For any real number x , let x denotes the largest integer less than or equal to x . If I = ∫ 0 10 10 x…

Given

$I={\int }_0^{10}\left[\sqrt{\frac{10x}{x+1}}\right]dx$

Check the values of $x$ at which the greatest integer function becomes integer.

$\frac{10x}{x+1}=1 \Rightarrow x=\frac{1}{9}$

$\frac{10x}{x+1}=4 \Rightarrow x=\frac{2}{3}$

$\frac{10x}{x+1}=9 \Rightarrow x=9$

$I={\int }_0^{\frac{1}{9}}0dx+{\int }_{\frac{1}{9}}^{\frac{2}{3}}1dx+{\int }_{\frac{2}{3}}^{9}2dx+{\int }_9^{10}3dx$

$I=0+[x{]}_{\frac{1}{9}}^{\frac{2}{3}}+[2x{]}_{\frac{2}{3}}^9+[3x{]}_9^{10}$

$I=\frac{2}{3}-\frac{1}{9}+18-\frac{4}{3}+30-27$

$I=21-\frac{2}{3}-\frac{1}{9}=\frac{182}{9}$

So, $9I=182$