The area of the region `{(x,y):x^2ley le|x^2-4|,,yge1}` is

Options

A. `3/4(4sqrt 2-1)`
B. `4/3(4sqrt2-1)`
C. `3/4(4sqrt 2+1)`
D. `4/3(4sqrt2+1)`

Answer

To find the area of the region defined by the inequalities \( x^2 \leq y \leq |x^2 - 4| \) and \( y \geq 1 \), we will follow these steps:

### Step 1: Determine the boundaries for \( y \)

The inequalities given are:
1. \( y \geq x^2 \)
2. \( y \leq |x^2 - 4| \)

To analyze the second inequality, we need to consider two cases for \( |x^2 - 4| \):
- Case 1: \( x^2 - 4 \geq 0 \) (i.e., \( x^2 \geq 4 \) or \( |x| \geq 2 \)), then \( |x^2 - 4| = x^2 - 4 \).
- Case 2: \( x^2 - 4 < 0 \) (i.e., \( -2 < x < 2 \)), then \( |x^2 - 4| = 4 - x^2 \).

### Step 2: Find the intersection points

We need to find the intersection points of \( y = x^2 \) and \( y = 4 - x^2 \):
\[ x^2 = 4 - x^2 \implies 2x^2 = 4 \implies x^2 = 2 \implies x = \pm \sqrt{2} \]

### Step 3: Set up the area calculation

The area will be calculated in two parts:
1. From \( x = -2 \) to \( x = -\sqrt{2} \) and from \( x = \sqrt{2} \) to \( x = 2 \), where \( y \) is bounded by \( x^2 \) and \( 4 - x^2 \).
2. From \( x = -\sqrt{2} \) to \( x = \sqrt{2} \), where \( y \) is bounded by \( 1 \) and \( 4 - x^2 \).

### Step 4: Calculate the area in the first region

For \( x \) in the intervals \( [-2, -\sqrt{2}] \) and \( [\sqrt{2}, 2] \):
\[ \text{Area}_1 = 2 \int_{\sqrt{2}}^{2} (4 - x^2 - x^2) \, dx = 2 \int_{\sqrt{2}}^{2} (4 - 2x^2) \, dx \]

Calculating the integral:
\[ = 2 \left[ 4x - \frac{2x^3}{3} \right]_{\sqrt{2}}^{2} \]

Evaluating the limits:
\[ = 2 \left[ (8 - \frac{16}{3}) - (4\sqrt{2} - \frac{4\sqrt{2}}{3}) \right] = 2 \left[ \frac{24}{3} - \frac{16}{3} - (4\sqrt{2} - \frac{4\sqrt{2}}{3}) \right] = 2 \left[ \frac{8}{3} - \frac{8\sqrt{2}}{3} \right] = \frac{16}{3} - \frac{16\sqrt{2}}{3} \]

### Step 5: Calculate the area in the second region

For \( x \) in the interval \( [-\sqrt{2}, \sqrt{2}] \):
\[ \text{Area}_2 = \int_{-\sqrt{2}}^{\sqrt{2}} (4 - x^2 - 1) \, dx = \int_{-\sqrt{2}}^{\sqrt{2}} (3 - x^2) \, dx \]

Calculating the integral:
\[ = \left[ 3x - \frac{x^3}{3} \right]_{-\sqrt{2}}^{\sqrt{2}} \]

Evaluating the limits:
\[ = \left[ 3\sqrt{2} - \frac{(\sqrt{2})^3}{3} \right] - \left[ -3\sqrt{2} + \frac{(-\sqrt{2})^3}{3} \right] = 6\sqrt{2} - \frac{2\sqrt{2}}{3} = \frac{18\sqrt{2}}{3} - \frac{2\sqrt{2}}{3} = \frac{16\sqrt{2}}{3} \]

### Step 6: Combine the areas

Finally, the total area \( A \) is:
\[ A = \text{Area}_1 + \text{Area}_2 = \left( \frac{16}{3} - \frac{16\sqrt{2}}{3} \right) + \frac{16\sqrt{2}}{3} = \frac{16}{3} \]

### Final Answer
The area of the region is \( \frac{16}{3} \).