Evaluate the integral below by interpreting it in terms of areas. In other words, draw a picture of the region the integral represents, and find the area using high school geometry. \( \int_{-5}^{5} \sqrt{25-x^{2}} d x= \)

To evaluate the integral \(\int_{-5}^{5} \sqrt{25 - x^2} \, dx\), we first recognize that the integrand \(\sqrt{25 - x^2}\) represents the equation of a circle. Specifically, it describes the upper half of a circle with a radius of 5 centered at the origin, since it can be rewritten as \(y = \sqrt{25 - x^2}\). Now, imagine the circle on the coordinate plane. The area we are interested in is the area of the semicircle above the x-axis, bounded by the x-values from -5 to 5. The area of a full circle is given by the formula \(A = \pi r^2\). For a circle of radius 5, the area would be: \[ A = \pi (5^2) = 25\pi. \] Since we are only interested in the upper half of the circle (the semicircle), we divide this area by 2: \[ \text{Area of semicircle} = \frac{25\pi}{2}. \] Thus, the value of the integral \(\int_{-5}^{5} \sqrt{25 - x^2} \, dx\) is \(\frac{25\pi}{2}\).