IB DP Mathematics: Applications and Interpretation – Further differentiation and integration, volumes of revolution and kinematics (HL) Revision Notes

For full explanations and longer worked examples, use the further calculus and kinematics study guide.

These notes cover the HL further calculus unit of IB Diploma Programme Mathematics: Applications and Interpretation. They are aligned to the IB Mathematics: applications and interpretation guide, first assessment 2021, syllabus sections 5.9–5.13, and everything here is HL only (AHL). They follow the IB guide for first assessment 2021, which remains the examined syllabus until the new course is first assessed in May 2029, so they apply to the May and November 2026, 2027 and 2028 HL sessions.

Check your coverage against the printable syllabus checklist, see the rest of the course on the IB DP Maths AI course hub, and test yourself with the practice questions.

Key definitions

• Second derivative: f′′(x) or d²y/dx², the derivative of the derivative.
• Concave-up: f′′(x) > 0. Concave-down: f′′(x) < 0.
• Point of inflexion: a point where the concavity changes.
• Related rates: rates of change of linked quantities, found by differentiating their connecting equation with respect to time.
• Displacement (s): position relative to a fixed origin; it can be negative.
• Velocity (v): rate of change of displacement. Speed: the magnitude of velocity, |v|.
• Acceleration (a): rate of change of velocity.
• ẋ and ẍ: dx/dt and d²x/dt².

Formulas

Topic	Result
Powers	d/dx(xⁿ) = nxⁿ⁻¹, n ∈ ℚ
Trigonometric	d/dx(sin x) = cos x; d/dx(cos x) = −sin x; d/dx(tan x) = 1/cos²x
Exponential and log	d/dx(eˣ) = eˣ; d/dx(ln x) = 1/x
Chain rule	dy/dx = (dy/du)(du/dx)
Product rule	d/dx(uv) = u(dv/dx) + v(du/dx)
Quotient rule	d/dx(u/v) = (v(du/dx) − u(dv/dx))/v²
Integrals	∫xⁿ dx = xⁿ⁺¹/(n + 1) + C (n ≠ −1); ∫(1/x) dx = ln|x| + C
Trig integrals	∫sin x dx = −cos x + C; ∫cos x dx = sin x + C; ∫(1/cos²x) dx = tan x + C
Exponential integral	∫eˣ dx = eˣ + C
Volume, x-axis	V = ∫ₐᵇ πy² dx
Volume, y-axis	V = ∫ₐᵇ πx² dy
Kinematics	v = ds/dt; a = dv/dt = d²s/dt² = v(dv/ds)
Displacement	∫ v(t) dt from t₁ to t₂
Total distance	∫ |v(t)| dt from t₁ to t₂

Work in radians for all calculus with trigonometric functions.

Method in steps

Related rates of change

1. Write the equation linking the quantities (for example V = (4/3)πr³).
2. Differentiate both sides with respect to t, using the chain rule.
3. Substitute the known rate and the values at that instant.
4. Solve for the unknown rate and give units. The sign tells you increasing or decreasing.

Classifying stationary points and inflexions

1. Solve f′(x) = 0.
2. Find f′′(x) at each solution: negative → maximum; positive → minimum; zero → test fails, so check the sign of f′(x) either side.
3. For inflexions, solve f′′(x) = 0 and check that f′′ changes sign.

Integration by substitution

1. Spot g(x) whose derivative g′(x) appears (up to a constant).
2. Let u = g(x); write du = g′(x) dx.
3. Rewrite the whole integral in u, including the limits if definite.
4. Integrate, then substitute back (indefinite) or evaluate with the u-limits (definite).

Area with negative parts

1. Find where the curve meets the axis inside the interval.
2. Integrate over each piece separately.
3. Add the absolute values, or use the GDC on ∫|f(x)| dx.

Volume about the y-axis

1. Rearrange to get x² in terms of y.
2. Convert the limits to y-values.
3. Evaluate π∫x² dy.

Total distance travelled

1. Solve v(t) = 0 to find where the direction changes.
2. Integrate v over each interval.
3. Add the absolute values, or use the GDC on ∫|v(t)| dt.

Small worked reminders

• d/dx(e⁻²ˣ cos x) = −2e⁻²ˣ cos x − e⁻²ˣ sin x = −e⁻²ˣ(2 cos x + sin x).
• ∫ sin(5x + 2) dx = −(1/5) cos(5x + 2) + C.
• ∫ 3x²/(x³ + 4) dx = ln|x³ + 4| + C, because the numerator is the derivative of the denominator.
• Region under y = √x for 0 ≤ x ≤ 4, rotated about the x-axis: V = π∫₀⁴ x dx = 8π.
• Region between y = x² and the y-axis for 0 ≤ y ≤ 4, rotated about the y-axis: x² = y, so V = π∫₀⁴ y dy = 8π.
• A circle’s radius grows at 0.5 cm s⁻¹. When r = 6: dA/dt = 2πr(dr/dt) = 2π(6)(0.5) = 6π ≈ 18.8 cm² s⁻¹.
• f(x) = x³ − 3x: f′′(x) = 6x, so the curve is concave-down for x < 0 and concave-up for x > 0, with an inflexion at (0, 0).
• If v = 3 − t, the particle is at rest at t = 3 and changes direction there. At t = 5, v = −2 and speed = 2 m s⁻¹.

