Let f be a differentiable function from

$\because f: R \rightarrow R$

and $|f(x)-f(y)| \leq 2 \cdot|x-y|^{3 / 2}$

$\Rightarrow \quad\left|\frac{f(x) \quad f(y)}{x \quad y}\right| \leq \sqrt{x-y}$

$\Rightarrow \quad \lim _{x \rightarrow y}\left|\frac{f(x)-f(y)}{x-y}\right| \leq \lim _{x \rightarrow y} 2 \sqrt{x-y}$

$\Rightarrow \quad\left|f^{\prime}(x)\right|=0$

$\therefore f(x)$ is a constant function.

Given $f(0)=1 \quad \Rightarrow \quad f(x)=1$

Hence, the integral

$\int_{0}^{1} f^{2}(x) d x=\int_{0}^{1} 1 d x=[x]_{0}^{1}=1$