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

Answer

To evaluate the integral \( I = \int_{0}^{2} |x - 1| \, dx \), we first need to analyze the absolute value function \( |x - 1| \).

### Step 1: Determine the intervals for the absolute value
The expression \( |x - 1| \) changes at the point where \( x = 1 \). Therefore, we will split the integral into two parts:
- From \( 0 \) to \( 1 \), where \( x - 1 < 0 \) and thus \( |x - 1| = -(x - 1) = 1 - x \).
- From \( 1 \) to \( 2 \), where \( x - 1 \geq 0 \) and thus \( |x - 1| = x - 1 \).

### Step 2: Rewrite the integral
We can rewrite the integral as:
\[ I = \int_{0}^{1} (1 - x) \, dx + \int_{1}^{2} (x - 1) \, dx \]

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

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

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

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