| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Exponential times polynomial |
| 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 specific values. While it involves Further Maths content and multiple parts, the techniques are routine for FP2 students: deriving the reduction formula is straightforward integration by parts, and parts (ii)-(iii) are mechanical applications. The final part requires some algebraic manipulation but no novel insight. |
| Spec | 1.08i Integration by parts4.08f Integrate using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| \(I_n = \int_0^1 x^n e^{-x}\,dx\), \(u=x^n\), \(dv = e^{-x}\,dx\), \(du = nx^{n-1}\,dx\), \(v = -e^{-x}\) | M1 | By parts |
| \(I_n = \left[-e^{-x}x^n\right]_0^1 + n\int_0^1 x^{n-1}e^{-x}\,dx = (-e^{-1}-0) + nI_{n-1}\) | A1 | Both terms before limits applied, soi |
| \(I_n = nI_{n-1} - e^{-1}\) | A1 | Or \(k = \dfrac{-1}{e}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(I_0 = \int_0^1 e^{-x}\,dx = \left[-e^{-x}\right]_0^1 = 1-e^{-1}\) | B1 | Or finding \(I_1\) |
| \(I_3 = 3I_2 - e^{-1} = 3(2I_1 - e^{-1}) - e^{-1} = 6I_1 - 4e^{-1} = 6(I_0 - e^{-1}) - 4e^{-1} = 6I_0 - 10e^{-1}\) | M1 | Complete method even if \(k\) is wrong |
| \(I_3 = 6 - 16e^{-1}\) | A1 | SC3 by parts 2 or 3 times |
| Answer | Marks | Guidance |
|---|---|---|
| \(I_{11} = 11I_{10} - e^{-1} = 11(10I_9 - e^{-1}) - e^{-1} = 110I_9 - 12e^{-1}\) | M1, A1 | Complete method; for \(I_{11}\) in terms of \(I_{10}\) or \(I_9\) in terms of \(I_8\) soi |
| Answer | Marks |
|---|---|
| \(990I_8 - I_{11} = 122e^{-1}\) | A1 |
# Question 4(i):
$I_n = \int_0^1 x^n e^{-x}\,dx$, $u=x^n$, $dv = e^{-x}\,dx$, $du = nx^{n-1}\,dx$, $v = -e^{-x}$ | M1 | By parts
$I_n = \left[-e^{-x}x^n\right]_0^1 + n\int_0^1 x^{n-1}e^{-x}\,dx = (-e^{-1}-0) + nI_{n-1}$ | A1 | Both terms before limits applied, soi
$I_n = nI_{n-1} - e^{-1}$ | A1 | Or $k = \dfrac{-1}{e}$
## Question 4(ii):
$I_0 = \int_0^1 e^{-x}\,dx = \left[-e^{-x}\right]_0^1 = 1-e^{-1}$ | B1 | Or finding $I_1$
$I_3 = 3I_2 - e^{-1} = 3(2I_1 - e^{-1}) - e^{-1} = 6I_1 - 4e^{-1} = 6(I_0 - e^{-1}) - 4e^{-1} = 6I_0 - 10e^{-1}$ | M1 | Complete method even if $k$ is wrong
$I_3 = 6 - 16e^{-1}$ | A1 | SC3 by parts 2 or 3 times
## Question 4(iii):
$I_{11} = 11I_{10} - e^{-1} = 11(10I_9 - e^{-1}) - e^{-1} = 110I_9 - 12e^{-1}$ | M1, A1 | Complete method; for $I_{11}$ in terms of $I_{10}$ or $I_9$ in terms of $I_8$ soi
$= 110(9I_8 - e^{-1}) - 12e^{-1} = 990I_8 - 122e^{-1}$
$990I_8 - I_{11} = 122e^{-1}$ | A1 |
---4 It is given that $I _ { n } = \int _ { 0 } ^ { 1 } x ^ { n } \mathrm { e } ^ { - x } \mathrm {~d} x$ for $n \geqslant 0$.\\
(i) Show that $I _ { n } = n I _ { n - 1 } + k$ for $n \geqslant 1$, where $k$ is a constant to be determined.\\
(ii) Find the exact value of $I _ { 3 }$.\\
(iii) Find the exact value of $990 I _ { 8 } - I _ { 11 }$.
\hfill \mbox{\textit{OCR FP2 2015 Q4 [9]}}