CAIE P3 2013 November — Question 2 4 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2013
SessionNovember
Marks4
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TopicIntegration by Substitution
TypeAlgebraic manipulation before substitution
DifficultyModerate -0.3 This is a straightforward substitution question requiring algebraic manipulation to express x in terms of u before integrating. While it requires recognizing that x = (u-1)/3 and rewriting the integrand, it's a standard technique tested routinely in P3. The manipulation is mechanical once the substitution is made, making it slightly easier than average but not trivial since students must handle the algebraic rearrangement correctly.
Spec1.08h Integration by substitution
2 Use the substitution \(u = 3 x + 1\) to find \(\int \frac { 3 x } { 3 x + 1 } \mathrm {~d} x\).
AnswerMarks Guidance
Carry out complete substitution including the use of \(\frac{du}{dx} = 3\)M1
Obtain \(\int\left(\frac{1}{3} - \frac{1}{3u}\right) du\)A1
Integrate to obtain form \(k_1u + k_2 \ln u\) or \(k_1u + k_2 \ln 3u\) where \(k_1k_2 \neq 0\)M1
Obtain \(\frac{1}{3}(3x+1) - \frac{1}{3}\ln(3x+1)\) or equivalent, condoning absence of modulus signs and \(+ c\)A1 [4] 
Carry out complete substitution including the use of $\frac{du}{dx} = 3$ | M1 |
Obtain $\int\left(\frac{1}{3} - \frac{1}{3u}\right) du$ | A1 |
Integrate to obtain form $k_1u + k_2 \ln u$ or $k_1u + k_2 \ln 3u$ where $k_1k_2 \neq 0$ | M1 |
Obtain $\frac{1}{3}(3x+1) - \frac{1}{3}\ln(3x+1)$ or equivalent, condoning absence of modulus signs and $+ c$ | A1 | [4]
2 Use the substitution $u = 3 x + 1$ to find $\int \frac { 3 x } { 3 x + 1 } \mathrm {~d} x$.

\hfill \mbox{\textit{CAIE P3 2013 Q2 [4]}}
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