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IB DP Maths AI · Topic 5 Calculus · SL 5.5, 5.8 · AHL 5.11–5.13 · ~7 min read

Integration, areas and kinematics

Integration as anti-differentiation, boundary conditions, definite integrals and areas with the GDC, the trapezoidal rule for data and functions, and for HL: integrating sin, cos, eˣ and 1/x, integration by inspection, areas between curves and the y-axis, volumes of revolution, and kinematics with displacement and distance.

Anti-differentiation+ C and boundary conditionsDefinite integralsArea under a curveTrapezoidal ruleIntegrating sin, cos, eˣ, 1/x (HL)Integration by inspection (HL)Area with the y-axis (HL)
RSby Riyaz Sir
mentorriyaz.in
No tag = SL and HLHL ONLY HL students onlyFORMULA BOOKLET given in the examM1A1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • Integration reverses differentiation: ∫axn dx = axn + 1n + 1 + C (n ≠ −1). A boundary condition (a known point) fixes C.
  • Definite integrals ∫ab f(x) dx give the area between the curve and the x-axis when f(x) ≥ 0. In AI you evaluate them on the GDC.
  • Trapezoidal rule: split the area into n strips of width h and add trapezia: A ≈ h2[(y0 + yn) + 2(y1 + … + yn − 1)]. It works on data tables too.
  • HL: standard integrals, volumes of revolution V = π∫y2 dx, and kinematics: displacement = ∫v dt, distance = ∫ǀvǀ dt.

By the end you can

  • Find indefinite integrals of axⁿ and use boundary conditions to find C.
  • Find areas under curves using definite integrals and the GDC.
  • Use the trapezoidal rule to approximate areas from functions and data.
  • Integrate sin x, cos x, eˣ, 1/x and composites by inspection. HL
  • Find areas between curves and the x- or y-axis, and volumes of revolution. HL
  • Use calculus in kinematics: displacement, velocity, acceleration and distance. HL
Syllabus noteFrom first exams in 2029, the trapezoidal rule leaves the AI SL syllabus. SL students sitting exams up to November 2028 still need it.

1Anti-differentiation

∫xn dx = xn + 1n + 1 + C, n ≠ −1 FORMULA BOOKLET
add 1 to the power, divide by the new power
Worked example

Find ∫(3x2 − 4x + 5) dx.

  1. x3 − 2x2 + 5x + C A1A1
Worked example

The gradient of a curve is f′(x) = 6x − 2, and the curve passes through (1, 4). Find f(x).

  1. f(x) = 3x2 − 2x + C A1
  2. 3 − 2 + C = 4 ⇒ C = 3 M1
  3. f(x) = 3x2 − 2x + 3 A1
Common mistakeForgetting + C on an indefinite integral. It costs a mark, and without it you can't use the boundary condition.

2Definite integrals and area

xy12341234area = 32/3 ≈ 10.7y = 4x − x²
The area between y = 4x − x² and the x-axis, from one root to the other.
Worked example

Find the area enclosed by y = 4x − x2 and the x-axis.

  1. roots x = 0 and x = 4 A1
  2. area = ∫04 (4x − x2) dx M1
  3. = 323 ≈ 10.7 (GDC) A1
  • Below the axis: an integral is negative where the curve is below the x-axis. For a total area, use ∫ǀf(x)ǀ dx on the GDC.
GDC (calculator) methodUse the GDC's ∫f(x)dx (fnInt) tool. Write the integral with its limits in your working, then the value: "∫04 (4x − x2) dx = 10.7".

3The trapezoidal rule

∫ab y dx ≈ h2[(y0 + yn) + 2(y1 + y2 + … + yn − 1)], h = b − an FORMULA BOOKLET
n strips of equal width h
xy012312y = √(x + 1)3 strips, h = 1trapezia ≈ 4.646exact = 4.667
Three trapezia under y = √(x + 1). They sit just under the curve, so the estimate is a little too small.
Worked example

Use the trapezoidal rule with 3 strips to estimate ∫03 √(x + 1) dx, and find the percentage error.

  1. h = 1; y-values 1, 1.414, 1.732, 2 A1
  2. 12[(1 + 2) + 2(1.414 + 1.732)] = 4.65 (4.6463) M1A1
  3. exact = 4.667 (GDC), so ε = ǀ4.6463 − 4.6667ǀ4.6667 × 100% = 0.437% M1A1
Worked example

A river is 4 m wide. Its depths at 1 m intervals across are 0, 1.2, 2.1, 1.8 and 0 m. Estimate the cross-sectional area.

