What is the area under the curve `f(x)=xe^(x)` above the X-axis and between the lines x=0 and x=1?

Options

A. `1/2` sq. unit
B. `1` sq unit
C. `3/2` sq. units
D. `2` sq. units

Answer

To find the area under the curve \( f(x) = x e^x \) above the x-axis and between the lines \( x = 0 \) and \( x = 1 \), we will use integration. The area can be calculated using the definite integral of the function from 0 to 1.

### Step-by-Step Solution:

1. **Set up the integral**:
The area \( A \) under the curve from \( x = 0 \) to \( x = 1 \) is given by:
\[ A = \int_{0}^{1} x e^x \, dx \]

2. **Use integration by parts**:
We will apply the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
Here, we can choose:
- \( u = x \) (thus \( du = dx \))
- \( dv = e^x \, dx \) (thus \( v = e^x \))

3. **Apply the integration by parts**:
Substituting into the integration by parts formula:
\[ \int x e^x \, dx = x e^x - \int e^x \, dx \]
Now, we compute the integral of \( e^x \):
\[ \int e^x \, dx = e^x \]
Therefore, we have:
\[ \int x e^x \, dx = x e^x - e^x \]

4. **Evaluate the definite integral**:
Now we need to evaluate the expression from 0 to 1:
\[ A = \left[ x e^x - e^x \right]_{0}^{1} \]
Plugging in the limits:
\[ A = \left( 1 \cdot e^1 - e^1 \right) - \left( 0 \cdot e^0 - e^0 \right) \]
Simplifying this:
\[ A = (e - e) - (0 - 1) = 0 + 1 = 1 \]

5. **Final result**:
Thus, the area under the curve \( f(x) = x e^x \) from \( x = 0 \) to \( x = 1 \) is:
\[ A = 1 \text{ square unit} \]