Area of region bounded by `x=0, y=0, x=2, y=2, y<=e^x & y>=lnx` is

Options

A. `6-4ln2`
B. `4 ln 2-2`
C. `2 ln 2-4`
D. `6 - 2 ln 2`

Answer

To find the area of the region bounded by the curves and lines given in the question, we will follow these steps:

### Step 1: Understand the Boundaries
The area is bounded by:
- \( x = 0 \) (y-axis)
- \( y = 0 \) (x-axis)
- \( x = 2 \) (vertical line)
- \( y = 2 \) (horizontal line)
- \( y = e^x \) (exponential curve)
- \( y = \ln x \) (natural logarithm curve)

### Step 2: Sketch the Graph
Draw the coordinate axes and plot the lines and curves:
- The line \( y = 2 \) is a horizontal line at \( y = 2 \).
- The line \( x = 2 \) is a vertical line at \( x = 2 \).
- The curve \( y = e^x \) starts at \( (0, 1) \) and increases rapidly.
- The curve \( y = \ln x \) starts at \( (1, 0) \) and increases slowly.

### Step 3: Find Points of Intersection
To find the area, we need to determine where the curves \( y = e^x \) and \( y = \ln x \) intersect within the bounds \( x = 0 \) and \( x = 2 \).

1. **Set the equations equal to find intersections**:
\[ e^x = \ln x \]
This can be solved graphically or numerically. We find that \( x = 1 \) is a point of intersection since \( e^1 = 2.718 \) and \( \ln(1) = 0 \).

### Step 4: Set Up the Integrals
The area can be calculated using definite integrals. The area can be found by integrating the top curve minus the bottom curve from \( x = 0 \) to \( x = 2 \).

1. **Area under \( y = e^x \)** from \( x = 0 \) to \( x = 1 \):
\[ A_1 = \int_0^1 e^x \, dx \]

2. **Area under \( y = \ln x \)** from \( x = 1 \) to \( x = 2 \):
\[ A_2 = \int_1^2 \ln x \, dx \]

### Step 5: Calculate the Integrals
1. **Calculate \( A_1 \)**:
\[ A_1 = \left[ e^x \right]_0^1 = e^1 - e^0 = e - 1 \]

2. **Calculate \( A_2 \)**:
\[ A_2 = \int \ln x \, dx = x \ln x - x \quad \text{(using integration by parts)} \]
Evaluate from \( 1 \) to \( 2 \):
\[ A_2 = \left[ x \ln x - x \right]_1^2 = (2 \ln 2 - 2) - (1 \cdot 0 - 1) = 2 \ln 2 - 2 + 1 = 2 \ln 2 - 1 \]

### Step 6: Calculate the Total Area
The total area \( A \) is given by:
\[ A = A_1 + A_2 = (e - 1) + (2 \ln 2 - 1) = e + 2 \ln 2 - 2 \]

### Final Answer
Thus, the area of the region bounded by the given curves and lines is:
\[ A = e + 2 \ln 2 - 2 \]