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

Practise 17 exam-style questions on Integration by parts 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.

17 questions
3–9 marks each
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17 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

4 marks

Find \(\displaystyle\int x\mathrm{e}^{3x}\,dx\)

[Integration by parts: \(\displaystyle\int u\dfrac{dv}{dx}\,dx = uv - \int v\dfrac{du}{dx}\,dx\)]

Model answer

\(u = x\), \(\dfrac{dv}{dx} = \mathrm{e}^{3x}\), so \(\dfrac{du}{dx} = 1\), \(v = \dfrac{1}{3}\mathrm{e}^{3x}\)
\(\displaystyle\int x\mathrm{e}^{3x}\,dx = \dfrac{1}{3}x\mathrm{e}^{3x} - \int \dfrac{1}{3}\mathrm{e}^{3x}\,dx\)
\(= \dfrac{1}{3}x\mathrm{e}^{3x} - \dfrac{1}{9}\mathrm{e}^{3x} + c\)

Explanation

Choose \(u = x\), because it becomes 1 when differentiated, and \(\dfrac{dv}{dx} = \mathrm{e}^{3x}\), which integrates to \(\dfrac{1}{3}\mathrm{e}^{3x}\). The formula gives \(\dfrac{1}{3}x\mathrm{e}^{3x} - \int \dfrac{1}{3}\mathrm{e}^{3x}\,dx\), and the remaining integral is \(\dfrac{1}{9}\mathrm{e}^{3x}\). Common errors are choosing \(u = \mathrm{e}^{3x}\), which makes the integral harder, and multiplying by 3 instead of dividing.

Mark scheme

1 mark
M1 Integrates by parts the right way round (\(u = x\), \(\dfrac{dv}{dx} = \mathrm{e}^{3x}\)) to reach \(Ax\mathrm{e}^{3x} - \displaystyle\int B\mathrm{e}^{3x}\,dx\); a correct final answer of \(\dfrac{1}{3}x\mathrm{e}^{3x} - \dfrac{1}{9}\mathrm{e}^{3x} + c\) with no working scores 2 marks (special case)
An answer of \(\dfrac{1}{3}x\mathrm{e}^{3x} - \dfrac{1}{9}\mathrm{e}^{3x} + c\) with no working scores 2 marks: a special case, as answers without working may not gain full credit when integration by parts is needed. Choosing \(u = \mathrm{e}^{3x}\) leads to \(\int \dfrac{x^2}{2}\,3\mathrm{e}^{3x}\,dx\), a harder integral: M0.
1 mark
A1 \(\dfrac{1}{3}x\mathrm{e}^{3x} - \displaystyle\int \dfrac{1}{3}\mathrm{e}^{3x}\,dx\)
1 mark
M1 (dM1) Integrates \(k\mathrm{e}^{3x}\) to \(\dfrac{k}{3}\mathrm{e}^{3x}\)
Depends on the first M mark.
1 mark
A1 \(\dfrac{1}{3}x\mathrm{e}^{3x} - \dfrac{1}{9}\mathrm{e}^{3x} + c\)
Accept \(\dfrac{1}{9}\mathrm{e}^{3x}(3x - 1) + c\). The \(+c\) is needed.

Question 2

6 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
\[\int_{1}^{9} \dfrac{\ln x}{\sqrt{x}}\,dx\]
giving your answer in the form \(a\ln 3 + b\), where \(a\) and \(b\) are integers.

