The shape of the lesson
By the end of the lesson, learners can
- The rule: widths, ordinates, weights: Split \([a, b]\) into \(n\) equal intervals (strips).
- Over-estimate or under -estimate?: Sketch the curve over the interval and look at the chords.
- Radians and more strips: For trigonometric functions every \(x\)-value is in radians, so set the calculator to radian mode first.
- Width: \(h = \dfrac{b-a}{n}\) with \(n\) intervals.
- Table: \(n + 1\) ordinates to 4 or 5 significant figures; radians for trig.
- Weights: ends once, middles twice, times \(\dfrac{h}{2}\).
- Sketch: bends up gives an over-estimate, bends down an under-estimate.
- Round only at the end, to the accuracy asked.
One picture every idea returns to
With \(n\) intervals of width \(h = \dfrac{b-a}{n}\) and ordinates \(y_0, y_1, \ldots, y_n\): \(\displaystyle\int_a^b y\,dx \approx \dfrac{h}{2}\left[y_0 + 2(y_1 + \cdots + y_{n-1}) + y_n\right]\). If the curve bends upwards (lies below its chords) the rule over-estimates; if it bends downwards it under-estimates.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Straight lines under a curve.. Paper 2 only (not examined in Paper 3). Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: The rule: widths, ordinates, weights | Split \([a, b]\) into \(n\) equal intervals (strips). Model the worked example A table of values on the board, one line at a time. Your turn: Use the trapezium rule with 3 intervals to estimate \(\displaystyle\int_0^3 \sqrt{x+1}\,dx\), to 3 significant figures. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Over-estimate or under -estimate? | Sketch the curve over the interval and look at the chords. Model the worked example Compare with the exact value on the board, one line at a time. | Pairs |
| 21 to 30 min | Teach and check: Radians and more strips | For trigonometric functions every \(x\)-value is in radians, so set the calculator to radian mode first. Model the worked example Halving the width on the board, one line at a time. | Frayer Model |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 29 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Radians and more strips, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Frayer Model
Pairs fill a four-box grid for the key term: definition, characteristics, examples and non-examples. The non-examples box is where the misconceptions surface, so give it the most time.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| \(n\) intervals have \(n + 1\) ordinates. | With 4 intervals on \([0, 2]\), \(h = 0.5\), not \(\dfrac{2}{5}\). Equally, do not forget the \(\dfrac{h}{2}\) at the end, or the answer comes out twice as big. |
| Do not judge by whether the curve is going up or down: \(y = \ln x\) is increasing, but it bends downwards, so the rule under-estimates. | It is the bending (concavity) that matters, not the slope. |
| In degree mode \(\sin(0.785\ldots)\) is about 0.0137, not 0.707, and the estimate collapses. | If a trigonometric estimate looks tiny, check the calculator mode. |
The trapezium rule: worksheet
6 questions, 29 marks. Written in the style of the papers; no past paper question is reproduced.
Use the trapezium rule with 3 intervals to estimate \(\displaystyle\int_0^3 \sqrt{x+1}\,dx\), giving your answer to 3 significant figures.
(a) Use the trapezium rule with 4 intervals to estimate \(\displaystyle\int_0^1 e^{x^2}\,dx\), giving your answer to 3 significant figures.
(b) State, with a reason, whether your estimate is an over-estimate or an under-estimate.
(a) Use the trapezium rule with 2 intervals to estimate \(\displaystyle\int_1^2 \ln x\,dx\), giving your answer to 3 decimal places.
(b) The exact value is \(2\ln 2 - 1\). Find the percentage error in your estimate and explain why the estimate is too small.
(a) Use the trapezium rule with 4 intervals to estimate \(\displaystyle\int_0^1 \dfrac{4}{1+x^2}\,dx\), giving your answer to 3 significant figures.
(b) (Paper 3) Show that the exact value of the integral is \(\pi\), and hence comment on the accuracy of your estimate.
