The value of `int_(0)^(1) "max" (e^(x) , e^(1-x) ) dx` equals

Options

A. `2(e-1)`
B. `2(e- sqrte)`
C. `2 ( e + sqrte)`
D. `2( e+ sqrte)`

Answer

To solve the integral \( \int_{0}^{1} \max(e^x, e^{1-x}) \, dx \), we first need to determine where the two functions \( e^x \) and \( e^{1-x} \) intersect within the interval \([0, 1]\).

### Step 1: Find the intersection points
To find the intersection points, we set \( e^x = e^{1-x} \).

Taking the natural logarithm on both sides gives:
\[ x = 1 - x \]
Solving for \( x \):
\[ 2x = 1 \implies x = \frac{1}{2} \]

### Step 2: Determine which function is greater in each interval
Now we will evaluate the functions at the intersection point and at the endpoints of the interval:
- At \( x = 0 \):
\[ e^0 = 1 \quad \text{and} \quad e^{1-0} = e \quad \Rightarrow \quad \max(1, e) = e \]

- At \( x = 1 \):
\[ e^1 = e \quad \text{and} \quad e^{1-1} = 1 \quad \Rightarrow \quad \max(e, 1) = e \]

- At \( x = \frac{1}{2} \):
\[ e^{1/2} = \sqrt{e} \quad \text{and} \quad e^{1-1/2} = \sqrt{e} \quad \Rightarrow \quad \max(\sqrt{e}, \sqrt{e}) = \sqrt{e} \]

### Step 3: Set up the integral
From \( x = 0 \) to \( x = \frac{1}{2} \), \( e^{1-x} \) is greater than \( e^x \), and from \( x = \frac{1}{2} \) to \( x = 1 \), \( e^x \) is greater than \( e^{1-x} \). Therefore, we can express the integral as:
\[ \int_{0}^{1} \max(e^x, e^{1-x}) \, dx = \int_{0}^{\frac{1}{2}} e^{1-x} \, dx + \int_{\frac{1}{2}}^{1} e^x \, dx \]

### Step 4: Calculate the integrals
1. **First integral**:
\[ \int_{0}^{\frac{1}{2}} e^{1-x} \, dx = e \int_{0}^{\frac{1}{2}} e^{-x} \, dx = e \left[-e^{-x}\right]_{0}^{\frac{1}{2}} = e \left[-e^{-\frac{1}{2}} + 1\right] = e \left(1 - \frac{1}{\sqrt{e}}\right) = e - \sqrt{e} \]

2. **Second integral**:
\[ \int_{\frac{1}{2}}^{1} e^x \, dx = \left[e^x\right]_{\frac{1}{2}}^{1} = e - e^{\frac{1}{2}} = e - \sqrt{e} \]

### Step 5: Combine the results
Now we add the results of the two integrals:
\[ \int_{0}^{1} \max(e^x, e^{1-x}) \, dx = (e - \sqrt{e}) + (e - \sqrt{e}) = 2e - 2\sqrt{e} \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{0}^{1} \max(e^x, e^{1-x}) \, dx = 2(e - \sqrt{e}) \]