Let `f : R to R` is a function defined as `f(x)` where `= {(|x-[x]| ,:[x] "is odd"),(|x - [x + 1]| ,:[x] "is even"):}`
[.] denotes the greatest integer function, then `int_(-2)^(4) dx` is equal to

Answer

To solve the problem, we need to analyze the function \( f(x) \) defined as follows:

\[ f(x) = \begin{cases} |x - [x]| & \text{if } [x] \text{ is odd} \\ |x - [x + 1]| & \text{if } [x] \text{ is even} \end{cases} \]

Where \([x]\) denotes the greatest integer function (also known as the floor function). We need to compute the integral:

\[ \int_{-2}^{4} f(x) \, dx \]

### Step 1: Understand the Function \( f(x) \)

1. **When \([x]\) is odd**: The function gives the fractional part of \( x \), which is \( |x - [x]| \).
2. **When \([x]\) is even**: The function gives \( |x - [x + 1]| \), which is equivalent to \( 1 - |x - [x]| \) since \([x + 1] = [x] + 1\).

### Step 2: Determine the Intervals

Next, we determine the intervals for \( x \) from \(-2\) to \(4\) and find the values of \([x]\):

- For \( -2 \leq x < -1 \): \([x] = -2\) (even)
- For \( -1 \leq x < 0 \): \([x] = -1\) (odd)
- For \( 0 \leq x < 1 \): \([x] = 0\) (even)
- For \( 1 \leq x < 2 \): \([x] = 1\) (odd)
- For \( 2 \leq x < 3 \): \([x] = 2\) (even)
- For \( 3 \leq x < 4 \): \([x] = 3\) (odd)

### Step 3: Evaluate \( f(x) \) on Each Interval

Now we compute \( f(x) \) on each interval:

1. **Interval \([-2, -1)\)**:
- \([x] = -2\) (even), so \( f(x) = 1 - |x - (-2)| = 1 - (x + 2) = -x - 1 \)

2. **Interval \([-1, 0)\)**:
- \([x] = -1\) (odd), so \( f(x) = |x - (-1)| = x + 1 \)

3. **Interval \([0, 1)\)**:
- \([x] = 0\) (even), so \( f(x) = 1 - |x - 0| = 1 - x \)

4. **Interval \([1, 2)\)**:
- \([x] = 1\) (odd), so \( f(x) = |x - 1| = x - 1 \)

5. **Interval \([2, 3)\)**:
- \([x] = 2\) (even), so \( f(x) = 1 - |x - 2| = 1 - (x - 2) = 3 - x \)

6. **Interval \([3, 4)\)**:
- \([x] = 3\) (odd), so \( f(x) = |x - 3| = x - 3 \)

### Step 4: Set Up the Integral

Now we can set up the integral by breaking it into parts:

\[ \int_{-2}^{4} f(x) \, dx = \int_{-2}^{-1} (-x - 1) \, dx + \int_{-1}^{0} (x + 1) \, dx + \int_{0}^{1} (1 - x) \, dx + \int_{1}^{2} (x - 1) \, dx + \int_{2}^{3} (3 - x) \, dx + \int_{3}^{4} (x - 3) \, dx \]

### Step 5: Calculate Each Integral

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

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

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

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

5. **For \([2, 3)\)**:
\[ \int_{2}^{3} (3 - x) \, dx = \left[3x - \frac{x^2}{2}\right]_{2}^{3} = \left[9 - \frac{9}{2}\right] - \left[6 - 2\right] = \left[\frac{9}{2} - 6\right] = \frac{9}{2} - \frac{12}{2} = -\frac{3}{2} \]

6. **For \([3, 4)\)**:
\[ \int_{3}^{4} (x - 3) \, dx = \left[\frac{x^2}{2} - 3x\right]_{3}^{4} = \left[8 - 12\right] - \left[\frac{9}{2} - 9\right] = -4 + \frac{9}{2} - 9 = -4 + \frac{9}{2} + \frac{18}{2} = \frac{9}{2} - 4 = \frac{1}{2} \]

### Step 6: Sum All the Areas

Now, we sum all the areas:

\[ \int_{-2}^{4} f(x) \, dx = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} - \frac{3}{2} + \frac{1}{2} = 3 \]

### Final Answer

Thus, the value of the integral is:

\[ \int_{-2}^{4} f(x) \, dx = 3 \]