The area bounded by the curve `y=xe^(-x);xy=0`and `x=c` where c is the x-coordinate of the curve's inflection point, is

Options

A. `1-3e^(-2)`
B. `1-2e^(-2)`
C. `1-e^(-2)`
D. none of these

Explanation

To find the area bounded by the curve \( y = x e^{-x} \), the lines \( xy = 0 \) (which corresponds to the x-axis and y-axis), and the line \( x = c \) where \( c \) is the x-coordinate of the curve's inflection point, we will follow these steps:

### Step 1: Find the inflection point of the curve
To find the inflection point, we need to compute the second derivative of the function \( y = x e^{-x} \).

1. **First Derivative**:
\[ y' = e^{-x} - x e^{-x} = e^{-x}(1 - x) \]

2. **Second Derivative**:
\[ y'' = \frac{d}{dx}(e^{-x}(1 - x)) = -e^{-x}(1 - x) + e^{-x}(-1) = e^{-x}(x - 2) \]

3. **Set the second derivative to zero**:
\[ e^{-x}(x - 2) = 0 \]
This gives us \( x - 2 = 0 \) or \( x = 2 \). Thus, the inflection point is at \( x = 2 \).

### Step 2: Set up the integral for the area
The area \( A \) bounded by the curve from \( x = 0 \) to \( x = 2 \) is given by the integral:
\[ A = \int_0^2 x e^{-x} \, dx \]

### Step 3: Solve the integral using integration by parts
We will use integration by parts where we let:
- \( u = x \) and \( dv = e^{-x} dx \)
- Then, \( du = dx \) and \( v = -e^{-x} \)

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ \int x e^{-x} \, dx = -x e^{-x} - \int -e^{-x} \, dx \]
\[ = -x e^{-x} + e^{-x} \]

### Step 4: Evaluate the definite integral
Now we evaluate the integral from 0 to 2:
\[ A = \left[-x e^{-x} + e^{-x}\right]_0^2 \]
Calculating at the limits:
1. At \( x = 2 \):
\[ -2 e^{-2} + e^{-2} = -2 e^{-2} + e^{-2} = -e^{-2} \]
2. At \( x = 0 \):
\[ -0 \cdot e^{0} + e^{0} = 1 \]

Thus, the area becomes:
\[ A = \left(-e^{-2} - 1\right) = 1 - 3e^{-2} \]

### Final Answer
The area bounded by the curve \( y = x e^{-x} \), the axes, and the line \( x = 2 \) is:
\[ A = 1 - 3e^{-2} \]