The value of `int_(-6)^(6) "max"(|2-|x||,4-|x|,3)dx` ,is

Options

A. 40
B. 50
C. 1
D. `(2)/(5)`

Answer

To solve the integral \( \int_{-6}^{6} \max(|2 - |x||, 4 - |x|, 3) \, dx \), we will first analyze the functions involved in the maximum and determine the intervals where each function is the maximum.

### Step 1: Identify the Functions
We have three functions:
1. \( f_1(x) = |2 - |x|| \)
2. \( f_2(x) = 4 - |x| \)
3. \( f_3(x) = 3 \)

### Step 2: Analyze Each Function
1. **For \( f_1(x) = |2 - |x|| \)**:
- This function changes at \( |x| = 2 \), giving us points \( x = -2 \) and \( x = 2 \).
- For \( |x| < 2 \), \( f_1(x) = 2 - |x| \).
- For \( |x| \geq 2 \), \( f_1(x) = |x| - 2 \).

2. **For \( f_2(x) = 4 - |x| \)**:
- This function changes at \( |x| = 4 \), giving us points \( x = -4 \) and \( x = 4 \).
- For \( |x| < 4 \), \( f_2(x) = 4 - |x| \).
- For \( |x| \geq 4 \), \( f_2(x) \) becomes negative.

3. **For \( f_3(x) = 3 \)**:
- This is a constant function.

### Step 3: Determine Intervals
We will analyze the maximum function in the intervals determined by the critical points \( -6, -4, -2, 2, 4, 6 \).

1. **Interval \( [-6, -4] \)**:
- Here, \( |x| \geq 4 \), thus \( f_1(x) = |x| - 2 \), \( f_2(x) < 0 \), and \( f_3(x) = 3 \).
- Maximum: \( \max(f_1(x), f_2(x), f_3(x)) = f_1(x) \).

2. **Interval \( [-4, -2] \)**:
- Here, \( |x| < 4 \) and \( |x| \geq 2 \), thus \( f_1(x) = |x| - 2 \), \( f_2(x) = 4 - |x| \), and \( f_3(x) = 3 \).
- Maximum: Compare \( f_1 \) and \( f_2 \):
- At \( x = -4 \): \( f_1(-4) = 2 \), \( f_2(-4) = 0 \), \( f_3 = 3 \) → Maximum is \( 3 \).
- At \( x = -2 \): \( f_1(-2) = 0 \), \( f_2(-2) = 2 \), \( f_3 = 3 \) → Maximum is \( 3 \).
- Maximum: \( f_3(x) = 3 \).

3. **Interval \( [-2, 2] \)**:
- Here, \( |x| < 2 \), thus \( f_1(x) = 2 - |x| \), \( f_2(x) = 4 - |x| \), and \( f_3(x) = 3 \).
- Compare \( f_1 \) and \( f_2 \):
- At \( x = 0 \): \( f_1(0) = 2 \), \( f_2(0) = 4 \), \( f_3 = 3 \) → Maximum is \( 4 \).
- Maximum: \( f_2(x) = 4 \).

4. **Interval \( [2, 4] \)**:
- Here, \( |x| < 4 \) and \( |x| \geq 2 \), thus \( f_1(x) = 2 - |x| \), \( f_2(x) = 4 - |x| \), and \( f_3(x) = 3 \).
- Compare \( f_1 \) and \( f_2 \):
- At \( x = 2 \): \( f_1(2) = 0 \), \( f_2(2) = 2 \), \( f_3 = 3 \) → Maximum is \( 3 \).
- At \( x = 4 \): \( f_1(4) = 2 \), \( f_2(4) = 0 \), \( f_3 = 3 \) → Maximum is \( 3 \).
- Maximum: \( f_3(x) = 3 \).

5. **Interval \( [4, 6] \)**:
- Here, \( |x| \geq 4 \), thus \( f_1(x) = |x| - 2 \), \( f_2(x) < 0 \), and \( f_3(x) = 3 \).
- Maximum: \( \max(f_1(x), f_2(x), f_3(x)) = f_1(x) \).

### Step 4: Set Up the Integral
Now we can set up the integral based on the maximum function in each interval:

\[ \int_{-6}^{-4} (|x| - 2) \, dx + \int_{-4}^{-2} 3 \, dx + \int_{-2}^{2} (4 - |x|) \, dx + \int_{2}^{4} 3 \, dx + \int_{4}^{6} (|x| - 2) \, dx \]

### Step 5: Calculate Each Integral
1. **Integral from \(-6\) to \(-4\)**:
\[ \int_{-6}^{-4} (|x| - 2) \, dx = \int_{-6}^{-4} (6 - 2) \, dx = \int_{-6}^{-4} 4 \, dx = 4 \times (2) = 8 \]

2. **Integral from \(-4\) to \(-2\)**:
\[ \int_{-4}^{-2} 3 \, dx = 3 \times (2) = 6 \]

3. **Integral from \(-2\) to \(2\)**:
\[ \int_{-2}^{2} (4 - |x|) \, dx = \int_{-2}^{0} (4 + x) \, dx + \int_{0}^{2} (4 - x) \, dx \]
- First part:
\[ \int_{-2}^{0} (4 + x) \, dx = \left[ 4x + \frac{x^2}{2} \right]_{-2}^{0} = (0) - ( -8 + 2) = 6 \]
- Second part:
\[ \int_{0}^{2} (4 - x) \, dx = \left[ 4x - \frac{x^2}{2} \right]_{0}^{2} = (8 - 2) - 0 = 6 \]
- Total:
\[ 6 + 6 = 12 \]

4. **Integral from \(2\) to \(4\)**:
\[ \int_{2}^{4} 3 \, dx = 3 \times (2) = 6 \]

5. **Integral from \(4\) to \(6\)**:
\[ \int_{4}^{6} (|x| - 2) \, dx = \int_{4}^{6} (x - 2) \, dx = \left[ \frac{x^2}{2} - 2x \right]_{4}^{6} = \left( 18 - 12 \right) - \left( 8 - 8 \right) = 6 \]

### Step 6: Combine All Integrals
Now, we can combine all the results:
\[ 8 + 6 + 12 + 6 + 6 = 38 \]

### Final Answer
The value of the integral is \( \boxed{38} \).