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(D) Given $f(x)=\frac{x}{(1+x^4)^{1/4}}$.First,calculate $f(f(x)) = \frac{f(x)}{(1+f(x)^4)^{1/4}} = \frac{\frac{x}{(1+x^4)^{1/4}}}{(1+\frac{x^4}{1+x^4

Vedclass · Accepted Answer

$39$

Vedclass · Answer

(D) Given $f(x)=\frac{x}{(1+x^4)^{1/4}}$.First,calculate $f(f(x)) = \frac{f(x)}{(1+f(x)^4)^{1/4}} = \frac{\frac{x}{(1+x^4)^{1/4}}}{(1+\frac{x^4}{1+x^4})^{1/4}} = \frac{x}{(1+2x^4)^{1/4}}$.By induction,$g(x) = f(f(f(f(x)))) = \frac{x}{(1+4x^4)^{1/4}}$.We need to evaluate $I = 18 \int_0^{\sqrt{2\sqrt{5}}} \frac{x^4}{(1+4x^4)^{1/4}} dx$. Note: The integral is $x^3 g(x) dx$.Let $1+4x^4 = t^4$,then $16x^3 dx = 4t^3 dt$,which implies $x^3 dx = \frac{1}{4} t^3 dt$.When $x=0$,$t=1$. When $x=\sqrt{2\sqrt{5}}$,$x^4 = 4(5) = 20$,so $t^4 = 1+4(20) = 81$,$t=3$.$I = 18 \int_1^3 \frac{1}{t} \cdot \frac{1}{4} t^3 dt = \frac{18}{4} \int_1^3 t^2 dt = \frac{9}{2} [\frac{t^3}{3}]_1^3 = \frac{9}{2} \cdot \frac{1}{3} (27-1) = \frac{3}{2} (26) = 39$.

Let $f: R \rightarrow R$ be a function defined by $f(x)=\frac{x}{(1+x^4)^{1/4}}$ and $g(x)=f(f(f(f(x))))$. Then find the value of $18 \int_0^{\sqrt{2\sqrt{5}}} x^3 g(x) dx$.

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