Integrating standard functions - Edexcel A-Level Maths (Paper 1) Exam Questions

Practise 18 exam-style questions on Integrating standard functions 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.

18 questions
2–9 marks each
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18 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

Find
\[\int \left(5\mathrm{e}^{-2x} - \dfrac{3}{2x}\right)dx, \quad x > 0\]

Model answer

\(\displaystyle\int 5\mathrm{e}^{-2x}\,dx = \dfrac{5}{-2}\mathrm{e}^{-2x} = -\dfrac{5}{2}\mathrm{e}^{-2x}\)
\(\displaystyle\int \dfrac{3}{2x}\,dx = \dfrac{3}{2}\int \dfrac{1}{x}\,dx = \dfrac{3}{2}\ln x\)
\(\displaystyle\int \left(5\mathrm{e}^{-2x} - \dfrac{3}{2x}\right)dx = -\dfrac{5}{2}\mathrm{e}^{-2x} - \dfrac{3}{2}\ln x + c\)

Explanation

Use \(\displaystyle\int \mathrm{e}^{kx}\,dx = \dfrac{1}{k}\mathrm{e}^{kx} + c\) with \(k = -2\), and write \(\dfrac{3}{2x}\) as \(\dfrac{3}{2} \times \dfrac{1}{x}\) so that \(\displaystyle\int \dfrac{1}{x}\,dx = \ln x\) can be used. Because \(\ln(2x) = \ln 2 + \ln x\), the answer \(-\dfrac{3}{2}\ln(2x) + c\) is also correct. Common errors are multiplying by \(-2\) instead of dividing, and treating \(\dfrac{1}{x}\) as a power to be increased.

Mark scheme

1 mark
M1 Integrates one term to the correct form: \(\mathrm{e}^{-2x} \to k\mathrm{e}^{-2x}\) or \(\dfrac{1}{x} \to k\ln x\); or a correct final answer of \(-\dfrac{5}{2}\mathrm{e}^{-2x} - \dfrac{3}{2}\ln x + c\) with no working
An answer of \(-\dfrac{5}{2}\mathrm{e}^{-2x} - \dfrac{3}{2}\ln x + c\) with no working scores 3 marks. \(\dfrac{1}{x} \to x^0\) or \(\dfrac{x^0}{0}\) is not a valid method for that term.
1 mark
A1 One term correct: \(-\dfrac{5}{2}\mathrm{e}^{-2x}\) or \(-\dfrac{3}{2}\ln x\)
1 mark
A1 \(-\dfrac{5}{2}\mathrm{e}^{-2x} - \dfrac{3}{2}\ln x + c\)
Accept \(\ln\lvert x\rvert\). Also accept \(-\dfrac{3}{2}\ln(2x)\) for the second term: it differs from \(-\dfrac{3}{2}\ln x\) by a constant, so it is a correct integral. The \(+ c\) is needed. Common errors are \(-10\mathrm{e}^{-2x}\) (multiplying by \(-2\) instead of dividing) and \(-3\ln(2x)\) or \(-\dfrac{3}{2x^2}\).

Question 2

4 marks

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

Show that
\[\dfrac{2}{1 + \cos 2x} \equiv \sec^2 x, \quad -\dfrac{\pi}{2} < x < \dfrac{\pi}{2}\]
Hence find the exact value of
\[\int_0^{\frac{\pi}{3}} \dfrac{1}{1 + \cos 2x}\,dx\]
[You may use \(\displaystyle\int \sec^2 kx\,dx = \dfrac{1}{k}\tan kx + c\)]

Model answer

\(1 + \cos 2x = 1 + (2\cos^2 x - 1) = 2\cos^2 x\)
\(\dfrac{2}{1 + \cos 2x} = \dfrac{2}{2\cos^2 x} = \dfrac{1}{\cos^2 x} = \sec^2 x\) as required
So \(\dfrac{1}{1 + \cos 2x} = \dfrac{1}{2}\sec^2 x\)
\(\displaystyle\int_0^{\frac{\pi}{3}} \dfrac{1}{2}\sec^2 x\,dx = \left[\dfrac{1}{2}\tan x\right]_0^{\frac{\pi}{3}} = \dfrac{1}{2}\tan\dfrac{\pi}{3} - \dfrac{1}{2}\tan 0 = \dfrac{\sqrt{3}}{2}\)

Explanation

The double-angle formula \(\cos 2x = 2\cos^2 x - 1\) turns the denominator into \(2\cos^2 x\), so the fraction is \(\sec^2 x\). The integrand is then half of \(\sec^2 x\), which integrates to \(\dfrac{1}{2}\tan x\), giving \(\dfrac{\sqrt{3}}{2}\). Common errors are using the wrong form of \(\cos 2x\), losing the factor \(\dfrac{1}{2}\) and giving a decimal instead of the exact value.

