For any real number x,let [x] denote the larg | vedclass

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(D) Let $f(x) = \frac{10x}{x+1}$.Then $f'(x) = \frac{10(x+1) - 10x}{(x+1)^2} = \frac{10}{(x+1)^2} > 0$ for all $x \in [0, 10]$.Thus,$f(x)$ is an incre

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$182$

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(D) Let $f(x) = \frac{10x}{x+1}$.Then $f'(x) = \frac{10(x+1) - 10x}{(x+1)^2} = \frac{10}{(x+1)^2} > 0$ for all $x \in [0, 10]$.Thus,$f(x)$ is an increasing function.We want to find the values of $x$ where $\sqrt{f(x)} = k$ for integer $k$.$\sqrt{\frac{10x}{x+1}} = k \implies \frac{10x}{x+1} = k^2 \implies 10x = k^2x + k^2 \implies x(10 - k^2) = k^2 \implies x = \frac{k^2}{10 - k^2}$.For $k=1$,$x = \frac{1}{9}$. For $k=2$,$x = \frac{4}{6} = \frac{2}{3}$. For $k=3$,$x = \frac{9}{1} = 9$.Since $f(x)$ is increasing,$\left[ \sqrt{f(x)} \right] = k$ for $x \in [x_k, x_{k+1})$.$I = \int_0^{1/9} 0 dx + \int_{1/9}^{2/3} 1 dx + \int_{2/3}^{9} 2 dx + \int_{9}^{10} 3 dx$.$I = 0 + (\frac{2}{3} - \frac{1}{9}) + 2(9 - \frac{2}{3}) + 3(10 - 9)$.$I = \frac{5}{9} + 2(\frac{25}{3}) + 3 = \frac{5}{9} + \frac{50}{3} + 3 = \frac{5 + 150 + 27}{9} = \frac{182}{9}$.Therefore,$9I = 182$.

$t \ (\text{in second})$	$0$	$2$	$4$	$6$	$8$	$10$
$v \ (\text{in m/s})$	$0$	$12$	$16$	$20$	$35$	$60$

For any real number $x$,let $[x]$ denote the largest integer less than or equal to $x$. If $I = \int_0^{10} \left[ \sqrt{\frac{10x}{x+1}} \right] dx$,then the value of $9I$ is . . . . . .

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