CAIE FP1 2012 November — Question 5 8 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2012
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicReduction Formulae
TypeProve formula by induction
DifficultyChallenging +1.2 This is a standard reduction formula question with straightforward integration by parts followed by routine induction. The integration by parts is mechanical (standard technique for x^n e^(-ax)), and the induction proof requires only basic algebraic manipulation using the given recurrence relation. While it involves improper integrals and is from Further Maths, the techniques are well-practiced and the question follows a predictable template.
Spec1.08i Integration by parts4.01a Mathematical induction: construct proofs4.08c Improper integrals: infinite limits or discontinuous integrands
5 Let \(I _ { n }\) denote \(\int _ { 0 } ^ { \infty } x ^ { n } \mathrm { e } ^ { - 2 x } \mathrm {~d} x\). Show that \(I _ { n } = \frac { 1 } { 2 } n I _ { n - 1 }\), for \(n \geqslant 1\). Prove by mathematical induction that, for all positive integers \(n , I _ { n } = \frac { n ! } { 2 ^ { n + 1 } }\).
Question 5:
AnswerMarks Guidance
Working/AnswerMark Guidance
\(\displaystyle\int_0^{\infty} x^n e^{-2x}\,dx = \left[x^n\cdot\frac{-e^{-2x}}{2}\right]_0^{\infty} + \int_0^{\infty} nx^{n-1}\frac{e^{-2x}}{2}\,dx\)M1 Integrates by parts
\(= \frac{n}{2}I_{n-1}\) (AG)A1 Part total 2
\(P_n\): \(I_n = \dfrac{n!}{2^{n+1}}\)— States proposition
\(n=1\): \(I_0 = \displaystyle\int_0^{\infty} e^{-2x}\,dx = \left[-\frac{1}{2}e^{-2x}\right]_0^{\infty} = \frac{1}{2}\)B1 Proves base case
\(I_1 = \frac{1}{2}\times\frac{1}{2} = \frac{1}{4} = \frac{1!}{2^2}\), \(\therefore P_1\) trueB1
\(P_k\): \(I_k = \dfrac{k!}{2^{k+1}}\) for some integer \(k\)B1 Shows \(P_k \Rightarrow P_{k+1}\)
\(\therefore I_{k+1} = \frac{k+1}{2}\times\frac{k!}{2^{k+1}} = \frac{(k+1)!}{2^{k+2}}\), \(\therefore P_k \Rightarrow P_{k+1}\)M1A1
Hence by PMI \(P_n\) is true for all positive integers \(n\)A1 States conclusion; part total 6; Total [8] 
## Question 5:

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\displaystyle\int_0^{\infty} x^n e^{-2x}\,dx = \left[x^n\cdot\frac{-e^{-2x}}{2}\right]_0^{\infty} + \int_0^{\infty} nx^{n-1}\frac{e^{-2x}}{2}\,dx$ | M1 | Integrates by parts |
| $= \frac{n}{2}I_{n-1}$ (AG) | A1 | Part total 2 |
| $P_n$: $I_n = \dfrac{n!}{2^{n+1}}$ | — | States proposition |
| $n=1$: $I_0 = \displaystyle\int_0^{\infty} e^{-2x}\,dx = \left[-\frac{1}{2}e^{-2x}\right]_0^{\infty} = \frac{1}{2}$ | B1 | Proves base case |
| $I_1 = \frac{1}{2}\times\frac{1}{2} = \frac{1}{4} = \frac{1!}{2^2}$, $\therefore P_1$ true | B1 | |
| $P_k$: $I_k = \dfrac{k!}{2^{k+1}}$ for some integer $k$ | B1 | Shows $P_k \Rightarrow P_{k+1}$ |
| $\therefore I_{k+1} = \frac{k+1}{2}\times\frac{k!}{2^{k+1}} = \frac{(k+1)!}{2^{k+2}}$, $\therefore P_k \Rightarrow P_{k+1}$ | M1A1 | |
| Hence by PMI $P_n$ is true for all positive integers $n$ | A1 | States conclusion; part total 6; **Total [8]** |

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5 Let $I _ { n }$ denote $\int _ { 0 } ^ { \infty } x ^ { n } \mathrm { e } ^ { - 2 x } \mathrm {~d} x$. Show that $I _ { n } = \frac { 1 } { 2 } n I _ { n - 1 }$, for $n \geqslant 1$.

Prove by mathematical induction that, for all positive integers $n , I _ { n } = \frac { n ! } { 2 ^ { n + 1 } }$.

\hfill \mbox{\textit{CAIE FP1 2012 Q5 [8]}}
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