Question:

Step 1: Define the fractional part function and its properties. The fractional part of a real number $x$, denoted by $\{x\}$, is defined as $x - [x]$, where $[x]$ is the greatest integer less than or equal to $x$. By definition, the fractional part always satisfies $0 \le \{x\} < 1$. Step 2: Simplify the integrand using the definition of the fractional part and the signum function. The integrand is $\text{sgn}(x - [x])$. Using the definition from Step 1, we can replace $x - [x]$ with $\{x\}$. So, the integrand becomes $\text{sgn}(\{x\})$. Now, let's analyze the value of $\text{sgn}(\{x\})$: Since $0 \le \{x\} < 1$: - If $\{x\} > 0$, then $\text{sgn}(\{x\}) = 1$. This occurs when $x$ is not an integer. - If $\{x\} = 0$, then $\text{sgn}(\{x\}) = 0$. This occurs when $x$ is an integer. Step 3: Decompose the definite integral into a sum of integrals over unit intervals. The integral range is from $0$ to $10$. We can break this integral into a sum of integrals over unit intervals, from $[0,1)$, $[1,2)$, up to $[9,10)$. $$ \int_0^{10} \text{sgn}(x - [x]) \, dx = \int_0^{10} \text{sgn}(\{x\}) \, dx $$ This can be written as: $$ \sum_{n=0}^{9} \int_n^{n+1} \text{sgn}(\{x\}) \, dx = \int_0^1 \text{sgn}(\{x\}) \, dx + \int_1^2 \text{sgn}(\{x\}) \, dx + \dots + \int_9^{10} \text{sgn}(\{x\}) \, dx $$ Step 4: Evaluate the integral over a generic unit interval $[n, n+1)$. Consider a general interval $[n, n+1)$ where $n$ is an integer. In this interval, for any $x$ such that $n \le x < n+1$, the fractional part $\{x\}$ is greater than $0$. The only point where $\{x\} = 0$ is at $x=n$, which is a single point and does not affect the value of the definite integral. Thus, for $x \in [n, n+1)$, $\{x\} > 0$, which implies $\text{sgn}(\{x\}) = 1$. Therefore, the integral over each unit interval is: $$ \int_n^{n+1} \text{sgn}(\{x\}) \, dx = \int_n^{n+1} 1 \, dx $$ Evaluating this integral: $$ [x]_n^{n+1} = (n+1) - n = 1 $$ So, the contribution from each unit interval is $1$. Step 5: Sum the results from each subinterval to find the total value of the integral. There are $10$ such unit intervals from $n=0$ to $n=9$: $[0,1)$, $[1,2)$, $\dots$, $[9,10)$. Each interval contributes a value of $1$ to the total integral. Summing these contributions: $$ \int_0^{10} \text{sgn}(\{x\}) \, dx = \underbrace{1 + 1 + \dots + 1}_{10 \text{ times}} = 10 $$ The final answer is $\boxed{10}$.