Question

For any real number $x$, let $[x]$ denote the largest integer less than or equal to $x$ If
$I=\int\limits_{0}^{10}\left[\sqrt{\frac{10 x}{x+1}}\right] d x,$
then the value of $9I$ is _____

Answer 1

Correct Answer: 182

Step 1: Analyze the Function Inside the Integral
The expression inside the floor function is $$ f(x) = \sqrt{\frac{10x}{x+1}}. $$
Let's examine its behavior as a function of $x$. First, observe that:
- As $x → 0$, we have $$ \sqrt{\frac{10x}{x+1}} \to \sqrt{\frac{0}{1}} = 0. $$
- As $x → \infty$, we have $$ \sqrt{\frac{10x}{x+1}} \to \sqrt{10} \quad \text{because} \quad \frac{10x}{x+1} \to 10. $$
Thus, the function $f(x)$ ranges from $0$ at $x = 0$ to $\sqrt{10}$ as $x$ increases towards 10.
Step 2: Determine the Floor Function's Behavior
The floor function $[f(x)]$ takes integer values depending on the value of $f(x)$. We need to determine intervals where $f(x)$ lies between integers. Let's approximate $\sqrt{10}$:
$$ \sqrt{10} \approx 3.162. $$
Thus, the floor function $[f(x)]$ will take integer values between 0 and 3.
Step 3: Find the Intervals for Each Value of $[f(x)]$
1. For $[f(x)] = 0$:
We want to find the values of $x$ for which $$ 0 ≤ \sqrt{\frac{10x}{x+1}} < 1. $$
Squaring both sides gives $$ 0 ≤ \frac{10x}{x+1} < 1, $$ or equivalently, $$ 0 ≤ 10x < x + 1, \quad \text{which simplifies to} \quad 9x < 1. $$
Therefore, $$ x < \frac{1}{9}. $$
Thus, for $x \in [0, \frac{1}{9})$, $[f(x)] = 0$.
2. For $[f(x)] = 1$:
We want to find the values of $x$ for which $$ 1 ≤ \sqrt{\frac{10x}{x+1}} < 2. $$
Squaring both sides gives $$ 1 ≤ \frac{10x}{x+1} < 4, $$ which simplifies to $$ 10x ≥ x + 1 \quad \text{and} \quad 10x < 4x + 4. $$
Solving these inequalities:
- From $10x ≥ x + 1$, we get $9x ≥ 1$, or $x ≥ \frac{1}{9}$.
- From $10x < 4x + 4$, we get $6x < 4$, or $x < \frac{2}{3}$.
Therefore, for $x \in [\frac{1}{9}, \frac{2}{3})$, $[f(x)] = 1$.
3. For $[f(x)] = 2$:
We want to find the values of $x$ for which $$ 2 ≤ \sqrt{\frac{10x}{x+1}} < 3. $$
Squaring both sides gives $$ 4 ≤ \frac{10x}{x+1} < 9, $$ which simplifies to $$ 4x + 4 ≤ 10x \quad \text{and} \quad 10x < 9x + 9. $$
Solving these inequalities:
- From $4x + 4 ≤ 10x$, we get $6x ≥ 4$, or $x ≥ \frac{2}{3}$.
- From $10x < 9x + 9$, we get $x < 9$.
Therefore, for $x \in [\frac{2}{3}, 9)$, $[f(x)] = 2$.
4. For $[f(x)] = 3$:
We want to find the values of $x$ for which $$ 3 ≤ \sqrt{\frac{10x}{x+1}} < 4. $$
Squaring both sides gives $$ 9 ≤ \frac{10x}{x+1} < 16, $$ which simplifies to $$ 9x + 9 ≤ 10x \quad \text{and} \quad 10x < 16x + 16. $$
Solving these inequalities:
- From $9x + 9 ≤ 10x$, we get $x ≥ 9$.
Thus, for $x \in [9, 10]$, $[f(x)] = 3$.
Step 4: Set Up the Integral
We now break the integral into parts based on the intervals:
- From $0$ to $\frac{1}{9}$, $[f(x)] = 0$.
- From $\frac{1}{9}$ to $\frac{2}{3}$, $[f(x)] = 1$.
- From $\frac{2}{3}$ to $9$, $[f(x)] = 2$.
- From $9$ to $10$, $[f(x)] = 3$.
Therefore, we can express the integral as: $$ I = \int_0^{\frac{1}{9}} 0 \, dx + \int_{\frac{1}{9}}^{\frac{2}{3}} 1 \, dx + \int_{\frac{2}{3}}^9 2 \, dx + \int_9^{10} 3 \, dx. $$
Step 5: Compute Each Integral
1. $\int_0^{\frac{1}{9}} 0 \, dx = 0$.
2. $\int_{\frac{1}{9}}^{\frac{2}{3}} 1 \, dx = \frac{2}{3} - \frac{1}{9} = \frac{5}{9}$.
3. $\int_{\frac{2}{3}}^9 2 \, dx = 2 \times (9 - \frac{2}{3}) = 2 \times \frac{25}{3} = \frac{50}{3}$.
4. $\int_9^{10} 3 \, dx = 3 \times (10 - 9) = 3$.
Thus, $$ I = 0 + \frac{5}{9} + \frac{50}{3} + 3 = \frac{5}{9} + \frac{150}{9} + \frac{27}{9} = \frac{182}{9}. $$
Step 6: Compute $9I$
Finally, we compute $$ 9I = 9 \times \frac{182}{9} = 182. $$
Thus, the value of $9I$ is $\boxed{182}$.