Evaluate
`int_(1)^(3)|x^(2)-2x|dx`

Answer

To evaluate the integral \( \int_{1}^{3} |x^2 - 2x| \, dx \), we will follow these steps:

### Step 1: Identify the expression inside the absolute value
The expression inside the absolute value is \( x^2 - 2x \). We can factor this expression:
\[ x^2 - 2x = x(x - 2) \]
This expression will change sign at the roots, which are \( x = 0 \) and \( x = 2 \).

### Step 2: Determine the intervals for the absolute value
We need to analyze the sign of \( x^2 - 2x \) in the intervals defined by the roots:
- For \( x < 0 \): \( x^2 - 2x > 0 \)
- For \( 0 < x < 2 \): \( x^2 - 2x < 0 \)
- For \( x > 2 \): \( x^2 - 2x > 0 \)

Since we are integrating from \( 1 \) to \( 3 \), we only need to consider the intervals \( [1, 2] \) and \( [2, 3] \).

### Step 3: Rewrite the integral without the absolute value
In the interval \( [1, 2] \), \( x^2 - 2x < 0 \), so \( |x^2 - 2x| = -(x^2 - 2x) = -x^2 + 2x \).
In the interval \( [2, 3] \), \( x^2 - 2x > 0 \), so \( |x^2 - 2x| = x^2 - 2x \).

Thus, we can split the integral as follows:
\[ \int_{1}^{3} |x^2 - 2x| \, dx = \int_{1}^{2} (-(x^2 - 2x)) \, dx + \int_{2}^{3} (x^2 - 2x) \, dx \]

### Step 4: Evaluate the first integral
Evaluate \( \int_{1}^{2} (-x^2 + 2x) \, dx \):
\[ \int_{1}^{2} (-x^2 + 2x) \, dx = \left[-\frac{x^3}{3} + x^2\right]_{1}^{2} \]
Calculating the limits:
\[ = \left[-\frac{2^3}{3} + 2^2\right] - \left[-\frac{1^3}{3} + 1^2\right] \]
\[ = \left[-\frac{8}{3} + 4\right] - \left[-\frac{1}{3} + 1\right] \]
\[ = \left[-\frac{8}{3} + \frac{12}{3}\right] - \left[-\frac{1}{3} + \frac{3}{3}\right] \]
\[ = \left[\frac{4}{3}\right] - \left[\frac{2}{3}\right] = \frac{4}{3} - \frac{2}{3} = \frac{2}{3} \]

### Step 5: Evaluate the second integral
Evaluate \( \int_{2}^{3} (x^2 - 2x) \, dx \):
\[ \int_{2}^{3} (x^2 - 2x) \, dx = \left[\frac{x^3}{3} - x^2\right]_{2}^{3} \]
Calculating the limits:
\[ = \left[\frac{3^3}{3} - 3^2\right] - \left[\frac{2^3}{3} - 2^2\right] \]
\[ = \left[9 - 9\right] - \left[\frac{8}{3} - 4\right] \]
\[ = 0 - \left[\frac{8}{3} - \frac{12}{3}\right] = 0 - \left[-\frac{4}{3}\right] = \frac{4}{3} \]

### Step 6: Combine the results
Now, we combine the results of both integrals:
\[ \int_{1}^{3} |x^2 - 2x| \, dx = \frac{2}{3} + \frac{4}{3} = \frac{6}{3} = 2 \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{1}^{3} |x^2 - 2x| \, dx = 2 \]