Integrating standard functions - Edexcel A-Level Maths (Paper 2) Exam Questions

Practise 18 exam-style questions on Integrating standard functions for Edexcel A-Level Maths (Paper 2). 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
\[\int \dfrac{(3x + 2)(x - 4)}{2x}\,dx, \quad x > 0\]
giving each term in its simplest form.

Model answer

\((3x + 2)(x - 4) = 3x^2 - 10x - 8\)
\(\dfrac{3x^2 - 10x - 8}{2x} = \dfrac{3}{2}x - 5 - \dfrac{4}{x}\)
\(\displaystyle\int \left(\dfrac{3}{2}x - 5 - \dfrac{4}{x}\right) dx = \dfrac{3}{4}x^2 - 5x - 4\ln x + c\)

Explanation

Multiply out the numerator and divide each term by \(2x\), so that the integrand is \(\dfrac{3}{2}x - 5 - \dfrac{4}{x}\). The power rule integrates the first two terms, and \(-\dfrac{4}{x}\) integrates to \(-4\ln x\) because the power rule does not work for \(x^{-1}\). Common errors are dividing only the first term by \(2x\), treating \(\dfrac{1}{x}\) with the power rule and leaving out the constant of integration.

Mark scheme

1 mark
M1 Expands the numerator and divides each term by \(2x\), giving at least two of \(\dfrac{3}{2}x\), \(-5\) and \(-\dfrac{4}{x}\) correct; or a correct final answer of \(\dfrac{3}{4}x^2 - 5x - 4\ln x + c\) with no working
An answer of \(\dfrac{3}{4}x^2 - 5x - 4\ln x + c\) with no working scores 3 marks. The numerator is \(3x^2 - 10x - 8\). Integrating the numerator and the denominator separately, or dividing only the first term by \(2x\), is M0.
1 mark
A1 Two of \(\dfrac{3}{4}x^2\), \(-5x\) and \(-4\ln x\) correct (unsimplified is fine)
E.g. \(\dfrac{3}{2} \times \dfrac{x^2}{2}\). \(-\dfrac{4}{x} \to \dfrac{4}{x^2}\) or \(-\dfrac{4x^0}{0}\) is not a valid integral of \(-\dfrac{4}{x}\).
1 mark
A1 \(\dfrac{3}{4}x^2 - 5x - 4\ln x + c\)
Depends on the M mark. All three terms simplified, with \(+ c\). Accept \(0.75x^2\), \(-4\ln\lvert x\rvert\) or \(-\ln x^4\). Also accept \(-4\ln 2x\) (from \(-\dfrac{8}{2x}\)), which differs from \(-4\ln x\) by a constant. Condone spurious integral signs.

Question 2

4 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
\[\int_{\frac{\pi}{8}}^{\frac{\pi}{4}} \left(2\operatorname{cosec} 2x + 1\right) dx\]
giving your answer in the form \(\ln\left(p + \sqrt{q}\right) + r\pi\), where \(p\) and \(q\) are integers and \(r\) is a rational number.
[You may use \(\displaystyle\int \operatorname{cosec} kx\,dx = -\dfrac{1}{k}\ln\lvert\operatorname{cosec} kx + \cot kx\rvert + c\)]

Model answer

\(\displaystyle\int \left(2\operatorname{cosec} 2x + 1\right) dx = 2 \times \left(-\dfrac{1}{2}\right)\ln\lvert\operatorname{cosec} 2x + \cot 2x\rvert + x = -\ln\lvert\operatorname{cosec} 2x + \cot 2x\rvert + x\)
At \(x = \dfrac{\pi}{4}\): \(2x = \dfrac{\pi}{2}\), \(\operatorname{cosec}\dfrac{\pi}{2} + \cot\dfrac{\pi}{2} = 1 + 0 = 1\)
At \(x = \dfrac{\pi}{8}\): \(2x = \dfrac{\pi}{4}\), \(\operatorname{cosec}\dfrac{\pi}{4} + \cot\dfrac{\pi}{4} = \sqrt{2} + 1\)
\(\left[-\ln\lvert\operatorname{cosec} 2x + \cot 2x\rvert + x\right]_{\frac{\pi}{8}}^{\frac{\pi}{4}} = \left(-\ln 1 + \dfrac{\pi}{4}\right) - \left(-\ln\left(\sqrt{2} + 1\right) + \dfrac{\pi}{8}\right)\)
\(= \ln\left(1 + \sqrt{2}\right) + \dfrac{\pi}{8}\)

