The area (in sq. units) bounded by `y={{:(e^(x),":",xge0),(e^(-x),":",xle0):}` with the axis from `x=-1" to "x=1` is

Options

A. e
B. 2e
C. `2e-2`
D. `2e+2`

Answer

To find the area bounded by the curves \( y = e^x \) for \( x \geq 0 \) and \( y = e^{-x} \) for \( x \leq 0 \) with the x-axis from \( x = -1 \) to \( x = 1 \), we can follow these steps:

### Step 1: Identify the area to be calculated
The area can be split into two parts:
1. From \( x = -1 \) to \( x = 0 \) under the curve \( y = e^{-x} \).
2. From \( x = 0 \) to \( x = 1 \) under the curve \( y = e^x \).

### Step 2: Set up the integral for the area
The total area \( A \) can be expressed as:
\[ A = \int_{-1}^{0} e^{-x} \, dx + \int_{0}^{1} e^{x} \, dx \]

### Step 3: Calculate the first integral
Calculate \( \int_{-1}^{0} e^{-x} \, dx \):
\[ \int e^{-x} \, dx = -e^{-x} + C \]
Now, evaluate the definite integral:
\[ \int_{-1}^{0} e^{-x} \, dx = \left[-e^{-x}\right]_{-1}^{0} = \left[-e^{0}\right] - \left[-e^{1}\right] = -1 + e = e - 1 \]

### Step 4: Calculate the second integral
Calculate \( \int_{0}^{1} e^{x} \, dx \):
\[ \int e^{x} \, dx = e^{x} + C \]
Now, evaluate the definite integral:
\[ \int_{0}^{1} e^{x} \, dx = \left[e^{x}\right]_{0}^{1} = e^{1} - e^{0} = e - 1 \]

### Step 5: Combine the areas
Now, combine the results from both integrals:
\[ A = (e - 1) + (e - 1) = 2e - 2 \]

### Final Answer
Thus, the area bounded by the curves and the x-axis from \( x = -1 \) to \( x = 1 \) is:
\[ \boxed{2e - 2} \text{ square units} \]