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

Options

A. `int_(2)^(4)f(x)dx=1`
B. `int_(2)^(4)f(x)dx=3`
C. `int_(-10)^(10)f(x)dx=10`
D. `int_(-1)^(1)f(x)dx=0`

Explanation

To solve the problem, we need to evaluate the integral of the function \( f(x) \) defined as:

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

where \([x]\) denotes the greatest integer function (also known as the floor function).

### Step 1: Determine the behavior of \( f(x) \) from \(-2\) to \(4\)

First, identify the integer values in the interval \([-2, 4]\):
- The integers are \(-2, -1, 0, 1, 2, 3, 4\).

Now, we will evaluate \( f(x) \) in each sub-interval defined by these integers.

### Step 2: Evaluate \( f(x) \) in each interval

1. **For \( x \in [-2, -1) \)**:
- Here, \([x] = -2\) (even).
- Thus, \( f(x) = |x - [-x + 1]| = |x - (-1)| = |x + 1| \).

2. **For \( x \in [-1, 0) \)**:
- Here, \([x] = -1\) (odd).
- Thus, \( f(x) = |x - [-x]| = |x + 1| \).

3. **For \( x \in [0, 1) \)**:
- Here, \([x] = 0\) (even).
- Thus, \( f(x) = |x - [x + 1]| = |x - 1| \).

4. **For \( x \in [1, 2) \)**:
- Here, \([x] = 1\) (odd).
- Thus, \( f(x) = |x - [x]| = |x - 1| \).

5. **For \( x \in [2, 3) \)**:
- Here, \([x] = 2\) (even).
- Thus, \( f(x) = |x - [x + 1]| = |x - 3| \).

6. **For \( x \in [3, 4) \)**:
- Here, \([x] = 3\) (odd).
- Thus, \( f(x) = |x - [x]| = |x - 3| \).

### Step 3: Integrate \( f(x) \) over each interval

Now we will calculate the integral over each interval:

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

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

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

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

5. **Integral from \(2\) to \(3\)**:
\[ \int_{2}^{3} (3 - x) \, dx = \left[ 3x - \frac{x^2}{2} \right]_{2}^{3} = \left( 9 - \frac{9}{2} \right) - \left( 6 - 2 \right) = \frac{9}{2} - 4 = \frac{1}{2} \]

6. **Integral from \(3\) to \(4\)**:
\[ \int_{3}^{4} (x - 3) \, dx = \left[ \frac{x^2}{2} - 3x \right]_{3}^{4} = \left( 8 - 12 \right) - \left( \frac{9}{2} - 9 \right) = -4 + \frac{9}{2} = -4 + 4.5 = 0.5 \]

### Step 4: Sum the integrals

Now, we sum the integrals from all intervals:
\[ \text{Total} = -\frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = 3 \]

### Final Answer

Thus, the value of the integral \( \int_{-2}^{4} f(x) \, dx \) is equal to \( 3 \).

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