What is the area bounded by the curves `y=e^(x),y=e^(-x)` and the straight line x=1?

Options

A. `(e+1/e)` sq. units
B. `(e-1/e)` sq. unit
C. `(e+1/e-2)` sq. units
D. `(e-1/e-2)`sq units

Answer

To find the area bounded by the curves \( y = e^x \), \( y = e^{-x} \), and the line \( x = 1 \), we can follow these steps:

### Step 1: Identify the curves and the area of interest
We have two curves:
- \( y = e^x \) (exponential growth curve)
- \( y = e^{-x} \) (exponential decay curve)

The area we are interested in is bounded by these curves and the vertical line \( x = 1 \).

### Step 2: Find the points of intersection
To find the area between the curves, we first need to determine where they intersect. Setting \( e^x = e^{-x} \):

\[ e^x = e^{-x} \implies e^{2x} = 1 \implies 2x = 0 \implies x = 0 \]

Thus, the curves intersect at the point \( (0, 1) \).

### Step 3: Set up the integral for the area
The area \( A \) between the curves from \( x = 0 \) to \( x = 1 \) can be calculated using the integral:

\[ A = \int_{0}^{1} (e^x - e^{-x}) \, dx \]

### Step 4: Calculate the integral
Now we will compute the integral:

\[ A = \int_{0}^{1} (e^x - e^{-x}) \, dx \]

This can be split into two separate integrals:

\[ A = \int_{0}^{1} e^x \, dx - \int_{0}^{1} e^{-x} \, dx \]

Calculating each integral:

1. For \( \int e^x \, dx \):

\[ \int e^x \, dx = e^x \bigg|_{0}^{1} = e^1 - e^0 = e - 1 \]

2. For \( \int e^{-x} \, dx \):

\[ \int e^{-x} \, dx = -e^{-x} \bigg|_{0}^{1} = -e^{-1} - (-e^{0}) = -\frac{1}{e} + 1 = 1 - \frac{1}{e} \]

### Step 5: Combine the results
Now substitute back into the area formula:

\[ A = (e - 1) - \left(1 - \frac{1}{e}\right) \]

Simplifying this gives:

\[ A = e - 1 - 1 + \frac{1}{e} = e - 2 + \frac{1}{e} \]

### Final Result
Thus, the area bounded by the curves \( y = e^x \), \( y = e^{-x} \), and the line \( x = 1 \) is:

\[ A = e - 2 + \frac{1}{e} \]