The value of integrals `int_(-2)^(2)max {x+|x|, x-[x]}dx` where [.] represents the greatest integer function is

Options

A. 4
B. 5
C. `(7)/(2)`
D. `(9)/(4)`

Answer

To solve the integral \( \int_{-2}^{2} \max \{ x + |x|, x - [x] \} \, dx \), where \([x]\) is the greatest integer function, we will break down the problem step by step.

### Step 1: Analyze the Functions
1. **Evaluate \(x + |x|\)**:
- For \(x < 0\): \( |x| = -x \) so \( x + |x| = x - x = 0 \).
- For \(x \geq 0\): \( |x| = x \) so \( x + |x| = x + x = 2x \).

Thus, we have:
\[ x + |x| = \begin{cases} 0 & \text{if } x < 0 \\ 2x & \text{if } x \geq 0 \end{cases} \]

2. **Evaluate \(x - [x]\)**:
- The greatest integer function \([x]\) gives the largest integer less than or equal to \(x\).
- For intervals:
- From \(-2\) to \(-1\): \([x] = -2\) so \(x - [x] = x + 2\).
- From \(-1\) to \(0\): \([x] = -1\) so \(x - [x] = x + 1\).
- From \(0\) to \(1\): \([x] = 0\) so \(x - [x] = x\).
- From \(1\) to \(2\): \([x] = 1\) so \(x - [x] = x - 1\).

### Step 2: Determine the Maximum Function
Now we can find the maximum of the two functions in each interval:
- **Interval \([-2, -1]\)**:
- \(x + |x| = 0\)
- \(x - [x] = x + 2\)
- Maximum: \( \max(0, x + 2) = x + 2 \) (since \(x + 2 > 0\) in this interval).

- **Interval \([-1, 0]\)**:
- \(x + |x| = 0\)
- \(x - [x] = x + 1\)
- Maximum: \( \max(0, x + 1) = x + 1 \) (since \(x + 1 > 0\) in this interval).

- **Interval \([0, 1]\)**:
- \(x + |x| = 2x\)
- \(x - [x] = x\)
- Maximum: \( \max(2x, x) = 2x \) (since \(2x > x\) for \(x > 0\)).

- **Interval \([1, 2]\)**:
- \(x + |x| = 2x\)
- \(x - [x] = x - 1\)
- Maximum: \( \max(2x, x - 1) = 2x \) (since \(2x > x - 1\) for \(x > 1\)).

### Step 3: Set Up the Integral
Now we can set up the integral:
\[ \int_{-2}^{2} \max \{ x + |x|, x - [x] \} \, dx = \int_{-2}^{-1} (x + 2) \, dx + \int_{-1}^{0} (x + 1) \, dx + \int_{0}^{1} (2x) \, dx + \int_{1}^{2} (2x) \, dx \]

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

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

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

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

### Step 5: Sum All Integrals
Now, we sum all the results:
\[ \frac{1}{2} + \frac{1}{2} + 1 + 3 = 5 \]

### Final Answer
Thus, the value of the integral is:
\[ \boxed{5} \]