(a) Use the trapezium rule with 4 intervals to estimate \(\displaystyle\int_0^{\pi} \sin x\,dx\), giving your answer in the form \(k\pi\left(1 + \sqrt{2}\right)\).
(b) Explain, with a sketch, why the estimate is less than the exact value 2.
A cyclist starts from rest. Her speed \(v\) m s-1 at time \(t\) seconds is recorded:
| \(t\) | 0 | 5 | 10 | 15 | 20 | 25 | 30 |
|---|---|---|---|---|---|---|---|
| \(v\) | 0 | 4.2 | 6.8 | 8.0 | 8.6 | 8.9 | 9.0 |
(a) Use the trapezium rule with all the data to estimate the distance she travels in the 30 seconds.
(b) The graph of \(v\) against \(t\) bends downwards throughout. State whether your answer is likely to be an over-estimate or an under-estimate.
Answer key and marking notes
- \(h = 1\); \(y = 1,\ 1.41421,\ 1.73205,\ 2\).
- \(\dfrac{1}{2}\left[1 + 2 + 2(1.41421 + 1.73205)\right] = \dfrac{1}{2}\times 9.29253 = 4.646\).
- Estimate: 4.65 (3 s.f.).
- (a) \(h = 0.25\); \(y = 1,\ 1.06449,\ 1.28403,\ 1.75505,\ 2.71828\).
- \(0.125\left[1 + 2.71828 + 2(1.06449 + 1.28403 + 1.75505)\right] = 0.125\times 11.92542 = 1.4907\). Estimate: 1.49.
- (b) The curve \(y = e^{x^2}\) bends upwards on \([0, 1]\), so the chords lie above it: an over-estimate.
- (a) \(h = 0.5\); \(y = 0,\ 0.40547,\ 0.69315\).
- \(0.25\left[0 + 0.69315 + 2(0.40547)\right] = 0.25\times 1.50409 = 0.376\).
- (b) Exact \(= 0.38629\). Error \(= \dfrac{0.38629 - 0.37602}{0.38629}\times 100\% = 2.66\%\).
- \(y = \ln x\) bends downwards, so each chord lies below the curve and the trapezia miss some area.
- (a) \(h = 0.25\); \(y = 4,\ 3.76471,\ 3.2,\ 2.56,\ 2\).
- \(0.125\left[4 + 2 + 2(3.76471 + 3.2 + 2.56)\right] = 0.125\times 25.04941 = 3.1312\). Estimate: 3.13.
- (b) \(\displaystyle\int_0^1 \dfrac{4}{1+x^2}\,dx = \left[4\tan^{-1}x\right]_0^1 = 4\times\dfrac{\pi}{4} = \pi\).
- \(\pi = 3.14159\), so the estimate is too small by about 0.010, an error of about 0.33%.
- (a) \(h = \dfrac{\pi}{4}\); \(y = 0,\ \dfrac{\sqrt{2}}{2},\ 1,\ \dfrac{\sqrt{2}}{2},\ 0\).
- \(\dfrac{\pi}{8}\left[0 + 0 + 2\left(\dfrac{\sqrt{2}}{2} + 1 + \dfrac{\sqrt{2}}{2}\right)\right] = \dfrac{\pi}{8}\times 2\left(1 + \sqrt{2}\right) = \dfrac{\pi}{4}\left(1 + \sqrt{2}\right)\), so \(k = \dfrac{1}{4}\) (value 1.90).
- (b) On \([0, \pi]\) the sine curve bends downwards, so every chord lies below the arch: the trapezia leave out the area between chord and curve, giving an under-estimate.
- (a) Distance \(= \displaystyle\int_0^{30} v\,dt\); 6 intervals, \(h = 5\).
- Ends: \(0 + 9.0 = 9.0\). Middles doubled: \(2(4.2 + 6.8 + 8.0 + 8.6 + 8.9) = 73\).
