Which courses does AscentIQ cover?

IB Mathematics: Analysis and Approaches SL, Applications and Interpretation SL, and Applications and Interpretation HL — exercises generated for each course's syllabus and marked against the IB mark scheme. Analysis and Approaches HL isn't available yet; it's on our roadmap. We add courses one at a time rather than several superficially.

How does the marking actually work?

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← IB Math AA SL

IB Math AA SL · Topic 5

Calculus

The largest topic of the course by recommended teaching time. Differentiation finds gradients, turning points and rates of change; integration reverses it and measures area; kinematics puts both to work on a moving particle. Unlike the other SL course, the functions here go beyond polynomials — powers, trigonometric, exponential and logarithmic functions — and the working is symbolic.

What the syllabus covers

Eleven assessed sections, SL 5.1 to 5.11 — from the idea of a limit to areas between curves.

SL 5.1Limits and the derivative as a rate of changeThe limit introduced informally; the derivative read as a gradient function and as a rate of change.

SL 5.2Increasing and decreasing functionsWhat f′(x) > 0, f′(x) = 0 and f′(x) < 0 say about the function, read from its graph.

SL 5.3Derivatives of functions with integer exponentsDifferentiating sums of terms axⁿ where every exponent is an integer.

SL 5.4Tangents and normalsTheir equations at a given point: gradient from the derivative, line through the point.

SL 5.5Introduction to integrationAntiderivatives of axⁿ-type functions, boundary conditions to fix the constant, and areas under a curve with technology.

SL 5.6Derivatives of xⁿ, sin x, cos x, eˣ and ln x; chain, product and quotient rulesRational exponents, the standard derivatives, and the three rules for combining functions.

SL 5.7The second derivativeConcavity, and the relationship between the graphs of f, f′ and f″.

SL 5.8Maxima, minima, optimisation and points of inflexionLocal maxima and minima and how to test for them; optimisation in context; points of inflexion.

SL 5.9KinematicsDisplacement, velocity and acceleration as derivatives and integrals of one another.

SL 5.10Indefinite integrals; inspection and substitutionIntegrals of xⁿ, sin x, cos x, 1/x and eˣ and their linear composites; reversing the chain rule, and substitution of the form ∫f(g(x))g′(x) dx.

SL 5.11Definite integrals and areas between curvesDefinite integrals evaluated analytically or with technology; areas enclosed by curves.

How it is examined

Calculus is examined on both papers and in a different register on each: exact and symbolic on Paper 1, numerical and in context on Paper 2.

Paper 1 (no calculator) asks for derivatives and integrals by hand — chain, product and quotient rules, integration by inspection or substitution — with exact answers such as 2e² − 1 or ln 3. A decimal is not what the question asks for.
Paper 2 (technology) brings contexts: a rate of change, an optimisation, a definite integral evaluated by the GDC. The calculus is a few lines; building the model, justifying the answer and putting it back in context are the rest.
Trigonometric calculus is done in radians. A GDC left in degree mode quietly gives wrong values for every derivative and integral that involves sin or cos.
Stationary points have to be classified. Solving f′(x) = 0 earns the method mark; stating that the point is a maximum or minimum needs a justification — the sign of f′ either side, or the sign of f″.
An indefinite integral needs its constant. When a boundary condition is given — a point the antiderivative passes through — the constant has to be found, not left as + C.
In kinematics, displacement is the integral of velocity, but distance travelled is the integral of its absolute value. When the particle turns round, the two differ, and the question usually arranges for that.
The formula booklet prints the standard derivatives and the product, quotient and chain rules. What is marked is how accurately you apply them — and whether the inner derivative was remembered.

A marked question, mark by mark

A kinematics question where the integration is correct and the question being answered is not the one that was asked.

Get the finishing habit, on fresh exercises

From SL 5.1 gradients to SL 5.11 areas between curves, every generated exercise comes back mark by mark — so the missing constant, classification or sign analysis shows up immediately, not on results day.

Start your trial

New to the codes? M1, A1, R1 explained →

Kinematics · SL 5.9

Seed 2H76

A particle moves in a straight line with velocity v(t) = 3t² − 12t + 9 m s⁻¹, for 0 ≤ t ≤ 4.

(a)Find the displacement of the particle from t = 0 to t = 4.
(b)Find the total distance travelled in the first 4 seconds.

Your work

s = ∫ v dt = t³ − 6t² + 9t

s(4) − s(0) = 64 − 96 + 36 = 4 m

distance = 4 m

Marks awarded

M1Mark awarded. Integrated the velocity correctly to obtain t³ − 6t² + 9t.
A1Mark awarded. Displacement = 4 m.
R1Mark not awarded. Distance given as 4 m: v changes sign at t = 1 and t = 3, so the distance is found over three intervals and comes to 12 m.

Where marks are actually lost

Calculus answers are finished by a sentence or a constant, not by the last number on the page.

Leaving out the constant of integration on an indefinite integral.

CostThe answer mark goes. When a boundary condition was meant to fix the constant, so do the marks that depended on it.

Stopping at the solution of f′(x) = 0 without saying whether it is a maximum or a minimum.

CostThe reasoning mark exists to separate the two. It is not earned by assertion, only by the sign of f′ or f″.

Forgetting the inner derivative when applying the chain rule.

CostThe derivative is wrong from that line on. Follow-through protects later marks only if the written method is consistent and readable.

Finding distance as a plain integral of velocity when the particle changes direction.

CostThe integral gives displacement. Distance needs the intervals split at the sign changes, and the marks are for that step.