks, where you may be asked to derive new rules or apply calculus to real-world contexts like population growth or economic modeling.

If you have any gaps in your understanding, or want to practice more exam-style questions tailored to your weak spots, you can ask Ollie, our AI tutor, for personalized help at any time. Ollie can walk you through step-by-step solutions, explain tricky concepts in simpler terms, and generate custom practice sets to help you master Calculus for your IB Math AA HL exams.