IB Maths AA calculus: differentiation, integration and kinematics

What calculus covers at SL and HL

This follows the current Analysis and Approaches guide (first assessment 2021). A new maths course is first assessed in 2029, so students sitting exams in 2027 and 2028 use this content.

The key ideas

The derivative f'(x) is the gradient of the curve at a point and the instantaneous rate of change. Almost every calculus question in AA comes back to one of three readings of it: the gradient of a tangent, where the function increases or decreases (the sign of f'(x)), and where it turns (f'(x) = 0). The second derivative tells you about concavity: f''(x) > 0 means concave up, f''(x) < 0 means concave down, and a point of inflection is where the concavity actually changes, not simply where f''(x) = 0.

Integration reverses differentiation. An indefinite integral needs + C; a definite integral gives a number, which is the signed area between the curve and the x-axis. Area below the axis comes out negative, so when a region crosses the axis you split the integral at the root, or on Paper 2 integrate |f(x)| on the GDC.

Kinematics is the same calculus in a physical setting: velocity v is the derivative of displacement s, and acceleration a is the derivative of v. Displacement is the integral of v; distance travelled is the integral of |v|. The two only agree if the particle never turns round.

Worked example 1: stationary points of f(x) = x^3 - 6x^2 + 9x + 1

1. Differentiate: f'(x) = 3x^2 - 12x + 9.
2. Set f'(x) = 0 and factorise: 3(x^2 - 4x + 3) = 3(x - 1)(x - 3) = 0, so x = 1 or x = 3.
3. Find the y-values: f(1) = 1 - 6 + 9 + 1 = 5 and f(3) = 27 - 54 + 27 + 1 = 1.
4. Classify with the second derivative: f''(x) = 6x - 12. f''(1) = -6 < 0, so (1, 5) is a local maximum. f''(3) = 6 > 0, so (3, 1) is a local minimum.
5. Check: the point of inflection is where f''(x) = 0, at x = 2, which lies between the two turning points, exactly as it should for a cubic.

Worked example 2: differentiate y = x^2 e^(3x) and find where the gradient is zero

1. This is a product u × v with u = x^2 and v = e^(3x). Then u' = 2x and, by the chain rule, v' = 3e^(3x).
2. Product rule: dy/dx = u'v + uv' = 2x e^(3x) + 3x^2 e^(3x).
3. Factorise: dy/dx = x e^(3x)(2 + 3x).
4. Since e^(3x) is never zero, dy/dx = 0 when x = 0 or x = -2/3.
5. Check: at x = 0 the function is y = 0 and is never negative (x^2 and e^(3x) are both non-negative), so x = 0 must be a minimum, which agrees with a zero gradient there.

Worked example 3: area and kinematics

1. Area: find the area enclosed by y = 4x - x^2 and the x-axis. The curve meets the axis where x(4 - x) = 0, at x = 0 and x = 4.
2. Integrate: the integral from 0 to 4 of (4x - x^2) dx = [2x^2 - x^3/3] from 0 to 4 = 32 - 64/3 = 32/3 square units. Check: the area under a parabolic arch is two thirds of base × height, and (2/3) × 4 × 4 = 32/3.
3. Kinematics: a particle has velocity v(t) = 6t - 3t^2 m/s for 0 <= t <= 3. Displacement = integral from 0 to 3 of v dt = [3t^2 - t^3] from 0 to 3 = 27 - 27 = 0 m, so it ends where it started.
4. Distance: v = 3t(2 - t) changes sign at t = 2. From 0 to 2 the integral is 12 - 8 = 4; from 2 to 3 it is 0 - 4 = -4. Distance travelled = 4 + 4 = 8 m.
5. On Paper 2 you can check the distance directly by integrating |6t - 3t^2| from 0 to 3 on the GDC, which also gives 8.

Common mistakes that cost marks

• Forgetting the inner derivative in the chain rule: the derivative of (2x + 1)^5 is 10(2x + 1)^4, not 5(2x + 1)^4.
• Classifying a stationary point by its y-value instead of by f''(x) or a sign change in f'(x).
• Calling every root of f''(x) a point of inflection. Check that the concavity really changes.
• Leaving out + C, or adding it to a definite integral.
• Adding signed areas when a region crosses the x-axis, so parts cancel. Split the integral at the root.
• Confusing displacement with distance in kinematics questions.
• Using degrees on the GDC in calculus. Derivatives of trig functions assume radians.

Exam technique and how a tutor helps

On Paper 1 calculus questions are usually broken into parts that build on each other: differentiate, find the stationary point, then use it to find an area or a range. If you get stuck on part (a), the question often gives you the result to show, so use it and carry on. On Paper 2, decide early whether the GDC can do the work: numerical integrals, intersection points and maximum values of awkward functions are all quicker on the calculator, and the markscheme expects you to use it.

Write the derivative or integral you are evaluating before you press any buttons, because method marks depend on it. Give answers to three significant figures unless the question says otherwise.

In one-to-one lessons a tutor watches the student differentiate on the shared whiteboard and catches the exact step that slips, usually the chain rule or a sign. The fix is a short run of targeted questions repeated over a few lessons until it is automatic, then mixed past-paper questions where calculus is hidden inside a longer problem. HL students also need regular practice on integration by parts and differential equations, which take time to become fluent.

Self-check: can you do these?

• Differentiate y = (2x + 1)^5. (Answer: 10(2x + 1)^4)
• Differentiate y = (ln x)/x. (Answer: (1 - ln x)/x^2)
• Evaluate the integral from 1 to e of 1/x dx. (Answer: 1)
• Find the equation of the tangent to y = x^2 + 3x at x = 1. (Answer: y = 5x - 1)
• Find the integral of cos(2x) dx. (Answer: (1/2) sin(2x) + C)