| Exam Board | AQA |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2021 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Reduction formula or recurrence |
| Difficulty | Challenging +1.2 This is a standard reduction formula question requiring integration by parts twice with sinh x, followed by routine application to find a specific value. The technique is well-practiced in Further Maths, though the hyperbolic function and multi-step algebraic manipulation elevate it slightly above average difficulty. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07d Differentiate/integrate: hyperbolic functions4.08f Integrate using partial fractions8.06a Reduction formulae: establish, use, and evaluate recursively |
| 12 | Hence show that \(\int _ { 0 } ^ { 1 } x ^ { 4 } \sinh x d x = \frac { 9 } { 2 } e + \frac { 65 } { 2 } e ^ { - 1 } - 24\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Selects integration by parts method | M1 (AO1.1a) | Method must be appropriate |
| \(u = x^n\), \(v' = \sinh x\), \(u' = nx^{n-1}\), \(v = \cosh x\) | A1 (AO1.1b) | Correct expressions for \(u'\) and \(v\) |
| \(S_n = [x^n \cosh x]_0^a - \int_0^a nx^{n-1} \cosh x \, dx\) | A1 (AO1.1b) | Correctly applies IBP first time |
| \(= a^n \cosh a - n\int_0^a x^{n-1} \cosh x \, dx\) | M1 (AO3.1a) | Applies IBP to integral of form \(\int_0^a x^r \cosh x \, dx\) |
| Second IBP: \(u = x^{n-1}\), \(v' = \cosh x\), \(u' = (n-1)x^{n-2}\), \(v = \sinh x\) | A1 (AO1.1b) | Correct result from 2nd IBP, limits substituted correctly in all terms |
| \(S_n = a^n \cosh a - n\left([x^{n-1}\sinh x]_0^a - \int_0^a (n-1)x^{n-2}\sinh x \, dx\right)\) | ||
| \(S_n = a^n \cosh a - n(a^{n-1}\sinh a - (n-1)S_{n-2})\) | M1 (AO1.1a) | Obtains expression for \(S_n\) in terms of \(S_{n-2}\) |
| \(S_n = n(n-1)S_{n-2} + a^n\cosh a - na^{n-1}\sinh a\) | R1 (AO2.1) | Completes rigorous argument including correct use of limits throughout |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Required integral is \(S_4\) with \(a=1\), requires \(S_2\) and \(S_0\) | M1 (AO2.2a) | Deduces integral is \(S_4\) with \(a=1\) |
| \(S_0 = \int_0^1 \sinh x \, dx = [\cosh x]_0^1 = \cosh 1 - 1\) | M1 (AO3.1a) | Uses reduction formula once |
| \(S_2 = (2)(1)S_0 + \cosh 1 - 2\sinh 1 = 2(\cosh 1 - 1) + \cosh 1 - 2\sinh 1 = 3\cosh 1 - 2\sinh 1 - 2\) | M1 (AO3.1a) | Uses reduction formula second time and finds \(S_0\) |
| \(S_4 = (4)(3)S_2 + \cosh 1 - 4\sinh 1\) | M1 (AO1.1a) | Converts both hyperbolic expressions to exponentials |
| \(= 12(3\cosh 1 - 2\sinh 1 - 2) + \cosh 1 - 4\sinh 1 = 37\cosh 1 - 28\sinh 1 - 24\) | ||
| \(= \frac{37}{2}(e + e^{-1}) - 14(e - e^{-1}) - 24 = \frac{9}{2}e + \frac{65}{2}e^{-1} - 24\) | R1 (AO2.1) | Completes rigorous argument |
## Question 12(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Selects integration by parts method | M1 (AO1.1a) | Method must be appropriate |
| $u = x^n$, $v' = \sinh x$, $u' = nx^{n-1}$, $v = \cosh x$ | A1 (AO1.1b) | Correct expressions for $u'$ and $v$ |
| $S_n = [x^n \cosh x]_0^a - \int_0^a nx^{n-1} \cosh x \, dx$ | A1 (AO1.1b) | Correctly applies IBP first time |
| $= a^n \cosh a - n\int_0^a x^{n-1} \cosh x \, dx$ | M1 (AO3.1a) | Applies IBP to integral of form $\int_0^a x^r \cosh x \, dx$ |
| Second IBP: $u = x^{n-1}$, $v' = \cosh x$, $u' = (n-1)x^{n-2}$, $v = \sinh x$ | A1 (AO1.1b) | Correct result from 2nd IBP, limits substituted correctly in all terms |
| $S_n = a^n \cosh a - n\left([x^{n-1}\sinh x]_0^a - \int_0^a (n-1)x^{n-2}\sinh x \, dx\right)$ | | |
| $S_n = a^n \cosh a - n(a^{n-1}\sinh a - (n-1)S_{n-2})$ | M1 (AO1.1a) | Obtains expression for $S_n$ in terms of $S_{n-2}$ |
| $S_n = n(n-1)S_{n-2} + a^n\cosh a - na^{n-1}\sinh a$ | R1 (AO2.1) | Completes rigorous argument including correct use of limits throughout |
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## Question 12(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Required integral is $S_4$ with $a=1$, requires $S_2$ and $S_0$ | M1 (AO2.2a) | Deduces integral is $S_4$ with $a=1$ |
| $S_0 = \int_0^1 \sinh x \, dx = [\cosh x]_0^1 = \cosh 1 - 1$ | M1 (AO3.1a) | Uses reduction formula once |
| $S_2 = (2)(1)S_0 + \cosh 1 - 2\sinh 1 = 2(\cosh 1 - 1) + \cosh 1 - 2\sinh 1 = 3\cosh 1 - 2\sinh 1 - 2$ | M1 (AO3.1a) | Uses reduction formula second time and finds $S_0$ |
| $S_4 = (4)(3)S_2 + \cosh 1 - 4\sinh 1$ | M1 (AO1.1a) | Converts both hyperbolic expressions to exponentials |
| $= 12(3\cosh 1 - 2\sinh 1 - 2) + \cosh 1 - 4\sinh 1 = 37\cosh 1 - 28\sinh 1 - 24$ | | |
| $= \frac{37}{2}(e + e^{-1}) - 14(e - e^{-1}) - 24 = \frac{9}{2}e + \frac{65}{2}e^{-1} - 24$ | R1 (AO2.1) | Completes rigorous argument |
---12 The integral $S _ { n }$ is defined by
$$S _ { n } = \int _ { 0 } ^ { a } x ^ { n } \sinh x \mathrm {~d} x \quad ( n \geq 0 )$$
12
\begin{enumerate}[label=(\alph*)]
\item Show that for $n \geq 2$
$$S _ { n } = n ( n - 1 ) S _ { n - 2 } + a ^ { n } \cosh a - n a ^ { n - 1 } \sinh a$$
\begin{center}
\begin{tabular}{|l|l|}
\hline
12
\item & Hence show that \(\int _ { 0 } ^ { 1 } x ^ { 4 } \sinh x d x = \frac { 9 } { 2 } e + \frac { 65 } { 2 } e ^ { - 1 } - 24\) \\
\hline
\end{tabular}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 2 2021 Q12 [12]}}