Let `f : R->R` be given by `f(x) = {|x-[x]|`, when [x] is odd and `|x-[x]-1|` , when [x] is even ,where [.] denotes the greatest integer function , then `int_-2^4 f(x) dx` is equal to (a) `(5)/(2)` (b) `(3)/(2)` (c) `5` (d) `3`

Options

A. `(5)/(2)`
B. `(3)/(2)`
C. 5
D. 3

Answer

To solve the integral \( \int_{-2}^{4} f(x) \, dx \) where the function \( f(x) \) is defined piecewise based on the greatest integer function \( [x] \), we need to analyze the function over the intervals determined by the integer values from \(-2\) to \(4\).

### Step 1: Determine the intervals and the function definition

The greatest integer function \( [x] \) takes on integer values, and we will evaluate \( f(x) \) in the following intervals:
- From \(-2\) to \(-1\) (where \( [x] = -2 \), which is even)
- From \(-1\) to \(0\) (where \( [x] = -1 \), which is odd)
- From \(0\) to \(1\) (where \( [x] = 0 \), which is even)
- From \(1\) to \(2\) (where \( [x] = 1 \), which is odd)
- From \(2\) to \(3\) (where \( [x] = 2 \), which is even)
- From \(3\) to \(4\) (where \( [x] = 3 \), which is odd)

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

1. For \( -2 \leq x < -1 \) (where \( [x] = -2 \)):
\[ f(x) = |x - (-2)| = |x + 2| \]
This simplifies to \( f(x) = x + 2 \) (since \( x + 2 \) is non-negative in this interval).

2. For \( -1 \leq x < 0 \) (where \( [x] = -1 \)):
\[ f(x) = |x - (-1)| = |x + 1| \]
This simplifies to \( f(x) = x + 1 \).

3. For \( 0 \leq x < 1 \) (where \( [x] = 0 \)):
\[ f(x) = |x - 0| = |x| = x. \]

4. For \( 1 \leq x < 2 \) (where \( [x] = 1 \)):
\[ f(x) = |x - 1| = x - 1. \]

5. For \( 2 \leq x < 3 \) (where \( [x] = 2 \)):
\[ f(x) = |x - 2| = x - 2. \]

6. For \( 3 \leq x < 4 \) (where \( [x] = 3 \)):
\[ f(x) = |x - 3| = x - 3. \]

### Step 3: Calculate the integral over each interval

Now we compute the integral over each interval:

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

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

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

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

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

6. **Interval \([3, 4]\)**:
\[ \int_{3}^{4} (x - 3) \, dx = \left[ \frac{x^2}{2} - 3x \right]_{3}^{4} = \left( \frac{16}{2} - 12 \right) - \left( \frac{9}{2} - 9 \right) = (8 - 12) - \left( \frac{9}{2} - \frac{18}{2} \right) = -4 - (-\frac{9}{2}) = -4 + \frac{9}{2} = -\frac{8}{2} + \frac{9}{2} = \frac{1}{2} \]

### Step 4: Sum the integrals

Now we sum the results from all intervals:
\[ \int_{-2}^{4} f(x) \, dx = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} - \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = 3 \]

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