  1. h = 1 A1
  2. 12[(0 + 0) + 2(1.2 + 2.1 + 1.8)] M1
  3. = 5.1 m2 A1
Exam tipOver or under? If the curve is concave down (bulging upwards), the trapezia lie under it and underestimate the area; if it is concave up, they overestimate. Use more strips for a better estimate.

4Standard integrals and integration by inspection HL

f(x)∫ f(x) dx
xn (n ≠ −1)xn + 1n + 1 + C
1xln x + C (x > 0)
sin x−cos x + C
cos xsin x + C
exex + C
1cos2xtan x + C
  • By inspection (reverse chain rule): ∫e3x dx = ⅓e3x + C; ∫cos(2x) dx = ½ sin(2x) + C; ∫2xx2 + 1 dx = ln(x2 + 1) + C, because the top is the derivative of the bottom.

5Areas and volumes HL

V = ∫ab πy2 dx (about the x-axis), V = ∫ab πx2 dy (about the y-axis) FORMULA BOOKLET
volume of revolution
Worked example

The region under y = x2 + 1 from x = 0 to x = 2 is rotated 360° about the x-axis. Find the volume.

  1. V = π∫02 (x2 + 1)2 dx M1A1
  2. = 43.1 (GDC; exactly 206π15) A1
  • Area with the y-axis: write x in terms of y and integrate with respect to y between y-limits: A = ∫cd x dy.

6Kinematics HL

v = dsdt, a = dvdt; displacement = ∫t_{1}t2 v dt, distance = ∫t_{1}t2 ǀvǀ dt FORMULA BOOKLET
differentiate down, integrate up
t (s)v (m/s)1234-3369+4−4+4v = 3t² − 12t + 9: displacement = 4 + (−4) + 4 = 4 mdistance = 4 + 4 + 4 = 12 m
Area above the t-axis is motion forward; below is motion backward. Distance counts both as positive.
Worked example

A particle has velocity v = 3t2 − 12t + 9 m s−1 for 0 ≤ t ≤ 4. Find its displacement, the total distance travelled, and its acceleration at t = 3.

  1. ∫04 v dt = 4 m M1A1
  2. ∫04 ǀvǀ dt = 12 m (v = 0 at t = 1 and 3) M1A1
  3. a = 6t − 12, so a(3) = 6 m s−2 A1
Linking questionIntegration reverses Differentiation and optimisation, and it is the method for solving Differential equations and phase portraits. The percentage error formula comes from Approximation, standard form and errors. For integration by hand in Maths AA, see Integration and kinematics.
Idea for your IA / explorationEstimate the area of a real shape (a lake on a map, a leaf, a running track) with the trapezoidal rule, then improve it by fitting a function and integrating, and compare the two methods.

Test yourself — cover the answers first

Quick check

  1. Find ∫6x² dx.
  2. Find ∫(2x + 1) dx.
  3. Evaluate ∫13 2x dx.
  4. With h = 0.5 and 4 strips, what are a and b if a = 0?
  5. Is the trapezoidal rule an over- or underestimate for a concave-down curve?
  6. HL Find ∫e2x dx.

Exam-style

  1. The rate of flow of water into a tank is r(t) = 40 − 2t litres per minute for 0 ≤ t ≤ 20. The tank starts with 50 litres. Find the volume after 10 minutes.
  2. A car's speed (m s⁻¹) is recorded every 2 s: 0, 6, 10, 13, 15. Use the trapezoidal rule to estimate the distance travelled in 8 s.
Show answers
  1. 2x³ + C
  2. x² + x + C
  3. 8
  4. b = 2
  5. an underestimate
  6. ½e2x + C
  7. V = 50 + ∫010 (40 − 2t) dt = 50 + [40t − t2]010
    = 50 + 300 = 350 litres
  8. h = 2
    22[(0 + 15) + 2(6 + 10 + 13)] = 15 + 58
    ≈ 73 m
Practise this chapter
IB Maths practice papers
Next chapterDifferential equations and phase portraits →
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Independent notes written by Riyaz Sir. Not produced, endorsed or approved by the International Baccalaureate Organization. Syllabus codes refer to the IB DP subject guides.