`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 values and break the integral into parts based on the critical points where the expressions inside the absolute values change sign. The critical points for this problem are \( x = 1 \) and \( x = 3 \).

### Step 1: Identify the intervals
We will break the integral into three intervals based on the critical points:
1. \( [0, 1] \)
2. \( [1, 3] \)
3. \( [3, 4] \)

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

#### Interval 1: \( [0, 1] \)
In this interval, both \( x-1 \) and \( x-3 \) are negative:
- \( |x-1| = 1-x \)
- \( |x-3| = 3-x \)

Thus, the integral becomes:
\[ \int_{0}^{1} (|x-1| + |x-3|) \, dx = \int_{0}^{1} ((1-x) + (3-x)) \, dx = \int_{0}^{1} (4 - 2x) \, dx \]

Calculating this integral:
\[ \int_{0}^{1} (4 - 2x) \, dx = \left[ 4x - x^2 \right]_{0}^{1} = (4 \cdot 1 - 1^2) - (4 \cdot 0 - 0^2) = 4 - 1 = 3 \]

#### Interval 2: \( [1, 3] \)
In this interval, \( x-1 \) is positive and \( x-3 \) is negative:
- \( |x-1| = x-1 \)
- \( |x-3| = 3-x \)

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

Calculating this integral:
\[ \int_{1}^{3} 2 \, dx = [2x]_{1}^{3} = 2 \cdot 3 - 2 \cdot 1 = 6 - 2 = 4 \]

#### Interval 3: \( [3, 4] \)
In this interval, both \( x-1 \) and \( x-3 \) are positive:
- \( |x-1| = x-1 \)
- \( |x-3| = x-3 \)

Thus, the integral becomes:
\[ \int_{3}^{4} (|x-1| + |x-3|) \, dx = \int_{3}^{4} ((x-1) + (x-3)) \, dx = \int_{3}^{4} (2x - 4) \, dx \]

Calculating this 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 3: Combine the results
Now, we add the results from all three intervals:
\[ 3 + 4 + 3 = 10 \]

Thus, the final answer is:
\[ \int_{0}^{4} (|x-1| + |x-3|) \, dx = 10 \]