- \(\dfrac{5}{2}(9.0 + 73) = 2.5\times 82 = 205\) m.
- (b) The curve bends downwards, so the chords lie below it: 205 m is an under-estimate.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
(Paper 2 only) Use the trapezium rule with 2 intervals to estimate \(\displaystyle\int_1^3 \dfrac{1}{x}\,dx\), giving your answer correct to 3 significant figures.
- A\(1.17\)
- B\(2.33\)
- C\(0.917\)
- D\(0.583\)
(Paper 2 only) Use the trapezium rule with 4 intervals to estimate \(\displaystyle\int_0^1 \sqrt{1 + x^3}\,dx\), giving your answer correct to 3 significant figures.
- A\(1.12\)
- B\(2.23\)
- C\(0.709\)
- D\(0.894\)
(Paper 2 only) The diagram shows \(y = \sqrt{x}\) with the 4 trapezia used to estimate \(\displaystyle\int_0^4 \sqrt{x}\,dx\). Which statement is correct?
- AThe estimate is an under-estimate, because each chord lies below the curve.
- BThe estimate is an over-estimate, because each chord lies above the curve.
- CThe estimate is exact, because \(\sqrt{x}\) is increasing.
- DThe estimate is an over-estimate, because \(\sqrt{x}\) is increasing.
(Paper 2 only) The speed of a cyclist, \(v\) m s-1, is recorded every 5 seconds: at \(t = 0, 5, 10, 15, 20\) the speeds are \(0, 8, 14, 18, 20\). Use the trapezium rule to estimate the distance travelled in the 20 seconds.
- A250 m
- B500 m
- C150 m
- D50 m
(Paper 2 only) Use the trapezium rule with 3 intervals to estimate \(\displaystyle\int_0^{\frac{\pi}{2}} \sqrt{1 + \cos x}\,dx\), giving your answer correct to 3 significant figures.
- A\(1.99\)
- B\(2.22\)
- C\(3.98\)
- D\(1.31\)
(Paper 2 only) Use the trapezium rule with 4 intervals to estimate \(\displaystyle\int_0^2 \dfrac{x}{1 + x}\,dx\), and state whether it is an over-estimate or an under-estimate. The curve \(y = \dfrac{x}{1 + x}\) bends downwards for \(x \ge 0\).
- A\(0.883\), an under-estimate
- B\(0.883\), an over-estimate
- C\(1.77\), an over-estimate
- D\(0.525\), an under-estimate
(Paper 2 only) Given that \(\displaystyle\int_0^1 \dfrac{1}{1 + x^2}\,dx = \dfrac{\pi}{4}\), use the trapezium rule with 2 intervals to find an approximation for \(\pi\), correct to 3 significant figures.
- A\(3.10\)
- B\(0.775\)
- C\(6.20\)
- D\(2.30\)
(Paper 2 only) Use the trapezium rule with 3 intervals to estimate \(\displaystyle\int_1^4 \ln x\,dx\). Hence estimate \(\displaystyle\int_1^4 \ln(x^2)\,dx\). Give both answers correct to 3 significant figures.
- A\(2.48\) and \(4.97\)
- B\(2.48\) and \(2.48\)
- C\(4.97\) and \(9.94\)
- D\(2.48\) and \(6.17\)
Quiz key
\(y_0 = 1\), \(y_1 = 0.5\), \(y_2 = \dfrac{1}{3}\). Estimate \(= \dfrac{1}{2}\left[1 + \dfrac{1}{3} + 2(0.5)\right] = \dfrac{1}{2} \times \dfrac{7}{3} = 1.17\). (The exact value is \(\ln 3 = 1.10\).) The tempting 2.33 forgets the \(\dfrac{h}{2}\). 0.917 does not double the middle ordinate, and 0.583 uses \(h = 0.5\).
