Calculus — IB Math AA SL AA SL Study Guide

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

Covers: Limits, derivatives (power and chain rule), stationary points and optimisation, definite and indefinite integrals, areas under and between curves, and kinematics applications aligned to the IB AA SL syllabus.

You should already know: IGCSE / pre-DP math.

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 SL 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 rates of change of functions) and integral calculus (which measures accumulated quantities and areas under curves). It makes up 30% of the IB AA SL final assessment, appearing in both Paper 1 (no calculator) and Paper 2 (calculator allowed) questions, typically worth 15 to 25 marks total across both papers. Examiners frequently combine calculus with functions, trigonometry, and real-world context questions to test application skills.

2. Limits, derivatives — power and chain rule

A limit describes the value a function approaches as its input approaches a given point, written as ${\lim }_{x\to a}f(x)=L$. The derivative of a function $f(x)$ is the instantaneous rate of change of the function at a point, defined formally as the limit of the difference quotient: $$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$$ You will not be required to use the limit definition to compute derivatives in exams, but you must understand it as the foundation of differential calculus.

The power rule is the core rule for differentiating polynomial terms: for any real number $n$, if $f(x)=x^n$, then: $$f^{\prime }(x)=n{x}^{n-1}$$ This works for all powers, including negative values (e.g, $f(x)=x^{-2}=\frac{1}{x^2}$, so $f^{\prime }(x)=-2x^{-3}=-\frac{2}{x^3}$) and fractional values (e.g, $f(x)=x^{1/2}=\sqrt{x}$, so $f^{\prime }(x)=\frac{1}{2}{x}^{-1/2}=\frac{1}{2\sqrt{x}}$).

The chain rule is used for differentiating composite functions ($y=f(g(x))$, where $g(x)$ is the inner function and $f(u)$ is the outer function with $u=g(x)$): $$\frac{dy}{dx}=f^{\prime }(g(x))\times g^{\prime }(x)$$ In simple terms: differentiate the outer function keeping the inner function unchanged, then multiply by the derivative of the inner function.

Worked Example

Differentiate $y=(2x^2+5x{)}^3$:

1. Identify outer function $u^3$ and inner function $u=2x^2+5x$
2. Differentiate outer function: $3u^2=3(2x^2+5x{)}^2$
3. Differentiate inner function: $4x+5$
4. Multiply the two results: $\frac{dy}{dx}=3(2x^2+5x{)}^2(4x+5)$ Examiners accept unsimplified derivatives unless explicitly asked for a simplified form.

3. Stationary points and optimisation

Stationary points are points on a curve where the gradient is zero, i.e, $f^{\prime }(x)=0$. There are three types of stationary points:

1. Local maximum: gradient changes from positive to negative as $x$ increases through the point
2. Local minimum: gradient changes from negative to positive as $x$ increases through the point
3. Horizontal point of inflection: gradient sign does not change (e.g, $y=x^3$ at $x=0$)

To classify stationary points, use either:

• First derivative test: test the sign of $f^{\prime }(x)$ for values just left and right of the stationary point, or
• Second derivative test: compute the second derivative $f^{\prime \prime }(x)$; if $f^{\prime \prime }(a)<0$ the point is a maximum, if $f^{\prime \prime }(a)>0$ it is a minimum, and if $f^{\prime \prime }(a)=0$ use the first derivative test.

Optimisation is the real-world application of stationary points, where you find the maximum or minimum value of a quantity (e.g, maximum volume of a container, minimum production cost). You will always need to justify that your solution is the required maximum or minimum to earn full marks.

Worked Example

An open-topped rectangular box has a square base of side length $x$ cm, and a total surface area of 96 cm². Find the maximum possible volume of the box.

