Numerical integration: 6 exam questions solved step by step

← Previous11 of 12Next →

What this lesson covers

  • When a function cannot be integrated, or only data is given, the area is estimated from ordinates.
  • The strip width h is the interval width divided by the number of strips, not the number of ordinates.
Key idea

6 questions, one method

  • 1. h = (b − a) ÷ number of strips.
  • 2. Every ordinate, to 4 d.p.
  • 3. The booklet rule, with weights.
  • 4. Round only at the end.
Your turn

Question 1 · Achieved

  • Use the trapezium rule with 4 strips
  • to estimate ∫₀² √(1 + x³) dx.
Worked example

Q1: h and the ordinates

  • 4 strips: 5 ordinates.
  • Write each one down.
  1. h = (2 − 0)/4 = 0.54 strips
  2. x = 0, 0.5, 1, 1.5, 25 ordinates
  3. y₀ = √1 = 1
  4. y₁ = √1.125 = 1.0607
  5. y₂ = √2 = 1.4142
  6. y₃ = √4.375 = 2.0917
  7. y₄ = √9 = 3
Worked example

Q1: the trapezium rule

  • Ends once, middle twice.
  • Then multiply by h/2.
  1. y: 1, 1.0607, 1.4142, 2.0917, 3
  2. A ≈ (h/2)[ends + 2 × middle]
  3. h/2 = 0.25
  4. ends: 1 + 3 = 4
  5. middle: 1.0607 + 1.4142 + 2.0917
  6. = 4.5666
  7. bracket = 4 + 2(4.5666) = 13.1332
  8. A ≈ 0.25 × 13.1332 = 3.2833
  9. ≈ 3.283 s.f.
Your turn

Question 2 · Achieved

  • Now use Simpson's rule with 4 strips
  • for the same ∫₀² √(1 + x³) dx.
Worked example

Q2: Simpson's rule

  • Odd ordinates × 4.
  • Even ordinates × 2.
  1. y: 1, 1.0607, 1.4142, 2.0917, 3
  2. A ≈ (h/3)[ends + 4 odd + 2 even]
  3. h/3 = 0.5/3 = 1/6
  4. ends: 1 + 3 = 4
  5. 4(1.0607 + 2.0917) = 12.6096y₁, y₃
  6. 2(1.4142) = 2.8284y₂
  7. bracket = 4 + 12.6096 + 2.8284
  8. A ≈ 19.4380 ÷ 6 = 3.2397
  9. ≈ 3.243 s.f.
Your turn

Question 3 · Merit

  • Use Simpson's rule with 5 ordinates
  • to estimate ∫₀¹ e^(−x²) dx.
  • Careful: 5 ordinates is not 5 strips.
Worked example

Q3: strips, not ordinates

  • Strips = ordinates − 1.
  • x² first, then e.
  1. h = (1 − 0)/5 = 0.25 is the ordinates
  2. h = (1 − 0)/4 = 0.254 strips
  3. x = 0, 0.25, 0.5, 0.75, 1
  4. y₀ = e⁰ = 1
  5. y₁ = e^(−0.0625) = 0.9394
  6. y₂ = e^(−0.25) = 0.7788
  7. y₃ = e^(−0.5625) = 0.5698
  8. y₄ = e^(−1) = 0.3679
Worked example

Q3: Simpson's rule

  • Weights 1, 4, 2, 4, 1.
  • Then multiply by h/3.
  1. h = 0.25, so h/3 = 1/12
  2. ends: 1 + 0.3679 = 1.3679y₀ + y₄
  3. 4(0.9394 + 0.5698) = 6.0368y₁, y₃
  4. 2(0.7788) = 1.5576y₂
  5. bracket = 8.9623
  6. A ≈ 8.9623 ÷ 12 = 0.7469
  7. ≈ 0.7473 s.f.
  8. h = 0.2 gives 0.597wrong h
Your turn

Question 4 · Merit, in context

  • Depths across a 12 m wide stream,
  • every 2 m. Use Simpson's rule to
  • estimate the cross-section area.
Across (m)Depth (m)
00
20.8
41.5
61.9
81.6
100.9
120
Worked example

