| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2019 |
| 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 Further Maths reduction formula question requiring integration by parts with substitution. Part (i) is straightforward substitution, part (ii) follows a predictable pattern for deriving reduction formulae, and part (iii) applies the formula recursively. While it requires careful algebraic manipulation and is harder than typical A-level questions due to being Further Maths content, it follows well-established techniques without requiring novel insight. |
| Spec | 1.08h Integration by substitution8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_0^1 x^2 e^{x^3}\, dx = \frac{1}{3}\left[e^{x^3}\right]_0^1 = \frac{1}{3}(e-1)\) | M1 A1 | Must show working. AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(I_n = \int_0^1 x^{n-2} \cdot x^2 e^{x^3}\, dx = \left[\frac{1}{3}x^{n-2}e^{x^3}\right]_0^1 - \frac{n-2}{3}\int_0^1 x^{n-3}e^{x^3}\, dx\) | M1 A1 | Integrates by parts |
| \(\frac{e}{3} - \frac{n-2}{3}I_{n-3} \Rightarrow 3I_n = e-(n-2)I_{n-3}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(I_5 = \frac{1}{3}(e - 3I_2) = \frac{1}{3}(e - e + 1) = \frac{1}{3}\) | M1 A1 | Applies reduction formula once and uses (i) |
| \(I_8 = \frac{1}{3}(e - 6I_5) = \frac{1}{3}(e-2)\) | A1 | Must be exact |
## Question 4(i):
| $\int_0^1 x^2 e^{x^3}\, dx = \frac{1}{3}\left[e^{x^3}\right]_0^1 = \frac{1}{3}(e-1)$ | M1 A1 | Must show working. AG |
|---|---|---|
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## Question 4(ii):
| $I_n = \int_0^1 x^{n-2} \cdot x^2 e^{x^3}\, dx = \left[\frac{1}{3}x^{n-2}e^{x^3}\right]_0^1 - \frac{n-2}{3}\int_0^1 x^{n-3}e^{x^3}\, dx$ | M1 A1 | Integrates by parts |
|---|---|---|
| $\frac{e}{3} - \frac{n-2}{3}I_{n-3} \Rightarrow 3I_n = e-(n-2)I_{n-3}$ | A1 | AG |
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## Question 4(iii):
| $I_5 = \frac{1}{3}(e - 3I_2) = \frac{1}{3}(e - e + 1) = \frac{1}{3}$ | M1 A1 | Applies reduction formula once and uses (i) |
|---|---|---|
| $I_8 = \frac{1}{3}(e - 6I_5) = \frac{1}{3}(e-2)$ | A1 | Must be exact |
---4 It is given that, for $n \geqslant 0$,
$$I _ { n } = \int _ { 0 } ^ { 1 } x ^ { n } \mathrm { e } ^ { x ^ { 3 } } \mathrm {~d} x$$
(i) Show that $I _ { 2 } = \frac { 1 } { 3 } ( \mathrm { e } - 1 )$.\\
(ii) Show that, for $n \geqslant 3$,
$$3 I _ { n } = \mathrm { e } - ( n - 2 ) I _ { n - 3 }$$
(iii) Hence find the exact value of $I _ { 8 }$.\\
\hfill \mbox{\textit{CAIE FP1 2019 Q4 [8]}}