For any real number x , let [x] denotes the largest integer less...

Given

$I = {∫}_{0}^{10}\left[\sqrt{\dfrac{10x}{x + 1}}\right]dx$

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

$\dfrac{10x}{x + 1} = 1 ⇒ x = \dfrac{1}{9}$

$\dfrac{10x}{x + 1} = 4 ⇒ x = \dfrac{2}{3}$

$\dfrac{10x}{x + 1} = 9 ⇒ x = 9$

$I = {∫}_{0}^{\dfrac{1}{9}}0dx + {∫}_{\dfrac{1}{9}}^{\dfrac{2}{3}}1dx + {∫}_{\dfrac{2}{3}}^{9}2dx + {∫}_{9}^{10}3dx$

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

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

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

So, $9I = 182$