Explanation

The integral of \(\operatorname{cosec} kx\) is in the formula booklet: with \(k = 2\), \(2\operatorname{cosec} 2x\) integrates to \(-\ln\lvert\operatorname{cosec} 2x + \cot 2x\rvert\), and the constant 1 integrates to \(x\). The exact values at \(2x = \dfrac{\pi}{2}\) and \(2x = \dfrac{\pi}{4}\) give \(-\ln 1 + \ln\left(\sqrt{2} + 1\right)\), so the answer is \(\ln\left(1 + \sqrt{2}\right) + \dfrac{\pi}{8}\). Examiners report candidates not using the formula booklet for these integrals, sign errors with the minus in front of the logarithm, and calculator decimals when an exact value is required.

Mark scheme

1 mark
M1 Integrates using the given result: \(2\operatorname{cosec} 2x \to \pm\alpha\ln\lvert\operatorname{cosec} 2x + \cot 2x\rvert\) (or uses the booklet's alternative form \(\alpha\ln\lvert\tan x\rvert\)). An answer with no working scores 0 marks.
An answer of \(\ln\left(1 + \sqrt{2}\right) + \dfrac{\pi}{8}\) with no working scores 0 marks: the question demands working (a calculator gives 1.274...). Alternative: \(\displaystyle\int \operatorname{cosec} kx\,dx = \dfrac{1}{k}\ln\left\lvert\tan\dfrac{kx}{2}\right\rvert + c\) (also in the formula booklet) gives \(\left[\ln\lvert\tan x\rvert + x\right]\) and \(-\ln\left(\sqrt{2} - 1\right) = \ln\left(\sqrt{2} + 1\right)\); it earns the same marks.
1 mark
A1 \(-\ln\lvert\operatorname{cosec} 2x + \cot 2x\rvert + x\)
Condone missing modulus signs. Coefficient: \(2 \times \left(-\dfrac{1}{2}\right) = -1\).
1 mark
M1 (dM1) Substitutes both limits and subtracts, using \(\operatorname{cosec}\dfrac{\pi}{2} = 1\), \(\cot\dfrac{\pi}{2} = 0\), \(\operatorname{cosec}\dfrac{\pi}{4} = \sqrt{2}\) and \(\cot\dfrac{\pi}{4} = 1\): \(\left(-\ln 1 + \dfrac{\pi}{4}\right) - \left(-\ln\left(\sqrt{2} + 1\right) + \dfrac{\pi}{8}\right)\)
Depends on the M mark at point 1 (not on the A mark just before). Limits in radians.
1 mark
A1 \(\ln\left(1 + \sqrt{2}\right) + \dfrac{\pi}{8}\)
\(p = 1\), \(q = 2\), \(r = \dfrac{1}{8}\). \(-\ln\left(\sqrt{2} - 1\right) + \dfrac{\pi}{8}\) is not in the form asked: A0. A decimal only is 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.

Use algebraic integration to find the exact value of
\[\int_4^9 \dfrac{\left(\sqrt{x} + 2\right)\left(\sqrt{x} - 3\right)}{x}\,dx\]
giving your answer in the form \(p\ln 3 + q\ln 2 + r\), where \(p\), \(q\) and \(r\) are integers.

