Integration by substitution - Edexcel A-Level Maths (Paper 1) Exam Questions

Practise 18 exam-style questions on Integration by substitution 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
2–8 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

Given that \(y = \left(1 + \mathrm{e}^{2x}\right)^3\), find \(\dfrac{dy}{dx}\).

Hence find \(\displaystyle\int \mathrm{e}^{2x}\left(1 + \mathrm{e}^{2x}\right)^2\,dx\)

Model answer

\(\dfrac{dy}{dx} = 3\left(1 + \mathrm{e}^{2x}\right)^2 \times 2\mathrm{e}^{2x} = 6\mathrm{e}^{2x}\left(1 + \mathrm{e}^{2x}\right)^2\)
So \(\displaystyle\int 6\mathrm{e}^{2x}\left(1 + \mathrm{e}^{2x}\right)^2\,dx = \left(1 + \mathrm{e}^{2x}\right)^3 + c\)
\(\displaystyle\int \mathrm{e}^{2x}\left(1 + \mathrm{e}^{2x}\right)^2\,dx = \dfrac{1}{6}\left(1 + \mathrm{e}^{2x}\right)^3 + c\)

Explanation

The chain rule gives \(\dfrac{dy}{dx} = 3\left(1 + \mathrm{e}^{2x}\right)^2 \times 2\mathrm{e}^{2x}\), which is 6 times the integrand. Integration reverses differentiation, so the integral is one-sixth of \(\left(1 + \mathrm{e}^{2x}\right)^3\), plus a constant. Common errors are missing the factor 2 from differentiating \(\mathrm{e}^{2x}\), which gives \(\dfrac{1}{3}\) instead of \(\dfrac{1}{6}\), and forgetting the constant of integration.

Mark scheme

1 mark
B1 \(\dfrac{dy}{dx} = 6\mathrm{e}^{2x}\left(1 + \mathrm{e}^{2x}\right)^2\); a correct derivative and a correct integral with no working score 3 marks
An answer of \(6\mathrm{e}^{2x}\left(1 + \mathrm{e}^{2x}\right)^2\) and \(\dfrac{1}{6}\left(1 + \mathrm{e}^{2x}\right)^3 + c\) with no working scores 3 marks. Accept the unsimplified \(3\left(1 + \mathrm{e}^{2x}\right)^2 \times 2\mathrm{e}^{2x}\). A common error is \(3\mathrm{e}^{2x}\left(1 + \mathrm{e}^{2x}\right)^2\) (the factor 2 from differentiating \(\mathrm{e}^{2x}\) missed): B0.
1 mark
M1 Uses their derivative to write \(\displaystyle\int \mathrm{e}^{2x}\left(1 + \mathrm{e}^{2x}\right)^2\,dx = k\left(1 + \mathrm{e}^{2x}\right)^3\) (reverses the chain rule)
Expanding to \(\mathrm{e}^{2x} + 2\mathrm{e}^{4x} + \mathrm{e}^{6x}\) and integrating term by term does not use the given result ('Hence'): M0A0.
1 mark
A1 \(\dfrac{1}{6}\left(1 + \mathrm{e}^{2x}\right)^3 + c\)
The \(+c\) is needed. Accept equivalent forms such as \(\dfrac{\left(1 + \mathrm{e}^{2x}\right)^3}{6} + c\).

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
\[\int_{0}^{\frac{\pi}{3}} \sin x\cos^4 x\,dx\]

Model answer

\(\dfrac{d}{dx}\left(\cos^5 x\right) = -5\cos^4 x\sin x\), so \(\displaystyle\int \sin x\cos^4 x\,dx = -\dfrac{1}{5}\cos^5 x + c\)
\(\left[-\dfrac{1}{5}\cos^5 x\right]_{0}^{\frac{\pi}{3}} = -\dfrac{1}{5}\left(\dfrac{1}{2}\right)^5 + \dfrac{1}{5}(1)^5\)
\(= -\dfrac{1}{160} + \dfrac{32}{160} = \dfrac{31}{160}\)

Explanation

Since \(\sin x\) is minus the derivative of \(\cos x\), the integral of \(\sin x\cos^4 x\) is \(-\dfrac{1}{5}\cos^5 x\) (reverse chain rule, or the substitution \(u = \cos x\)). The limits then give \(\cos\dfrac{\pi}{3} = \dfrac{1}{2}\) and \(\cos 0 = 1\). Common errors are the wrong sign and evaluating \(\left(\dfrac{1}{2}\right)^5\) incorrectly.

