The smallest interval $[a, b]$ such that $\int\limits_0^1 \frac{1}{\sqrt{1+x^4}} d x \in[a, b]$ is given by

$[1 / \sqrt{2}, 1]$

Let $I=\int\limits_0^1 \frac{1}{\sqrt{1+x^4}} d x$

We have,

$0 \leq x \leq 1 \Rightarrow 0 \leq x^4 \leq 1 \Rightarrow 1 \leq 1+x^4 \leq 2$

$\Rightarrow 1 \leq \sqrt{1+x^4} \leq \sqrt{2}$

$\Rightarrow \frac{1}{\sqrt{2}} \leq \frac{1}{\sqrt{1+x^4}} \leq 1$

$\Rightarrow \frac{1}{\sqrt{2}}(1-0) \leq \int\limits_0^1 \frac{1}{\sqrt{1+x^4}} d x \leq 1(1-0)$

$\Rightarrow \frac{1}{\sqrt{2}} \leq I \leq 1 \Rightarrow I \in\left[\frac{1}{\sqrt{2}}, 1\right]$