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 \( x \leq 0 \) along with the x-axis from \( x = -1 \) to \( x = 1 \), we can follow these steps:

### Step 1: Identify the curves and the area of interest
We have two curves:
- For \( x \geq 0 \): \( y = e^x \)
- For \( x \leq 0 \): \( y = e^{-x} \)

We need to find the area between these curves and the x-axis from \( x = -1 \) to \( x = 1 \).

### Step 2: Calculate the area for \( x \) from 0 to 1
The area \( A_2 \) under the curve \( y = e^x \) from \( x = 0 \) to \( x = 1 \) can be calculated using the definite integral:
\[ A_2 = \int_0^1 e^x \, dx \]

### Step 3: Evaluate the integral
The integral of \( e^x \) is \( e^x \). Therefore,
\[ A_2 = \left[ e^x \right]_0^1 = e^1 - e^0 = e - 1 \]

### Step 4: Calculate the area for \( x \) from -1 to 0
Similarly, the area \( A_1 \) under the curve \( y = e^{-x} \) from \( x = -1 \) to \( x = 0 \) can be calculated using the definite integral:
\[ A_1 = \int_{-1}^0 e^{-x} \, dx \]

### Step 5: Evaluate the integral for \( A_1 \)
The integral of \( e^{-x} \) is \( -e^{-x} \). Therefore,
\[ A_1 = \left[ -e^{-x} \right]_{-1}^0 = -e^0 - (-e^1) = -1 + e = e - 1 \]

### Step 6: Total area calculation
Since \( A_1 = A_2 \), the total area \( A \) from \( x = -1 \) to \( x = 1 \) is:
\[ A = A_1 + A_2 = (e - 1) + (e - 1) = 2(e - 1) \]

### Step 7: Final expression
Thus, the total area bounded by the curves and the x-axis from \( x = -1 \) to \( x = 1 \) is:
\[ A = 2(e - 1) \]

### Conclusion
The area (in square units) bounded by the curves and the x-axis from \( x = -1 \) to \( x = 1 \) is \( 2(e - 1) \).