Answer is: option2
\( \frac{25\pi}{4} \)Solution:
We are given the integral:
\[ \int_0^5 \sqrt{25 - x^2} \, dx \]
This integral represents the area under the upper semicircle of the circle \( x^2 + y^2 = 25 \) from \( x = 0 \) to \( x = 5 \).
The equation \( y = \sqrt{25 - x^2} \) is the upper half of the circle with radius 5 centered at the origin.
The full area of the circle is \( \pi r^2 = \pi \cdot 25 = 25\pi \).
The upper half of the circle (semicircle) has area \( \frac{25\pi}{2} \).
Since the bounds are from \( x = 0 \) to \( x = 5 \), we are taking half of the semicircle (i.e., one-quarter of the full circle).
\[ \frac{1}{4} \cdot 25\pi = \frac{25\pi}{4} \]
Answer: (B) \( \frac{25\pi}{4} \)