4 Let
$$I _ { n } = \int _ { 1 } ^ { \mathrm { e } } x ^ { 2 } ( \ln x ) ^ { n } \mathrm {~d} x$$
for \(n \geqslant 0\). Show that, for all \(n \geqslant 1\),
$$I _ { n } = \frac { 1 } { 3 } \mathrm { e } ^ { 3 } - \frac { 1 } { 3 } n I _ { n - 1 }$$
Find the exact value of \(I _ { 3 }\).
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Question 4:
Answer Marks
Guidance
Answer/Working Marks
Guidance
\(I_n = \int_1^e x^2(\ln x)^n\,dx = \left[(\ln x)^n\frac{x^3}{3}\right]_1^e - \int_1^e n(\ln x)^{n-1}\cdot\frac{1}{x}\cdot\frac{x^3}{3}\,dx\) M1A1, A1
Integrates by parts
\(= \frac{e^3}{3} - \frac{n}{3}I_{n-1}\) (AG) A1
Reduction formula; Part Mark: 4
\(I_0 = \int_1^e x^2\,dx = \left[\frac{x^3}{3}\right]_1^e = \frac{e^3-1}{3}\) B1
Finds \(I_0\)
\(\Rightarrow I_1 = \frac{e^3}{3} - \frac{1}{3}\left(\frac{e^3-1}{3}\right) = \frac{2e^3+1}{9}\) M1
Uses reduction formula
\(\Rightarrow I_2 = \frac{e^3}{3} - \frac{2}{3}\left(\frac{2e^3+1}{9}\right) = \frac{5e^3-2}{27}\) A1
Obtains \(I_2\)
\(\Rightarrow I_3 = \frac{e^3}{3} - \left(\frac{5e^3-2}{27}\right) = \frac{4e^3+2}{27}\) A1
Obtains \(I_3\); Part Mark: 4; Total: [8]
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## Question 4:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $I_n = \int_1^e x^2(\ln x)^n\,dx = \left[(\ln x)^n\frac{x^3}{3}\right]_1^e - \int_1^e n(\ln x)^{n-1}\cdot\frac{1}{x}\cdot\frac{x^3}{3}\,dx$ | M1A1, A1 | Integrates by parts |
| $= \frac{e^3}{3} - \frac{n}{3}I_{n-1}$ (AG) | A1 | Reduction formula; Part Mark: 4 |
| $I_0 = \int_1^e x^2\,dx = \left[\frac{x^3}{3}\right]_1^e = \frac{e^3-1}{3}$ | B1 | Finds $I_0$ |
| $\Rightarrow I_1 = \frac{e^3}{3} - \frac{1}{3}\left(\frac{e^3-1}{3}\right) = \frac{2e^3+1}{9}$ | M1 | Uses reduction formula |
| $\Rightarrow I_2 = \frac{e^3}{3} - \frac{2}{3}\left(\frac{2e^3+1}{9}\right) = \frac{5e^3-2}{27}$ | A1 | Obtains $I_2$ |
| $\Rightarrow I_3 = \frac{e^3}{3} - \left(\frac{5e^3-2}{27}\right) = \frac{4e^3+2}{27}$ | A1 | Obtains $I_3$; Part Mark: 4; Total: **[8]** |
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4 Let
$$I _ { n } = \int _ { 1 } ^ { \mathrm { e } } x ^ { 2 } ( \ln x ) ^ { n } \mathrm {~d} x$$
for $n \geqslant 0$. Show that, for all $n \geqslant 1$,
$$I _ { n } = \frac { 1 } { 3 } \mathrm { e } ^ { 3 } - \frac { 1 } { 3 } n I _ { n - 1 }$$
Find the exact value of $I _ { 3 }$.
\hfill \mbox{\textit{CAIE FP1 2012 Q4 [8]}}