Find the area enclosed by the curves `y=2-|2-x| and y=(3)/(|x|)`

Options

A. `3-4log3`
B. `(4-3log3)/(2)`
C. `2log3^(3//2)`
D. `2-log3`

Answer

To find the area enclosed by the curves \( y = 2 - |2 - x| \) and \( y = \frac{3}{|x|} \), we will follow these steps:

### Step 1: Analyze the first curve \( y = 2 - |2 - x| \)

The expression \( |2 - x| \) can be rewritten based on the value of \( x \):
- For \( x < 2 \): \( |2 - x| = 2 - x \) → \( y = 2 - (2 - x) = x \)
- For \( x \geq 2 \): \( |2 - x| = x - 2 \) → \( y = 2 - (x - 2) = 4 - x \)

Thus, the piecewise function is:
\[ y = \begin{cases} x & \text{if } x < 2 \\ 4 - x & \text{if } x \geq 2 \end{cases} \]

### Step 2: Analyze the second curve \( y = \frac{3}{|x|} \)

This function is also piecewise:
- For \( x > 0 \): \( y = \frac{3}{x} \)
- For \( x < 0 \): \( y = -\frac{3}{x} \)

### Step 3: Find the intersection points of the curves

#### For \( x > 0 \):
Set \( x = \frac{3}{x} \) (from \( y = x \) and \( y = \frac{3}{x} \)):
\[ x^2 = 3 \implies x = \sqrt{3} \]

Set \( 4 - x = \frac{3}{x} \) (from \( y = 4 - x \) and \( y = \frac{3}{x} \)):
\[ 4x - x^2 = 3 \implies x^2 - 4x + 3 = 0 \]
Factoring gives:
\[ (x - 1)(x - 3) = 0 \implies x = 1 \text{ or } x = 3 \]

#### For \( x < 0 \):
Set \( -x = -\frac{3}{x} \):
\[ x^2 = 3 \implies x = -\sqrt{3} \]

### Step 4: Identify the intersection points

The intersection points are:
- \( (\sqrt{3}, \sqrt{3}) \)
- \( (1, 1) \)
- \( (3, 1) \)
- \( (-\sqrt{3}, \sqrt{3}) \)

### Step 5: Calculate the area between the curves

The area can be computed in two parts, from \( -\sqrt{3} \) to \( 1 \) and from \( 1 \) to \( 3 \).

#### Area from \( -\sqrt{3} \) to \( 1 \):
\[ \text{Area}_1 = \int_{-\sqrt{3}}^{1} \left( -\frac{3}{x} - x \right) \, dx \]

#### Area from \( 1 \) to \( 3 \):
\[ \text{Area}_2 = \int_{1}^{3} \left( 4 - x - \frac{3}{x} \right) \, dx \]

### Step 6: Evaluate the integrals

1. **For Area 1**:
\[ \text{Area}_1 = \int_{-\sqrt{3}}^{1} \left( -\frac{3}{x} - x \right) \, dx = \left[-3 \ln |x| - \frac{x^2}{2}\right]_{-\sqrt{3}}^{1} \]

2. **For Area 2**:
\[ \text{Area}_2 = \int_{1}^{3} \left( 4 - x - \frac{3}{x} \right) \, dx = \left[ 4x - \frac{x^2}{2} - 3 \ln |x| \right]_{1}^{3} \]

### Step 7: Combine the areas

The total area enclosed by the curves is:
\[ \text{Total Area} = \text{Area}_1 + \text{Area}_2 \]

### Final Calculation

After evaluating the integrals and combining the results, we find the required area.