Mark scheme

1 mark
M1 Uses \(\cos 2x = 2\cos^2 x - 1\) in the denominator, so that \(1 + \cos 2x = 2\cos^2 x\). An answer with no working scores 0 marks.
An answer with no working scores 0 marks: the identity is a show that (restating it earns nothing) and the question demands all stages of working, so \(\dfrac{\sqrt{3}}{2}\) alone scores nothing. Using \(\cos 2x = \cos^2 x - \sin^2 x\) with \(1 = \cos^2 x + \sin^2 x\) is equally valid. Using \(\cos 2x = 1 - 2\sin^2 x\) and then \(1 - \sin^2 x = \cos^2 x\) to reach \(2\cos^2 x\) is also valid; \(\cos 2x = 1 - 2\sin^2 x\) with no conversion to \(\cos^2 x\) (so that \(\sec^2 x\) is never reached) is M0.
1 mark
A1* \(\dfrac{2}{1 + \cos 2x} = \dfrac{2}{2\cos^2 x} = \dfrac{1}{\cos^2 x} = \sec^2 x\) cso
Given answer: needs the line \(\dfrac{2}{2\cos^2 x}\) (or equivalent) and a conclusion.
1 mark
M1 Writes the integrand as \(\dfrac{1}{2}\sec^2 x\) and integrates to \(k\tan x\)
May use the given identity even if the show that was not completed. Integrating \(\sec^2 x\) without the factor \(\dfrac{1}{2}\) can score this M1 but not the A1.
1 mark
A1 \(\left[\dfrac{1}{2}\tan x\right]_0^{\frac{\pi}{3}} = \dfrac{\sqrt{3}}{2}\) cso
Uses \(\tan\dfrac{\pi}{3} = \sqrt{3}\) and \(\tan 0 = 0\). Accept \(\dfrac{1}{2}\sqrt{3}\) or \(\dfrac{\sqrt{3}}{2}\) typed as sqrt(3)/2; a decimal only (0.866) is A0. A common error is \(\sqrt{3}\), from losing the factor \(\dfrac{1}{2}\): A0.

Question 3

5 marks

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

Find the exact value of
\[\int_{-\mathrm{e}}^{-1} \dfrac{4x - 3}{x}\,dx\]

Model answer

\(\dfrac{4x - 3}{x} = 4 - \dfrac{3}{x}\)
\(\displaystyle\int \left(4 - \dfrac{3}{x}\right)dx = 4x - 3\ln\lvert x\rvert + c\)
\(\displaystyle\int_{-\mathrm{e}}^{-1} \left(4 - \dfrac{3}{x}\right)dx = \left[4x - 3\ln\lvert x\rvert\right]_{-\mathrm{e}}^{-1}\)
\(= \left(-4 - 3\ln 1\right) - \left(-4\mathrm{e} - 3\ln\mathrm{e}\right)\)
\(= -4 - 0 + 4\mathrm{e} + 3\)
\(= 4\mathrm{e} - 1\)

Explanation

Divide each term by \(x\) to get \(4 - \dfrac{3}{x}\), which integrates to \(4x - 3\ln\lvert x\rvert\). The modulus matters here because \(x\) is negative between the limits: \(\ln\lvert -1\rvert = 0\) and \(\ln\lvert -\mathrm{e}\rvert = 1\), so the value is \((-4) - (-4\mathrm{e} - 3) = 4\mathrm{e} - 1\). Examiners report candidates being careless with the modulus in \(\ln\lvert x\rvert\), and the logarithm of a negative number is not defined.

Mark scheme

1 mark
B1 Writes the integrand as \(4 - \dfrac{3}{x}\) (or \(4 - 3x^{-1}\)). An answer with no working scores 0 marks.
An answer with no working scores 0 marks: the question demands all stages of working (a calculator gives 9.873...). A correct answer needs at least the divided integrand, the integral with \(\ln\lvert x\rvert\) and the substituted limits.
1 mark
M1 Integrates: \(4 \to 4x\) or \(\dfrac{3}{x} \to 3\ln\lvert x\rvert\) (allow \(k\ln\lvert x\rvert\) or \(k\ln x\))
Independent of the B1. Applying the power rule to \(x^{-1}\), giving \(\dfrac{x^0}{0}\), is not a valid method for that term.
1 mark
A1 \(4x - 3\ln\lvert x\rvert\) (or \(4x - 3\ln(-x)\))
The modulus (or \(\ln(-x)\)) is needed here because \(x\) is negative on the interval: \(4x - 3\ln x\) is A0 unless the modulus appears when the limits are substituted. Alternative: the substitution \(u = -x\) gives \(\displaystyle\int_1^{\mathrm{e}} \dfrac{4u + 3}{u}\,du = \left[4u + 3\ln u\right]_1^{\mathrm{e}}\), scored in the same way.
1 mark
M1 (dM1) Substitutes the limits \(-1\) and \(-\mathrm{e}\) into their integrated expression and subtracts, using \(\ln\lvert -1\rvert = 0\) and \(\ln\lvert -\mathrm{e}\rvert = 1\)
Depends on the M mark for integrating (not only the point just before). The substitution must be seen, e.g. \((-4 - 3\ln 1) - (-4\mathrm{e} - 3\ln\mathrm{e})\).
1 mark
A1 \(4\mathrm{e} - 1\) cso
A decimal (9.87) only is A0. Working with logarithms of negative numbers, e.g. \(\ln(-\mathrm{e}) = 1\), is an error: A0 cso, although the method marks can still be scored. Examiners report candidates being careless with the modulus in \(\ln\lvert x\rvert\).

