The value of `int_(-1)^(1) max[2-x,2,1+x]` dx is

Options

A. 4
B. `9//2`
C. 2
D. none of these

Answer

To solve the integral \( \int_{-1}^{1} \max[2-x, 2, 1+x] \, dx \), we will break it down step by step.

### Step 1: Identify the intervals
We need to evaluate the maximum of the three functions \( 2-x \), \( 2 \), and \( 1+x \) over the interval from \(-1\) to \(1\). We can break this interval into two parts: from \(-1\) to \(0\) and from \(0\) to \(1\).

### Step 2: Analyze the interval \([-1, 0]\)
- For \( x = -1 \):
- \( 2 - (-1) = 3 \)
- \( 2 = 2 \)
- \( 1 + (-1) = 0 \)

The maximum value is \( 3 \) (from \( 2 - x \)).

- For \( x = 0 \):
- \( 2 - 0 = 2 \)
- \( 2 = 2 \)
- \( 1 + 0 = 1 \)

The maximum value is \( 2 \) (from \( 2 \)).

Since \( 2-x \) is a decreasing function and \( 1+x \) is an increasing function, we can conclude that in the interval \([-1, 0]\), the maximum function is \( 2 - x \).

### Step 3: Analyze the interval \([0, 1]\)
- For \( x = 0 \):
- \( 2 - 0 = 2 \)
- \( 2 = 2 \)
- \( 1 + 0 = 1 \)

The maximum value is \( 2 \) (from \( 2 \)).

- For \( x = 1 \):
- \( 2 - 1 = 1 \)
- \( 2 = 2 \)
- \( 1 + 1 = 2 \)

The maximum value is \( 2 \) (from \( 2 \)).

In this interval, the maximum function is also \( 2 \).

### Step 4: Set up the integral
Now we can set up the integral as follows:
\[ \int_{-1}^{1} \max[2-x, 2, 1+x] \, dx = \int_{-1}^{0} (2-x) \, dx + \int_{0}^{1} 2 \, dx \]

### Step 5: Calculate the integrals
1. Calculate \( \int_{-1}^{0} (2-x) \, dx \):
\[ \int (2-x) \, dx = 2x - \frac{x^2}{2} \]
Evaluating from \(-1\) to \(0\):
\[ \left[ 2(0) - \frac{0^2}{2} \right] - \left[ 2(-1) - \frac{(-1)^2}{2} \right] = 0 - \left[ -2 - \frac{1}{2} \right] = 0 + 2 + \frac{1}{2} = \frac{5}{2} \]

2. Calculate \( \int_{0}^{1} 2 \, dx \):
\[ \int 2 \, dx = 2x \]
Evaluating from \(0\) to \(1\):
\[ 2(1) - 2(0) = 2 \]

### Step 6: Combine the results
Now we combine the results of the two integrals:
\[ \int_{-1}^{1} \max[2-x, 2, 1+x] \, dx = \frac{5}{2} + 2 = \frac{5}{2} + \frac{4}{2} = \frac{9}{2} \]

### Final Answer
Thus, the value of the integral is:
\[ \frac{9}{2} \]