| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure (Further Additional Pure) |
| Year | 2018 |
| Session | December |
| Marks | 13 |
| Topic | Reduction Formulae |
| Type | Algebraic function with square root |
| Difficulty | Challenging +1.8 This is a Further Maths reduction formula question requiring integration by parts to establish the recurrence relation, then applying it twice. While technically demanding with multiple steps and algebraic manipulation of surds, it follows a standard reduction formula template that Further Maths students practice extensively. The verification in part (a) and systematic application in part (c) are routine for this level. |
| Spec | 8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| \(I_1 = \int_1^5 x\sqrt{2 + x^2} \, dx\) | M1 | 1.1 |
| \(= \left[\frac{1}{3}(2 + x^2)^{\frac{3}{2}}\right]_1^5\) | A1 | 1.1 |
| \(= \frac{1}{3}(81\sqrt{3} - 3\sqrt{3}) = 26\sqrt{3}\) | A1 | 2.2a |
| Total: [3] | Substitute \(n = 1\) and attempt integral by substitution or inspection; For \(k(2 + x^2)^{\frac{3}{2}}\), any non-zero \(k\); Correctly gained from correct \(k\) and use of (1, 5) limits |
| Answer | Marks | Guidance |
|---|---|---|
| \(I_n = \int_1^5 x^{n-1} \cdot x\sqrt{2 + x^2} \, dx\) | M1 | 1.1a |
| \(= x^{n-1} \cdot \frac{1}{3}(2 + x^2)^{\frac{3}{2}} - \int (n-1)x^{n-2} \cdot \frac{1}{3}(2 + x^2)^{\frac{3}{2}} \, dx\) | A1, A1 | 1.1, 1.1 |
| \(3I_n = 5^{n-1} \cdot 81\sqrt{3} - 3\sqrt{3} - (n-1)\int x^{n-2}(2 + x^2)\sqrt{2 + x^2} \, dx\) | M1 | 2.1 |
| \(= 5^{n-1} \cdot 81\sqrt{3} - 3\sqrt{3} - (n-1)[2I_{n-2} + I_n]\) | A1 | 1.1 |
| \(\Rightarrow (3+ n - 1)I_n = 3\sqrt{3}(27 \times 5^{n-1} - 1) - 2(n-1)I_{n-2} \Rightarrow\) Result | A1 | 1.1 |
| Total: [6] | AG Legitimately obtained; Correct splitting and attempt to integrate by parts; One for each term correctly integrated (ignore limits until the end); Splitting the second integral's terms, ready to get \(I\)'s; Correct second integral in terms of \(I\)'s |
| Answer | Marks | Guidance |
|---|---|---|
| Use of \(n = 3\) in Reduction Formula: \(5I_3 = 3\sqrt{3}(27 \times 25 - 1) - 2(2)I_1\) | M1 | 1.1a |
| \(I_3 = \frac{1}{5}(2022\sqrt{3} - 104\sqrt{3}) = \frac{1918}{5}\sqrt{3}\) | A1 | 1.1 |
| Use of \(n = 5\) in Reduction Formula: \(7I_5 = 3\sqrt{3}(27 \times 625 - 1) - 2(4)I_3\) | M1 | 1.1 |
| \(I_5 = \frac{1}{7}(50622\sqrt{3} - 8 \times \frac{1918}{5}\sqrt{3}) = \frac{237766}{35}\sqrt{3}\) | A1 | 2.2a |
| Total: [4] | Candidates may work downwards |
**Part (a)**
| $I_1 = \int_1^5 x\sqrt{2 + x^2} \, dx$ | M1 | 1.1 |
| $= \left[\frac{1}{3}(2 + x^2)^{\frac{3}{2}}\right]_1^5$ | A1 | 1.1 |
| $= \frac{1}{3}(81\sqrt{3} - 3\sqrt{3}) = 26\sqrt{3}$ | A1 | 2.2a |
| Total: [3] | Substitute $n = 1$ and attempt integral by substitution or inspection; For $k(2 + x^2)^{\frac{3}{2}}$, any non-zero $k$; Correctly gained from correct $k$ and use of (1, 5) limits | |
**Part (b)**
| $I_n = \int_1^5 x^{n-1} \cdot x\sqrt{2 + x^2} \, dx$ | M1 | 1.1a |
| $= x^{n-1} \cdot \frac{1}{3}(2 + x^2)^{\frac{3}{2}} - \int (n-1)x^{n-2} \cdot \frac{1}{3}(2 + x^2)^{\frac{3}{2}} \, dx$ | A1, A1 | 1.1, 1.1 |
| $3I_n = 5^{n-1} \cdot 81\sqrt{3} - 3\sqrt{3} - (n-1)\int x^{n-2}(2 + x^2)\sqrt{2 + x^2} \, dx$ | M1 | 2.1 |
| $= 5^{n-1} \cdot 81\sqrt{3} - 3\sqrt{3} - (n-1)[2I_{n-2} + I_n]$ | A1 | 1.1 |
| $\Rightarrow (3+ n - 1)I_n = 3\sqrt{3}(27 \times 5^{n-1} - 1) - 2(n-1)I_{n-2} \Rightarrow$ Result | A1 | 1.1 |
| Total: [6] | AG Legitimately obtained; Correct splitting and attempt to integrate by parts; One for each term correctly integrated (ignore limits until the end); Splitting the second integral's terms, ready to get $I$'s; Correct second integral in terms of $I$'s | |
**Part (c)**
| Use of $n = 3$ in Reduction Formula: $5I_3 = 3\sqrt{3}(27 \times 25 - 1) - 2(2)I_1$ | M1 | 1.1a |
| $I_3 = \frac{1}{5}(2022\sqrt{3} - 104\sqrt{3}) = \frac{1918}{5}\sqrt{3}$ | A1 | 1.1 |
| Use of $n = 5$ in Reduction Formula: $7I_5 = 3\sqrt{3}(27 \times 625 - 1) - 2(4)I_3$ | M1 | 1.1 |
| $I_5 = \frac{1}{7}(50622\sqrt{3} - 8 \times \frac{1918}{5}\sqrt{3}) = \frac{237766}{35}\sqrt{3}$ | A1 | 2.2a |
| Total: [4] | Candidates may work downwards | |6 For positive integers $n$, the integrals $I _ { n }$ are given by $I _ { n } = \int _ { 1 } ^ { 5 } x ^ { n } \sqrt { 2 + x ^ { 2 } } \mathrm {~d} x$.
\begin{enumerate}[label=(\alph*)]
\item Show that $I _ { 1 } = 26 \sqrt { 3 }$.
\item Prove that, for $n \geqslant 3 , ( n + 2 ) I _ { n } = 3 \sqrt { 3 } \left( 27 \times 5 ^ { n - 1 } - 1 \right) - 2 ( n - 1 ) I _ { n - 2 }$.
\item Determine the exact value of $I _ { 5 }$ as a rational multiple of $\sqrt { 3 }$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure 2018 Q6 [13]}}