The area below `y=e^(x)` and above y = x between x = 0 and x = 2 is

Options

A. `e^(2)-1`
B. `e^(2)+e`
C. `e^(2)-3`
D. `e+2`

Answer

To find the area between the curves \( y = e^x \) and \( y = x \) from \( x = 0 \) to \( x = 2 \), we will follow these steps:

### Step 1: Identify the curves and the area to be calculated
We need to calculate the area between the curve \( y = e^x \) (which is above) and the line \( y = x \) (which is below) from \( x = 0 \) to \( x = 2 \).

### Step 2: Set up the integral
The area \( A \) between the two curves can be expressed as:
\[ A = \int_{0}^{2} (e^x - x) \, dx \]
Here, \( e^x \) is the upper function and \( x \) is the lower function in the interval from 0 to 2.

### Step 3: Calculate the integral
Now we will compute the integral:
\[ A = \int_{0}^{2} e^x \, dx - \int_{0}^{2} x \, dx \]

#### Step 3.1: Calculate \( \int_{0}^{2} e^x \, dx \)
The integral of \( e^x \) is \( e^x \):
\[ \int e^x \, dx = e^x + C \]
Thus,
\[ \int_{0}^{2} e^x \, dx = e^2 - e^0 = e^2 - 1 \]

#### Step 3.2: Calculate \( \int_{0}^{2} x \, dx \)
The integral of \( x \) is \( \frac{x^2}{2} \):
\[ \int x \, dx = \frac{x^2}{2} + C \]
Thus,
\[ \int_{0}^{2} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{2} = \frac{2^2}{2} - \frac{0^2}{2} = \frac{4}{2} = 2 \]

### Step 4: Combine the results
Now we can substitute the results back into the area formula:
\[ A = (e^2 - 1) - 2 = e^2 - 3 \]

### Final Answer
The area below \( y = e^x \) and above \( y = x \) from \( x = 0 \) to \( x = 2 \) is:
\[ \boxed{e^2 - 3} \]