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

Options

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

Answer

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

### Step-by-Step Solution:

1. **Identify the curves and the line**: We have two curves: \( y = e^x \) and \( y = e^{-x} \), and a vertical line \( x = 2 \). The area we want to find is between these curves from \( x = 0 \) to \( x = 2 \).

2. **Find the points of intersection**: The curves \( y = e^x \) and \( y = e^{-x} \) intersect at \( x = 0 \) because:
\[ e^0 = e^{-0} = 1 \]
Thus, the intersection point is \( (0, 1) \).

3. **Set up the integral for the area**: The area \( A \) between the curves from \( x = 0 \) to \( x = 2 \) can be expressed as:
\[ A = \int_{0}^{2} (e^x - e^{-x}) \, dx \]
Here, \( e^x \) is the upper curve and \( e^{-x} \) is the lower curve in the interval from \( 0 \) to \( 2 \).

4. **Evaluate the integral**:
- The integral of \( e^x \) is \( e^x \).
- The integral of \( e^{-x} \) is \( -e^{-x} \).
Therefore, we have:
\[ A = \left[ e^x + e^{-x} \right]_{0}^{2} \]

5. **Calculate the definite integral**:
- Substitute the limits into the expression:
\[ A = \left( e^2 + e^{-2} \right) - \left( e^0 + e^{-0} \right) \]
- Simplifying this gives:
\[ A = \left( e^2 + \frac{1}{e^2} \right) - (1 + 1) = e^2 + \frac{1}{e^2} - 2 \]

6. **Final result**: The area bounded by the curves and the line is:
\[ A = e^2 + \frac{1}{e^2} - 2 \]