| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure (Further Additional Pure) |
| Year | 2018 |
| Session | March |
| Marks | 14 |
| Topic | Integration by Parts |
| Type | Reduction formula or recurrence |
| Difficulty | Challenging +1.8 This is a challenging Further Maths question requiring derivation of a reduction formula via integration by parts, then applying it to a surface of revolution problem. While the techniques are standard for Further Maths students (reduction formulas, parametric surface area), the multi-stage nature, algebraic manipulation required in part (ii), and the need to connect the reduction formula to the geometric application make this substantially harder than typical A-level questions but not exceptionally difficult for Further Additional Pure. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian4.08d Volumes of revolution: about x and y axes8.06a Reduction formulae: establish, use, and evaluate recursively8.06b Arc length and surface area: of revolution, cartesian or parametric |
| Answer | Marks | Guidance |
|---|---|---|
| \(I_1 = \int_0^{\sqrt{3}} t\sqrt{1+t^2} \, dt = \left[\frac{1}{3}(1+t^2)^{\frac{3}{2}}\right]_0^{\sqrt{3}} = \frac{1}{3}(8-1) = \frac{7}{3}\) | M1, A1 [2] | By "recognition" (*reverse Chain Rule*) integration or any other suitable method (e.g. by substitution): MUST have \(k(1+t^2)^{1.5}\); AG from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| \(I_n = \int_0^{\sqrt{3}} t^n\sqrt{1+t^2} \, dt = \int_0^{\sqrt{3}} t^{n-1} \cdot t\sqrt{1+t^2} \, dt\) | M1 | Correct splitting and attempt at *integration by parts* |
| \(= \left[t^{n-1} \cdot \frac{1}{3}(1+t^2)^{\frac{3}{2}}\right]_0^{\sqrt{3}} - \int_0^{\sqrt{3}} (n-1)t^{n-2} \cdot \frac{1}{3}(1+t^2)^{\frac{3}{2}} \, dt\) | A1, A1 | One for each correct part |
| \(\Rightarrow 3I_e = (\sqrt{3})^{.8} - 0 - (n-1)\int_0^{\sqrt{3}} t^{n-2}(1+t^2)\sqrt{1+t^2} \, dt\) | M1 | Splitting up the power of \((1+t^2)\) suitably |
| \(\Rightarrow 3I_e = 8(\sqrt{3})^{n-1} - (n-1)[I_{n-2} + I_n]\) | A1 | 2nd integral correctly identified in terms of \(I\)'s |
| \(\Rightarrow (n+2)I_n = 8(\sqrt{3})^{n-1} - (n-1)I_{n-2}\) | A1 [6] | AG Legitimately obtained from correct rearrangement of \(I\) terms |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = (30t)^2 + (30t)^2 = 900t^2(1+t^2)\) soi | M1, A1 | Attempted use of formula; Correct, any form |
| \(A = 2\pi \int_0^{\sqrt{3}} 15t^2 \cdot 30t\sqrt{1+t^2} \, dt\) | M1 | Formula used correctly with relevant terms in \(t\) |
| \(= 900\pi I_3\) i.e. \(k = 900, m = 3\) | A1 | |
| Use of \(n = 3\) in Reduction Formula: \(5I_3 = 8(\sqrt{3})^5 - 2 \times \frac{7}{3} \Rightarrow I_3 = \frac{58}{15}\) | M1, A1 [6] | cao |
| so that \(A = 3480\pi\) | A1 |
## (i)
| $I_1 = \int_0^{\sqrt{3}} t\sqrt{1+t^2} \, dt = \left[\frac{1}{3}(1+t^2)^{\frac{3}{2}}\right]_0^{\sqrt{3}} = \frac{1}{3}(8-1) = \frac{7}{3}$ | M1, A1 [2] | By "recognition" (*reverse Chain Rule*) integration or any other suitable method (e.g. by substitution): MUST have $k(1+t^2)^{1.5}$; AG from correct working |
## (ii)
| $I_n = \int_0^{\sqrt{3}} t^n\sqrt{1+t^2} \, dt = \int_0^{\sqrt{3}} t^{n-1} \cdot t\sqrt{1+t^2} \, dt$ | M1 | Correct splitting and attempt at *integration by parts* |
| $= \left[t^{n-1} \cdot \frac{1}{3}(1+t^2)^{\frac{3}{2}}\right]_0^{\sqrt{3}} - \int_0^{\sqrt{3}} (n-1)t^{n-2} \cdot \frac{1}{3}(1+t^2)^{\frac{3}{2}} \, dt$ | A1, A1 | One for each correct part |
| $\Rightarrow 3I_e = (\sqrt{3})^{.8} - 0 - (n-1)\int_0^{\sqrt{3}} t^{n-2}(1+t^2)\sqrt{1+t^2} \, dt$ | M1 | Splitting up the power of $(1+t^2)$ suitably |
| $\Rightarrow 3I_e = 8(\sqrt{3})^{n-1} - (n-1)[I_{n-2} + I_n]$ | A1 | 2nd integral correctly identified in terms of $I$'s |
| $\Rightarrow (n+2)I_n = 8(\sqrt{3})^{n-1} - (n-1)I_{n-2}$ | A1 [6] | AG Legitimately obtained from correct rearrangement of $I$ terms |
## (iii)
| $\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = (30t)^2 + (30t)^2 = 900t^2(1+t^2)$ **soi** | M1, A1 | Attempted use of formula; Correct, any form |
| $A = 2\pi \int_0^{\sqrt{3}} 15t^2 \cdot 30t\sqrt{1+t^2} \, dt$ | M1 | Formula used correctly with relevant terms in $t$ |
| $= 900\pi I_3$ i.e. $k = 900, m = 3$ | A1 | |
| Use of $n = 3$ in Reduction Formula: $5I_3 = 8(\sqrt{3})^5 - 2 \times \frac{7}{3} \Rightarrow I_3 = \frac{58}{15}$ | M1, A1 [6] | cao |
| so that $A = 3480\pi$ | A1 | |
---6 In this question you must show detailed reasoning.
It is given that $I _ { n } = \int _ { 0 } ^ { \sqrt { 3 } } t ^ { n } \sqrt { 1 + t ^ { 2 } } \mathrm {~d} t$ for integers $n \geqslant 0$.\\
(i) Show that $I _ { 1 } = \frac { 7 } { 3 }$.\\
(ii) Prove that, for $n \geqslant 2 , ( n + 2 ) I _ { n } = 8 ( \sqrt { 3 } ) ^ { n - 1 } - ( n - 1 ) I _ { n - 2 }$.
The curve $C$ is defined parametrically by
$$x = 10 t ^ { 3 } , y = 15 t ^ { 2 } \text { for } 0 \leqslant t \leqslant \sqrt { 3 }$$
When the curve $C$ is rotated through $2 \pi$ radians about the $x$-axis, a surface of revolution is formed with surface area $A$.\\
(iii) Determine
\begin{itemize}
\item the values of the integers $k$ and $m$ such that $A = k \pi I _ { m }$,
\item the exact value of $A$.
\end{itemize}
\hfill \mbox{\textit{OCR Further Additional Pure 2018 Q6 [14]}}