Find the area of the region bounded by
y = ln x between the ordinates x = 1 and x = e

Answer

To find the area of the region bounded by the curve \( y = \ln x \) between the ordinates \( x = 1 \) and \( x = e \), we can follow these steps:

### Step 1: Understand the Problem
We need to find the area under the curve \( y = \ln x \) from \( x = 1 \) to \( x = e \). The area can be calculated using the definite integral.

### Step 2: Set Up the Integral
The area \( A \) can be expressed as:
\[ A = \int_{1}^{e} \ln x \, dx \]

### Step 3: Use Integration by Parts
To evaluate the integral \( \int \ln x \, dx \), we will use integration by parts. Recall the formula for integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
Let:
- \( u = \ln x \) (thus, \( du = \frac{1}{x} \, dx \))
- \( dv = dx \) (thus, \( v = x \))

### Step 4: Apply Integration by Parts
Using the integration by parts formula:
\[ \int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx \]
This simplifies to:
\[ \int \ln x \, dx = x \ln x - \int 1 \, dx = x \ln x - x + C \]

### Step 5: Evaluate the Definite Integral
Now we evaluate the definite integral from \( 1 \) to \( e \):
\[ A = \left[ x \ln x - x \right]_{1}^{e} \]
Calculating at the upper limit \( x = e \):
\[ = e \ln e - e = e \cdot 1 - e = e - e = 0 \]
Calculating at the lower limit \( x = 1 \):
\[ = 1 \ln 1 - 1 = 1 \cdot 0 - 1 = -1 \]
Now, substituting these limits into the expression:
\[ A = (0) - (-1) = 1 \]

### Final Answer
Thus, the area of the region bounded by \( y = \ln x \) between \( x = 1 \) and \( x = e \) is:
\[ \boxed{1} \text{ square unit} \]