AS & A Level · Practice questions · The trapezium rule

Chords, strips and a sketch.

Six original Cambridge-style questions on the trapezium rule: tables of values, deciding whether an estimate is too big or too small, and data from a context.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Width, table, weights, then a sketch.

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.

Recall

[3 marks]

Use the trapezium rule with 3 intervals to estimate \(\displaystyle\int_0^3 \sqrt{x+1}\,dx\), giving your answer to 3 significant figures.

\(h = 1\); \(y = 1,\ 1.41421,\ 1.73205,\ 2\). ✓
\(\dfrac{1}{2}\left[1 + 2 + 2(1.41421 + 1.73205)\right] = \dfrac{1}{2}\times 9.29253 = 4.646\). ✓
Estimate: 4.65 (3 s.f.). ✓

Calculation

[5 marks]

(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) \(h = 0.25\); \(y = 1,\ 1.06449,\ 1.28403,\ 1.75505,\ 2.71828\). ✓
\(0.125\left[1 + 2.71828 + 2(1.06449 + 1.28403 + 1.75505)\right] = 0.125\times 11.92542 = 1.4907\). Estimate: 1.49. ✓
(b) The curve \(y = e^{x^2}\) bends upwards on \([0, 1]\), so the chords lie above it: an over-estimate. ✓

Analysis

[5 marks]

(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) \(h = 0.5\); \(y = 0,\ 0.40547,\ 0.69315\). ✓
\(0.25\left[0 + 0.69315 + 2(0.40547)\right] = 0.25\times 1.50409 = 0.376\). ✓
(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. ✓

Analysis

[6 marks]

(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) \(h = 0.25\); \(y = 4,\ 3.76471,\ 3.2,\ 2.56,\ 2\). ✓
\(0.125\left[4 + 2 + 2(3.76471 + 3.2 + 2.56)\right] = 0.125\times 25.04941 = 3.1312\). 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%. ✓

Calculation

[5 marks]

(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) \(h = \dfrac{\pi}{4}\); \(y = 0,\ \dfrac{\sqrt{2}}{2},\ 1,\ \dfrac{\sqrt{2}}{2},\ 0\). ✓
\(\dfrac{\pi}{8}\left[0 + 0 + 2\left(\dfrac{\sqrt{2}}{2} + 1 + \dfrac{\sqrt{2}}{2}\right)\right] = \dfrac{\pi}{8}\times 2\left(1 + \sqrt{2}\right) = \dfrac{\pi}{4}\left(1 + \sqrt{2}\right)\), 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. ✓

Analysis

[5 marks]

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.

(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. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.