The area enclosed between the curve `y = log_e (x +e )` and the coordinate axes is

Options

A. 4
B. 3
C. 2
D. 1

Explanation

To find the area enclosed between the curve \( y = \log_e(x + e) \) and the coordinate axes, we can follow these steps:

### Step 1: Identify the points of intersection
We need to find where the curve intersects the x-axis and y-axis.

1. **For the x-axis**: Set \( y = 0 \):
\[ 0 = \log_e(x + e) \]
This implies:
\[ x + e = 1 \implies x = 1 - e \]

2. **For the y-axis**: Set \( x = 0 \):
\[ y = \log_e(0 + e) = \log_e(e) = 1 \]

Thus, the points of intersection are \( (1 - e, 0) \) and \( (0, 1) \).

### Step 2: Set up the integral for the area
The area \( A \) enclosed between the curve and the axes can be calculated using the integral:
\[ A = \int_{0}^{1} (e^y - e) \, dy \]
where \( e^y = x + e \) rearranged gives \( x = e^y - e \).

### Step 3: Evaluate the integral
Now we compute the integral:
\[ A = \int_{0}^{1} (e^y - e) \, dy \]

This can be split into two separate integrals:
\[ A = \int_{0}^{1} e^y \, dy - \int_{0}^{1} e \, dy \]

Calculating the first integral:
\[ \int e^y \, dy = e^y \bigg|_0^1 = e^1 - e^0 = e - 1 \]

Calculating the second integral:
\[ \int e \, dy = e \cdot y \bigg|_0^1 = e \cdot 1 - e \cdot 0 = e \]

Putting it all together:
\[ A = (e - 1) - e = -1 \]

Since area cannot be negative, we take the absolute value:
\[ A = 1 \text{ square unit} \]

### Final Answer
The area enclosed between the curve \( y = \log_e(x + e) \) and the coordinate axes is \( 1 \) square unit.

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