The area bounded by the curves `y=x e^x ,y=x e^(-x)` and the line `x=1` is `2/e s qdotu n i t s` (b) `1-2/e s qdotu n i t s` `1/e s qdotu n i t s` (d) `1-1/e s qdotu n i t s`

Options

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

Explanation

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

### Step-by-Step Solution

1. **Identify the curves and the area of interest**:
We have two curves: \( y = x e^x \) and \( y = x e^{-x} \). We need to find the area between these two curves from \( x = 0 \) to \( x = 1 \).

2. **Set up the integral for the area**:
The area \( A \) between the curves from \( x = 0 \) to \( x = 1 \) can be expressed as:
\[ A = \int_{0}^{1} (x e^x - x e^{-x}) \, dx \]

3. **Split the integral**:
We can separate the integral into two parts:
\[ A = \int_{0}^{1} x e^x \, dx - \int_{0}^{1} x e^{-x} \, dx \]

4. **Integrate \( \int x e^x \, dx \)**:
We will use integration by parts for \( \int x e^x \, dx \):
- Let \( u = x \) and \( dv = e^x \, dx \).
- Then, \( du = dx \) and \( v = e^x \).
- Applying integration by parts:
\[ \int x e^x \, dx = x e^x - \int e^x \, dx = x e^x - e^x + C \]
Evaluating from 0 to 1:
\[ \left[ x e^x - e^x \right]_{0}^{1} = \left[ 1 \cdot e^1 - e^1 \right] - \left[ 0 \cdot e^0 - e^0 \right] = (e - e) - (0 - 1) = 1 \]

5. **Integrate \( \int x e^{-x} \, dx \)**:
Again, we will use integration by parts:
- Let \( u = x \) and \( dv = e^{-x} \, dx \).
- Then, \( du = dx \) and \( v = -e^{-x} \).
- Applying integration by parts:
\[ \int x e^{-x} \, dx = -x e^{-x} - \int -e^{-x} \, dx = -x e^{-x} + e^{-x} + C \]
Evaluating from 0 to 1:
\[ \left[ -x e^{-x} + e^{-x} \right]_{0}^{1} = \left[ -1 \cdot e^{-1} + e^{-1} \right] - \left[ -0 \cdot e^{0} + e^{0} \right] = (-\frac{1}{e} + \frac{1}{e}) - (0 + 1) = -1 \]

6. **Combine the results**:
Now we can substitute back into our area formula:
\[ A = 1 - (-1) = 1 + 1 = 2 \]

7. **Final area calculation**:
Since we need to express the area in terms of \( e \):
\[ A = 2 \cdot \frac{1}{e} = \frac{2}{e} \]

### Conclusion
The area bounded by the curves \( y = x e^x \), \( y = x e^{-x} \), and the line \( x = 1 \) is:
\[ \boxed{\frac{2}{e}} \text{ square units} \]