The area bounded by the curves `y=e^(x),y=e^(-x)` and `y=2`, is

Options

A. `log (16/e)`
B. log(4/e)
C. 2log(4/e)
D. log(8/e)

Answer

To find the area bounded by the curves \(y = e^x\), \(y = e^{-x}\), and \(y = 2\), we will follow these steps:

### Step 1: Find Points of Intersection
First, we need to find the points where the curves intersect.

1. **Intersection of \(y = e^x\) and \(y = 2\)**:
\[ e^x = 2 \implies x = \ln(2) \]

2. **Intersection of \(y = e^{-x}\) and \(y = 2\)**:
\[ e^{-x} = 2 \implies e^x = \frac{1}{2} \implies x = -\ln(2) \]

3. **Intersection of \(y = e^x\) and \(y = e^{-x}\)**:
\[ e^x = e^{-x} \implies e^{2x} = 1 \implies x = 0 \]

The points of intersection are \((\ln(2), 2)\), \((- \ln(2), 2)\), and \((0, 1)\).

### Step 2: Set Up the Integral
The area we need to calculate is bounded between the curves from \(x = -\ln(2)\) to \(x = \ln(2)\). The area \(A\) can be expressed as:
\[ A = \int_{-\ln(2)}^{\ln(2)} (e^{-x} - e^x) \, dx \]

### Step 3: Calculate the Integral
Now, we will compute the integral:
\[ A = \int_{-\ln(2)}^{\ln(2)} (e^{-x} - e^x) \, dx \]

1. **Integrate \(e^{-x}\)**:
\[ \int e^{-x} \, dx = -e^{-x} \]

2. **Integrate \(e^x\)**:
\[ \int e^x \, dx = e^x \]

Thus, we have:
\[ A = \left[-e^{-x} - e^x\right]_{-\ln(2)}^{\ln(2)} \]

### Step 4: Evaluate the Integral
Now we evaluate the definite integral:
\[ A = \left[-e^{-\ln(2)} - e^{\ln(2)}\right] - \left[-e^{\ln(2)} - e^{-\ln(2)}\right] \]

Calculating each term:
1. \(e^{-\ln(2)} = \frac{1}{2}\)
2. \(e^{\ln(2)} = 2\)

Substituting these values in:
\[ A = \left[-\frac{1}{2} - 2\right] - \left[-2 - \frac{1}{2}\right] \]
\[ = \left[-\frac{1}{2} - 2\right] + \left[2 + \frac{1}{2}\right] \]
\[ = -\frac{1}{2} - 2 + 2 + \frac{1}{2} = 0 \]

### Step 5: Final Calculation
The area simplifies to:
\[ A = 2 - \frac{1}{2} - 2 + \frac{1}{2} = 2 - 2 = 0 \]

However, we need to calculate the absolute area between the curves:
\[ A = 2 - 1 = \frac{3}{2} \]

### Conclusion
The area bounded by the curves \(y = e^x\), \(y = e^{-x}\), and \(y = 2\) is:
\[ \boxed{\frac{3}{2}} \]