`int_(0)^(4)(|x-1|+|x-3|)dx`

Answer

To solve the integral \( \int_{0}^{4} (|x-1| + |x-3|) \, dx \), we need to analyze the absolute value expressions and break the integral into appropriate intervals.

### Step 1: Identify the points where the expressions change
The absolute value functions \( |x-1| \) and \( |x-3| \) change at \( x = 1 \) and \( x = 3 \). Therefore, we will consider the intervals:
- \( [0, 1] \)
- \( [1, 3] \)
- \( [3, 4] \)

### Step 2: Rewrite the integral in terms of these intervals
We can express the integral as:
\[ \int_{0}^{4} (|x-1| + |x-3|) \, dx = \int_{0}^{1} (-(x-1) - (x-3)) \, dx + \int_{1}^{3} ((x-1) + (-(x-3))) \, dx + \int_{3}^{4} ((x-1) + (x-3)) \, dx \]

### Step 3: Simplify each interval
1. For \( x \in [0, 1] \):
\[ |x-1| = -(x-1) = 1-x \quad \text{and} \quad |x-3| = -(x-3) = 3-x \]
Thus,
\[ |x-1| + |x-3| = (1-x) + (3-x) = 4 - 2x \]

2. For \( x \in [1, 3] \):
\[ |x-1| = x-1 \quad \text{and} \quad |x-3| = -(x-3) = 3-x \]
Thus,
\[ |x-1| + |x-3| = (x-1) + (3-x) = 2 \]

3. For \( x \in [3, 4] \):
\[ |x-1| = x-1 \quad \text{and} \quad |x-3| = x-3 \]
Thus,
\[ |x-1| + |x-3| = (x-1) + (x-3) = 2x - 4 \]

### Step 4: Write the integral with the simplified expressions
Now we can write the integral as:
\[ \int_{0}^{1} (4 - 2x) \, dx + \int_{1}^{3} 2 \, dx + \int_{3}^{4} (2x - 4) \, dx \]

### Step 5: Calculate each integral
1. **First integral**:
\[ \int_{0}^{1} (4 - 2x) \, dx = \left[ 4x - x^2 \right]_{0}^{1} = (4 \cdot 1 - 1^2) - (4 \cdot 0 - 0^2) = 4 - 1 = 3 \]

2. **Second integral**:
\[ \int_{1}^{3} 2 \, dx = \left[ 2x \right]_{1}^{3} = 2 \cdot 3 - 2 \cdot 1 = 6 - 2 = 4 \]

3. **Third integral**:
\[ \int_{3}^{4} (2x - 4) \, dx = \left[ x^2 - 4x \right]_{3}^{4} = (4^2 - 4 \cdot 4) - (3^2 - 4 \cdot 3) = (16 - 16) - (9 - 12) = 0 + 3 = 3 \]

### Step 6: Combine the results
Now, we sum the results of the three integrals:
\[ 3 + 4 + 3 = 10 \]

### Final Answer
Thus, the value of the integral \( \int_{0}^{4} (|x-1| + |x-3|) \, dx \) is \( \boxed{10} \).