| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2010 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Derive reduction formula by integration by parts |
| Difficulty | Challenging +1.2 This is a standard reduction formula question requiring integration by parts with clear substitutions. Part (i) involves routine application of IBP with u=(1-2x)^n, dv=e^x dx, followed by algebraic manipulation. Part (ii) is mechanical recursion using the derived formula. While it requires careful algebra and is from Further Maths FP2, the technique is well-practiced and follows a predictable pattern, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 1.08i Integration by parts8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Reasonable attempt at integration by parts | M1 | Leading to second integral |
| \(\frac{(1-2x)^{n+1}}{-2(n+1)}e^x - \int \frac{(1-2x)^{n+1}}{-2(n+1)}e^x\,dx\) | A1 | Or equivalent form |
| Evidence of limits used in integrated part | M1 | Should show \(\pm 1\) |
| Tidy to give \(I_{n+1} = 2(n+1)I_n - 1\) | A1 | Allow \(I_{n+1} = 2(n+1)I_n - 1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Show any one of \(I_3=6I_2-1\), \(I_2=4I_1-1\) | B1 | May be implied |
| Get \(I_0(=e^{\frac{1}{2}}-1)\) or \(I_1(=2e^{\frac{1}{2}}-3)\) | B1 | |
| Substitute values back for \(I_3\) | M1 | Not involving \(n\) |
| Get \(48e^{\frac{1}{2}} - 79\) | A1 |
# Question 5:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Reasonable attempt at integration by parts | M1 | Leading to second integral |
| $\frac{(1-2x)^{n+1}}{-2(n+1)}e^x - \int \frac{(1-2x)^{n+1}}{-2(n+1)}e^x\,dx$ | A1 | Or equivalent form |
| Evidence of limits used in integrated part | M1 | Should show $\pm 1$ |
| Tidy to give $I_{n+1} = 2(n+1)I_n - 1$ | A1 | Allow $I_{n+1} = 2(n+1)I_n - 1$ |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Show any one of $I_3=6I_2-1$, $I_2=4I_1-1$ | B1 | May be implied |
| Get $I_0(=e^{\frac{1}{2}}-1)$ or $I_1(=2e^{\frac{1}{2}}-3)$ | B1 | |
| Substitute values back for $I_3$ | M1 | Not involving $n$ |
| Get $48e^{\frac{1}{2}} - 79$ | A1 | |
---5 It is given that, for $n \geqslant 0$,
$$I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 2 } } ( 1 - 2 x ) ^ { n } \mathrm { e } ^ { x } \mathrm {~d} x$$
(i) Prove that, for $n \geqslant 1$,
$$I _ { n } = 2 n I _ { n - 1 } - 1$$
(ii) Find the exact value of $I _ { 3 }$.
\hfill \mbox{\textit{OCR FP2 2010 Q5 [8]}}