| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2007 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Logarithmic power integrals |
| 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 I₃. While it involves Further Maths content (FP2), the technique is routine: choose u = (ln x)^n and dv = dx, apply the formula, and evaluate limits. The iteration for part (ii) is mechanical calculation requiring I₀, I₁, I₂, then I₃. More challenging than average A-level due to being Further Maths and requiring careful algebraic manipulation, but follows a well-practiced template without novel insight. |
| Spec | 1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Attempt at parts on \(\int 1 \cdot (\ln x)^n dx\) | M1 | Two terms seen |
| Get \(x(\ln x)^n - \int n(\ln x)^{n-1} dx\) | A1 | |
| Put in limits correctly in line above | M1 | |
| Clearly get A.G. | A1 | \(\ln e = 1\), \(\ln 1 = 0\) seen or implied |
| (ii) Attempt \(I_5\) to \(I_2\) as \(I_1 = e - 3I_2\) | M1 | |
| Continue sequence in terms of \(\ln\) | A1 | \(I_2 = e-2I_1\) and/or \(I_1 = e-I_0\) |
| Attempt \(I_0\) or \(I_1\) | M1 | \((I_0 = e-1, I_1 = 1)\) |
| Get \(6 - 2e\) | A1 | cao |
**(i)** Attempt at parts on $\int 1 \cdot (\ln x)^n dx$ | M1 | Two terms seen
Get $x(\ln x)^n - \int n(\ln x)^{n-1} dx$ | A1 |
Put in limits correctly in line above | M1 |
Clearly get A.G. | A1 | $\ln e = 1$, $\ln 1 = 0$ seen or implied
**(ii)** Attempt $I_5$ to $I_2$ as $I_1 = e - 3I_2$ | M1 |
Continue sequence in terms of $\ln$ | A1 | $I_2 = e-2I_1$ and/or $I_1 = e-I_0$
Attempt $I_0$ or $I_1$ | M1 | $(I_0 = e-1, I_1 = 1)$
Get $6 - 2e$ | A1 | cao5 It is given that, for non-negative integers $n$,
$$I _ { n } = \int _ { 1 } ^ { \mathrm { e } } ( \ln x ) ^ { n } \mathrm {~d} x$$
(i) Show that, for $n \geqslant 1$,
$$I _ { n } = \mathrm { e } - n I _ { n - 1 } .$$
(ii) Find $I _ { 3 }$ in terms of e.
\hfill \mbox{\textit{OCR FP2 2007 Q5 [8]}}