| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Partial fractions after substitution |
| Difficulty | Challenging +1.2 This is a structured multi-part integration question requiring substitution followed by partial fractions. While it involves several techniques (substitution, partial fractions, definite integration), the question provides the substitution and guides students through each step. The algebraic manipulation is moderately involved but follows standard A-level procedures without requiring novel insight. Slightly above average difficulty due to the multi-step nature and combination of techniques, but well within the scope of typical A-level Pure Mathematics questions. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| \(du = e^x dx\) | B1 | Correct statement linking \(du\) and \(dx\); or \(dx = \dfrac{1}{u+2}du\) |
| \(\displaystyle\int \frac{7(u+2)-8}{u^2} \cdot \frac{1}{u+2}du\) | M1 | Use \(e^x = u+2\) to attempt integrand in terms of \(u\); must see clear evidence of substitution including how \(e^x dx\) is dealt with; M0 for going straight from \(7e^x - 8\) to \(7u+6\) with no justification; must include \(du\) |
| \(= \displaystyle\int \frac{7u+14-8}{u^2(u+2)}du = \int \frac{7u+6}{u^2(u+2)}du\) | A1 | Correct integrand; including both integral sign and \(du\) throughout, as AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(\dfrac{A}{u} + \dfrac{B}{u^2} + \dfrac{C}{u+2} = \dfrac{7u+6}{u^2(u+2)}\); \(Au(u+2)+B(u+2)+Cu^2 = 7u+6\) | M1 | Attempt correct partial fractions; may have \(\dfrac{Au+B}{u^2} + \dfrac{C}{u+2}\) but M0 for just \(\dfrac{B}{u^2}\) with no \(\dfrac{A}{u}\) |
| \(\dfrac{2}{u} + \dfrac{3}{u^2} - \dfrac{2}{u+2}\) | A1 | Correct partial fractions; \(A=2, B=3, C=-2\); possibly implied by values |
| \(2\ln | u | - 2\ln |
| A1FT | FT on their two or three fractions as long as \(ku^{-2}\) and one or two fractions each with a linear denominator; condone no brackets as long as implied by later working | |
| \((2\ln4 - 2\ln6 - \tfrac{3}{4}) - (2\ln2 - 2\ln4 - \tfrac{3}{2})\) | M1 | Attempt use of correct limits; correct order and subtraction; \(u\) or \(x\) but commensurate with their integral |
| \(\left(\dfrac{3}{2} - \dfrac{3}{4}\right) + \ln\left(\dfrac{4\times4}{6\times2}\right)^2\) | M1 | Attempt to rearrange correct numerical integral to required form; correct attempt to combine ln terms |
| \(\dfrac{3}{4} + \ln\dfrac{16}{9}\) | A1 | Obtain \(\dfrac{3}{4} + \ln\dfrac{16}{9}\); condone \(\dfrac{3}{4} + 2\ln\dfrac{4}{3}\); fractions must be simplified; A0 if further work done e.g. multiplying by constant to clear fractions |
## Question 12(a):
$du = e^x dx$ | **B1** | Correct statement linking $du$ and $dx$; or $dx = \dfrac{1}{u+2}du$
$\displaystyle\int \frac{7(u+2)-8}{u^2} \cdot \frac{1}{u+2}du$ | **M1** | Use $e^x = u+2$ to attempt integrand in terms of $u$; must see clear evidence of substitution including how $e^x dx$ is dealt with; **M0** for going straight from $7e^x - 8$ to $7u+6$ with no justification; must include $du$
$= \displaystyle\int \frac{7u+14-8}{u^2(u+2)}du = \int \frac{7u+6}{u^2(u+2)}du$ | **A1** | Correct integrand; including both integral sign and $du$ throughout, as **AG**
**[3]**
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## Question 12(b):
$\dfrac{A}{u} + \dfrac{B}{u^2} + \dfrac{C}{u+2} = \dfrac{7u+6}{u^2(u+2)}$; $Au(u+2)+B(u+2)+Cu^2 = 7u+6$ | **M1** | Attempt correct partial fractions; may have $\dfrac{Au+B}{u^2} + \dfrac{C}{u+2}$ but **M0** for just $\dfrac{B}{u^2}$ with no $\dfrac{A}{u}$
$\dfrac{2}{u} + \dfrac{3}{u^2} - \dfrac{2}{u+2}$ | **A1** | Correct partial fractions; $A=2, B=3, C=-2$; possibly implied by values
$2\ln|u| - 2\ln|u+2| - 3u^{-1}$ | **M1** | Attempt integration of $\dfrac{B}{u^2}$ and at least one of $\dfrac{A}{u}$ or $\dfrac{C}{u+2}$
| **A1FT** | FT on their two or three fractions as long as $ku^{-2}$ and one or two fractions each with a linear denominator; condone no brackets as long as implied by later working
$(2\ln4 - 2\ln6 - \tfrac{3}{4}) - (2\ln2 - 2\ln4 - \tfrac{3}{2})$ | **M1** | Attempt use of correct limits; correct order and subtraction; $u$ or $x$ but commensurate with their integral
$\left(\dfrac{3}{2} - \dfrac{3}{4}\right) + \ln\left(\dfrac{4\times4}{6\times2}\right)^2$ | **M1** | Attempt to rearrange correct numerical integral to required form; correct attempt to combine ln terms
$\dfrac{3}{4} + \ln\dfrac{16}{9}$ | **A1** | Obtain $\dfrac{3}{4} + \ln\dfrac{16}{9}$; condone $\dfrac{3}{4} + 2\ln\dfrac{4}{3}$; fractions must be simplified; **A0** if further work done e.g. multiplying by constant to clear fractions
**[7]**12
\begin{enumerate}[label=(\alph*)]
\item Use the substitution $u = \mathrm { e } ^ { x } - 2$ to show that
$$\int \frac { 7 \mathrm { e } ^ { x } - 8 } { \left( \mathrm { e } ^ { x } - 2 \right) ^ { 2 } } \mathrm {~d} x = \int \frac { 7 u + 6 } { u ^ { 2 } ( u + 2 ) } \mathrm { d } u$$
\item Hence show that
$$\int _ { \ln 4 } ^ { \ln 6 } \frac { 7 \mathrm { e } ^ { x } - 8 } { \left( \mathrm { e } ^ { x } - 2 \right) ^ { 2 } } \mathrm {~d} x = a + \ln b$$
where $a$ and $b$ are rational numbers to be determined.
\section*{END OF QUESTION PAPER}
\end{enumerate}
\hfill \mbox{\textit{OCR H240/01 2023 Q12 [10]}}