Model answer

\(\dfrac{\left(\sqrt{x} + 2\right)\left(\sqrt{x} - 3\right)}{x} = \dfrac{x - \sqrt{x} - 6}{x} = 1 - x^{-\frac{1}{2}} - \dfrac{6}{x}\)
\(\displaystyle\int_4^9 \left(1 - x^{-\frac{1}{2}} - \dfrac{6}{x}\right) dx = \left[x - 2x^{\frac{1}{2}} - 6\ln x\right]_4^9\)
\(= (9 - 6 - 6\ln 9) - (4 - 4 - 6\ln 4)\)
\(= 3 - 6\ln 9 + 6\ln 4\)
\(= -12\ln 3 + 12\ln 2 + 3\)

Explanation

Multiply out the brackets and divide each term by \(x\), giving \(1 - x^{-\frac{1}{2}} - \dfrac{6}{x}\); the last term integrates to a logarithm and the middle term by the power rule. After substituting the limits, \(6\ln 9 = 12\ln 3\) and \(6\ln 4 = 12\ln 2\) put the answer in the form asked, with \(p = -12\), \(q = 12\) and \(r = 3\). Common errors are a wrong middle term when multiplying out (it is \(-\sqrt{x}\)) and leaving \(-6\ln 9 + 6\ln 4\).

Mark scheme

1 mark
M1 Multiplies out the brackets and divides by \(x\) to obtain terms that can be integrated, at least two of \(1\), \(-x^{-\frac{1}{2}}\) and \(-\dfrac{6}{x}\) correct. An answer with no working scores 0 marks.
An answer of \(-12\ln 3 + 12\ln 2 + 3\) with no working scores 0 marks: the question demands working (a calculator gives \(-1.8656...\)).
1 mark
M1 Integrates: \(\dfrac{1}{x} \to \ln x\) and at least one power of \(x\) increased by 1
May be awarded for their terms even if the expansion contains a slip.
1 mark
A1 \(x - 2x^{\frac{1}{2}} - 6\ln x\)
1 mark
M1 (dM1) Substitutes both limits and subtracts: \((9 - 6 - 6\ln 9) - (4 - 4 - 6\ln 4)\)
Depends on the second M mark (not the point just before).
1 mark
A1 \(-12\ln 3 + 12\ln 2 + 3\), i.e. \(p = -12\), \(q = 12\), \(r = 3\)
\(3 - 6\ln 9 + 6\ln 4\), \(3 + 6\ln\dfrac{4}{9}\) or \(3 + 12\ln\dfrac{2}{3}\) is not in the form asked: A0. A decimal only is A0.

Question 4

8 marks

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

A curve \(C\) has equation \(y = \mathrm{f}(x)\), where
\[\dfrac{dy}{dx} = a\sin 2x + 3\cos x\]
and \(a\) is a constant.
\(C\) has a stationary point where \(x = \dfrac{\pi}{6}\), and \(C\) passes through the point \((0, 1)\).
Find \(\mathrm{f}(x)\).
Hence find the greatest value of \(y\) for \(0 \leqslant x \leqslant \pi\).

Model answer

Stationary at \(x = \dfrac{\pi}{6}\): \(a\sin\dfrac{\pi}{3} + 3\cos\dfrac{\pi}{6} = 0\), so \(\dfrac{\sqrt{3}}{2}a + \dfrac{3\sqrt{3}}{2} = 0\) and \(a = -3\)
\(y = \displaystyle\int (-3\sin 2x + 3\cos x)\,dx = \dfrac{3}{2}\cos 2x + 3\sin x + c\)
At \((0, 1)\): \(1 = \dfrac{3}{2} + 0 + c\), so \(c = -\dfrac{1}{2}\)
\(\mathrm{f}(x) = \dfrac{3}{2}\cos 2x + 3\sin x - \dfrac{1}{2}\)
Stationary points: \(-3\sin 2x + 3\cos x = 0\), \(-6\sin x\cos x + 3\cos x = 0\), \(3\cos x(1 - 2\sin x) = 0\)
\(\cos x = 0\): \(x = \dfrac{\pi}{2}\); \(\sin x = \dfrac{1}{2}\): \(x = \dfrac{\pi}{6}, \dfrac{5\pi}{6}\)
\(\mathrm{f}\left(\dfrac{\pi}{6}\right) = \dfrac{3}{4} + \dfrac{3}{2} - \dfrac{1}{2} = \dfrac{7}{4}\), \(\mathrm{f}\left(\dfrac{\pi}{2}\right) = -\dfrac{3}{2} + 3 - \dfrac{1}{2} = 1\), \(\mathrm{f}\left(\dfrac{5\pi}{6}\right) = \dfrac{7}{4}\)
End points: \(\mathrm{f}(0) = 1\), \(\mathrm{f}(\pi) = 1\)
The greatest value of \(y\) for \(0 \leqslant x \leqslant \pi\) is \(\dfrac{7}{4}\)

