CAIE FP1 2011 November — Question 6 8 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2011
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicReduction Formulae
TypeCompound expressions with binomial expansion
DifficultyChallenging +1.2 This is a standard reduction formula question requiring integration by parts to establish the recurrence relation, followed by iterative application to find I_3. The integration by parts is straightforward with clear choice of u and dv, and the algebraic manipulation is routine. While it requires multiple steps and careful algebra, it follows a well-practiced template that Further Maths students drill extensively. The exact value calculation is mechanical once the relation is established.
Spec4.08f Integrate using partial fractions
6 Let \(I _ { n } = \int _ { 0 } ^ { 1 } x ^ { n } ( 1 - x ) ^ { \frac { 1 } { 2 } } \mathrm {~d} x\), for \(n \geqslant 0\). Show that, for \(n \geqslant 1\), $$( 3 + 2 n ) I _ { n } = 2 n I _ { n - 1 }$$ Hence find the exact value of \(I _ { 3 }\).
Question 6:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(I_n = \int_0^1 x^n(1-x)^{\frac{1}{2}}\,dx\)
\(= \left[-\frac{2}{3}x^n(1-x)^{\frac{3}{2}}\right]_0^1 + \frac{2}{3}\int_0^1 nx^{n-1}(1-x)(1-x)^{\frac{1}{2}}\,dx\)M1A1 Integrates by parts
\(= 0 + \frac{2n}{3}\int_0^1 x^{n-1}(1-x)^{\frac{1}{2}}\,dx - \frac{2n}{3}\int_0^1 x^n(1-x)^{\frac{1}{2}}\,dx\)M1A1 Rearranges
\(= \frac{2n}{3}I_{n-1} - \frac{2n}{3}I_n\)
\(\Rightarrow (2n+3)I_n = 2nI_{n-1}\) (AG)A1 Obtains printed result
\(I_0 = \int_0^1(1-x)^{\frac{1}{2}}\,dx = \left[-\frac{2}{3}(1-x)^{\frac{3}{2}}\right]_0^1 = \frac{2}{3}\)B1 Evaluates \(I_0\)
\(I_3 = \frac{6}{9}\times\frac{4}{7}\times\frac{2}{5}\times\frac{2}{3} = \frac{32}{315}\)M1A1 Uses reduction formula 
# Question 6:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $I_n = \int_0^1 x^n(1-x)^{\frac{1}{2}}\,dx$ | | |
| $= \left[-\frac{2}{3}x^n(1-x)^{\frac{3}{2}}\right]_0^1 + \frac{2}{3}\int_0^1 nx^{n-1}(1-x)(1-x)^{\frac{1}{2}}\,dx$ | M1A1 | Integrates by parts |
| $= 0 + \frac{2n}{3}\int_0^1 x^{n-1}(1-x)^{\frac{1}{2}}\,dx - \frac{2n}{3}\int_0^1 x^n(1-x)^{\frac{1}{2}}\,dx$ | M1A1 | Rearranges |
| $= \frac{2n}{3}I_{n-1} - \frac{2n}{3}I_n$ | | |
| $\Rightarrow (2n+3)I_n = 2nI_{n-1}$ (AG) | A1 | Obtains printed result | **Part total: 5** |
| $I_0 = \int_0^1(1-x)^{\frac{1}{2}}\,dx = \left[-\frac{2}{3}(1-x)^{\frac{3}{2}}\right]_0^1 = \frac{2}{3}$ | B1 | Evaluates $I_0$ |
| $I_3 = \frac{6}{9}\times\frac{4}{7}\times\frac{2}{5}\times\frac{2}{3} = \frac{32}{315}$ | M1A1 | Uses reduction formula | **Part total: 3** |

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6 Let $I _ { n } = \int _ { 0 } ^ { 1 } x ^ { n } ( 1 - x ) ^ { \frac { 1 } { 2 } } \mathrm {~d} x$, for $n \geqslant 0$. Show that, for $n \geqslant 1$,

$$( 3 + 2 n ) I _ { n } = 2 n I _ { n - 1 }$$

Hence find the exact value of $I _ { 3 }$.

\hfill \mbox{\textit{CAIE FP1 2011 Q6 [8]}}
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