The trapezium rule - Edexcel A-Level Maths (Paper 1) Exam Questions

Practise 16 exam-style questions on The trapezium rule for Edexcel A-Level Maths (Paper 1). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

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16 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

3 marks

A student is given the table of values of \(y = \dfrac{2^x}{x}\) below, where the values of \(y\) are rounded to 4 decimal places where they are not exact.

The student uses the five values to estimate \(\displaystyle\int_1^3 \dfrac{2^x}{x}\,dx\). The student's working is
\[h = \dfrac{3 - 1}{5} = 0.4\]
\[\int_1^3 \dfrac{2^x}{x}\,dx \approx \dfrac{1}{2}(0.4)\{(2 + 2.6667) + 2(1.8856 + 2 + 2.2627)\} = 3.39\]
Identify the mistake in the student's working, and find a correct estimate for the integral, giving your answer to 3 significant figures.

[The trapezium rule: \(\displaystyle\int_a^b y\,dx \approx \dfrac{1}{2}h\{(y_0 + y_n) + 2(y_1 + y_2 + \ldots + y_{n-1})\}\), where \(h = \dfrac{b - a}{n}\)]

Table with columns "x", "1", "1.5", "2", "2.5", "3". x: y; 1: 2; 1.5: 1.8856; 2: 2; 2.5: 2.2627; 3: 2.6667.

Model answer

There are 5 values (ordinates), so there are 4 strips, not 5. The student should have used \(h = \dfrac{3 - 1}{4} = 0.5\).
\(\displaystyle\int_1^3 \dfrac{2^x}{x}\,dx \approx \dfrac{1}{2}(0.5)\{(2 + 2.6667) + 2(1.8856 + 2 + 2.2627)\}\)
\(= 0.25\{4.6667 + 12.2966\}\)
\(= 4.24\) (3 s.f.)

Explanation

In the formula \(h = \dfrac{b - a}{n}\), n is the number of strips, which is one less than the number of values in the table. Five values from x = 1 to x = 3 give four strips of width 0.5, and redoing the calculation with h = 0.5 gives 4.24. Using the number of values instead of the number of strips is one of the commonest trapezium rule errors in the examiners' reports.

Mark scheme

1 mark
B1 Identifies the mistake: the five values in the table give 4 strips, not 5, so \(h = \dfrac{3 - 1}{4} = 0.5\) (the student divided by the number of values instead of the number of strips); a correct final answer of 4.24 with no working scores the M1 and A1 only
An answer of 4.24 with no working scores 2 marks: the B1 needs the mistake to be stated. Accept 'h should be 0.5' or 'there are only 4 intervals'. Do not accept vague answers such as 'h is wrong' with no correct value or reason.
1 mark
M1 A full attempt at the trapezium rule with \(h = 0.5\): \(\dfrac{0.5}{2}\{(2 + 2.6667) + 2(1.8856 + 2 + 2.2627)\}\)
Condone copying slips. Allow the sum of the areas of four trapezia. Accept values to 3 or more decimal places.
1 mark
A1 awrt 4.24, i.e. in [4.235, 4.245)
The calculator value of the integral is 4.21, which scores A0.

Question 2

5 marks

The table shows corresponding values of \(x\) and \(y\) for \(y = \dfrac{1}{1 + \ln x}\), with the values of \(y\) given to 4 decimal places.

Find the values of \(a\) and \(b\), giving each to 4 decimal places.

Hence use the trapezium rule, with strips of width 0.2, to find an estimate for
\[\int_1^2 \dfrac{1}{1 + \ln x}\,dx\]
Give your answer to 3 significant figures.

[The trapezium rule: \(\displaystyle\int_a^b y\,dx \approx \dfrac{1}{2}h\{(y_0 + y_n) + 2(y_1 + y_2 + \ldots + y_{n-1})\}\), where \(h = \dfrac{b - a}{n}\)]

Table with columns "x", "1", "1.2", "1.4", "1.6", "1.8", "2". x: y; 1: 1; 1.2: a; 1.4: 0.7482; 1.6: 0.6803; 1.8: b; 2: 0.5906.

Model answer

\(a = \dfrac{1}{1 + \ln 1.2} = 0.8458\), \(b = \dfrac{1}{1 + \ln 1.8} = 0.6298\)
\(h = 0.2\)
\(\displaystyle\int_1^2 \dfrac{1}{1 + \ln x}\,dx \approx \dfrac{1}{2}(0.2)\{(1 + 0.5906) + 2(0.8458 + 0.7482 + 0.6803 + 0.6298)\}\)
\(= 0.1\{1.5906 + 5.8082\}\)
\(= 0.740\) (3 s.f.)

Explanation

The missing values come from substituting x = 1.2 and x = 1.8 into the equation of the curve. Strips of width 0.2 from 1 to 2 give five strips and six values, so the bracket is the two end values plus twice the four middle values, multiplied by \(\dfrac{0.2}{2}\). A common error is to give the calculator's value of the integral (0.737) instead of the trapezium estimate.