1. Surface area formula: $x^2+4xh=96$, so rearrange for height $h=\frac{96-x^2}{4x}$
2. Volume formula: $V=x^{2}h=x^2\times \frac{96-x^2}{4x}=24x-\frac{1}{4}{x}^3$
3. Differentiate volume: $\frac{dV}{dx}=24-\frac{3}{4}{x}^2$
4. Set derivative to zero to find stationary points: $24=\frac{3}{4}{x}^2\to x^2=32\to x=4\sqrt{2}$ (length cannot be negative)
5. Second derivative: $\frac{d^{2}V}{d{x}^2}=-\frac{3}{2}x$, which is negative for $x>0$, so this is a maximum
6. Calculate maximum volume: $V=24(4\sqrt{2})-\frac{1}{4}(4\sqrt{2}{)}^3=96\sqrt{2}-32\sqrt{2}=64\sqrt{2}\approx 90.5$ cm³

4. Definite and indefinite integrals

Integration is the reverse of differentiation. An indefinite integral has no upper or lower bounds, and returns a family of functions that share the same derivative. You must always add a constant of integration $C$ to indefinite integrals, as the derivative of any constant is zero.

The power rule for integration is the reverse of the differentiation power rule, for $n\ne -1$: $$\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$$ For composite functions with a linear inner term $(ax+b)$, the integration rule is: $$\int (ax+b{)}^{n}dx=\frac{(ax+b{)}^{n+1}}{a(n+1)}+C,n\ne -1$$

A definite integral has upper and lower bounds $a$ and $b$, and returns a numerical value equal to the net area between the curve and the x-axis between $a$ and $b$, as defined by the Fundamental Theorem of Calculus: $${\int }_a^{b}f(x)dx=F(b)-F(a)$$ where $F(x)$ is the antiderivative of $f(x)$. You do not need to include the constant $C$ for definite integrals, as it cancels out when you subtract $F(a)$ from $F(b)$.

Worked Examples

1. Compute the indefinite integral $\int (3x^2+2\sqrt{x}-x^{-3})dx$: $$\int (3x^2+2x^{1/2}-x^{-3})dx=\frac{3x^3}{3}+\frac{2x^{3/2}}{3/2}-\frac{x^{-2}}{-2}+C=x^3+\frac{4}{3}{x}^{3/2}+\frac{1}{2x^2}+C$$
2. Compute the definite integral ${\int }_1^4(2x+1)dx$: Antiderivative $F(x)=x^2+x$, so ${\int }_1^4(2x+1)dx=F(4)-F(1)=(16+4)-(1+1)=20-2=18$

5. Areas under and between curves

The area under a curve $f(x)$ between $x=a$ and $x=b$ is calculated using definite integrals, with two key rules:

1. If $f(x)\ge 0$ for all $x$ in $[a,b]$, area = ${\int }_a^{b}f(x)dx$
2. If $f(x)<0$ for any $x$ in $[a,b]$, the integral returns a negative value, so you take the absolute value of the integral for that region. If the curve crosses the x-axis between $a$ and $b$, split the integral at the root, compute each segment separately, take absolute values of negative segments, then sum all results.

The area between two curves $f(x)$ and $g(x)$ between $x=a$ and $x=b$, where $f(x)\ge g(x)$ for all $x$ in $[a,b]$, is: $$Area={\int }_a^b[f(x)-g(x)]dx$$ If the curves cross each other in the interval, split the integral at the intersection points, take the absolute value of each segment, then sum.

Worked Example

Find the area between $y=x^2$ and $y=x+6$:

1. Find intersection points: $x^2=x+6\to x^2-x-6=0\to (x-3)(x+2)=0\to x=-2,x=3$
2. Between $x=-2$ and $x=3$, $y=x+6$ is the upper function, so area = ${\int }_{-2}^3(x+6-x^2)dx$
3. Antiderivative: $\frac{1}{2}{x}^2+6x-\frac{1}{3}{x}^3$
4. Evaluate at bounds: $F(3)=4.5+18-9=13.5$, $F(-2)=2-12+\frac{8}{3}=-\frac{22}{3}$
5. Area = $13.5-(-\frac{22}{3})=\frac{125}{6}\approx 20.83$ square units

