| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2021 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Reduction formula or recurrence |
| Difficulty | Challenging +1.8 This is a multi-part FP2 reduction formula question requiring integration by parts to derive a recurrence relation, then applying it twice with parametric surface area. While technically demanding with multiple steps, it follows a standard FP2 template: derive the reduction formula (routine but algebraically careful), verify the surface area integral setup (standard parametric formula), then apply the recurrence twice with boundary evaluation. The algebraic manipulation is substantial but methodical rather than requiring novel insight. Harder than typical A-level due to Further Maths content and length, but standard for FP2 reduction formula questions. |
| Spec | 8.06a Reduction formulae: establish, use, and evaluate recursively8.06b Arc length and surface area: of revolution, cartesian or parametric |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_n = t^{n-1} \times K(4+5t^2)^{\frac{3}{2}} - \int(n-1)t^{n-2} \times K(4+5t^2)^{\frac{3}{2}}\,dt\) | M1 | Splits integrand correctly and applies integration by parts in correct direction |
| \(I_n = t^{n-1} \times \frac{2}{3\times10}(4+5t^2)^{\frac{3}{2}} - \int(n-1)t^{n-2} \times \frac{2}{3\times10}(4+5t^2)^{\frac{3}{2}}\,dt\) | A1 | Correct result of applying parts, need not be simplified |
| \(= t^{n-1} \times \frac{1}{15}(4+5t^2)^{\frac{3}{2}} - \frac{(n-1)}{15}\int t^{n-2}(4+5t^2)^{\frac{1}{2}}\times(4+5t^2)\,dt\) then splitting to identify \(I_n\) and \(I_{n-2}\) | M1 | Splits integrand to identify \(I_n\) and \(I_{n-2}\) (allow if \(I_{n-1}\) appears due to error) |
| \(\Rightarrow 15I_n = t^{n-1}(4+5t^2)^{\frac{3}{2}} - 4(n-1)I_{n-2} - 5(n-1)I_n\) | M1 | Rearranges to make \(I_n\) the subject from an equation in \(I_n\) and \(I_{n-2}\) |
| \(I_n = \dfrac{t^{n-1}}{5(n+2)}(4+5t^2)^{\frac{3}{2}} - \dfrac{4(n-1)}{5(n+2)}I_{n-2}\) | A1* | Correct completion to given result |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Surface area \(= 2\pi\int_0^1 y\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt\) | B1 | Correct parametric formula; must include \(2\pi\) and limits (may be added later, \(2\pi\) must be seen before cancelling) |
| \(\frac{dx}{dt} = \frac{5}{\sqrt{5}}t^4\) and \(\frac{dy}{dt} = 2t^3\) | B1 | Correct derivatives of \(x\) and \(y\) w.r.t. \(t\) |
| \(\int y\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt = \int \frac{1}{2}t^4\sqrt{\left(\frac{5}{\sqrt{5}}t^4\right)^2+(2t^3)^2}\,dt\) | M1 | Applies derivatives and \(y\) to the integral formula |
| \(= \int \frac{1}{2}t^4\sqrt{5t^8+4t^6}\,dt = \frac{1}{2}\int t^7\sqrt{4+5t^2}\,dt\) | M1 | Squares derivatives and takes common factor \(t^3\) from square root |
| Hence surface area \(= \pi\int_0^1 t^7\sqrt{4+5t^2}\,dt\) | A1* | Correct answer with limits, \(dt\) present, \(2\pi\) correctly processed |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([I_1]_0^1 = \left[\frac{1}{15}(4+5t^2)^{\frac{3}{2}}\right]_0^1 = \frac{27}{15}-\frac{8}{15}=\frac{19}{15}\) | B1 | Correct value for \(I_1\) between the limits |
| \(\int_0^1 t^7\sqrt{4+5t^2}\,dt = \left[\frac{t^6}{5\times9}(4+5t^2)^{\frac{3}{2}}\right]_0^1 - \frac{4\times6}{5\times9}[I_5]_0^1\) | M1 | Applies reduction formula from (a) |
| \(= \frac{3}{5} - \frac{8}{15}\left(\left[\frac{t^4}{5\times7}(4+5t^2)^{\frac{3}{2}}\right]_0^1 - \frac{4\times4}{5\times7}[I_3]_0^1\right)\) | M1 | Applies reduction formula two more times to link \(I_1\) and \(I_7\) |
| \(= \frac{3}{5} - \frac{8}{15}\left(\frac{27}{35} - \frac{16}{35}\left(\left[\frac{t^2}{5\times5}(4+5t^2)^{\frac{3}{2}}\right]_0^1 - \frac{4\times2}{5\times5}[I_1]_0^1\right)\right)\) | M1 | Applies limits to reach a value (complete process) |
| Total surface area \(= (\pi)\left[\frac{3}{5}-\frac{8}{15}\left(\frac{27}{35}-\frac{16}{35}\left(\frac{27}{25}-\frac{8}{25}\times\frac{19}{15}\right)\right)\right] = \frac{69509\pi}{196875}\) | A1 | awrt \(1.11\) (3 s.f.); must have all three method marks |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_n = t^{n-1} \times K(4+5t^2)^{\frac{3}{2}} - \int(n-1)t^{n-2} \times K(4+5t^2)^{\frac{3}{2}}\,dt$ | M1 | Splits integrand correctly and applies integration by parts in correct direction |