Mark scheme

1 mark
B1 \(a = 0.8458\); a correct final answer of a = 0.8458, b = 0.6298 and 0.740 with no working scores 3 marks (see comments)
An answer of a = 0.8458, b = 0.6298 and 0.740 with no working scores 3 marks (B1 for a, B1 for b, and the final A1 as a special case). Values of a and b may be written down from a calculator.
1 mark
B1 \(b = 0.6298\)
Accept values that are correct to 4 decimal places only.
1 mark
B1 Uses \(h = 0.2\) with 5 strips (6 values of y)
May be implied by the factor 0.1 in front of the bracket. h = 0.2 is exact.
1 mark
M1 A full attempt at the trapezium rule with their a and b: \(\dfrac{0.2}{2}\{(1 + 0.5906) + 2(a + 0.7482 + 0.6803 + b)\}\)
Condone copying slips. Allow the sum of five trapezia. Accept values to 3 or more decimal places.
1 mark
A1 awrt 0.740, i.e. in [0.7395, 0.7405)
The calculator value of the integral is 0.737, which scores A0. Do not accept 0.74 unless 0.740 (or 0.7399) has been seen.

Question 3

6 marks

With strips of width 1, the trapezium rule gives an estimate for
\[\int_0^3 x\mathrm{e}^{-\frac{x}{3}}\,dx\]

Find this estimate to 3 significant figures.

Given that the exact value of the integral is \(9 - \dfrac{18}{\mathrm{e}}\), find the percentage error in your estimate. Use your unrounded estimate and give your answer to 2 significant figures.

Explain how the trapezium rule could be used to give a more accurate estimate of the integral.

[The trapezium rule: \(\displaystyle\int_a^b y\,dx \approx \dfrac{1}{2}h\{(y_0 + y_n) + 2(y_1 + y_2 + \ldots + y_{n-1})\}\), where \(h = \dfrac{b - a}{n}\)]

Model answer

\(h = 1\); y-values at x = 0, 1, 2, 3: 0, \(\mathrm{e}^{-\frac{1}{3}} = 0.7165\), \(2\mathrm{e}^{-\frac{2}{3}} = 1.0268\), \(3\mathrm{e}^{-1} = 1.1036\)
\(\displaystyle\int_0^3 x\mathrm{e}^{-\frac{x}{3}}\,dx \approx \dfrac{1}{2}\{(0 + 1.1036) + 2(0.7165 + 1.0268)\} = 2.2952\ldots = 2.30\) (3 s.f.)
Exact value \(= 9 - \dfrac{18}{\mathrm{e}} = 2.3782\)
Percentage error \(= \dfrac{2.3782 - 2.2952}{2.3782} \times 100 = 3.49\%\), so 3.5% (2 s.f.)
To improve the estimate, use more strips of smaller width, e.g. 6 strips with h = 0.5.

Explanation

With h = 1 the four values of \(x\mathrm{e}^{-\frac{x}{3}}\) at x = 0, 1, 2, 3 give an estimate of 2.295. The percentage error compares the difference with the exact value \(9 - \dfrac{18}{\mathrm{e}}\), so divide by the exact value, not the estimate. The estimate improves as the number of strips increases, because narrower trapezia follow the curve more closely; a common error is to suggest more decimal places instead.

Mark scheme

1 mark
B1 States or uses \(h = 1\) with the values \(0, \mathrm{e}^{-\frac{1}{3}}, 2\mathrm{e}^{-\frac{2}{3}}, 3\mathrm{e}^{-1}\) (0, 0.7165, 1.0268, 1.1036); answers of 2.30 and 3.5% with no working score 2 marks (see comments)
An answer of 2.30 and 3.5% with no working scores 2 marks (special case: the two A marks); the final B1 needs an explanation. Accept the values to 3 or more decimal places, e.g. awrt 0.717, i.e. in [0.7165, 0.7175), for the second.
1 mark
M1 A full attempt at the trapezium rule: \(\dfrac{1}{2}\{(0 + 1.1036) + 2(0.7165 + 1.0268)\}\)
Condone copying slips. Allow the sum of three trapezia (the first is a triangle). Accept values to 3 or more decimal places.
1 mark
A1 awrt 2.30, i.e. in [2.295, 2.305)
Unrounded value 2.2952. The calculator value of the integral, 2.38, scores A0.
1 mark
M1 Percentage error \(= \dfrac{\left(9 - \frac{18}{\mathrm{e}}\right) - \text{their estimate}}{9 - \frac{18}{\mathrm{e}}} \times 100\)
\(9 - \dfrac{18}{\mathrm{e}} = 2.3782\). Allow the difference in either order. Dividing by their estimate instead of the exact value is M0.
1 mark
A1 3.5% (awrt 3.5, i.e. in [3.45, 3.55))
From 3.489%. Allow more accurate answers in the range, e.g. 3.49.
1 mark
B1 Use more strips (narrower strips, i.e. a smaller value of h), e.g. 6 strips of width 0.5
Do not accept 'use more decimal places' or 'use a calculator' or 'integrate exactly'.

Question 4

9 marks

In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.

Use the trapezium rule with 3 strips of equal width to estimate
\[\int_1^4 \dfrac{x + 2}{\sqrt{x}}\,dx\]
giving your answer to 4 significant figures.

