CAIE FP1 2013 June — Question 5 8 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2013
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Substitution
TypeShow substitution transforms integral, then apply integration by parts or further substitution
DifficultyChallenging +1.2 This is a structured Further Maths question requiring substitution (u = x²) for the first part, then integration by parts to derive a reduction formula, followed by recursive application. While it involves multiple techniques and careful algebraic manipulation, the path is clearly signposted through the parts, and reduction formulas are standard FP1 fare. The computational steps are moderate but methodical rather than requiring novel insight.
Spec1.08h Integration by substitution8.06a Reduction formulae: establish, use, and evaluate recursively

5 Show that \(\int _ { 0 } ^ { 1 } x \mathrm { e } ^ { - x ^ { 2 } } \mathrm {~d} x = \frac { 1 } { 2 } - \frac { 1 } { 2 \mathrm { e } }\). Let \(I _ { n } = \int _ { 0 } ^ { 1 } x ^ { n } \mathrm { e } ^ { - x ^ { 2 } } \mathrm {~d} x\). Show that \(I _ { 2 n + 1 } = n I _ { 2 n - 1 } - \frac { 1 } { 2 \mathrm { e } }\) for \(n \geqslant 1\). Find the exact value of \(I _ { 7 }\).

Question 5:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(I_1 = \int_0^1 xe^{-x^2}dx = \left[-\frac{e^{-x^2}}{2}\right]_0^1 = \frac{1}{2} - \frac{1}{2e}\) (AG)M1A1 Finds \(I_1\); 2 marks
\(I_{2n+1} = \int_0^1 x^{2n+1}e^{-x^2}dx = \left[-x^{2n}\frac{e^{-x^2}}{2}\right]_0^1 + \int_0^1 2nx^{2n-1}\frac{e^{-x^2}}{2}dx\)M1A1 Integrates by parts
\(= \left[-\frac{1}{2e}\right] - [0] + nI_{2n-1} = nI_{2n-1} - \frac{1}{2e}\) (AG)A1 Obtains reduction formula; 3 marks
\(I_3 = \frac{1}{2} - \frac{1}{2e} - \frac{1}{2e} = \frac{1}{2} - \frac{1}{e}\)M1 Attempts to use reduction formula at least once
\(I_5 = 2\left(\frac{1}{2} - \frac{1}{e}\right) - \frac{1}{2e} = 1 - \frac{5}{2e}\)A1 Obtains \(I_5\) or some intermediate result correctly
\(I_7 = 3\left(1 - \frac{5}{2e}\right) - \frac{1}{2e} = 3 - \frac{8}{e}\)A1 3 marks; total [8]
## Question 5:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $I_1 = \int_0^1 xe^{-x^2}dx = \left[-\frac{e^{-x^2}}{2}\right]_0^1 = \frac{1}{2} - \frac{1}{2e}$ (AG) | M1A1 | Finds $I_1$; 2 marks |
| $I_{2n+1} = \int_0^1 x^{2n+1}e^{-x^2}dx = \left[-x^{2n}\frac{e^{-x^2}}{2}\right]_0^1 + \int_0^1 2nx^{2n-1}\frac{e^{-x^2}}{2}dx$ | M1A1 | Integrates by parts |
| $= \left[-\frac{1}{2e}\right] - [0] + nI_{2n-1} = nI_{2n-1} - \frac{1}{2e}$ (AG) | A1 | Obtains reduction formula; 3 marks |
| $I_3 = \frac{1}{2} - \frac{1}{2e} - \frac{1}{2e} = \frac{1}{2} - \frac{1}{e}$ | M1 | Attempts to use reduction formula at least once |
| $I_5 = 2\left(\frac{1}{2} - \frac{1}{e}\right) - \frac{1}{2e} = 1 - \frac{5}{2e}$ | A1 | Obtains $I_5$ or some intermediate result correctly |
| $I_7 = 3\left(1 - \frac{5}{2e}\right) - \frac{1}{2e} = 3 - \frac{8}{e}$ | A1 | 3 marks; total [8] |

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5 Show that $\int _ { 0 } ^ { 1 } x \mathrm { e } ^ { - x ^ { 2 } } \mathrm {~d} x = \frac { 1 } { 2 } - \frac { 1 } { 2 \mathrm { e } }$.

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

Find the exact value of $I _ { 7 }$.

\hfill \mbox{\textit{CAIE FP1 2013 Q5 [8]}}