Question 4

7 marks

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

Find the exact value of
\[\int_0^{\frac{\pi}{4}} \sin^2 x\cos 2x\,dx\]
giving your answer in the form \(\dfrac{a + b\pi}{c}\), where \(a\), \(b\) and \(c\) are integers.

Model answer

\(\sin^2 x = \dfrac{1}{2}(1 - \cos 2x)\), so \(\sin^2 x\cos 2x = \dfrac{1}{2}\cos 2x - \dfrac{1}{2}\cos^2 2x\)
\(\cos^2 2x = \dfrac{1}{2}(1 + \cos 4x)\), so \(\sin^2 x\cos 2x = \dfrac{1}{2}\cos 2x - \dfrac{1}{4} - \dfrac{1}{4}\cos 4x\)
\(\displaystyle\int_0^{\frac{\pi}{4}} \left(\dfrac{1}{2}\cos 2x - \dfrac{1}{4} - \dfrac{1}{4}\cos 4x\right)dx = \left[\dfrac{1}{4}\sin 2x - \dfrac{1}{4}x - \dfrac{1}{16}\sin 4x\right]_0^{\frac{\pi}{4}}\)
\(= \left(\dfrac{1}{4}\sin\dfrac{\pi}{2} - \dfrac{\pi}{16} - \dfrac{1}{16}\sin\pi\right) - 0 = \dfrac{1}{4} - \dfrac{\pi}{16}\)
\(= \dfrac{4 - \pi}{16}\)

Explanation

Replace \(\sin^2 x\) by \(\dfrac{1}{2}(1 - \cos 2x)\), and then the resulting \(\cos^2 2x\) by \(\dfrac{1}{2}(1 + \cos 4x)\), so that the integrand is a sum of a constant and cosines of multiple angles. Each cosine integrates to a sine divided by its coefficient, and the limits give \(\dfrac{1}{4} - \dfrac{\pi}{16} = \dfrac{4 - \pi}{16}\). Common errors are sign slips in the double-angle formulae, multiplying instead of dividing by 2 or 4 when integrating, and not writing the answer in the form asked for.

Mark scheme

1 mark
M1 Uses \(\sin^2 x = \dfrac{1}{2}(1 - \cos 2x)\) (from \(\cos 2x = 1 - 2\sin^2 x\)) to write the integrand in terms of \(\cos 2x\). An answer with no working scores 0 marks.
An answer of \(\dfrac{4 - \pi}{16}\) with no working scores 0 marks: the question demands all stages of working. Alternative: \(\cos 2x = 1 - 2\sin^2 x\) gives \(\sin^2 x - 2\sin^4 x\), which needs \(\sin^4 x\) in terms of \(\cos 2x\) and \(\cos 4x\); mark equivalently.
1 mark
M1 Uses \(\cos^2 2x = \dfrac{1}{2}(1 + \cos 4x)\) for the \(\cos^2 2x\) term
Awarded for a correct use of this identity on the \(\cos^2 2x\) term of their expression.
1 mark
A1 Integrand \(=\dfrac{1}{2}\cos 2x - \dfrac{1}{4} - \dfrac{1}{4}\cos 4x\) oe
From \(\dfrac{1}{2}\cos 2x - \dfrac{1}{2}\cos^2 2x\).
1 mark
M1 Integrates their expression: \(\cos kx \to \pm\dfrac{1}{k}\sin kx\) for at least one term
Follow through their integrand if it contains \(\cos 2x\) and \(\cos 4x\) terms.
1 mark
A1 \(\dfrac{1}{4}\sin 2x - \dfrac{1}{4}x - \dfrac{1}{16}\sin 4x\) oe
1 mark
M1 (dM1) Substitutes the limits \(\dfrac{\pi}{4}\) and 0 and subtracts, using \(\sin\dfrac{\pi}{2} = 1\) and \(\sin\pi = 0\)
Depends on the M mark for integrating (the fourth point), not on the point just before.
1 mark
A1 \(\dfrac{4 - \pi}{16}\) cso
Accept \(a = 4\), \(b = -1\), \(c = 16\) or an equivalent with integers, e.g. \(\dfrac{8 - 2\pi}{32}\); typed (4 - pi)/16. \(\dfrac{1}{4} - \dfrac{\pi}{16}\) not combined into the form asked is A0, as is a decimal only (0.0537).

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