6. Kinematics applications

Kinematics is the study of motion of objects in a straight line, and you will use differentiation and integration to relate three core quantities: displacement $s$ (position relative to a fixed origin), velocity $v$ (rate of change of displacement), and acceleration $a$ (rate of change of velocity):

• Differentiation rules: $v=\frac{ds}{dt}$, $a=\frac{dv}{dt}=\frac{d^{2}s}{d{t}^2}$
• Integration rules: $s=\int vdt$, $v=\int adt$

Key exam notes:

• Speed is the magnitude of velocity, so $speed=|v|$
• Displacement is the net change in position, calculated as ${\int }_a^{b}v(t)dt$
• Total distance traveled is the total path length, calculated as ${\int }_a^b|v(t)|dt$; if velocity changes sign in the interval, split the integral at the time $t$ where $v(t)=0$, take absolute values of each segment, then sum.

Worked Example

A particle moves in a straight line with velocity $v(t)=3t-6$ m/s, and initial displacement $s(0)=4$ m. Find (a) displacement at $t=3$, (b) total distance traveled between $t=0$ and $t=3$.

1. Part (a): Compute displacement function $s(t)=\int (3t-6)dt=\frac{3}{2}{t}^2-6t+C$. Use $s(0)=4$ so $C=4$, so $s(t)=1.5t^2-6t+4$. At $t=3$, $s(3)=13.5-18+4=-0.5$ m.
2. Part (b): Find when velocity is zero: $3t-6=0\to t=2$. Split integral into $[0,2]$ and $[2,3]$:

• Distance $[0,2]$: $\left|{\int }_0^2(3t-6)dt\right|=|(6-12)-0|=|-6|=6$ m
• Distance $[2,3]$: $\left|{\int }_2^3(3t-6)dt\right|=|(13.5-18)-(6-12)|=|-4.5+6|=1.5$ m
• Total distance: $6+1.5=7.5$ m

7. Common Pitfalls (and how to avoid them)

• Forgetting the constant of integration $C$ in indefinite integrals: Students rush through integration problems and only add $C$ when prompted, losing 1 mark per missing $C$. Correct move: write $+C$ immediately after computing any indefinite integral, before simplifying.
• Mixing up differentiation and integration power rules: Students reverse the multiply/divide and subtract/add steps when working quickly. Correct move: memorize the difference: differentiation = multiply by power, subtract 1; integration = add 1 to power, divide by new power.
• Confusing displacement and total distance in kinematics: Students assume integrating velocity always gives distance, but displacement is net change, distance is total path length. Correct move: if asked for distance, first find all times $t$ where $v(t)=0$ in the interval, split the integral, take absolute values of each segment before summing.
• Getting negative area values for curve area questions: Students integrate functions that dip below the x-axis without adjusting for sign, or subtract the upper function from the lower function. Correct move: always sketch the curve first if possible, split integrals at crossing points with the x-axis or between two curves, take absolute value of all integral results before summing.
• Skipping stationary point classification in optimisation problems: Students find the $x$ value where the derivative is zero and jump straight to the final answer, losing 1 mark for missing justification. Correct move: always do either the first or second derivative test, and write a 1-sentence justification (e.g, "$f^{\prime \prime }(2)=-4<0$, so this is a local maximum").

8. Practice Questions (IB Math: Analysis & Approaches SL Style)

Question 1 (No calculator, 5 marks)

a) Differentiate $f(x)=(3x^2-4x{)}^4$ with respect to $x$. (3 marks) b) Find the gradient of the curve at $x=1$. (2 marks)

Solution

a) Use the chain rule: outer function $u^4$, inner function $u=3x^2-4x$.

