The area of the region bounded between the curves `y=e||x|In|x||,x^2+y^2-2(|x|+|y|)+1>=0` and X-axis where `|x|<=1`, if `alpha` is the x-coordinate of the point of intersection of curves in 1st quadrant, is

Options

A. `4[int_(0)^alpha ex In x dx+int_(alpha)^(1)(1-sqrt(1-(x-1)^2))dx]`
B. `4[int_(0)^alpha ex In x dx+int_(1)^(alpha)(1-sqrt(1-(x-1)^2))dx]`
C. `4[-int_(0)^alpha ex In x dx+int_(alpha)^(1)(1-sqrt(1-(x-1)^2))dx]`
D. `2[int_(0)^alpha ex In x dx+int_(alpha)^(1)(1-sqrt(1-(x-1)^2))dx]`

Answer

To solve the problem of finding the area of the region bounded between the curves \( y = e^{|x| \ln |x|} \) and the circle defined by \( x^2 + y^2 - 2(|x| + |y|) + 1 \geq 0 \) within the bounds \( |x| \leq 1 \), we will follow these steps:

### Step 1: Understand the curves
The first curve is given by:
\[ y = e^{|x| \ln |x|} \]
The second curve can be rewritten as:
\[ x^2 + y^2 - 2|x| - 2|y| + 1 \geq 0 \]
This represents the area outside a circle centered at \( (1, 1) \) with a radius of 1.

### Step 2: Identify the area of interest
We need to find the area bounded by these curves above the x-axis, specifically for \( |x| \leq 1 \). This means we will consider the first quadrant where \( x \geq 0 \) and \( y \geq 0 \).

### Step 3: Find the intersection points
To find the intersection points of the two curves in the first quadrant, we set:
\[ e^{x \ln x} = 1 - \sqrt{1 - (x - 1)^2} \]
This will give us the x-coordinate \( \alpha \) of the intersection point.

### Step 4: Set up the area calculation
The area \( A \) can be calculated as the sum of two integrals:
1. From \( 0 \) to \( \alpha \) for the curve \( y = e^{x \ln x} \)
2. From \( \alpha \) to \( 1 \) for the circle.

The area can be expressed as:
\[ A = \int_0^{\alpha} e^{x \ln x} \, dx + \int_{\alpha}^{1} \left(1 - \sqrt{1 - (x - 1)^2}\right) \, dx \]

### Step 5: Calculate the integrals
1. The first integral can be simplified:
\[ \int_0^{\alpha} e^{x \ln x} \, dx = \int_0^{\alpha} x^x \, dx \]
This integral needs to be evaluated numerically or using special functions.

2. The second integral represents the area under the circle:
\[ \int_{\alpha}^{1} \left(1 - \sqrt{1 - (x - 1)^2}\right) \, dx \]
This can be computed using trigonometric substitution or numerical methods.

### Step 6: Combine the areas
Finally, we combine the results from both integrals to get the total area.