Explanation

The stationary point gives an equation for \(a\) (\(a = -3\)), integration with \(+ c\) gives \(y\), and the point \((0, 1)\) fixes \(c = -\dfrac{1}{2}\). To find the greatest value, factorise \(\dfrac{dy}{dx} = 3\cos x(1 - 2\sin x)\) rather than dividing by \(\cos x\), and compare the values at all the stationary points and the end points; the greatest is \(\dfrac{7}{4}\). Examiners report candidates losing solutions by dividing by a trigonometric factor and forgetting the constant of integration.

Mark scheme

1 mark
M1 Sets \(\dfrac{dy}{dx} = 0\) at \(x = \dfrac{\pi}{6}\): \(a\sin\dfrac{\pi}{3} + 3\cos\dfrac{\pi}{6} = 0\). An answer with no working scores 0 marks.
An answer of \(\mathrm{f}(x) = \dfrac{3}{2}\cos 2x + 3\sin x - \dfrac{1}{2}\) and a greatest value of \(\dfrac{7}{4}\) with no working scores 0 marks: the question demands working.
1 mark
A1 \(a = -3\)
From \(\dfrac{\sqrt{3}}{2}a + \dfrac{3\sqrt{3}}{2} = 0\).
1 mark
M1 Integrates: \(\sin 2x \to \pm\dfrac{1}{2}\cos 2x\) or \(\cos x \to \pm\sin x\)
May be done with \(a\) still unknown.
1 mark
A1 \(y = \dfrac{3}{2}\cos 2x + 3\sin x + c\) (with their \(a\): \(-\dfrac{a}{2}\cos 2x + 3\sin x + c\))
1 mark
M1 (dM1) Uses \((0, 1)\) to find \(c\): \(1 = \dfrac{3}{2} + 0 + c\)
Depends on the M mark at point 3 (not on the A mark just before). The integrated expression must include a constant.
1 mark
A1 \(\mathrm{f}(x) = \dfrac{3}{2}\cos 2x + 3\sin x - \dfrac{1}{2}\)
Must follow \(a = -3\).
1 mark
M1 Finds all the stationary points in the interval by solving \(-3\sin 2x + 3\cos x = 0\) with factorisation, \(3\cos x(1 - 2\sin x) = 0\), so \(x = \dfrac{\pi}{6}, \dfrac{\pi}{2}, \dfrac{5\pi}{6}\), and evaluates \(\mathrm{f}\) at these points (and the end points)
Depends on the second M mark. Dividing by \(\cos x\) and losing \(x = \dfrac{\pi}{2}\) can still score this mark if \(\mathrm{f}\left(\dfrac{\pi}{6}\right)\) and \(\mathrm{f}\left(\dfrac{5\pi}{6}\right)\) are compared with the end-point values. Alternative: \(\mathrm{f}(x) = 1 + 3\sin x - 3\sin^2 x = \dfrac{7}{4} - 3\left(\sin x - \dfrac{1}{2}\right)^2\) by completing the square.
1 mark
A1 Greatest value \(\dfrac{7}{4}\)
Exact; 1.75 is acceptable. Giving the \(x\)-value only is A0.

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