`int_(-1)^(3) {|x-1|+ [x]} dx` with usual notations is

Options

A. 3
B. 4
C. 5
D. 6

Answer

To solve the integral \( I = \int_{-1}^{3} \left( |x-1| + [x] \right) dx \), we will break it down into two parts: the absolute value function \( |x-1| \) and the greatest integer function \( [x] \).

### Step 1: Break the integral into two parts
We can express the integral as:
\[ I = \int_{-1}^{3} |x-1| \, dx + \int_{-1}^{3} [x] \, dx \]
Let \( I_1 = \int_{-1}^{3} |x-1| \, dx \) and \( I_2 = \int_{-1}^{3} [x] \, dx \).

### Step 2: Evaluate \( I_1 = \int_{-1}^{3} |x-1| \, dx \)
The absolute value function \( |x-1| \) changes at \( x = 1 \):
- For \( x < 1 \), \( |x-1| = -(x-1) = 1-x \)
- For \( x \geq 1 \), \( |x-1| = x-1 \)

Thus, we can split the integral at \( x = 1 \):
\[ I_1 = \int_{-1}^{1} (1-x) \, dx + \int_{1}^{3} (x-1) \, dx \]

### Step 3: Compute \( \int_{-1}^{1} (1-x) \, dx \)
\[ \int_{-1}^{1} (1-x) \, dx = \left[ x - \frac{x^2}{2} \right]_{-1}^{1} = \left( 1 - \frac{1^2}{2} \right) - \left( -1 - \frac{(-1)^2}{2} \right) \]
Calculating this gives:
\[ = \left( 1 - \frac{1}{2} \right) - \left( -1 - \frac{1}{2} \right) = \frac{1}{2} + \frac{3}{2} = 2 \]

### Step 4: Compute \( \int_{1}^{3} (x-1) \, dx \)
\[ \int_{1}^{3} (x-1) \, dx = \left[ \frac{x^2}{2} - x \right]_{1}^{3} = \left( \frac{3^2}{2} - 3 \right) - \left( \frac{1^2}{2} - 1 \right) \]
Calculating this gives:
\[ = \left( \frac{9}{2} - 3 \right) - \left( \frac{1}{2} - 1 \right) = \left( \frac{9}{2} - \frac{6}{2} \right) - \left( \frac{1}{2} - \frac{2}{2} \right) = \frac{3}{2} + \frac{1}{2} = 2 \]

### Step 5: Combine results for \( I_1 \)
Thus, we have:
\[ I_1 = 2 + 2 = 4 \]

### Step 6: Evaluate \( I_2 = \int_{-1}^{3} [x] \, dx \)
The greatest integer function \( [x] \) is piecewise constant. We break it at the integers:
\[ I_2 = \int_{-1}^{0} (-1) \, dx + \int_{0}^{1} (0) \, dx + \int_{1}^{2} (1) \, dx + \int_{2}^{3} (2) \, dx \]

### Step 7: Compute each part of \( I_2 \)
1. \( \int_{-1}^{0} (-1) \, dx = -1 \cdot (0 - (-1)) = -1 \)
2. \( \int_{0}^{1} (0) \, dx = 0 \)
3. \( \int_{1}^{2} (1) \, dx = 1 \cdot (2 - 1) = 1 \)
4. \( \int_{2}^{3} (2) \, dx = 2 \cdot (3 - 2) = 2 \)

### Step 8: Combine results for \( I_2 \)
Thus, we have:
\[ I_2 = -1 + 0 + 1 + 2 = 2 \]

### Step 9: Combine \( I_1 \) and \( I_2 \)
Finally, we find:
\[ I = I_1 + I_2 = 4 + 2 = 6 \]

### Final Answer
The value of the integral is:
\[ \boxed{6} \]