\(y = 1, 1.00778, 1.06066, 1.19242, 1.41421\). Estimate \(= \dfrac{0.25}{2}\left[1 + 1.41421 + 2(1.00778 + 1.06066 + 1.19242)\right] = 0.125 \times 8.93594 = 1.12\). The tempting 2.23 uses \(h\) in place of \(\dfrac{h}{2}\) and then doubles again (it is twice the answer). 0.709 does not double the middle ordinates, and 0.894 uses \(h = 0.2\) by dividing by the 5 ordinates instead of the 4 intervals.
\(y = \sqrt{x}\) bends downwards, so each chord joining two points of the curve lies BELOW the curve. The trapezia miss the thin slivers between the chords and the curve, so the total area of the trapezia is smaller than the true area: an under-estimate. (Exact value \(\dfrac{16}{3} = 5.33\); the rule gives \(\dfrac{1}{2}[0 + 2 + 2(1 + \sqrt{2} + \sqrt{3})] = 5.15\).) The tempting "over-estimate" reasons from whether the curve is rising, but what matters is which way it bends, not whether it increases.
Distance \(\approx \dfrac{5}{2}\left[0 + 20 + 2(8 + 14 + 18)\right] = 2.5 \times 100 = 250\) m. The tempting 500 m forgets the \(\dfrac{1}{2}\). 150 m adds all the speeds once each without doubling the middle ones, and 50 m uses \(h = 1\) instead of the 5-second interval.
Ordinates: \(\sqrt{2} = 1.41421\), \(1.36603\), \(1.22474\), \(1\). Estimate \(= \dfrac{\pi}{12}\left[1.41421 + 1 + 2(1.36603 + 1.22474)\right] = \dfrac{\pi}{12} \times 7.59575 = 1.99\). (The exact value is 2, so this is a slight under-estimate.) The tempting 2.22 evaluates \(\cos x\) in degrees. 3.98 uses \(h\) in place of \(\dfrac{h}{2}\), and 1.31 does not double the middle ordinates.
\(y = 0, \dfrac{1}{3}, \dfrac{1}{2}, \dfrac{3}{5}, \dfrac{2}{3}\). Estimate \(= \dfrac{0.5}{2}\left[0 + \dfrac{2}{3} + 2\left(\dfrac{1}{3} + \dfrac{1}{2} + \dfrac{3}{5}\right)\right] = 0.25 \times 3.5333 = 0.883\). A downward-bending curve lies ABOVE its chords, so the trapezia miss area: an under-estimate (the exact value is \(2 - \ln 3 = 0.901\)). The tempting "over-estimate" gets the bending rule the wrong way round. 1.77 forgets the \(\dfrac{1}{2}\), and 0.525 does not double the middle ordinates.
\(y = 1, 0.8, 0.5\) at \(x = 0, 0.5, 1\). Estimate \(= \dfrac{0.5}{2}[1 + 0.5 + 2(0.8)] = 0.25 \times 3.1 = 0.775\). So \(\dfrac{\pi}{4} \approx 0.775\) and \(\pi \approx 3.10\). The tempting 0.775 stops at the integral without multiplying by 4. 6.20 forgets the \(\dfrac{1}{2}\) in \(\dfrac{h}{2}\), and 2.30 does not double the middle ordinate (\(4 \times 0.25 \times 2.3\)).
\(\dfrac{1}{2}\left[0 + \ln 4 + 2(\ln 2 + \ln 3)\right] = \dfrac{1}{2}(1.38629 + 3.58352) = 2.48\). Since \(\ln(x^2) = 2\ln x\), the second integral is twice the first: \(4.97\). (Exact first value: \(4\ln 4 - 3 = 2.55\), so 2.48 is an under-estimate, as \(\ln x\) bends downwards.) The tempting "2.48 and 2.48" ignores the power. "4.97 and 9.94" forgets the \(\dfrac{1}{2}\) in the rule, and 6.17 treats \(\ln(x^2)\) as \((\ln x)^2\) and squares the estimate.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.