[Integration by parts: \(\displaystyle\int u\dfrac{dv}{dx}\,dx = uv - \int v\dfrac{du}{dx}\,dx\)]

Model answer

\(u = \ln x\), \(\dfrac{dv}{dx} = x^{-\frac{1}{2}}\), so \(\dfrac{du}{dx} = \dfrac{1}{x}\), \(v = 2x^{\frac{1}{2}}\)
\(\displaystyle\int \dfrac{\ln x}{\sqrt{x}}\,dx = 2x^{\frac{1}{2}}\ln x - \int 2x^{\frac{1}{2}} \times \dfrac{1}{x}\,dx = 2\sqrt{x}\ln x - \int 2x^{-\frac{1}{2}}\,dx = 2\sqrt{x}\ln x - 4\sqrt{x}\)
\(\left[2\sqrt{x}\ln x - 4\sqrt{x}\right]_{1}^{9} = (6\ln 9 - 12) - (0 - 4) = 6\ln 9 - 8\)
\(= 12\ln 3 - 8\)

Explanation

Take \(u = \ln x\), since \(\ln x\) is hard to integrate but simple to differentiate, and \(\dfrac{dv}{dx} = x^{-\frac{1}{2}}\). The second integral simplifies to \(\int 2x^{-\frac{1}{2}}\,dx = 4\sqrt{x}\), and the limits use \(\ln 1 = 0\) and \(\ln 9 = 2\ln 3\). Common errors are taking \(u = x^{-\frac{1}{2}}\) and not writing the answer in the form asked for.

Mark scheme

1 mark
M1 Integrates by parts with \(u = \ln x\), \(\dfrac{dv}{dx} = x^{-\frac{1}{2}}\), to reach \(Ax^{\frac{1}{2}}\ln x - \displaystyle\int Bx^{\frac{1}{2}} \times \dfrac{1}{x}\,dx\). An answer with no working scores 0 marks.
An answer of \(12\ln 3 - 8\) with no working scores 0 marks: the question demands working.
1 mark
A1 \(2x^{\frac{1}{2}}\ln x - \displaystyle\int 2x^{-\frac{1}{2}}\,dx\)
1 mark
M1 (dM1) Integrates \(x^{-\frac{1}{2}}\) to \(kx^{\frac{1}{2}}\)
Depends on the first M mark.
1 mark
A1 \(2\sqrt{x}\ln x - 4\sqrt{x}\)
1 mark
M1 (ddM1) Substitutes 9 and 1 and subtracts, using \(\ln 1 = 0\)
Depends on both previous M marks.
1 mark
A1 \(12\ln 3 - 8\) cso
From \((6\ln 9 - 12) - (0 - 4)\). Do not accept \(6\ln 9 - 8\) as the final answer (not in the form asked) or 5.18 alone.

Question 3

6 marks

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

Figure 1 shows a sketch of part of the curve \(C\) with equation
\[y = (x - 1)\mathrm{e}^{-x}\]
The curve meets the \(x\)-axis at the point \(A\). The region \(R\), shown shaded in Figure 1, is bounded by \(C\), the \(x\)-axis and the line with equation \(x = 3\).

Find the exact area of \(R\), giving your answer in the form \(\dfrac{a\mathrm{e}^2 + b}{\mathrm{e}^3}\), where \(a\) and \(b\) are integers.

[Integration by parts: \(\displaystyle\int u\dfrac{dv}{dx}\,dx = uv - \int v\dfrac{du}{dx}\,dx\)]

Figure 1: a sketch of part of the curve C and the vertical line x = 3. On the left the curve is below the x-axis; it rises to cross the x-axis at A, reaches a maximum and then decreases slowly, staying above the x-axis. The region R between C, the x-axis and the line x = 3 is shaded.

Model answer

\(C\) meets the \(x\)-axis where \(x - 1 = 0\) (since \(\mathrm{e}^{-x} > 0\)), so \(A\) is at \(x = 1\)
\(u = x - 1\), \(\dfrac{dv}{dx} = \mathrm{e}^{-x}\), so \(\dfrac{du}{dx} = 1\), \(v = -\mathrm{e}^{-x}\)
\(\displaystyle\int (x - 1)\mathrm{e}^{-x}\,dx = -(x - 1)\mathrm{e}^{-x} + \int \mathrm{e}^{-x}\,dx = -(x - 1)\mathrm{e}^{-x} - \mathrm{e}^{-x} = -x\mathrm{e}^{-x}\)
Area \(= \left[-x\mathrm{e}^{-x}\right]_{1}^{3} = -3\mathrm{e}^{-3} + \mathrm{e}^{-1} = \dfrac{\mathrm{e}^2 - 3}{\mathrm{e}^3}\)