Mark scheme

1 mark
M1 Integrates to the form \(k\cos^5 x\), by inspection or with the substitution \(u = \cos x\). An answer with no working scores 0 marks.
An answer of \(\dfrac{31}{160}\) with no working scores 0 marks: the question demands working.
1 mark
A1 \(-\dfrac{1}{5}\cos^5 x\) (or \(-\dfrac{1}{5}u^5\) with \(u = \cos x\))
1 mark
M1 (dM1) Substitutes the limits, using \(\cos\dfrac{\pi}{3} = \dfrac{1}{2}\) and \(\cos 0 = 1\) (or uses the changed limits \(u = \dfrac{1}{2}\) and \(u = 1\)), and subtracts
Depends on the first M mark.
1 mark
A1 \(\dfrac{31}{160}\) cso
Accept 0.19375 (exact).

Question 3

6 marks

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

Use the substitution \(u = 1 + \ln x\) to show that
\[\int_{1}^{\mathrm{e}} \dfrac{\ln x}{x(1 + \ln x)^2}\,dx = \ln 2 - \dfrac{1}{2}\]

Model answer

\(u = 1 + \ln x \Rightarrow \dfrac{du}{dx} = \dfrac{1}{x}\), so \(dx = x\,du\), and \(\ln x = u - 1\)
When \(x = 1\), \(u = 1\); when \(x = \mathrm{e}\), \(u = 2\)
\(\displaystyle\int_{1}^{\mathrm{e}} \dfrac{\ln x}{x(1 + \ln x)^2}\,dx = \int_{1}^{2} \dfrac{u - 1}{x\,u^2}\,x\,du = \int_{1}^{2} \left(\dfrac{1}{u} - \dfrac{1}{u^2}\right) du\)
\(= \left[\ln u + \dfrac{1}{u}\right]_{1}^{2} = \left(\ln 2 + \dfrac{1}{2}\right) - (0 + 1) = \ln 2 - \dfrac{1}{2}\)

Explanation

With \(u = 1 + \ln x\), \(du = \dfrac{1}{x}\,dx\), so the \(x\) in the denominator cancels, and \(\ln x = u - 1\) turns the integrand into \(\dfrac{u - 1}{u^2}\). Splitting this into \(\dfrac{1}{u} - \dfrac{1}{u^2}\) gives \(\ln u + \dfrac{1}{u}\), evaluated between the new limits 1 and 2. Common errors are keeping the limits 1 and \(\mathrm{e}\) after substituting and integrating \(-u^{-2}\) as \(-u^{-1}\).

Mark scheme

1 mark
B1 \(\dfrac{du}{dx} = \dfrac{1}{x}\) (so \(dx = x\,du\)) and \(\ln x = u - 1\). An answer with no working scores 0 marks.
An answer with no working scores 0 marks: this is a 'show that' question and the given substitution must be used.
1 mark
M1 Substitutes fully to obtain \(\displaystyle\int \dfrac{u - 1}{u^2}\,du\)
\(x\), \(\ln x\) and \(dx\) must all be replaced; the factor \(x\) cancels.
1 mark
B1 Limits \(u = 1\) (when \(x = 1\)) and \(u = 2\) (when \(x = \mathrm{e}\))
Or returns to \(x\) and uses the limits 1 and \(\mathrm{e}\).
1 mark
M1 Writes \(\dfrac{u - 1}{u^2} = \dfrac{1}{u} - \dfrac{1}{u^2}\) and integrates to the form \(a\ln u + bu^{-1}\)
Depends on the first M mark. Integrating the numerator and the denominator separately is M0.
1 mark
A1 \(\ln u + \dfrac{1}{u}\)
Or, in \(x\), \(\ln(1 + \ln x) + \dfrac{1}{1 + \ln x}\).
1 mark
A1* \(\ln 2 - \dfrac{1}{2}\) cso, from \(\left(\ln 2 + \dfrac{1}{2}\right) - (\ln 1 + 1)\)
\(\ln 1 = 0\) must be seen or clearly implied.

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 seedling is 2 cm tall when it is planted. Its height, \(h\) cm, \(t\) days after it is planted is modelled by
\[\dfrac{dh}{dt} = \dfrac{24t}{\left(t^2 + 16\right)^{\frac{3}{2}}}, \quad t \geqslant 0\]
Using algebraic integration, show that
\[h = 8 - \dfrac{24}{\sqrt{t^2 + 16}}\]
Find the height of the seedling, as predicted by the model, 3 days after it is planted.
Using calculus, find the exact value of \(t\) at which the seedling is growing fastest, and find the exact height of the seedling at this time.

