Evaluating a semicircle integral geometrically

Squaring both sides of $y=\sqrt{4-(x-2)^2}$ gives $(x-2)^2+y^2=4$, which is a circle centered at $(2,0)$ with radius $2$. Since $y=\sqrt{4-(x-2)^2}\ge 0$, the graph is only the upper half of this circle, i.e., a semicircle sitting on the $x$-axis. The interval $[0,4]$ matches exactly the diameter of this circle (from $x=0$ to $x=4$), so the region under $f(x)$ on $[0,4]$ is precisely this upper semicircular region.

The area of a full circle of radius $2$ is $\pi r^2 = 4\pi$, so the area of the upper semicircle is half of that: $2\pi$. Since $f(x)\ge 0$ on $[0,4]$, the definite integral equals this area directly, giving $\int_0^4 \sqrt{4-(x-2)^2}\,dx = 2\pi$.