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 \(I _ { 7 }\).
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Question 5:
Answer Marks
Guidance
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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## 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]}}