Evaluating a definite integral as a quarter-circle area

The graph of $y = \sqrt{25 - x^2}$ is the upper half of the circle $x^2 + y^2 = 25$, which has radius $5$. Over the interval $[-5, 0]$, the curve sweeps out exactly one quarter of the full disk of radius 5, since $x=-5$ corresponds to the leftmost point $(-5,0)$ and $x=0$ corresponds to the top point $(0,5)$.

Since $f(x) \ge 0$ on $[-5,0]$, the definite integral equals the area of this quarter-circle region. The area of a full circle of radius 5 is $\pi (5)^2 = 25\pi$, so a quarter of that area is $\frac{25\pi}{4}$. Therefore $\int_{-5}^{0}\sqrt{25-x^2}\,dx = \dfrac{25\pi}{4}$.