Edexcel C4 2014 January — Question 2 10 marks

Exam BoardEdexcel
ModuleC4 (Core Mathematics 4)
Year2014
SessionJanuary
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Parts
TypeIndependent multi-part (different techniques)
DifficultyStandard +0.3 Part (i) is a standard integration by parts application with a straightforward substitution. Part (ii) involves routine partial fractions with a repeated linear factor and simple integration of the resulting terms. Both are textbook-standard techniques with no novel insight required, making this slightly easier than average for C4.
Spec1.02y Partial fractions: decompose rational functions1.08i Integration by parts

2. (i) Find $$\int x \cos \left( \frac { x } { 2 } \right) \mathrm { d } x$$ (ii) (a) Express \(\frac { 1 } { x ^ { 2 } ( 1 - 3 x ) }\) in partial fractions.
(b) Hence find, for \(0 < x < \frac { 1 } { 3 }\) $$\int \frac { 1 } { x ^ { 2 } ( 1 - 3 x ) } \mathrm { d } x$$

(i)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\int x\cos\left(\frac{x}{2}\right)dx\), with \(u=x \Rightarrow \frac{du}{dx}=1\); \(\frac{dv}{dx}=\cos\left(\frac{x}{2}\right) \Rightarrow v=2\sin\left(\frac{x}{2}\right)\)M1 A1
\(=2x\sin\left(\frac{x}{2}\right)-\int 2\sin\left(\frac{x}{2}\right)\{dx\}\)A1
\(=2x\sin\left(\frac{x}{2}\right)+4\cos\left(\frac{x}{2}\right)\{+c\}\)A1
[3]
(ii)(a)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{1}{x^2(1-3x)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{(1-3x)}\)B1 At least one of "\(B\)" or "\(C\)" correct. Breaks up their partial fraction correctly into three terms and both "\(B\)"\(=1\) and "\(C\)"\(=9\).
\(B=1, C=9\)B1 cso See notes below.
\(1 \equiv Ax(1-3x)+B(1-3x)+Cx^2\); \(x=0, 1=B\); \(x=\frac{1}{3}, 1=\frac{1}{9}C \Rightarrow C=9\); \(x^2\) terms: \(0=-3A+C \Rightarrow A=3\)M1 Writes down a correct identity and attempts to find the value of either one of "\(A\)", "\(B\)" or "\(C\)".
A1Correct value for "\(A\)" which is found using a correct identity and follows from their partial fraction decomposition.
[4]
(ii)(b)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\int \frac{1}{x^2(1-3x)}dx=\int\left[\frac{3}{x}+\frac{1}{x^2}+\frac{9}{(1-3x)}\right]dx\)M1 Either \(\pm\frac{P}{x}\to\pm a\ln x\) or \(\pm a\ln kx\); or \(\pm\frac{Q}{x^2}\to\pm b x^{-1}\) or \(\frac{R}{(1-3x)}\to\pm c\ln(1-3x)\)
\(=3\ln x+\frac{x^{-1}}{(-1)}+\frac{9}{(-3)}\ln(1-3x)\{+c\}\)A1ft At least two terms correctly integrated; All three terms correctly integrated. Ignore absence of "\(+c\)"
[3]
[10]
Notes for 2.(i)
M1: Integration by parts is applied in the form \(\pm \lambda x\sin\left(\frac{x}{2}\right)-\int \mu\sin\left(\frac{x}{2}\right)\{dx\}\) (where \(\lambda \neq 0, \mu \neq 0\))
A1: \(2x\sin\left(\frac{x}{2}\right)-\int 2\sin\left(\frac{x}{2}\right)\{dx\}\) or equivalent. Can be un-simplified.
**(i)**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int x\cos\left(\frac{x}{2}\right)dx$, with $u=x \Rightarrow \frac{du}{dx}=1$; $\frac{dv}{dx}=\cos\left(\frac{x}{2}\right) \Rightarrow v=2\sin\left(\frac{x}{2}\right)$ | M1 A1 | |
| $=2x\sin\left(\frac{x}{2}\right)-\int 2\sin\left(\frac{x}{2}\right)\{dx\}$ | A1 | |
| $=2x\sin\left(\frac{x}{2}\right)+4\cos\left(\frac{x}{2}\right)\{+c\}$ | A1 | |
| | | [3] |

**(ii)(a)**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{x^2(1-3x)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{(1-3x)}$ | B1 | At least one of "$B$" or "$C$" correct. Breaks up their partial fraction correctly into **three terms** and both "$B$"$=1$ and "$C$"$=9$. |
| $B=1, C=9$ | B1 cso | See notes below. |
| $1 \equiv Ax(1-3x)+B(1-3x)+Cx^2$; $x=0, 1=B$; $x=\frac{1}{3}, 1=\frac{1}{9}C \Rightarrow C=9$; $x^2$ terms: $0=-3A+C \Rightarrow A=3$ | M1 | Writes down a **correct identity** and attempts to find the value of either one of "$A$", "$B$" or "$C$". |
| | A1 | Correct value for "$A$" which is found using a correct identity and follows from their partial fraction decomposition. |
| | | [4] |

**(ii)(b)**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int \frac{1}{x^2(1-3x)}dx=\int\left[\frac{3}{x}+\frac{1}{x^2}+\frac{9}{(1-3x)}\right]dx$ | M1 | Either $\pm\frac{P}{x}\to\pm a\ln x$ or $\pm a\ln kx$; or $\pm\frac{Q}{x^2}\to\pm b x^{-1}$ or $\frac{R}{(1-3x)}\to\pm c\ln(1-3x)$ |
| $=3\ln x+\frac{x^{-1}}{(-1)}+\frac{9}{(-3)}\ln(1-3x)\{+c\}$ | A1ft | **At least two terms correctly integrated**; **All three terms correctly integrated**. Ignore absence of "$+c$" |
| | | [3] |
| | | [10] |

**Notes for 2.(i)**

M1: Integration by parts is applied in the form $\pm \lambda x\sin\left(\frac{x}{2}\right)-\int \mu\sin\left(\frac{x}{2}\right)\{dx\}$ (where $\lambda \neq 0, \mu \neq 0$)

A1: $2x\sin\left(\frac{x}{2}\right)-\int 2\sin\left(\frac{x}{2}\right)\{dx\}$ or equivalent. Can be un-simplified.

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2. (i) Find

$$\int x \cos \left( \frac { x } { 2 } \right) \mathrm { d } x$$

(ii) (a) Express $\frac { 1 } { x ^ { 2 } ( 1 - 3 x ) }$ in partial fractions.\\
(b) Hence find, for $0 < x < \frac { 1 } { 3 }$

$$\int \frac { 1 } { x ^ { 2 } ( 1 - 3 x ) } \mathrm { d } x$$

\hfill \mbox{\textit{Edexcel C4 2014 Q2 [10]}}