| $I_n = t^{n-1} \times \frac{2}{3\times10}(4+5t^2)^{\frac{3}{2}} - \int(n-1)t^{n-2} \times \frac{2}{3\times10}(4+5t^2)^{\frac{3}{2}}\,dt$ | A1 | Correct result of applying parts, need not be simplified |
| $= t^{n-1} \times \frac{1}{15}(4+5t^2)^{\frac{3}{2}} - \frac{(n-1)}{15}\int t^{n-2}(4+5t^2)^{\frac{1}{2}}\times(4+5t^2)\,dt$ then splitting to identify $I_n$ and $I_{n-2}$ | M1 | Splits integrand to identify $I_n$ and $I_{n-2}$ (allow if $I_{n-1}$ appears due to error) |
| $\Rightarrow 15I_n = t^{n-1}(4+5t^2)^{\frac{3}{2}} - 4(n-1)I_{n-2} - 5(n-1)I_n$ | M1 | Rearranges to make $I_n$ the subject from an equation in $I_n$ and $I_{n-2}$ |
| $I_n = \dfrac{t^{n-1}}{5(n+2)}(4+5t^2)^{\frac{3}{2}} - \dfrac{4(n-1)}{5(n+2)}I_{n-2}$ | A1* | Correct completion to given result |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Surface area $= 2\pi\int_0^1 y\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt$ | B1 | Correct parametric formula; must include $2\pi$ and limits (may be added later, $2\pi$ must be seen before cancelling) |
| $\frac{dx}{dt} = \frac{5}{\sqrt{5}}t^4$ and $\frac{dy}{dt} = 2t^3$ | B1 | Correct derivatives of $x$ and $y$ w.r.t. $t$ |
| $\int y\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt = \int \frac{1}{2}t^4\sqrt{\left(\frac{5}{\sqrt{5}}t^4\right)^2+(2t^3)^2}\,dt$ | M1 | Applies derivatives and $y$ to the integral formula |
| $= \int \frac{1}{2}t^4\sqrt{5t^8+4t^6}\,dt = \frac{1}{2}\int t^7\sqrt{4+5t^2}\,dt$ | M1 | Squares derivatives and takes common factor $t^3$ from square root |
| Hence surface area $= \pi\int_0^1 t^7\sqrt{4+5t^2}\,dt$ | A1* | Correct answer with limits, $dt$ present, $2\pi$ correctly processed |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[I_1]_0^1 = \left[\frac{1}{15}(4+5t^2)^{\frac{3}{2}}\right]_0^1 = \frac{27}{15}-\frac{8}{15}=\frac{19}{15}$ | B1 | Correct value for $I_1$ between the limits |
| $\int_0^1 t^7\sqrt{4+5t^2}\,dt = \left[\frac{t^6}{5\times9}(4+5t^2)^{\frac{3}{2}}\right]_0^1 - \frac{4\times6}{5\times9}[I_5]_0^1$ | M1 | Applies reduction formula from (a) |
| $= \frac{3}{5} - \frac{8}{15}\left(\left[\frac{t^4}{5\times7}(4+5t^2)^{\frac{3}{2}}\right]_0^1 - \frac{4\times4}{5\times7}[I_3]_0^1\right)$ | M1 | Applies reduction formula two more times to link $I_1$ and $I_7$ |
| $= \frac{3}{5} - \frac{8}{15}\left(\frac{27}{35} - \frac{16}{35}\left(\left[\frac{t^2}{5\times5}(4+5t^2)^{\frac{3}{2}}\right]_0^1 - \frac{4\times2}{5\times5}[I_1]_0^1\right)\right)$ | M1 | Applies limits to reach a value (complete process) |
| Total surface area $= (\pi)\left[\frac{3}{5}-\frac{8}{15}\left(\frac{27}{35}-\frac{16}{35}\left(\frac{27}{25}-\frac{8}{25}\times\frac{19}{15}\right)\right)\right] = \frac{69509\pi}{196875}$ | A1 | awrt $1.11$ (3 s.f.); must have all three method marks |
---\begin{enumerate}
\item In this question you must show all stages of your working.
\end{enumerate}
You must not use the integration facility on your calculator.
$$I _ { n } = \int t ^ { n } \sqrt { 4 + 5 t ^ { 2 } } \mathrm {~d} t \quad n \geqslant 0$$
(a) Show that, for $n > 1$
$$I _ { n } = \frac { t ^ { n - 1 } } { 5 ( n + 2 ) } \left( 4 + 5 t ^ { 2 } \right) ^ { \frac { 3 } { 2 } } - \frac { 4 ( n - 1 ) } { 5 ( n + 2 ) } I _ { n - 2 }$$
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{1241b133-4161-4c04-9b50-067904cc25c2-20_385_394_829_833}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The curve shown in Figure 1 is defined by the parametric equations
$$x = \frac { 1 } { \sqrt { 5 } } t ^ { 5 } \quad y = \frac { 1 } { 2 } t ^ { 4 } \quad 0 \leqslant t \leqslant 1$$
This curve is rotated through $2 \pi$ radians about the $x$-axis to form a hollow open shell.\\
(b) Show that the external surface area of the shell is given by
$$\pi \int _ { 0 } ^ { 1 } t ^ { 7 } \sqrt { 4 + 5 t ^ { 2 } } \mathrm {~d} t$$
Using the results in parts (a) and (b) and making each step of your working clear,\\
(c) determine the value of the external surface area of the shell, giving your answer to 3 significant figures.\\
\hfill \mbox{\textit{Edexcel FP2 2021 Q7 [15]}}