Key Concepts and Formulas:

• Greatest Integer Function Property: For any real number $x$ and integer $k$, $[x-k] = [x]-k$.
• Definition of Minimum: The minimum of a set of numbers is the smallest value in that set.
• Absolute Value Definition: $|y| = y$ if $y \ge 0$, and $|y| = -y$ if $y < 0$.
• Definite Integral of a Step Function: If a function is piecewise constant over an interval, its definite integral is the sum of the products of the constant value and the length of the sub-interval.
• Sum of Squares Formula: $\sum_{n=1}^{N} n^2 = \frac{N(N+1)(2N+1)}{6}$.

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Step-by-Step Solution:

1. Simplify the function $f(x)$

The function is given by $f(x)=\min \{[x-1],[x-2], \ldots,[x-10]\}$. We can use the property of the greatest integer function: $[t-k] = [t]-k$ for any integer $k$. Applying this to each term within the minimum function: $[x-1] = [x]-1$ $[x-2] = [x]-2$ $\vdots$ $[x-10] = [x]-10$

So, $f(x) = \min \{[x]-1, [x]-2, \ldots, [x]-10\}$. Let $I = [x]$. Then $f(x) = \min \{I-1, I-2, \ldots, I-10\}$. The values $I-1, I-2, \ldots, I-10$ are in decreasing order. Therefore, the minimum value is $I-10$. Thus, $f(x) = [x]-10$. Using the greatest integer function property in reverse, we can write $f(x) = [x-10]$.

2. Simplify the integral expression using the properties of $f(x)$ and absolute value

The expression to evaluate is $E = \int\limits_{0}^{10} f(x) \mathrm{d} x+\int\limits_{0}^{10}(f(x))^{2} \mathrm{~d} x+\int\limits_{0}^{10}|f(x)| \mathrm{d} x$. We established that $f(x) = [x-10]$. We need to determine the sign of $f(x)$ for $x \in [0, 10]$. For $x \in [0, 10]$, the term $x-10$ ranges from $0-10 = -10$ to $10-10 = 0$. So, $x-10 \in [-10, 0]$. This implies that $f(x) = [x-10]$ will always be a non-positive integer. Specifically, $f(x) \le 0$ for all $x \in [0, 10]$.

Since $f(x) \le 0$ for $x \in [0, 10]$, we can simplify $|f(x)|$. According to the definition of absolute value, if $y \le 0$, then $|y| = -y$. Therefore, $|f(x)| = -f(x)$ for $x \in [0, 10]$.

Substitute this into the expression for $E$: $E = \int\limits_{0}^{10} f(x) \mathrm{d} x+\int\limits_{0}^{10}(f(x))^{2} \mathrm{~d} x+\int\limits_{0}^{10}(-f(x)) \mathrm{d} x$ Using the linearity of definite integrals, we can combine the first and third integrals: $E = \left(\int\limits_{0}^{10} f(x) \mathrm{d} x - \int\limits_{0}^{10} f(x) \mathrm{d} x\right) + \int\limits_{0}^{10}(f(x))^{2} \mathrm{~d} x$ The first two terms cancel out: $E = \int\limits_{0}^{10}(f(x))^{2} \mathrm{~d} x$

3. Evaluate the remaining integral

We need to compute $E = \int\limits_{0}^{10} ([x-10])^{2} \mathrm{~d} x$. We know $f(x) = [x-10] = [x]-10$. The greatest integer function $[x]$ is a step function, changing its value at integer points. We can evaluate the integral by breaking the interval $[0, 10]$ into unit sub-intervals: $[0, 1), [1, 2), \ldots, [9, 10)$.

In each sub-interval $[k, k+1)$ for $k \in \{0, 1, \ldots, 9\}$, the value of $[x]$ is $k$. Therefore, for $x \in [k, k+1)$, $f(x) = [x]-10 = k-10$.

The integral can be written as a sum of integrals over these sub-intervals: $E = \sum_{k=0}^{9} \int_{k}^{k+1} ([x]-10)^2 \mathrm{d} x$ Substituting the constant value of $([x]-10)$ in each interval: $E = \int_{0}^{1} (0-10)^2 \mathrm{d} x + \int_{1}^{2} (1-10)^2 \mathrm{d} x + \int_{2}^{3} (2-10)^2 \mathrm{d} x + \ldots + \int_{9}^{10} (9-10)^2 \mathrm{d} x$ $E = \int_{0}^{1} (-10)^2 \mathrm{d} x + \int_{1}^{2} (-9)^2 \mathrm{d} x + \int_{2}^{3} (-8)^2 \mathrm{d} x + \ldots + \int_{9}^{10} (-1)^2 \mathrm{d} x$ For a constant $C$, $\int_{a}^{b} C^2 \mathrm{d} x = C^2 (b-a)$. Since the length of each interval is 1, each integral evaluates to the square of the constant: $E = (-10)^2 \cdot (1-0) + (-9)^2 \cdot (2-1) + (-8)^2 \cdot (3-2) + \ldots + (-1)^2 \cdot (10-9)$ $E = (-10)^2 + (-9)^2 + (-8)^2 + \ldots + (-1)^2$ $E = 10^2 + 9^2 + 8^2 + \ldots + 1^2$ This is the sum of the squares of the first 10 natural numbers. Using the sum of squares formula $\sum_{n=1}^{N} n^2 = \frac{N(N+1)(2N+1)}{6}$ with $N=10$: $E = \frac{10(10+1)(2 \cdot 10+1)}{6}$ $E = \frac{10 \cdot 11 \cdot 21}{6}$ $E = \frac{2310}{6}$ $E = 385$

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Common Mistakes & Tips:

• Simplifying $f(x)$: Ensure correct application of the $[x-k] = [x]-k$ property. Misinterpreting the minimum of the set $\{[x]-1, \ldots, [x]-10\}$ can lead to an incorrect $f(x)$.
• Absolute Value Analysis: Carefully determine the sign of $f(x)$ over the integration interval. Incorrectly simplifying $|f(x)|$ will alter the integral.
• Integral of Step Functions: Break the integral at integer points where the greatest integer function changes. The integral of a constant over an interval is that constant multiplied by the interval's length.

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Summary:

The problem involves simplifying a function defined using the minimum of greatest integer functions, and then evaluating a definite integral expression. By using the property $[x-k] = [x]-k$, we simplified $f(x)$ to $[x-10]$. Over the interval $[0, 10]$, $f(x)$ is always non-positive, which simplified the absolute value term $|f(x)|$ to $-f(x)$. This led to the cancellation of the $\int f(x) dx$ and $\int -f(x) dx$ terms, leaving only $\int (f(x))^2 dx$. The integral of $([x-10])^2$ over $[0, 10]$ was then evaluated by splitting the integral into unit intervals, resulting in the sum of squares of integers from 1 to 10, which was calculated using the standard formula.

The final answer is $\boxed{385}$.