To solve the given integral:

$\int_{-2}^{2} \sqrt{4 - x^2} \, dx$

we can recognize that the function $y = \sqrt{4 - x^2}$ describes a semicircle with radius $r = 2$ and center at the origin (0, 0). The integral from $-2$ to $2$ calculates the area of this semicircle.

Solution:

1. Identify the Shape: Since $y = \sqrt{4 - x^2}$ is the top half of a circle with radius 2, the region described by this integral is a semicircle.

2. Area of a Semicircle: The area of a full circle is $\pi r^2$. For a semicircle, we take half of that: $\text{Area} = \frac{1}{2} \pi r^2$

3. Substitute the Radius: With $r = 2$, $\text{Area} = \frac{1}{2} \pi (2)^2 = \frac{1}{2} \pi \cdot 4 = 2\pi$

Therefore, the value of the integral is:

$\int_{-2}^{2} \sqrt{4 - x^2} \, dx = 2\pi$

Would you like more details or have any questions?

Related Questions:

1. What would the integral represent if the limits were from $-1$ to $1$?
2. How would the result change if the equation were $\sqrt{9 - x^2}$ instead?
3. Can you generalize the result for any radius $r$ in the function $\sqrt{r^2 - x^2}$?
4. How does the symmetry of the function simplify the integration process?
5. How would you approach finding the area if the integrand were not a perfect semicircle function?

Tip:

When dealing with integrals that represent areas of well-known shapes (like semicircles), using geometric formulas can save time over direct integration.