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?

Your work is transcribed, then evaluated against the exercise's mark scheme one mark at a time. Each Method, Answer, and Reasoning mark is awarded or withheld with a written reason. Follow-through is applied, so an early error doesn't automatically cost you the marks that depend on it.

Can the AI get the maths wrong?

The maths isn't generated by AI. Every number in every exercise is computed deterministically before the AI is involved — the AI only writes the wording around values it's been given, and is blocked from producing numbers of its own.

Do I have to write by hand?

You don't have to, but the whiteboard is where the marking is worth most. Writing out formula, substitution, and answer lets each method mark be awarded or withheld on its own. Typing takes final answers only — quicker, but there's no working to mark, and nothing to review afterwards.

What do I need to use it?

A tablet with a stylus is the best experience — the whiteboard is built for it. A laptop or desktop works fully too. We wouldn't recommend a phone: the whiteboard needs more room than a phone screen gives it.

Seven days of full access — every chapter of your course, unlimited generated exercises, and per-mark feedback on the work you write out. No card required to start.

What happens when the trial ends?

If you'd like to continue afterwards, leave your details at ascentiq.io/continue or write to renzo@ascentiq.io, and you'll hear back by email — including about the asynchronous tutoring we're building on top of your results.

What is the asynchronous tutoring?

It's a tutoring method built on your results in the exercise mode — where your marks are lost, which topics are slipping, what to work on next — followed up at your own pace, without having to coordinate a live session. It's still in development, so there's no fixed format yet. If you'd like to hear how it will work, write to renzo@ascentiq.io.

← IB Math AI HL

IB Math AI HL · Topic 5

Calculus

At SL calculus is applied to polynomials: optimisation and area. At HL it is deeper and wider — the chain, product and quotient rules, integration by substitution, volumes of revolution — and then it opens onto differential equations: separating variables, slope fields, Euler's method and coupled systems whose behaviour is read from eigenvalues.

What the syllabus covers

Eighteen assessed sections: SL 5.1 to 5.8, shared with AI SL, and AHL 5.9 to 5.18, which are HL only.

SL 5.1Limits and the derivativeThe limit concept introduced informally; the derivative interpreted as a gradient function and as a rate of change.

SL 5.2Increasing and decreasingReading f′(x) > 0, f′(x) = 0 and f′(x) < 0 off a graph, and what each means for the function itself.

SL 5.3Differentiating polynomialsFunctions of the form f(x) = axⁿ + bxⁿ⁻¹ + …, where all exponents are integers.

SL 5.4Tangents and normalsTheir equations at a given point — gradient from the derivative, line through the point with substitution shown.

SL 5.5IntegrationAnti-differentiation of axⁿ-type functions (n ≠ −1); boundary conditions to pin down the constant; definite integrals by technology; area under a curve above the x-axis.

SL 5.6Stationary pointsSolving f′(x) = 0; identifying local maxima and minima among the solutions.

SL 5.7Optimisation in contextWord problems that run model → differentiate → solve → interpret, with units and a conclusion in terms of the scenario.

SL 5.8Trapezoidal ruleApproximating areas under curves when integration is not available — table of values, interval width h, weighted sum.

AHL 5.9Chain, product and quotient rules; related ratesDerivatives of powers, sine, cosine, tangent, eˣ and ln x by hand, and rates of change linked through a relationship.

AHL 5.10The second derivativeConcavity, points of inflexion, and the second-derivative test.

AHL 5.11Integration by inspection and substitutionReversing the chain rule, and substitution of the form ∫f(g(x))g′(x) dx.

AHL 5.12Areas and volumes of revolutionAreas against the y-axis, and volumes of revolution about the x-axis and the y-axis.

AHL 5.13KinematicsDisplacement, velocity and acceleration by differentiation and integration, and total distance.

AHL 5.14Differential equations and separation of variablesSetting up an equation from a context and solving it by separating the variables.

AHL 5.15Slope fieldsReading the direction of solution curves, and the equation of an isocline.

AHL 5.16Euler's methodApproximating a solution step by step, for one equation and for coupled systems.

AHL 5.17Phase portraits of coupled equationsEquilibrium points, and their type — stable or unstable node, saddle, spiral, centre — from the eigenvalues.

AHL 5.18Second-order equations with Euler's methodRewriting a second-order equation as a pair of first-order equations, then stepping forward.

How it is examined

Calculus at HL is examined as technique and as modelling. The technique is by hand where an exact answer is asked for.

From AHL 5.9 differentiation and integration go beyond polynomials: rational powers, sin, cos, tan, eˣ and ln x, done by hand. The GDC verifies; it does not replace the working when an exact answer is asked for.
Related rates connect a relationship to time: write the relationship, differentiate with respect to t, substitute. Units and the sign of the rate — increasing or decreasing — belong to the answer.
A point of inflexion needs a change of concavity, not just f″(x) = 0. The sign of f″ either side is the justification.
A volume of revolution is π∫y² dx about the x-axis, and π∫x² dy about the y-axis. Which variable you integrate with respect to, and the limits in that variable, are most of the working.
A differential equation from a context is built from a sentence: write dy/dx, separate, integrate both sides, and use the initial condition to find the constant.
Euler's method is a written table. Each step uses the previous values: xₙ₊₁ = xₙ + h and yₙ₊₁ = yₙ + h f(xₙ, yₙ). For a coupled system, both components are updated from the old values.
For coupled linear systems the equilibrium is where both derivatives are zero, and its type is read from the eigenvalues of the system's matrix.

A marked question, mark by mark

An Euler's-method question where the steps are right and the percentage error is measured against the wrong value.

Practise the whole chain, on fresh exercises

From SL 5.1 to AHL 5.18, every generated exercise comes back mark by mark, with a reason for each — so a quietly shifted step in Euler's method shows up now, not on results day.

Start your trial

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

Euler's method · AHL 5.16

Seed 1V63

Consider dy/dx = x + y with y(0) = 1. Use Euler's method with step h = 0.1 to estimate y(0.2). The exact value is y(0.2) = 1.24281, to five decimal places.

(a)Find the Euler estimate of y(0.2).
(b)Find the percentage error of the estimate, correct to three significant figures.

Your work

y₁ = 1 + 0.1(0 + 1) = 1.1

y₂ = 1.1 + 0.1(0.1 + 1.1) = 1.22

error = (1.24281 − 1.22) / 1.22 × 100 = 1.87%

Marks awarded

M1Mark awarded. Euler's formula applied with the previous x and y values at each step.
A1Mark awarded. y(0.2) ≈ 1.22.
A1Mark not awarded. The error was divided by the estimate, 1.22. A percentage error is relative to the exact value: 0.02281 / 1.24281 = 1.84%.

Where marks are actually lost

At HL the working is longer, and the marks are lost where one line of it quietly changes.

Dividing by the approximation instead of the exact value when finding a percentage error.

CostThe percentage is for a different quantity. The method mark may stay; the accuracy mark does not.

Using the new x, or the wrong step, inside the slope in an Euler step.

CostEuler's method evaluates the slope at the start of each step. A shifted value changes every later line.

Calling a point an inflexion because f″(x) = 0.

CostA zero of the second derivative is only a candidate. The reasoning mark is for the change of concavity either side.

Dropping the constant of integration after separating the variables.

CostThe general solution is incomplete, so there is nothing for the initial condition to fix, and the marks that follow are lost.