Syllabus map
AA SL Calculus syllabus
The checklist below names the Standard Level syllabus points for this topic. Higher Level extensions are not included on this page.
- 5.1Introduction to limits and derivatives
- 5.2Increasing and decreasing functions
- 5.3Derivative of f(x) = axⁿ
- 5.4Tangents and normals
- 5.5Definite integrals
- 5.6Product, chain and quotient rules
- 5.7The second derivative
- 5.8Maximum, minimum and inflection points; optimization
- 5.9Kinematic problems
- 5.10Indefinite integrals, including inspection and substitution
- 5.11Definite integrals and area under and between curves
Study by topic
Calculus lessons
Work through focused lessons mapped to the syllabus points. Each lesson includes concise notes, an original worked example, practice and common mistakes.
Learn the ideas
Key ideas and methods
Limits, derivatives and gradients
- The derivative f′(x) is the limiting gradient of a secant as its two points approach one another. At a point where f is differentiable, f′(x) is the gradient of the tangent and the instantaneous rate of change.
- For f(x) = axⁿ, f′(x) = anxⁿ⁻¹. Use the product, chain and quotient rules when the expression is a product, a composite or a quotient of functions.
- A tangent with gradient m at (a, f(a)) has equation y − f(a) = m(x − a). A normal is perpendicular to the tangent, so its gradient is −1/m when m ≠ 0.
Monotonicity and stationary points
- Where f′(x) > 0, the function is increasing; where f′(x) < 0, it is decreasing. Solve f′(x) = 0 to find stationary points, then use sign changes or the second derivative to classify them.
- The second derivative f″ describes how the gradient changes. A change in concavity may indicate an inflection point, but check the function’s behaviour around the candidate value.
- For optimization, define the quantity to optimize, express it in one variable, state any domain restrictions, find critical points and test which value gives the required maximum or minimum.
Integrals and accumulation
- An indefinite integral reverses differentiation and includes a constant: ∫f(x)dx = F(x) + C where F′(x) = f(x). Substitution can simplify a composite expression by replacing its inner function with a new variable.
- A definite integral ∫ₐᵇ f(x)dx is signed area. For geometric area below the x-axis, reverse the sign; for area between curves, integrate upper function minus lower function over intervals where their order is fixed.
- The trapezoidal rule estimates area from equally spaced ordinates; it is an approximation. In kinematics, if s(t) is position, v(t) = s′(t) and a(t) = v′(t). Track units and distinguish displacement from distance travelled.
Follow the reasoning
Worked examples
These original examples show common reasoning steps. They are not copied from IB examination papers.
Worked example 1
Classify stationary points
Question. For f(x) = x³ − 3x, find and classify the stationary points.
- f′(x) = 3x² − 3 = 3(x − 1)(x + 1), so stationary points occur at x = −1 and x = 1.
- f(−1) = 2 and f(1) = −2. Also f″(x) = 6x.
- At x = −1, f″(−1) < 0, giving a local maximum (−1, 2). At x = 1, f″(1) > 0, giving a local minimum (1, −2).
Answer: Local maximum (−1, 2); local minimum (1, −2).
Worked example 2
Find a tangent and normal
Question. Find the tangent and normal to y = x² + 1 at x = 2.
- The point is (2, 5). The derivative is dy/dx = 2x, so the tangent gradient is 4.
- Tangent: y − 5 = 4(x − 2), or y = 4x − 3.
- The normal gradient is −1/4. Normal: y − 5 = −¼(x − 2).
Answer: Tangent y = 4x − 3; normal y − 5 = −¼(x − 2).
Worked example 3
Integrate by substitution and evaluate
Question. Evaluate ∫₀¹ 2x(x² + 1)³ dx.
- Let u = x² + 1, so du = 2x dx. When x = 0, u = 1; when x = 1, u = 2.
- The integral becomes ∫₁² u³ du = [u⁴/4]₁².
- Evaluate: (16 − 1)/4 = 15/4.
Answer: 15/4.
Try it yourself
Practice with worked answers
Question 1. Differentiate y = (x² + 1)³.
Show worked answer
By the chain rule, dy/dx = 3(x² + 1)²(2x) = 6x(x² + 1)².
Question 2. Find the stationary point of f(x) = x² − 6x + 5 and classify it.
Show worked answer
f′(x) = 2x − 6 = 0 gives x = 3. f(3) = −4 and f″(x) = 2 > 0, so the point (3, −4) is a minimum.
Question 3. Find the area between y = 4 and y = x² from x = 0 to x = 2.
Show worked answer
On this interval 4 ≥ x². Area = ∫₀²(4 − x²)dx = [4x − x³/3]₀² = 8 − 8/3 = 16/3 square units.
Avoid these slips
Common mistakes
- Forgetting the chain-rule factor from the inner function.
- Calling every solution of f′(x) = 0 a maximum or minimum without checking the behaviour.
- Using the same gradient for the tangent and normal; perpendicular non-vertical lines have gradients whose product is −1.
- Treating a definite integral as unsigned geometric area when the curve is below the axis.
- Omitting the constant of integration from an indefinite integral.
Before you move on
Revision checklist
- Can I interpret a derivative as a gradient or rate in context?
- Can I choose and apply the power, product, chain or quotient rule?
- Can I classify stationary points and set up an optimization model?
- Can I distinguish signed integral, geometric area, displacement and distance?
About this guide
Independent SJMaths resource; not an official IB publication. Follows the AA syllabus first assessed in 2021 (final assessment: November 2028). Use the separate 2029 guide for the new course.
Check the IB Mathematics programme page and the official AA subject guide for the authoritative syllabus and examination-session requirements.