| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Compound expressions with binomial expansion |
| Difficulty | Challenging +1.2 This is a standard reduction formula question requiring integration by parts to establish the recurrence relation, then iterative application to find I_3. While it involves Further Maths content (reduction formulae with fractional powers), the technique is routine and the algebraic manipulation straightforward. The question follows a predictable template with clear guidance, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 1.08i Integration by parts4.08f Integrate using partial fractions8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_n = \int_0^1 x^n(1-x)^{\frac{3}{2}}\,dx\) | M1 | For integrating by parts (correct way round) |
| \(= \left[-\frac{2}{5}x^n(1-x)^{\frac{5}{2}}\right]_0^1 + \frac{2}{5}n\int_0^1 x^{n-1}(1-x)^{\frac{5}{2}}\,dx\) | A1 | For correct first stage |
| \(\Rightarrow I_n = \frac{2}{5}n\int_0^1 x^{n-1}(1-x)^{\frac{5}{2}}\,dx\) | A1 | |
| \(\Rightarrow I_n = \frac{2}{5}n\int_0^1 x^{n-1}(1-x)(1-x)^{\frac{3}{2}}\,dx\) | M1 | For splitting \((1-x)^{\frac{5}{2}}\) suitably |
| \(\Rightarrow I_n = \frac{2}{5}nI_{n-1} - \frac{2}{5}nI_n\) | A1 | For obtaining correct relation between \(I_n\) and \(I_{n-1}\) |
| \(\Rightarrow I_n = \frac{2n}{2n+5}I_{n-1}\) | A1 | For correct result (answer given) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_0 = \left[-\frac{2}{5}(1-x)^{\frac{5}{2}}\right]_0^1 = \frac{2}{5}\) | M1 | For evaluating \(I_0\) [OR \(I_1\) by parts] |
| M1 | For using recurrence relation 3 [OR 2] times (may be combined together) | |
| \(I_3 = \frac{6}{11}I_2 = \frac{6}{11}\times\frac{4}{9}I_1 = \frac{6}{11}\times\frac{4}{9}\times\frac{2}{7}I_0\) | A1 | For 3 [OR 2] correct fractions |
| \(I_3 = \frac{32}{1155}\) | A1 | For correct exact result |
## Question 6:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_n = \int_0^1 x^n(1-x)^{\frac{3}{2}}\,dx$ | M1 | For integrating by parts (correct way round) |
| $= \left[-\frac{2}{5}x^n(1-x)^{\frac{5}{2}}\right]_0^1 + \frac{2}{5}n\int_0^1 x^{n-1}(1-x)^{\frac{5}{2}}\,dx$ | A1 | For correct first stage |
| $\Rightarrow I_n = \frac{2}{5}n\int_0^1 x^{n-1}(1-x)^{\frac{5}{2}}\,dx$ | A1 | |
| $\Rightarrow I_n = \frac{2}{5}n\int_0^1 x^{n-1}(1-x)(1-x)^{\frac{3}{2}}\,dx$ | M1 | For splitting $(1-x)^{\frac{5}{2}}$ suitably |
| $\Rightarrow I_n = \frac{2}{5}nI_{n-1} - \frac{2}{5}nI_n$ | A1 | For obtaining correct relation between $I_n$ and $I_{n-1}$ |
| $\Rightarrow I_n = \frac{2n}{2n+5}I_{n-1}$ | A1 | For correct result (**answer given**) |
**Total: 6 marks**
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_0 = \left[-\frac{2}{5}(1-x)^{\frac{5}{2}}\right]_0^1 = \frac{2}{5}$ | M1 | For evaluating $I_0$ [OR $I_1$ by parts] |
| | M1 | For using recurrence relation 3 [OR 2] times (may be combined together) |
| $I_3 = \frac{6}{11}I_2 = \frac{6}{11}\times\frac{4}{9}I_1 = \frac{6}{11}\times\frac{4}{9}\times\frac{2}{7}I_0$ | A1 | For 3 [OR 2] correct fractions |
| $I_3 = \frac{32}{1155}$ | A1 | For correct exact result |
**Total: 4 marks**
---6 It is given that $I _ { n } = \int _ { 0 } ^ { 1 } x ^ { n } ( 1 - x ) ^ { \frac { 3 } { 2 } } \mathrm {~d} x$, for $n \geqslant 0$.\\
(i) Show that $I _ { n } = \frac { 2 n } { 2 n + 5 } I _ { n - 1 }$, for $n \geqslant 1$.\\
(ii) Hence find the exact value of $I _ { 3 }$.
\hfill \mbox{\textit{OCR FP2 2011 Q6 [10]}}