`int_0^(n+1) min {(|x-1|,|x-2|,|x-3|,............,|x-n|) }dx` equais to (where n be a positive integer)

Options

A. `((n +1))/(4 )`
B. `((n +2))/(4)`
C. `((n +3))/(4)`
D. `((n +4))/(4)`

Answer

To solve the integral \( I = \int_0^{n+1} \min \{ |x-1|, |x-2|, |x-3|, \ldots, |x-n| \} \, dx \), we will analyze the function inside the integral step by step.

### Step 1: Understand the function \( \min \{ |x-1|, |x-2|, |x-3|, \ldots, |x-n| \} \)

The function \( |x-k| \) represents the distance from \( x \) to \( k \). The minimum of these absolute values will change at the points \( x = 1, 2, 3, \ldots, n \).

### Step 2: Identify intervals

We need to analyze the behavior of the function in the intervals:
- \( [0, 1] \)
- \( [1, 2] \)
- \( [2, 3] \)
- \( [3, 4] \)
- \( \ldots \)
- \( [n, n+1] \)

### Step 3: Calculate the minimum in each interval

1. **Interval \( [0, 1] \)**:
- Here, \( |x-k| \) for \( k = 1, 2, \ldots, n \) gives:
- \( |x-1| = 1-x \)
- \( |x-2| = 2-x \)
- \( |x-3| = 3-x \)
- ...
- \( |x-n| = n-x \)
- The minimum is \( |x-1| = 1-x \) since \( x \) is closer to 1.

2. **Interval \( [1, 2] \)**:
- The minimum is \( |x-1| \) for \( x \in [1, 2] \) since \( x \) is closer to 1 than to any other point.

3. **Interval \( [2, 3] \)**:
- The minimum is \( |x-2| \) for \( x \in [2, 3] \).

4. **Interval \( [3, 4] \)**:
- The minimum is \( |x-3| \) for \( x \in [3, 4] \).

5. **Continue this pattern until \( [n, n+1] \)**:
- The minimum is \( |x-n| \) for \( x \in [n, n+1] \).

### Step 4: Set up the integral

Now we can express the integral as a sum of integrals over each interval:

\[ I = \int_0^1 (1-x) \, dx + \int_1^2 (x-1) \, dx + \int_2^3 (2-x) \, dx + \int_3^4 (x-3) \, dx + \ldots + \int_n^{n+1} (x-n) \, dx \]

### Step 5: Calculate each integral

1. **For \( \int_0^1 (1-x) \, dx \)**:
\[ = \left[ x - \frac{x^2}{2} \right]_0^1 = 1 - \frac{1}{2} = \frac{1}{2} \]

2. **For \( \int_1^2 (x-1) \, dx \)**:
\[ = \left[ \frac{x^2}{2} - x \right]_1^2 = \left( \frac{4}{2} - 2 \right) - \left( \frac{1}{2} - 1 \right) = 2 - 2 + \frac{1}{2} = \frac{1}{2} \]

3. **For \( \int_2^3 (2-x) \, dx \)**:
\[ = \left[ 2x - \frac{x^2}{2} \right]_2^3 = \left( 6 - \frac{9}{2} \right) - \left( 4 - 2 \right) = 6 - 4.5 - 2 = -0.5 \text{ (but we take absolute value)} \]

Continuing this way, we find that each integral contributes \( \frac{1}{2} \) until the last one.

### Step 6: Sum the contributions

Since there are \( n \) intervals, and each contributes \( \frac{1}{2} \):

\[ I = \frac{1}{2} + \frac{1}{2} + \ldots + \frac{1}{2} \text{ (n times)} = \frac{n+1}{2} \]

### Final Result

Thus, the value of the integral is:

\[ I = \frac{(n+1)}{4} \]