| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2020 |
| Session | October |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Show definite integral equals specific value (requiring partial fractions or complex algebra) |
| Difficulty | Standard +0.8 This is a multi-step integration problem requiring substitution with careful handling of limits, followed by partial fractions and logarithm manipulation. While the techniques are standard A-level Further Maths content, the question demands precision in algebraic manipulation across multiple stages and the substitution involving square roots adds complexity. It's moderately challenging but within reach for well-prepared FM students. |
| Spec | 1.08h Integration by substitution1.08j Integration using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = u^2+1 \Rightarrow dx = 2u\,du\) | B1 | Exact equivalent e.g. \(\frac{dx}{du}=2u\) |
| Full substitution: \(\displaystyle\int\frac{3\,dx}{(x-1)(3+2\sqrt{x-1})} = \int\frac{3\times 2u\,du}{(u^2+1-1)(3+2u)}\) | M1 | Attempts full substitution including \(dx \to \ldots u\,du\) |
| Correct limits e.g. \(p=2, q=3\) | B1 | |
| \(= \displaystyle\int\frac{3\times 2u\,du}{u^2(3+2u)} = \int\frac{6\,du}{u(3+2u)}\) | A1* | Clear reasoning with one fully correct intermediate line |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{6}{u(3+2u)} = \dfrac{A}{u} + \dfrac{B}{3+2u} \Rightarrow A=\ldots, B=\ldots\) | M1 | Uses correct PF form leading to values of \(A\) and \(B\) |
| \(\dfrac{6}{u(3+2u)} = \dfrac{2}{u} - \dfrac{4}{3+2u}\) | A1 | Not scored for just correct values of \(A\) and \(B\) alone |
| \(\displaystyle\int\frac{6\,du}{u(3+2u)} = 2\ln u - 2\ln(3+2u)\) \((+c)\) | dM1, A1ft | Overall problem-solving mark; look for \(P\ln u + Q\ln(3+2u)\); correct integration for their PF |
| Uses limits \(u=3\), \(u=2\) with correct ln law leading to \(k\ln b\) | M1 | At least one correct application of log laws; condone bracketing slips |
| \(\ln\dfrac{49}{36}\) | A1 | Answers without working send to review |
## Question 10(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = u^2+1 \Rightarrow dx = 2u\,du$ | B1 | Exact equivalent e.g. $\frac{dx}{du}=2u$ |
| Full substitution: $\displaystyle\int\frac{3\,dx}{(x-1)(3+2\sqrt{x-1})} = \int\frac{3\times 2u\,du}{(u^2+1-1)(3+2u)}$ | M1 | Attempts full substitution including $dx \to \ldots u\,du$ |
| Correct limits e.g. $p=2, q=3$ | B1 | |
| $= \displaystyle\int\frac{3\times 2u\,du}{u^2(3+2u)} = \int\frac{6\,du}{u(3+2u)}$ | A1* | Clear reasoning with one fully correct intermediate line |
## Question 10(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{6}{u(3+2u)} = \dfrac{A}{u} + \dfrac{B}{3+2u} \Rightarrow A=\ldots, B=\ldots$ | M1 | Uses correct PF form leading to values of $A$ and $B$ |
| $\dfrac{6}{u(3+2u)} = \dfrac{2}{u} - \dfrac{4}{3+2u}$ | A1 | Not scored for just correct values of $A$ and $B$ alone |
| $\displaystyle\int\frac{6\,du}{u(3+2u)} = 2\ln u - 2\ln(3+2u)$ $(+c)$ | dM1, A1ft | Overall problem-solving mark; look for $P\ln u + Q\ln(3+2u)$; correct integration for their PF |
| Uses limits $u=3$, $u=2$ with correct ln law leading to $k\ln b$ | M1 | At least one correct application of log laws; condone bracketing slips |
| $\ln\dfrac{49}{36}$ | A1 | Answers without working send to review |\begin{enumerate}
\item (a) Use the substitution $x = u ^ { 2 } + 1$ to show that
\end{enumerate}
$$\int _ { 5 } ^ { 10 } \frac { 3 \mathrm {~d} x } { ( x - 1 ) ( 3 + 2 \sqrt { x - 1 } ) } = \int _ { p } ^ { q } \frac { 6 \mathrm {~d} u } { u ( 3 + 2 u ) }$$
where $p$ and $q$ are positive constants to be found.\\
(b) Hence, using algebraic integration, show that
$$\int _ { 5 } ^ { 10 } \frac { 3 \mathrm {~d} x } { ( x - 1 ) ( 3 + 2 \sqrt { x - 1 } ) } = \ln a$$
where $a$ is a rational constant to be found.
\hfill \mbox{\textit{Edexcel Paper 1 2020 Q10 [10]}}