Using your GDC in this unit

All three HL papers require technology, but the GDC supports the method rather than replacing it.

• Definite integrals: write the full integral with limits first, for example V = π∫₀² eˣ dx, then evaluate. The written integral usually carries the method mark.
• Total distance: enter ∫|v(t)| dt directly, or find the roots of v(t) from the graph and integrate piece by piece.
• Areas with negative parts: use the graph to find where the curve crosses the axis, then integrate |f(x)|.
• Stationary points and inflexions: the graph’s maximum and minimum tools confirm your answers, but a question asking you to “show” or “justify” needs f′′ and the sign argument written out.
• Checking derivatives: the numerical derivative at a point is a quick check of a chain-rule or quotient-rule answer.
• Accuracy: store intermediate values in the GDC memory and round only the final answer to 3 significant figures.

Must-know distinctions

• Displacement vs distance: displacement is ∫v dt and can be negative or zero; distance is ∫|v| dt and is never negative.
• Velocity vs speed: velocity has a sign for direction; speed is |v|.
• Definite integral vs area: a definite integral is negative where the curve is below the x-axis; an area is always positive.
• Stationary point vs point of inflexion: f′(x) = 0 at a stationary point; concavity changes at an inflexion. A point can be both (for example (0, 0) on y = x³).
• x-axis vs y-axis rotation: πy² dx with x-limits versus πx² dy with y-limits.
• a = dv/dt vs a = v(dv/ds): use the first when v is given in terms of t; use the second when v is given in terms of s.
• Speeding up vs slowing down: same signs of v and a → speeding up; opposite signs → slowing down.

Quick self-test

1. Differentiate y = 5x^(2/5).
2. Differentiate y = ln(3x² + 1).
3. Differentiate y = x²e⁻ˣ and factorise.
4. Differentiate y = (tan x)/x.
5. Find the x-coordinates of the points of inflexion of f(x) = x⁴ − 4x³.
6. Find ∫ 1/cos²(3x) dx.
7. Find ∫ x e^(x²) dx.
8. Find ∫₀² (x − 1)³ dx and the area enclosed by y = (x − 1)³ and the x-axis for 0 ≤ x ≤ 2.
9. The region under y = x for 0 ≤ x ≤ 3 is rotated about the x-axis. Find the exact volume.
10. v = 4 − t² m s⁻¹ for 0 ≤ t ≤ 3. Find the displacement and the total distance travelled.
11. v = 2s + 3. Find the acceleration when s = 1.
12. x = cos 3t. Show that ẍ = −9x.

Answers

1. dy/dx = 2x^(−3/5)
2. dy/dx = 6x/(3x² + 1) (chain rule)
3. dy/dx = 2xe⁻ˣ − x²e⁻ˣ = xe⁻ˣ(2 − x)
4. dy/dx = (x/cos²x − tan x)/x²
5. f′′(x) = 12x² − 24x = 12x(x − 2). f′′ changes sign at both roots, so x = 0 and x = 2.
6. (1/3) tan 3x + C
7. u = x², du = 2x dx, so the integral is (1/2)e^(x²) + C
8. ∫₀² (x − 1)³ dx = 0 (the two halves cancel). Area = 1/4 + 1/4 = 1/2
9. V = π∫₀³ x² dx = 9π
10. Displacement = ∫₀³ (4 − t²) dt = 3 m. v = 0 at t = 2; distance = 16/3 + 7/3 = 23/3 ≈ 7.67 m
11. a = v(dv/ds) = (2s + 3)(2) = 10 m s⁻²
12. ẋ = −3 sin 3t, ẍ = −9 cos 3t = −9x

Where marks are usually lost

• Leaving out the inner derivative in the chain rule, such as d/dx(e^(4x)) written as e^(4x).
• Writing the quotient rule numerator in the wrong order, which flips the sign of the answer.
• Substituting the particular value (for example r = 5) before differentiating in a related-rates question.
• Using f′′(x) = 0 as proof of an inflexion without showing the change in concavity.
• Omitting + C, or writing ln x instead of ln|x| when x could be negative.
• Keeping x-limits after a substitution changes the variable to u.
• Reporting a net definite integral as an area when part of the region is below the axis.
• Using x-limits, or πy², when the rotation is about the y-axis.
• Giving displacement when the question asks for distance travelled, or quoting a negative speed.
• Writing a GDC answer to fewer than 3 significant figures, or rounding early and carrying the error forward.

Official syllabus

International Baccalaureate Organization, Diploma Programme, Mathematics: applications and interpretation guide, first assessment 2021.