The integral

$\int_{-7}^{7} \sqrt{49 - x^2} \, dx$

represents the area of a semicircle with radius $r = 7$, centered at the origin, above the $x$-axis. This interpretation arises because the function $\sqrt{49 - x^2}$ is the upper half of the circle described by the equation

$x^2 + y^2 = 49.$

Step-by-step solution:

1. Recognize the geometry:

   • The given function is the equation of a semicircle with radius 7. The integral computes the total area under the curve from $x = -7$ to $x = 7$.

2. Area of the semicircle:

   • The area of a full circle is given by $\pi r^2$, where $r$ is the radius.
   • For a semicircle, the area is half of that: $\text{Area} = \frac{1}{2} \pi r^2.$

3. Substitute $r = 7$: $\text{Area} = \frac{1}{2} \pi (7)^2 = \frac{1}{2} \pi (49) = \frac{49\pi}{2}.$

Final Answer:

$\int_{-7}^{7} \sqrt{49 - x^2} \, dx = \frac{49\pi}{2}.$

This result is derived directly from the geometric interpretation of the integral.

---

Would you like a detailed verification of this using trigonometric substitution or further explanation?

Here are related questions to expand your understanding:

1. What is the integral of $\sqrt{a^2 - x^2}$ in general?
2. How does one compute similar integrals without geometric interpretation?
3. What is the area under a full circle curve $\sqrt{r^2 - x^2}$ over $x \in [-r, r]$?
4. How would this integral change if the limits were $\int_{0}^{7} \sqrt{49 - x^2} \, dx$?
5. Can this integral be evaluated without recognizing it as a semicircle?

Tip: When encountering a square root of the form $\sqrt{a^2 - x^2}$, always check if it corresponds to a geometric shape (a circle or semicircle).