| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2017 |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Logarithmic power integrals |
| Difficulty | Challenging +1.3 This is a standard reduction formula question requiring integration by parts to establish the recurrence relation, then applying it to find I_3. The integration by parts is straightforward with u=(ln x)^n, and the mean value calculation is routine. While it requires careful algebraic manipulation and is from Further Maths, it follows a well-established template without requiring novel insight. |
| Spec | 4.06b Method of differences: telescoping series4.08e Mean value of function: using integral |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_1^e \ln x \, dx = x\ln x - x\) | 1 | B1 |
| \(I_n = \int_1^e (\ln x)^{n-1} \cdot \ln x \, dx\) | 1 | M1 |
| \(= \left[(\ln x)^{n-1}(x\ln x - x)\right]_1^e - \int_1^e (n-1)(\ln x)^{n-2} \cdot \dfrac{1}{x}(x\ln x - x)\,dx\) | 2 | M1A1 |
| \(= 0 - \int_1^e (n-1)(\ln x)^{n-2}(\ln x - 1)\,dx = (n-1)\left[I_{n-2} - I_{n-1}\right]\) (AG) | 2 | M1A1 |
| Alternative for obtaining reduction formula: | ||
| \(I_n = \int_1^e (\ln x)^n \times 1\,dx = \left[x(\ln x)^n\right]_1^e - \int_1^e n(\ln x)^{n-1}\,dx\) | 2 | M1A1 |
| \(\Rightarrow I_n = e - nI_{n-1}\) | 1 | A1 |
| Similarly \(I_{n-1} = e - (n-1)I_{n-2}\) | 1 | B1 |
| \(\Rightarrow I_n + nI_{n-1} = I_{n-1} + (n-1)I_{n-2}\) | 1 | M1 |
| \(\Rightarrow I_n = (n-1)\left[I_{n-2} - I_{n-1}\right]\) (AG) | 1 | A1 |
| Subtotal | 6 | |
| \(I_0 = [x]_1^e = e - 1\) | 1 | B1 |
| \(I_1 = [x\ln x - x]_1^e = 1\) | 1 | B1 |
| \(I_2 = 1 \times (e - 1 - 1) = e - 2\) | 1 | M1 |
| \(I_3 = 2(I_1 - I_2) = 2(1 - [e-2]) = 6 - 2e\) | 1 | A1 |
| \(\text{MV} = \dfrac{I_3}{e-1} = \dfrac{6-2e}{e-1}\) | 2 | M1 A1\(\wedge\) |
| Subtotal | 6 |
## Question 9:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_1^e \ln x \, dx = x\ln x - x$ | 1 | B1 |
| $I_n = \int_1^e (\ln x)^{n-1} \cdot \ln x \, dx$ | 1 | M1 |
| $= \left[(\ln x)^{n-1}(x\ln x - x)\right]_1^e - \int_1^e (n-1)(\ln x)^{n-2} \cdot \dfrac{1}{x}(x\ln x - x)\,dx$ | 2 | M1A1 |
| $= 0 - \int_1^e (n-1)(\ln x)^{n-2}(\ln x - 1)\,dx = (n-1)\left[I_{n-2} - I_{n-1}\right]$ (AG) | 2 | M1A1 |
| **Alternative for obtaining reduction formula:** | | |
| $I_n = \int_1^e (\ln x)^n \times 1\,dx = \left[x(\ln x)^n\right]_1^e - \int_1^e n(\ln x)^{n-1}\,dx$ | 2 | M1A1 |
| $\Rightarrow I_n = e - nI_{n-1}$ | 1 | A1 |
| Similarly $I_{n-1} = e - (n-1)I_{n-2}$ | 1 | B1 |
| $\Rightarrow I_n + nI_{n-1} = I_{n-1} + (n-1)I_{n-2}$ | 1 | M1 |
| $\Rightarrow I_n = (n-1)\left[I_{n-2} - I_{n-1}\right]$ (AG) | 1 | A1 |
| **Subtotal** | **6** | |
| $I_0 = [x]_1^e = e - 1$ | 1 | B1 |
| $I_1 = [x\ln x - x]_1^e = 1$ | 1 | B1 |
| $I_2 = 1 \times (e - 1 - 1) = e - 2$ | 1 | M1 |
| $I_3 = 2(I_1 - I_2) = 2(1 - [e-2]) = 6 - 2e$ | 1 | A1 |
| $\text{MV} = \dfrac{I_3}{e-1} = \dfrac{6-2e}{e-1}$ | 2 | M1 A1$\wedge$ |
| **Subtotal** | **6** | |9 It is given that $I _ { n } = \int _ { 1 } ^ { \mathrm { e } } ( \ln x ) ^ { n } \mathrm {~d} x$ for $n \geqslant 0$.\\
(i) Show that
$$I _ { n } = ( n - 1 ) \left[ I _ { n - 2 } - I _ { n - 1 } \right] \text { for } n \geqslant 2 .$$
(ii) Hence find, in an exact form, the mean value of $( \ln x ) ^ { 3 }$ with respect to $x$ over the interval $1 \leqslant x \leqslant \mathrm { e }$. [6]\\
\hfill \mbox{\textit{CAIE FP1 2017 Q9 [12]}}