Sketch the region `{(x, 0):y=sqrt(4-x^(2))}` and X-axis. Find the area of the region using integration.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given function
The function given is \( y = \sqrt{4 - x^2} \). This represents the upper half of a circle with a radius of 2 centered at the origin.

### Step 2: Sketch the region
To sketch the region, we need to plot the function \( y = \sqrt{4 - x^2} \) along with the x-axis. The circle equation derived from squaring the function is \( x^2 + y^2 = 4 \). The region we are interested in is the area above the x-axis and below the curve from \( x = -2 \) to \( x = 2 \).

### Step 3: Identify the limits of integration
The limits of integration are determined by the points where the curve intersects the x-axis. This occurs at \( x = -2 \) and \( x = 2 \).

### Step 4: Set up the integral for area
The area \( A \) of the region can be expressed as:
\[ A = \int_{-2}^{2} \sqrt{4 - x^2} \, dx \]

### Step 5: Use symmetry to simplify the integral
Since the function \( \sqrt{4 - x^2} \) is even (i.e., \( f(x) = f(-x) \)), we can simplify the integral:
\[ A = 2 \int_{0}^{2} \sqrt{4 - x^2} \, dx \]

### Step 6: Evaluate the integral
To evaluate the integral \( \int \sqrt{4 - x^2} \, dx \), we can use the formula:
\[ \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) + C \]
Here, \( a = 2 \).

Thus, we have:
\[ \int \sqrt{4 - x^2} \, dx = \frac{x}{2} \sqrt{4 - x^2} + 2 \sin^{-1}\left(\frac{x}{2}\right) + C \]

### Step 7: Apply the limits
Now we apply the limits from 0 to 2:
\[ A = 2 \left[ \left( \frac{x}{2} \sqrt{4 - x^2} + 2 \sin^{-1}\left(\frac{x}{2}\right) \right) \bigg|_{0}^{2} \right] \]

Calculating at the upper limit \( x = 2 \):
\[ = 2 \left[ \left( \frac{2}{2} \sqrt{4 - 2^2} + 2 \sin^{-1}(1) \right) - \left( \frac{0}{2} \sqrt{4 - 0^2} + 2 \sin^{-1}(0) \right) \right] \]
\[ = 2 \left[ \left( 1 \cdot 0 + 2 \cdot \frac{\pi}{2} \right) - 0 \right] \]
\[ = 2 \left[ \pi \right] = 2\pi \]

### Final Area Calculation
Thus, the total area \( A \) is:
\[ A = 2 \cdot 2\pi = 2\pi \text{ square units} \]

### Summary
The area of the region bounded by the curve \( y = \sqrt{4 - x^2} \) and the x-axis from \( x = -2 \) to \( x = 2 \) is \( 2\pi \) square units.