7 Let \(I _ { n } = \int _ { 0 } ^ { 1 } \frac { 1 } { \left( 1 + x ^ { 4 } \right) ^ { n } } \mathrm {~d} x\). By considering \(\frac { \mathrm { d } } { \mathrm { d } x } \left( \frac { x } { \left( 1 + x ^ { 4 } \right) ^ { n } } \right)\), show that
$$4 n I _ { n + 1 } = \frac { 1 } { 2 ^ { n } } + ( 4 n - 1 ) I _ { n }$$
Given that \(I _ { 1 } = 0.86697\), correct to 5 decimal places, find \(I _ { 3 }\).
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\(D[x(1 + x^4)^n] = (1 + x^4)^n - 4nx^4(1 + x^4)^{n-1}\) M1
\(= (1 - 4n)(1 + x^4)^n + 4n(1 + x^4)^{n-1}\) A1
\(\Rightarrow \left[x(1 + x^4)^n\right]_0^1 = (1 - 4n)I_n + 4nI_{n-1}\) M1
\(\Rightarrow 4nI_{n+1} = 2^{-n} + (4n - 1)I_n\) (AG) A1
\(8I_3 = 1/4 + 7I_2; \quad 4I_2 = 1/2 + 3I_1\) B1B1
\(I_3 = 9/64 + (21/32)I_1 \approx 0.7096\) or \(0.710\) M1A1
OR \(n = 1 \quad 4I_2 = \frac{1}{2} + 3 \times 0.86697 \Rightarrow I_2 = 0.7752275\) \(n = 2 \quad 8I_3 = \frac{1}{4} + 7 \times 0.7752275 \Rightarrow I_3 = 0.7095740625\) \(\therefore I_3 = 0.7096\) or \(0.710\) (No penalty for correct 5 d.p. value.) M1 Use of formula, A1 Gets \(I_2\), A1ft Subs value for \(I_2\) in \(I_3\) formula, A1 obtains \(I_3\) correct (cao)
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$D[x(1 + x^4)^n] = (1 + x^4)^n - 4nx^4(1 + x^4)^{n-1}$ | M1 |
$= (1 - 4n)(1 + x^4)^n + 4n(1 + x^4)^{n-1}$ | A1 |
$\Rightarrow \left[x(1 + x^4)^n\right]_0^1 = (1 - 4n)I_n + 4nI_{n-1}$ | M1 |
$\Rightarrow 4nI_{n+1} = 2^{-n} + (4n - 1)I_n$ (AG) | A1 |
$8I_3 = 1/4 + 7I_2; \quad 4I_2 = 1/2 + 3I_1$ | B1B1 |
$I_3 = 9/64 + (21/32)I_1 \approx 0.7096$ or $0.710$ | M1A1 |
OR |
$n = 1 \quad 4I_2 = \frac{1}{2} + 3 \times 0.86697 \Rightarrow I_2 = 0.7752275$ | |
$n = 2 \quad 8I_3 = \frac{1}{4} + 7 \times 0.7752275 \Rightarrow I_3 = 0.7095740625$ | |
$\therefore I_3 = 0.7096$ or $0.710$ | |
(No penalty for correct 5 d.p. value.) | |
M1 Use of formula, A1 Gets $I_2$, A1ft Subs value for $I_2$ in $I_3$ formula, A1 obtains $I_3$ correct (cao) | |
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7 Let $I _ { n } = \int _ { 0 } ^ { 1 } \frac { 1 } { \left( 1 + x ^ { 4 } \right) ^ { n } } \mathrm {~d} x$. By considering $\frac { \mathrm { d } } { \mathrm { d } x } \left( \frac { x } { \left( 1 + x ^ { 4 } \right) ^ { n } } \right)$, show that
$$4 n I _ { n + 1 } = \frac { 1 } { 2 ^ { n } } + ( 4 n - 1 ) I _ { n }$$
Given that $I _ { 1 } = 0.86697$, correct to 5 decimal places, find $I _ { 3 }$.
\hfill \mbox{\textit{CAIE FP1 2008 Q7 [8]}}