Let f:R R be a function defined by f(x)=x(1+x^4)^1/4, and g(x)=f(f(f(f(x)))).

Question

Let $f:R	o R$ be a function defined by $f(x)=\frac{x}{{(1+{x}^{4})}^{1/4}}$, and $g(x)=f(f(f(f(x))))$. Then, $18{\int }_{\sqrt{2\sqrt{5}}}^{0}{x}^{2}g(x)dx$ is equal to

Accepted Answer

$$\begin{aligned}
&f(x)=\frac{x}{\left(1+x^4\right)^{1 / 4}}\
&f \circ f(x)=\frac{f(x)}{\left(1+f(x)^4\right)^{1 / 4}}=\frac{\frac{x}{\left(1+x^4\right)^{1 / 4}}}{\left(1+\frac{x^4}{1+x^4}\right)^{1 / 4}}=\frac{x}{\left(1+2 x^4\right)^{1 / 4}}\
&f(f(f(f(x))))=\frac{x}{\left(1+4 x^4\right)^{1 / 4}}
\end{aligned}$$

$$18 \int_\limits0^{\sqrt{2 \sqrt{5}}} \frac{x^3}{\left(1+4 x^4\right)^{1 / 4}} d x$$

$$\begin{aligned}
& \text { Let } 1+4 \mathrm{x}^4=\mathrm{t}^4 \
& 16 \mathrm{x}^3 \mathrm{dx}=4 \mathrm{t}^3 \mathrm{dt} \
& \frac{18}{4} \int_\limits1^3 \frac{\mathrm{t}^3 \mathrm{dt}}{\mathrm{t}} \
& =\frac{9}{2}\left(\frac{\mathrm{t}^3}{3}\right)_1^3 \
& =\frac{3}{2}[26]=39
\end{aligned}$$

Solutions · 1