Let f(x)= minimum `{e^(x),3//2,1+e^(-x)},0lexle1`. Find the area bounded by `y=f(x)`, X-axis and the line x=1.

Answer

To find the area bounded by the curve \( y = f(x) \), the x-axis, and the line \( x = 1 \), we will follow these steps:

### Step 1: Define the function \( f(x) \)
We have:
\[ f(x) = \min\{e^x, \frac{3}{2}, 1 + e^{-x}\} \]
for \( 0 \leq x \leq 1 \).

### Step 2: Determine the intervals for \( f(x) \)
We need to find where each function intersects within the interval \( [0, 1] \).

1. **Find intersection points of \( e^x \) and \( \frac{3}{2} \)**:
\[ e^x = \frac{3}{2} \implies x = \ln\left(\frac{3}{2}\right) \]

2. **Find intersection points of \( e^x \) and \( 1 + e^{-x} \)**:
\[ e^x = 1 + e^{-x} \implies e^{2x} - e^x - 1 = 0 \]
Let \( y = e^x \):
\[ y^2 - y - 1 = 0 \implies y = \frac{1 \pm \sqrt{5}}{2} \]
Taking the positive root:
\[ e^x = \frac{1 + \sqrt{5}}{2} \implies x = \ln\left(\frac{1 + \sqrt{5}}{2}\right) \]

### Step 3: Evaluate the function in the intervals
Now we evaluate \( f(x) \) in the intervals determined by the intersection points:
- For \( 0 \leq x < \ln\left(\frac{3}{2}\right) \), \( f(x) = e^x \).
- For \( \ln\left(\frac{3}{2}\right) \leq x < \ln\left(\frac{1 + \sqrt{5}}{2}\right) \), \( f(x) = \frac{3}{2} \).
- For \( \ln\left(\frac{1 + \sqrt{5}}{2}\right) \leq x \leq 1 \), \( f(x) = 1 + e^{-x} \).

### Step 4: Calculate the area under the curve
The area \( A \) can be calculated by integrating \( f(x) \) over the specified intervals:

1. From \( 0 \) to \( \ln\left(\frac{3}{2}\right) \):
\[ A_1 = \int_0^{\ln\left(\frac{3}{2}\right)} e^x \, dx = [e^x]_0^{\ln\left(\frac{3}{2}\right)} = \frac{3}{2} - 1 = \frac{1}{2} \]

2. From \( \ln\left(\frac{3}{2}\right) \) to \( \ln\left(\frac{1 + \sqrt{5}}{2}\right) \):
\[ A_2 = \int_{\ln\left(\frac{3}{2}\right)}^{\ln\left(\frac{1 + \sqrt{5}}{2}\right)} \frac{3}{2} \, dx = \frac{3}{2} \left(\ln\left(\frac{1 + \sqrt{5}}{2}\right) - \ln\left(\frac{3}{2}\right)\right) \]

3. From \( \ln\left(\frac{1 + \sqrt{5}}{2}\right) \) to \( 1 \):
\[ A_3 = \int_{\ln\left(\frac{1 + \sqrt{5}}{2}\right)}^1 (1 + e^{-x}) \, dx = [x - e^{-x}]_{\ln\left(\frac{1 + \sqrt{5}}{2}\right)}^1 \]

### Step 5: Combine the areas
The total area \( A \) is:
\[ A = A_1 + A_2 + A_3 \]

### Step 6: Simplify the expression
After calculating each area, combine them to get the final area bounded by the curve, the x-axis, and the line \( x = 1 \).