The value of 9 0 9 [ 10 x x + 1 ] d x , where [ t ] denotes the greatest integer less than or equal to t , is [2024]

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Solution

Q.
The value of $9 \int_{0}^{9} [\sqrt{\frac{10 x}{x + 1}}] d x,$ where $[t]$ denotes the greatest integer less than or equal to $t,$ is ______ . [2024]

Ans.
(155)

$I = 9 \int_{0}^{9} [\sqrt{\frac{10 x}{x + 1}}] d x$

Now, at $x = 0,$ $[\sqrt{\frac{10 x}{x + 1}}] = 0$ at $x = 9,$ $[\sqrt{\frac{10 x}{x + 1}}] = 3$

Integer values between 0 and 3 are 1 and 2

So $[\sqrt{\frac{10 x}{x + 1}}] = 1 and [\sqrt{\frac{10 x}{x + 1}}] = 2$

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

$\Rightarrow 10 x = x + 1 and 10 x = 4 x + 4$

$\Rightarrow x = \frac{1}{9} and x = \frac{2}{3}$

$I = 9 \int_{0}^{1 / 9} [\sqrt{\frac{10 x}{x + 1}}] d x + \int_{1 / 9}^{2 / 3} [\sqrt{\frac{10 x}{x + 1}}] d x + \int_{2 / 3}^{9} [\sqrt{\frac{10 x}{x + 1}}] d x$

$= 9 [\int_{0}^{1 / 9} 0 d x + \int_{1 / 9}^{2 / 3} 1 d x + \int_{2 / 3}^{9} 2 d x]$ [ $∵$ $[t]$ is a greatest integer function]

$= 9 [0 + \frac{2}{3} - \frac{1}{9} + 18 - \frac{4}{3}] = 9 [- \frac{2}{3} - \frac{1}{9} + 18]$

$= 162 - 6 - 1 = 155$

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