$7.4535$ is not the answer

A shorter version would be to consider the values of $\left \lfloor \sqrt{x} \right \rfloor$ for $x\in [1,5]$ between square points (where $x$ assumes a perfect square value). By definition, we have,

$\left \lfloor \sqrt{x} \right \rfloor= \begin{cases} 0 \quad \forall x\in [0,1) \\ 1 \quad \forall x\in [1,4) \\ 2 \quad \forall x\in [4,5] \end{cases}$

This is the same as your method but is a little bit quicker.