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Step-by-Step Solution
Step 1: Understand the Function f(x)
We are given
f(x) = \min\{[\,x - 1\,],[\,x - 2\,], \dots, [\,x - 10\,]\} ,
where [\,t\,] denotes the greatest integer less than or equal to t .
Observe that for any x in [\,0, 10\,] ,
x - 10 \leq x - k for 1 \leq k \leq 10 .
Hence, the smallest (minimum) among the values [\,x - 1\,], [\,x - 2\,], \ldots, [\,x - 10\,]
is always [\,x - 10\,].
Thus,
f(x) = [\,x - 10\,]\ \text{for}\ 0 \leq x \leq 10.
Step 2: Express the Required Sum of Integrals
We want to evaluate the following sum:
\[
\int_{0}^{10} f(x)\,dx \;+\; \int_{0}^{10} \bigl(f(x)\bigr)^2\,dx \;+\; \int_{0}^{10} \lvert f(x)\rvert\,dx.
\]
Step 3: Analyze the Sign of f(x)
For 0 \le x < 10, we have x - 10 < 0, so [\,x - 10\,] is a negative integer or zero (at x=10 exactly, x-10=0 but that is just a single point). Therefore, on the interval 0 \le x < 10, f(x) < 0.
When a number a is negative, a + |a| = 0 . Thus, for almost every x in [0,10) :
\[
f(x) + |f(x)| = 0.
\]
Step 4: Combine the Three Integrals
Observe:
\[
\int_{0}^{10} f(x)\,dx + \int_{0}^{10} |f(x)|\,dx
\;=\;
\int_{0}^{10} \Bigl(f(x) + |f(x)|\Bigr) dx
\;=\;
\int_{0}^{10} 0 \, dx
\;=\; 0.
\]
Therefore,
\[
\int_{0}^{10} f(x)\,dx \;+\; \int_{0}^{10} \bigl(f(x)\bigr)^2\,dx \;+\; \int_{0}^{10} \lvert f(x)\rvert\,dx
\;=\;
\int_{0}^{10} \bigl(f(x)\bigr)^2 \,dx.
\]
Step 5: Compute \displaystyle \int_{0}^{10} \bigl(f(x)\bigr)^2 \,dx
Since f(x) = [\,x - 10\,] , we note that on each integer interval [n, n+1) for n=0,1,\dots,9, the value of [\,x - 10\,] is a constant integer:
For 0 \le x < 1,\; x - 10 \in [-10, -9), \; f(x) = -10, so f(x)^2 = 100.
For 1 \le x < 2,\; x - 10 \in [-9, -8), \; f(x) = -9,\; f(x)^2 = 81.
For 2 \le x < 3,\; f(x) = -8,\; f(x)^2 = 64.
For 3 \le x < 4,\; f(x) = -7,\; f(x)^2 = 49.
For 4 \le x < 5,\; f(x) = -6,\; f(x)^2 = 36.
For 5 \le x < 6,\; f(x) = -5,\; f(x)^2 = 25.
For 6 \le x < 7,\; f(x) = -4,\; f(x)^2 = 16.
For 7 \le x < 8,\; f(x) = -3,\; f(x)^2 = 9.
For 8 \le x < 9,\; f(x) = -2,\; f(x)^2 = 4.
For 9 \le x < 10,\; f(x) = -1,\; f(x)^2 = 1.
The integral from 0 to 10 is just the sum of these constant values over each interval of length 1:
\[
\int_{0}^{10} [\,x - 10\,]^2 \, dx = 100 + 81 + 64 + 49 + 36 + 25 + 16 + 9 + 4 + 1.
\]
The sum of squares of the first 10 natural numbers is
\[
1^2 + 2^2 + \cdots + 10^2 \;=\; 385.
\]
Therefore,
\[
\int_{0}^{10} f(x)\,dx
\;+\;
\int_{0}^{10} \bigl(f(x)\bigr)^2 \,dx
\;+\;
\int_{0}^{10} |f(x)|\,dx
\;=\; 385.
\]
Final Answer
The required value of the given expression is \boxed{385}.