Q4: the stream

  • 7 depths: 6 strips, even.
  • Area in m².
  1. h = 2, so h/3 = 2/36 strips
  2. ends: 0 + 0 = 0banks
  3. 4(0.8 + 1.9 + 0.9) = 14.4x = 2, 6, 10
  4. 2(1.5 + 1.6) = 6.2x = 4, 8
  5. bracket = 0 + 14.4 + 6.2 = 20.6
  6. A ≈ (2/3)(20.6) = 13.733
  7. about 13.7 m²3 s.f.
Your turn

Question 5 · Excellence

  • Use the trapezium rule with 4 strips
  • to estimate ∫₀^π sin x dx.
  • Compare it with the exact value,
  • and explain the difference.
Question 5 · Excellence
Worked example

Q5: the estimate

  • h = π/4, in radians.
  • Exact values first.
  1. h = (π − 0)/4 = π/44 strips
  2. y: 0, √2/2, 1, √2/2, 0sin x, radians
  3. ends: 0 + 0 = 0
  4. middle: √2/2 + 1 + √2/2 = 1 + √2
  5. A ≈ (π/8)[0 + 2(1 + √2)]h/2 = π/8
  6. = π(1 + √2)/4 ≈ 1.8961
  7. ≈ 1.903 s.f.
Worked example

Q5: compare and explain

  • Integrate exactly.
  • Why under, not over?
  1. trapezium ≈ 1.8961
  2. ∫₀^π sin x dx = [−cos x]₀^πexact
  3. = −cos π − (−cos 0)
  4. = 1 + 1 = 2
  5. 1.8961 < 2: an under-estimate
  6. error ≈ 0.104, about 5.2%
  7. concave down on 0 < x < π(sin x)″ = −sin x < 0
  8. strip tops sit below the curve
Your turn

Question 6 · Excellence, in context

  • Water flows into a farm tank at r L/min.
  • Estimate the volume in 50 minutes,
  • using Simpson's rule where you can.
t (min)r (L/min)
00
1040
2064
3072
4060
5036
Worked example

Q6: count the strips

  • 6 readings: 5 strips.
  • Simpson's needs even.
  1. volume = area under the r graph
  2. 6 readings, so 5 stripsh = 10 min
  3. Simpson's on all 5 stripsneeds an even number
  4. Simpson's for t = 0 to 404 strips
  5. trapezium for t = 40 to 50the last strip
Worked example

Q6: two rules, then add

  • Weights 1, 4, 2, 4, 1.
  • Litres = L/min × min.
  1. h = 10, so h/3 = 10/3
  2. ends: 0 + 60 = 60t = 0, 40
  3. 4(40 + 72) = 448t = 10, 30
  4. 2(64) = 128t = 20
  5. (10/3)(636) = 2120 LSimpson's, 0 to 40
  6. (10/2)(60 + 36) = 480 Ltrapezium, 40 to 50
  7. V ≈ 2120 + 480 = 2600 L
  8. about 2600 litres in 50 min
Summary

The method that solved all six

  • h from the strips, every ordinate, the right weights.
StepHow
Strip widthh = (b − a) ÷ strips
Ordinatesone more than the strips
Trapezium(h/2)[ends + 2 × middle]
Simpson's(h/3)[ends + 4 odd + 2 even]
Odd stripsSimpson's on an even part
Accuracyround only at the end

More in AS 91579

The integration toolkit: surds, e, trig and the chain rule6 minIntegrating fractions: when it is ln, and when it is not6 minThe constant of integration: find c, never assume it6 minDefinite integrals and area, including between two curves6 minTrapezium and Simpson's rules: numerical areas6 minDifferential equations: growth, decay and Newton's cooling6 minIntegration toolkit: 7 exam questions solved step by step10 minIntegrating fractions: 7 exam questions solved step by step10 minConstant of integration: 7 exam questions solved step by step11 minDefinite integrals: 7 exam questions solved step by step11 minNumerical integration: 6 exam questions solved step by step11 minDifferential equations: 6 exam questions solved step by step11 min

Is your child on track?

Take the free placement test