Explanation

The curve crosses the axis where \(x - 1 = 0\), because \(\mathrm{e}^{-x}\) is never zero, so the limits are 1 and 3. Take \(u = x - 1\) and \(v = -\mathrm{e}^{-x}\); the two terms combine to \(-x\mathrm{e}^{-x}\), and \(\mathrm{e}^{-1} - 3\mathrm{e}^{-3}\) is written over \(\mathrm{e}^3\) to give the form asked for. Common errors are taking \(v = \mathrm{e}^{-x}\) and other sign slips from the \(-x\) in the exponent.

Mark scheme

1 mark
B1 \(A\) is at \(x = 1\), so the limits are 1 and 3. An answer with no working scores 0 marks.
An answer of \(\dfrac{\mathrm{e}^2 - 3}{\mathrm{e}^3}\) with no working scores 0 marks: the question demands working.
1 mark
M1 Integrates by parts with \(u = x - 1\), \(\dfrac{dv}{dx} = \mathrm{e}^{-x}\): \(\pm(x - 1)\mathrm{e}^{-x} \pm \displaystyle\int \mathrm{e}^{-x}\,dx\)
Alternative: expand to \(x\mathrm{e}^{-x} - \mathrm{e}^{-x}\) and integrate \(x\mathrm{e}^{-x}\) by parts; mark in the same way.
1 mark
A1 \(-(x - 1)\mathrm{e}^{-x} + \displaystyle\int \mathrm{e}^{-x}\,dx\)
1 mark
A1 \(-(x - 1)\mathrm{e}^{-x} - \mathrm{e}^{-x}\), i.e. \(-x\mathrm{e}^{-x}\)
1 mark
M1 (dM1) Substitutes 3 and 1 and subtracts
Depends on the M mark for integrating by parts.
1 mark
A1 \(\dfrac{\mathrm{e}^2 - 3}{\mathrm{e}^3}\) cso
From \(-3\mathrm{e}^{-3} + \mathrm{e}^{-1}\). Accept \(a = 1\), \(b = -3\). \(\mathrm{e}^{-1} - 3\mathrm{e}^{-3}\) must be written in the form asked for to score this mark. Not 0.219 alone.

Question 4

8 marks

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

The curve \(C\) shown in Figure 1 is defined by
\[x = \mathrm{e}^{t}, \quad y = t(2 - t), \quad 0 \leqslant t \leqslant 2\]
It meets the \(x\)-axis at \(A\) and \(B\), and \(R\) is the shaded region between \(C\) and the \(x\)-axis.

Show that the area of \(R\) equals
\[\int_{0}^{2} \left(2t - t^2\right)\mathrm{e}^{t}\,dt\]
Hence, using algebraic integration, find the exact area of \(R\).

[Integration by parts: \(\displaystyle\int u\dfrac{dv}{dx}\,dx = uv - \int v\dfrac{du}{dx}\,dx\)]

Figure 1: a sketch of the curve C. It starts at A on the positive x-axis, rises steeply to a maximum and then falls more gently to meet the x-axis again at B. The region R between C and the x-axis from A to B is shaded.

