Evaluate
`int_(-2)^(4)x[x]dx`
where [.] denotes the greatest integer function.

Answer

To evaluate the integral \( \int_{-2}^{4} x \lfloor x \rfloor \, dx \), where \( \lfloor x \rfloor \) denotes the greatest integer function, we first need to break the integral into segments where the greatest integer function is constant.

### Step 1: Identify the intervals
The greatest integer function \( \lfloor x \rfloor \) changes its value at integer points. Therefore, we will break the integral into the following intervals:
- From \(-2\) to \(-1\)
- From \(-1\) to \(0\)
- From \(0\) to \(1\)
- From \(1\) to \(2\)
- From \(2\) to \(3\)
- From \(3\) to \(4\)

### Step 2: Write the integral as a sum of integrals
We can express the integral as:
\[ \int_{-2}^{4} x \lfloor x \rfloor \, dx = \int_{-2}^{-1} x(-2) \, dx + \int_{-1}^{0} x(-1) \, dx + \int_{0}^{1} x(0) \, dx + \int_{1}^{2} x(1) \, dx + \int_{2}^{3} x(2) \, dx + \int_{3}^{4} x(3) \, dx \]

### Step 3: Evaluate each integral
1. **For \( \int_{-2}^{-1} x(-2) \, dx \)**:
\[ = -2 \int_{-2}^{-1} x \, dx = -2 \left[ \frac{x^2}{2} \right]_{-2}^{-1} = -2 \left( \frac{(-1)^2}{2} - \frac{(-2)^2}{2} \right) = -2 \left( \frac{1}{2} - 2 \right) = -2 \left( \frac{1}{2} - \frac{4}{2} \right) = -2 \left( -\frac{3}{2} \right) = 3 \]

2. **For \( \int_{-1}^{0} x(-1) \, dx \)**:
\[ = -\int_{-1}^{0} x \, dx = -\left[ \frac{x^2}{2} \right]_{-1}^{0} = -\left( 0 - \frac{(-1)^2}{2} \right) = -\left( -\frac{1}{2} \right) = \frac{1}{2} \]

3. **For \( \int_{0}^{1} x(0) \, dx \)**:
\[ = 0 \]

4. **For \( \int_{1}^{2} x(1) \, dx \)**:
\[ = \int_{1}^{2} x \, dx = \left[ \frac{x^2}{2} \right]_{1}^{2} = \left( \frac{2^2}{2} - \frac{1^2}{2} \right) = \left( \frac{4}{2} - \frac{1}{2} \right) = \frac{3}{2} \]

5. **For \( \int_{2}^{3} x(2) \, dx \)**:
\[ = 2 \int_{2}^{3} x \, dx = 2 \left[ \frac{x^2}{2} \right]_{2}^{3} = \left( 3^2 - 2^2 \right) = 9 - 4 = 5 \]

6. **For \( \int_{3}^{4} x(3) \, dx \)**:
\[ = 3 \int_{3}^{4} x \, dx = 3 \left[ \frac{x^2}{2} \right]_{3}^{4} = 3 \left( \frac{4^2}{2} - \frac{3^2}{2} \right) = 3 \left( \frac{16}{2} - \frac{9}{2} \right) = 3 \left( \frac{7}{2} \right) = \frac{21}{2} \]

### Step 4: Combine all the results
Now we can sum all the evaluated integrals:
\[ 3 + \frac{1}{2} + 0 + \frac{3}{2} + 5 + \frac{21}{2} \]
Converting everything to a common denominator (2):
\[ = \frac{6}{2} + \frac{1}{2} + 0 + \frac{3}{2} + \frac{10}{2} + \frac{21}{2} = \frac{6 + 1 + 0 + 3 + 10 + 21}{2} = \frac{41}{2} \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{-2}^{4} x \lfloor x \rfloor \, dx = \frac{41}{2} \]