Evaluate the integral `l=int_(0)^(2)|1-x| d x `.

Answer

To evaluate the integral \( l = \int_{0}^{2} |1 - x| \, dx \), we will break it down into manageable steps.

### Step 1: Analyze the Absolute Value
The expression \( |1 - x| \) can be rewritten based on the value of \( x \):
- For \( x < 1 \), \( |1 - x| = 1 - x \)
- For \( x \geq 1 \), \( |1 - x| = -(1 - x) = x - 1 \)

### Step 2: Break the Integral into Intervals
Since the expression changes at \( x = 1 \), we can split the integral into two parts:
\[ l = \int_{0}^{1} (1 - x) \, dx + \int_{1}^{2} (x - 1) \, dx \]

### Step 3: Evaluate the First Integral
Now, we evaluate the first integral:
\[ \int_{0}^{1} (1 - x) \, dx \]
Calculating this integral:
\[ = \left[ x - \frac{x^2}{2} \right]_{0}^{1} \]
Substituting the limits:
\[ = \left( 1 - \frac{1^2}{2} \right) - \left( 0 - 0 \right) = 1 - \frac{1}{2} = \frac{1}{2} \]

### Step 4: Evaluate the Second Integral
Next, we evaluate the second integral:
\[ \int_{1}^{2} (x - 1) \, dx \]
Calculating this integral:
\[ = \left[ \frac{x^2}{2} - x \right]_{1}^{2} \]
Substituting the limits:
\[ = \left( \frac{2^2}{2} - 2 \right) - \left( \frac{1^2}{2} - 1 \right) = \left( 2 - 2 \right) - \left( \frac{1}{2} - 1 \right) = 0 - \left( \frac{1}{2} - 1 \right) = 0 + \frac{1}{2} = \frac{1}{2} \]

### Step 5: Combine the Results
Now, we combine the results of both integrals:
\[ l = \frac{1}{2} + \frac{1}{2} = 1 \]

### Final Answer
Thus, the value of the integral is:
\[ \boxed{1} \]