Use algebraic integration to find the exact value of this integral, and hence find the percentage error in your trapezium rule estimate, giving your answer to 2 significant figures.

The curve \(y = \dfrac{x + 2}{\sqrt{x}}\) is convex for \(1 \leqslant x \leqslant 4\). Without further calculation, state, with a reason, whether the trapezium rule with 6 strips of equal width would give a larger or a smaller estimate than your estimate with 3 strips.

[The trapezium rule: \(\displaystyle\int_a^b y\,dx \approx \dfrac{1}{2}h\{(y_0 + y_n) + 2(y_1 + y_2 + \ldots + y_{n-1})\}\), where \(h = \dfrac{b - a}{n}\)]

Model answer

\(h = 1\); y at x = 1, 2, 3, 4: 3, 2.8284, 2.8868, 3
\(\displaystyle\int_1^4 \dfrac{x + 2}{\sqrt{x}}\,dx \approx \dfrac{1}{2}\{(3 + 3) + 2(2.8284 + 2.8868)\} = 8.715\) (4 s.f.)
\(\dfrac{x + 2}{\sqrt{x}} = x^{\frac{1}{2}} + 2x^{-\frac{1}{2}}\)
\(\displaystyle\int_1^4 \left(x^{\frac{1}{2}} + 2x^{-\frac{1}{2}}\right)dx = \left[\dfrac{2}{3}x^{\frac{3}{2}} + 4x^{\frac{1}{2}}\right]_1^4 = \left(\dfrac{16}{3} + 8\right) - \left(\dfrac{2}{3} + 4\right) = \dfrac{26}{3}\)
Percentage error \(= \dfrac{8.71518 - 8.66667}{8.66667} \times 100 = 0.56\%\) (2 s.f.)
Smaller: the curve is convex, so the tops of the trapezia lie above it and the estimate is too large; with 6 narrower strips the trapezia fit the curve more closely, so the estimate is closer to \(\dfrac{26}{3}\) and therefore smaller than 8.715.

Explanation

With h = 1 the four values give a trapezium estimate of 8.715, and splitting the integrand into \(x^{\frac{1}{2}} + 2x^{-\frac{1}{2}}\) lets it be integrated exactly to \(\dfrac{26}{3}\). The percentage error is the difference as a percentage of the exact value. Because the curve is convex, every trapezium estimate is too big, and more, narrower strips reduce the excess, so the 6-strip estimate is smaller; common errors are integrating the fraction without splitting it and saying 'larger' because more strips are 'more accurate'.

Mark scheme

1 mark
B1 The values of y at x = 1, 2, 3, 4 (h = 1): 3, \(2\sqrt{2}\) (2.8284), \(\dfrac{5}{\sqrt{3}}\) (2.8868), 3. An answer with no working scores 0 marks.
An answer of 8.715, \(\dfrac{26}{3}\), 0.56% and 'smaller' with no working scores 0 marks: the question demands working. Accept the values to 3 or more decimal places.
1 mark
M1 A full attempt at the trapezium rule: \(\dfrac{1}{2}\{(3 + 3) + 2(2.8284 + 2.8868)\}\)
Condone copying slips. Allow the sum of three trapezia. Accept values to 3 or more decimal places.
1 mark
A1 awrt 8.715, i.e. in [8.7145, 8.7155)
Unrounded value 8.71518. The exact value of the integral, 8.667, scores A0 here.
1 mark
M1 Writes the integrand as \(x^{\frac{1}{2}} + 2x^{-\frac{1}{2}}\) and integrates: at least one power of x increased by 1
The fraction must be split (or another valid method used) before integrating.
1 mark
A1 \(\left[\dfrac{2}{3}x^{\frac{3}{2}} + 4x^{\frac{1}{2}}\right]_1^4 = \left(\dfrac{16}{3} + 8\right) - \left(\dfrac{2}{3} + 4\right) = \dfrac{26}{3}\)
Exact: requires evidence of the limits being substituted. \(\dfrac{26}{3}\) from a calculator with no integration seen scores M0A0.
1 mark
M1 Percentage error \(= \dfrac{\text{their estimate} - \frac{26}{3}}{\frac{26}{3}} \times 100\)
Allow the difference in either order. Dividing by their estimate instead of the exact value is M0.
1 mark
A1 awrt 0.56%, i.e. in [0.555, 0.565)
From 0.5597...%. Using the rounded estimate 8.715 gives 0.558%, which is also awrt 0.56: accept.
1 mark
B1 Smaller
No working needed. 'Larger' scores B0B0.
1 mark
B1 (dB1) A correct reason: the curve is convex, so the chords (tops of the trapezia) lie above the curve and every trapezium estimate is too large; narrower strips fit the curve more closely, so the 6-strip estimate is closer to \(\dfrac{26}{3}\) from above, i.e. smaller (but still greater than \(\dfrac{26}{3}\))
Can only be awarded if 'smaller' is given. Also accept: each new ordinate at the middle of an old strip lies below the old chord because the curve is convex. For reference, the 6-strip estimate is 8.6794.

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