Model answer

Let \(u = t^2 + 16\), so \(\dfrac{du}{dt} = 2t\)
\(\displaystyle\int \dfrac{24t}{\left(t^2 + 16\right)^{\frac{3}{2}}}\,dt = \int 12u^{-\frac{3}{2}}\,du = -24u^{-\frac{1}{2}} + c\)
So \(h = c - \dfrac{24}{\sqrt{t^2 + 16}}\)
\(t = 0\), \(h = 2\): \(2 = c - \dfrac{24}{4}\), so \(c = 8\) and \(h = 8 - \dfrac{24}{\sqrt{t^2 + 16}}\) as required
When \(t = 3\): \(h = 8 - \dfrac{24}{\sqrt{25}} = 8 - 4.8 = 3.2\) cm
The growth rate is \(r = 24t\left(t^2 + 16\right)^{-\frac{3}{2}}\)
\(\dfrac{dr}{dt} = 24\left(t^2 + 16\right)^{-\frac{3}{2}} - 72t^2\left(t^2 + 16\right)^{-\frac{5}{2}} = 24\left(t^2 + 16\right)^{-\frac{5}{2}}\left(16 - 2t^2\right)\)
\(\dfrac{dr}{dt} = 0 \Rightarrow t^2 = 8 \Rightarrow t = 2\sqrt{2}\) (as \(t \geqslant 0\)); \(\dfrac{dr}{dt}\) changes from positive to negative there, so the rate is greatest
Height: \(h = 8 - \dfrac{24}{\sqrt{24}} = 8 - \sqrt{24} = 8 - 2\sqrt{6}\) cm

Explanation

The substitution \(u = t^2 + 16\) (or the reverse chain rule) gives \(\displaystyle\int 12u^{-\frac{3}{2}}\,du = -24u^{-\frac{1}{2}}\), and \(h = 2\) when \(t = 0\) fixes the constant as 8; then \(h = 3.2\) cm when \(t = 3\). The seedling grows fastest where the derivative of the rate \(\dfrac{dh}{dt}\) is zero, which needs the product (or quotient) rule and gives \(t = 2\sqrt{2}\) and \(h = 8 - 2\sqrt{6}\) cm. Common errors are forgetting the constant of integration, differentiating \(h\) instead of \(\dfrac{dh}{dt}\), and giving decimals where exact values are asked for.

Mark scheme

1 mark
M1 Integrates \(\dfrac{24t}{\left(t^2 + 16\right)^{\frac{3}{2}}}\) using a substitution such as \(u = t^2 + 16\) (or the reverse chain rule) to obtain \(k\left(t^2 + 16\right)^{-\frac{1}{2}}\). An answer with no working scores 0 marks.
An answer with no working scores 0 marks: the question demands working, so 3.2 cm, \(t = 2\sqrt{2}\) and \(8 - 2\sqrt{6}\) cm with no working score nothing.
1 mark
A1 \(-24\left(t^2 + 16\right)^{-\frac{1}{2}} + c\)
With \(u = t^2 + 16\): \(\displaystyle\int 12u^{-\frac{3}{2}}\,du = -24u^{-\frac{1}{2}}\).
1 mark
M1 Uses \(h = 2\) when \(t = 0\) to find their constant: \(2 = -\dfrac{24}{4} + c\)
May follow an incorrect integral. A missing constant of integration is M0.
1 mark
A1* \(c = 8\), so \(h = 8 - \dfrac{24}{\sqrt{t^2 + 16}}\) cso
Given answer: needs a correct integral and the constant found from \(h = 2\) at \(t = 0\).
1 mark
B1 \(h = 3.2\) cm when \(t = 3\)
From \(8 - \dfrac{24}{5}\). Accept \(\dfrac{16}{5}\).
1 mark
M1 Differentiates the growth rate \(\dfrac{24t}{\left(t^2 + 16\right)^{\frac{3}{2}}}\) with respect to \(t\) using the product or quotient rule (with the chain rule) and sets the result equal to zero
It is the rate \(\dfrac{dh}{dt}\), not \(h\), that is maximised: differentiating \(h\) and setting it to zero is M0. A correct derivative is \(24\left(t^2 + 16\right)^{-\frac{5}{2}}\left(16 - 2t^2\right)\) oe.
1 mark
A1 \(t^2 = 8\), so \(t = 2\sqrt{2}\) (days)
\(t = -2\sqrt{2}\) must be rejected if seen. A decimal (2.83) only is A0. A justification that this is a maximum is not required.
1 mark
A1 Height \(= 8 - \dfrac{24}{\sqrt{24}} = 8 - 2\sqrt{6}\) cm
Accept \(8 - \sqrt{24}\). A decimal (3.10) only is A0.

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