• Derivative of outer: $4u^3=4(3x^2-4x{)}^3$
• Derivative of inner: $6x-4$
• Final derivative: $f^{\prime }(x)=4(3x^2-4x{)}^3(6x-4)$ Award 1 mark for identifying composite structure, 1 mark for correct derivatives of inner and outer functions, 1 mark for final correct expression. b) Substitute $x=1$: $f^{\prime }(1)=4(3-4{)}^3(6-4)=4(-1{)}^3(2)=4(-1)(2)=-8$ Award 1 mark for correct substitution, 1 mark for final answer.

Question 2 (Calculator allowed, 6 marks)

Find the total area bounded by $y=x^3-9x$, the x-axis, $x=-2$ and $x=2$.

Solution

1. Find roots of the function in the interval: $x^3-9x=x(x^2-9)=x(x-3)(x+3)$. The only root in $[-2,2]$ is $x=0$.
2. Test sign: for $x\in [-2,0]$, $f(-1)=-1+9=8>0$ (above x-axis); for $x\in [0,2]$, $f(1)=1-9=-8<0$ (below x-axis).
3. Split integral: $$Area={\int }_{-2}^0(x^3-9x)dx+\left|{\int }_0^2(x^3-9x)dx\right|$$
4. Compute first integral: $F(x)=\frac{x^4}{4}-\frac{9x^2}{2}$ ${\int }_{-2}^{0}f(x)dx=0-(\frac{16}{4}-\frac{9\ast 4}{2})=-(4-18)=14$
5. Compute second integral: ${\int }_0^{2}f(x)dx=(\frac{16}{4}-\frac{9\ast 4}{2})-0=4-18=-14$, absolute value =14
6. Total area = $14+14=28$ square units Award 1 mark for finding root in interval, 1 mark for splitting integral, 1 mark for recognizing sign change, 2 marks for correct integral computations, 1 mark for final answer.

Question 3 (Calculator allowed, 7 marks)

A closed cylindrical water tank has a volume of 2000 m³. Find the minimum total surface area of the tank to minimize construction cost.

Solution

1. Let $r$ = radius of base, $h$ = height of cylinder. Volume: $\pi r^{2}h=2000\to h=\frac{2000}{\pi r^2}$
2. Total surface area: $S=2\pi r^2+2\pi rh$
3. Substitute $h$: $S(r)=2\pi r^2+2\pi r(\frac{2000}{\pi r^2})=2\pi r^2+\frac{4000}{r}$
4. Differentiate: $S^{\prime }(r)=4\pi r-\frac{4000}{r^2}$
5. Set to zero: $4\pi r=\frac{4000}{r^2}\to r^3=\frac{1000}{\pi }\to r=\sqrt[3]{\frac{1000}{\pi }}\approx 6.83$ m
6. Second derivative: $S^{\prime \prime }(r)=4\pi +\frac{8000}{r^3}$, which is positive for all $r>0$, so this is a minimum
7. Compute surface area: $S=2\pi (6.83{)}^2+\frac{4000}{6.83}\approx 293+586=879$ m² (3 significant figures) Award 1 mark for correct volume and surface area formulas, 1 mark for expressing $S$ in terms of $r$ only, 1 mark for correct derivative, 1 mark for solving for $r$, 1 mark for second derivative justification, 1 mark for correct substitution, 1 mark for final answer.

9. Quick Reference Cheatsheet

10. What's Next

Calculus is the highest-weighted topic in IB Math AA SL, and the foundational skills covered in this guide connect directly to more advanced syllabus content including derivatives and integrals of trigonometric, exponential, and logarithmic functions, curve sketching, and continuous probability distributions. Mastering these core rules and application techniques is critical for scoring a 6 or 7 on your final exam, as 25-30% of all marks across Paper 1 and Paper 2 are tied to calculus concepts.

If you struggle with any of the concepts in this guide, or want to practice more exam-style questions tailored to your weak areas, you can ask Ollie, our AI tutor, for personalized help at any time by visiting Ollie. You can also find more topic-specific study guides, timed quizzes, and full mock exams aligned to the IB Math AA SL syllabus on the homepage, designed to help you target your revision and maximize your exam score.