Let the domain of the function \(\mathrm{f}(\mathrm{x})={\log }_{2}{\log }_{4}{\log }_{6}\left(3+4\mathrm{x}-{\mathrm{x}…

\(f(x)={\log ⁡}_{2}{\log ⁡}_{4}{\log ⁡}_{6}(3+4x−{x}^{2})\)

\({\log ⁡}_{6}(3+4x−{x}^{2})>0\Rightarrow 3+4x−{x}^{2}>1\)

\(−{x}^{2}+4x+2>0\Rightarrow {x}^{2}−4x−2<0\)

Roots: \(x=2\pm \sqrt{6}\).

So, \(2−\sqrt{6}

\({\log ⁡}_{4}({\log ⁡}_{6}(3+4x−{x}^{2}))>0\)\(\Rightarrow {\log ⁡}_{6}(3+4x−{x}^{2})>1\)

\(3+4x−{x}^{2}>6\Rightarrow −{x}^{2}+4x−3>0\)\(\Rightarrow {x}^{2}−4x+3<0\)

so \(1

\({\log ⁡}_{2}({\log ⁡}_{4}{\log ⁡}_{6}(⋅))\text{ requires }{\log ⁡}_{4}{\log ⁡}_{6}(⋅)>0,\)

which is the same as step 2. So

\(\text{domain}=(a,b)=(1,3)\mathrm{.}\)

Thus \(b−a=2\) and we need

\({\int }_{0}^{2}⌊{x}^{2}⌋\text{ }dx\mathrm{.}\)

On \([0,2]\), \({x}^{2}\) runs from 0 to 4; break at where \({x}^{2}=k\):

• \(0\leq x<1\): \(⌊{x}^{2}⌋=0\)
• \(1\leq x<\sqrt{2}\): \(⌊{x}^{2}⌋=1\)
• \(\sqrt{2}\leq x<\sqrt{3}\): \(⌊{x}^{2}⌋=2\)
• \(\sqrt{3}\leq x\leq 2\): \(⌊{x}^{2}⌋=3\)

So

\({\int }_{0}^{2}⌊{x}^{2}⌋dx={\int }_{0}^{1}0\text{ }dx+{\int }_{1}^{\sqrt{2}}1\text{ }dx\)\(+{\int }_{\sqrt{2}}^{\sqrt{3}}2\text{ }dx+{\int }_{\sqrt{3}}^{2}3\text{ }dx\)

\(=(\sqrt{2}−1)+2(\sqrt{3}−\sqrt{2})+3(2−\sqrt{3})\mathrm{.}\)

Simplify:

\(=\sqrt{2}−1+2\sqrt{3}−2\sqrt{2}+6−3\sqrt{3}\)\(=5−\sqrt{2}−\sqrt{3}\mathrm{.}\)

Thus it matches \(p−\sqrt{q}−\sqrt{r}\) with

\(p+q+r=5+2+3=10.\)