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

Answer

To evaluate the integral \( I = \int_{0}^{2} |x - 1| \, dx \), we need to consider the behavior of the function \( |x - 1| \) over the interval from 0 to 2.

### Step 1: Identify the points where the expression inside the modulus changes sign.
The expression \( x - 1 \) changes sign at \( x = 1 \):
- For \( x < 1 \), \( x - 1 < 0 \) so \( |x - 1| = -(x - 1) = 1 - x \).
- For \( x \geq 1 \), \( x - 1 \geq 0 \) so \( |x - 1| = x - 1 \).

### Step 2: Break the integral into two parts based on the point where the modulus changes.
We can split the integral at \( x = 1 \):
\[ I = \int_{0}^{1} |x - 1| \, dx + \int_{1}^{2} |x - 1| \, dx \]
This gives us:
\[ I = \int_{0}^{1} (1 - x) \, dx + \int_{1}^{2} (x - 1) \, dx \]

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

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

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

### Final Answer:
Thus, the value of the integral \( I = \int_{0}^{2} |x - 1| \, dx \) is \( \boxed{1} \).