If `f(x)=x(x-1), 0 le x le 1, and f(x+1)=f(x) AA x in R`, then `|int_(2)^(4)f(x)dx|` is _____________

Answer

To solve the problem, we need to evaluate the integral \( | \int_{2}^{4} f(x) \, dx | \) where \( f(x) = x(x - 1) \) for \( 0 \leq x \leq 1 \) and \( f(x + 1) = f(x) \) for \( x \in \mathbb{R} \). This indicates that \( f(x) \) is a periodic function with a period of 1.

### Step-by-Step Solution:

1. **Identify the Periodicity of \( f(x) \)**:
Since \( f(x + 1) = f(x) \), we know that \( f(x) \) is periodic with a period of 1. Therefore, we can evaluate the integral from 2 to 4 by breaking it down into intervals of length 1.

2. **Break Down the Integral**:
We can express the integral as:
\[ \int_{2}^{4} f(x) \, dx = \int_{2}^{3} f(x) \, dx + \int_{3}^{4} f(x) \, dx \]
Because of the periodicity, we have:
\[ f(x) = f(x - 2) \text{ for } x \in [2, 3] \text{ and } f(x) = f(x - 3) \text{ for } x \in [3, 4] \]
Thus, we can rewrite:
\[ \int_{2}^{3} f(x) \, dx = \int_{0}^{1} f(x) \, dx \]
\[ \int_{3}^{4} f(x) \, dx = \int_{0}^{1} f(x) \, dx \]

3. **Combine the Integrals**:
Therefore, we can combine the integrals:
\[ \int_{2}^{4} f(x) \, dx = \int_{0}^{1} f(x) \, dx + \int_{0}^{1} f(x) \, dx = 2 \int_{0}^{1} f(x) \, dx \]

4. **Calculate \( \int_{0}^{1} f(x) \, dx \)**:
Now we need to calculate \( \int_{0}^{1} f(x) \, dx \):
\[ f(x) = x(x - 1) = x^2 - x \]
Thus,
\[ \int_{0}^{1} f(x) \, dx = \int_{0}^{1} (x^2 - x) \, dx \]
We can compute this integral:
\[ = \left[ \frac{x^3}{3} - \frac{x^2}{2} \right]_{0}^{1} = \left( \frac{1^3}{3} - \frac{1^2}{2} \right) - \left( 0 - 0 \right) \]
\[ = \frac{1}{3} - \frac{1}{2} = \frac{2}{6} - \frac{3}{6} = -\frac{1}{6} \]

5. **Final Calculation**:
Now substituting back, we have:
\[ \int_{2}^{4} f(x) \, dx = 2 \left(-\frac{1}{6}\right) = -\frac{2}{6} = -\frac{1}{3} \]
Therefore,
\[ | \int_{2}^{4} f(x) \, dx | = \left| -\frac{1}{3} \right| = \frac{1}{3} \]

### Final Answer:
\[ | \int_{2}^{4} f(x) \, dx | = \frac{1}{3} \]