The value of `int_(-1)^(3)(|x|+|x-1|) dx` is equal to

Options

A. 9
B. 6
C. 3
D. None of these

Answer

To solve the integral \( \int_{-1}^{3} (|x| + |x-1|) \, dx \), we need to break it down into intervals based on the points where the expressions inside the absolute values change sign. The critical points for the absolute values are \( x = 0 \) and \( x = 1 \).

### Step 1: Identify the intervals
The intervals based on the critical points are:
1. \( [-1, 0] \)
2. \( [0, 1] \)
3. \( [1, 3] \)

### Step 2: Evaluate the integral on each interval

#### Interval 1: \( [-1, 0] \)
In this interval, both \( |x| \) and \( |x-1| \) are negative:
- \( |x| = -x \)
- \( |x-1| = -(x-1) = -x + 1 \)

Thus, the expression becomes:
\[ |x| + |x-1| = -x + (-x + 1) = -2x + 1 \]
Now, we can set up the integral:
\[ \int_{-1}^{0} (-2x + 1) \, dx \]

#### Interval 2: \( [0, 1] \)
In this interval, \( |x| \) is positive and \( |x-1| \) is negative:
- \( |x| = x \)
- \( |x-1| = -(x-1) = -x + 1 \)

Thus, the expression becomes:
\[ |x| + |x-1| = x + (-x + 1) = 1 \]
Now, we can set up the integral:
\[ \int_{0}^{1} 1 \, dx \]

#### Interval 3: \( [1, 3] \)
In this interval, both \( |x| \) and \( |x-1| \) are positive:
- \( |x| = x \)
- \( |x-1| = x - 1 \)

Thus, the expression becomes:
\[ |x| + |x-1| = x + (x - 1) = 2x - 1 \]
Now, we can set up the integral:
\[ \int_{1}^{3} (2x - 1) \, dx \]

### Step 3: Calculate each integral

1. **Integral from \(-1\) to \(0\)**:
\[ \int_{-1}^{0} (-2x + 1) \, dx = \left[-x^2 + x\right]_{-1}^{0} = \left[0 - 0\right] - \left[-1 + 1\right] = 0 - 0 = 1 \]

2. **Integral from \(0\) to \(1\)**:
\[ \int_{0}^{1} 1 \, dx = [x]_{0}^{1} = 1 - 0 = 1 \]

3. **Integral from \(1\) to \(3\)**:
\[ \int_{1}^{3} (2x - 1) \, dx = \left[x^2 - x\right]_{1}^{3} = \left[9 - 3\right] - \left[1 - 1\right] = 6 - 0 = 6 \]

### Step 4: Add the results of the integrals
Now we sum the results from all three intervals:
\[ 1 + 1 + 6 = 8 \]

### Conclusion
Thus, the value of the integral \( \int_{-1}^{3} (|x| + |x-1|) \, dx \) is \( 8 \).