Model answer

\(\dfrac{dx}{dt} = \mathrm{e}^{t}\), so area \(= \displaystyle\int y\,\dfrac{dx}{dt}\,dt = \int t(2 - t)\mathrm{e}^{t}\,dt\)
\(y = 0\) when \(t = 0\) (\(A\), \(x = 1\)) and \(t = 2\) (\(B\), \(x = \mathrm{e}^2\)), so area \(= \displaystyle\int_{0}^{2} \left(2t - t^2\right)\mathrm{e}^{t}\,dt\)
\(\displaystyle\int \left(2t - t^2\right)\mathrm{e}^{t}\,dt = \left(2t - t^2\right)\mathrm{e}^{t} - \int (2 - 2t)\mathrm{e}^{t}\,dt\)
\(\displaystyle\int (2 - 2t)\mathrm{e}^{t}\,dt = (2 - 2t)\mathrm{e}^{t} + \int 2\mathrm{e}^{t}\,dt = (2 - 2t)\mathrm{e}^{t} + 2\mathrm{e}^{t} = (4 - 2t)\mathrm{e}^{t}\)
So the integral is \(\left(2t - t^2 - 4 + 2t\right)\mathrm{e}^{t} = -(t - 2)^2\mathrm{e}^{t}\)
Area \(= \left[-(t - 2)^2\mathrm{e}^{t}\right]_{0}^{2} = 0 - (-4) = 4\)

Explanation

For the parametric area use \(\int y\,\dfrac{dx}{dt}\,dt\) with the \(t\)-values where \(y = 0\), which gives \(\int_{0}^{2} (2t - t^2)\mathrm{e}^{t}\,dt\). The quadratic factor needs integration by parts twice, each time with \(u\) the polynomial and \(\dfrac{dv}{dt} = \mathrm{e}^{t}\), and the result simplifies to \(-(t - 2)^2\mathrm{e}^{t}\). Common errors are sign slips in the second application and using the \(x\)-limits 1 and \(\mathrm{e}^2\) with the \(t\)-integral.

Mark scheme

1 mark
M1 Uses area \(= \displaystyle\int y\,\dfrac{dx}{dt}\,dt\) with \(\dfrac{dx}{dt} = \mathrm{e}^{t}\): \(\displaystyle\int t(2 - t)\mathrm{e}^{t}\,dt\). An answer with no working scores 0 marks.
An answer with no working scores 0 marks: the first step is a 'show that' and the question demands working, so an area of 4 alone scores 0.
1 mark
B1 The limits: \(t = 0\) at \(A\) (\(x = 1\)) and \(t = 2\) at \(B\) (\(x = \mathrm{e}^2\)), where \(y = 0\)
1 mark
A1* \(\displaystyle\int_{0}^{2} \left(2t - t^2\right)\mathrm{e}^{t}\,dt\) with the integral sign, the limits and \(dt\)
cso. Requires the M1 and the B1.
1 mark
M1 First application of integration by parts with \(u = 2t - t^2\), \(\dfrac{dv}{dt} = \mathrm{e}^{t}\): \(\left(2t - t^2\right)\mathrm{e}^{t} - \displaystyle\int (2 - 2t)\mathrm{e}^{t}\,dt\)
Allow sign slips in the derivative of \(2t - t^2\). The tabular (DI) method is acceptable. Alternative: the Cartesian form \(y = \ln x\,(2 - \ln x)\) integrated from 1 to \(\mathrm{e}^2\) is valid ('Hence' refers to the integral); mark equivalent steps the same way.
1 mark
M1 (dM1) Second application: \(\displaystyle\int (2 - 2t)\mathrm{e}^{t}\,dt = (2 - 2t)\mathrm{e}^{t} + \int 2\mathrm{e}^{t}\,dt\)
Depends on the previous M mark.
1 mark
A1 \(\left(-t^2 + 4t - 4\right)\mathrm{e}^{t}\) oe, i.e. \(-(t - 2)^2\mathrm{e}^{t}\)
Accept unsimplified \(\left(2t - t^2\right)\mathrm{e}^{t} - (2 - 2t)\mathrm{e}^{t} - 2\mathrm{e}^{t}\).
1 mark
M1 (ddM1) Substitutes 2 and 0 and subtracts, using \(\mathrm{e}^{0} = 1\)
Depends on both M marks for integrating by parts.
1 mark
A1 4 cso
From \(0 - (-4)\).

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