` int_(-1)^(2)f(x)dx " where " f(x) = |x+1| +|x|+ |x-1|` is equal to

Options

A. `7/2`
B. `9/2`
C. `13/2`
D. `19/2`

Explanation

To solve the integral \( \int_{-1}^{2} f(x) \, dx \) where \( f(x) = |x+1| + |x| + |x-1| \), we will first analyze the function \( f(x) \) by breaking it down into intervals based on the points where the expressions inside the absolute values change sign.

### Step 1: Identify the critical points
The critical points for \( f(x) \) occur when the expressions inside the absolute values are zero:
- \( x + 1 = 0 \) → \( x = -1 \)
- \( x = 0 \)
- \( x - 1 = 0 \) → \( x = 1 \)

Thus, we will consider the intervals: \( [-1, 0] \), \( [0, 1] \), and \( [1, 2] \).

### Step 2: Define \( f(x) \) in each interval

1. **Interval \( [-1, 0] \)**:
- \( x + 1 \geq 0 \) → \( |x + 1| = x + 1 \)
- \( x < 0 \) → \( |x| = -x \)
- \( x - 1 < 0 \) → \( |x - 1| = -x + 1 \)

Therefore, in this interval:
\[ f(x) = (x + 1) + (-x) + (-x + 1) = 2 - x \]

2. **Interval \( [0, 1] \)**:
- \( x + 1 \geq 0 \) → \( |x + 1| = x + 1 \)
- \( x \geq 0 \) → \( |x| = x \)
- \( x - 1 < 0 \) → \( |x - 1| = -x + 1 \)

Therefore, in this interval:
\[ f(x) = (x + 1) + x + (-x + 1) = 2 + x \]

3. **Interval \( [1, 2] \)**:
- \( x + 1 \geq 0 \) → \( |x + 1| = x + 1 \)
- \( x \geq 0 \) → \( |x| = x \)
- \( x - 1 \geq 0 \) → \( |x - 1| = x - 1 \)

Therefore, in this interval:
\[ f(x) = (x + 1) + x + (x - 1) = 3x \]

### Step 3: Set up the integral
Now we can express the integral \( I \) as the sum of integrals over the three intervals:
\[ I = \int_{-1}^{0} (2 - x) \, dx + \int_{0}^{1} (2 + x) \, dx + \int_{1}^{2} (3x) \, dx \]

### Step 4: Calculate each integral

1. **First integral**:
\[ \int_{-1}^{0} (2 - x) \, dx = \left[ 2x - \frac{x^2}{2} \right]_{-1}^{0} = \left( 0 - 0 \right) - \left( -2 - \frac{1}{2} \right) = 2 + \frac{1}{2} = \frac{5}{2} \]

2. **Second integral**:
\[ \int_{0}^{1} (2 + x) \, dx = \left[ 2x + \frac{x^2}{2} \right]_{0}^{1} = \left( 2 + \frac{1}{2} \right) - (0) = 2 + \frac{1}{2} = \frac{5}{2} \]

3. **Third integral**:
\[ \int_{1}^{2} (3x) \, dx = \left[ \frac{3x^2}{2} \right]_{1}^{2} = \left( \frac{3(2^2)}{2} - \frac{3(1^2)}{2} \right) = \left( \frac{12}{2} - \frac{3}{2} \right) = 6 - \frac{3}{2} = \frac{12}{2} - \frac{3}{2} = \frac{9}{2} \]

### Step 5: Combine the results
Now, we add the results of the three integrals:
\[ I = \frac{5}{2} + \frac{5}{2} + \frac{9}{2} = \frac{19}{2} \]

### Final Answer
Thus, the value of the integral \( \int_{-1}^{2} f(x) \, dx \) is:
\[ \boxed{\frac{19}{2}} \]