\(\displaystyle\int_0^2\sqrt{4-x^2}\,\mathrm dx\)
Recall that a circle of radius 2 centered at the origin has equation
\(x^2+y^2=4\implies y=\pm\sqrt{4-x^2}\)
where the positive root gives the top half of the circle in the x-y plane. The definite integral corresponds to the area of the right half of this top half. Since the area of a circle with radius \(r\) is \(\pi r^2\), it follows that the area of a quarter-circle would be \(\dfrac{\pi r^2}4\).
You have \(r=2\), so the definite integral is equal to \(\dfrac{2^2\pi}4=\pi\).
Another way to verify this is to actually compute the integral. Let \(x=2\sin u\), so that \(\mathrm dx=2\cos u\,\mathrm du\). Now
\(\displaystyle\int_0^2\sqrt{4-x^2}\,\mathrm dx=\int_0^{\pi/2}\sqrt{4-(2\sin u)^2}(2\cos u)\,\mathrm du=4\int_0^{\pi/2}\cos^2u\,\mathrm du\)
Recall the half-angle identity for cosine:
\(\cos^2u=\dfrac{1+\cos2u}2\)
This means the integral is equivalent to
\(\displaystyle2\int_0^{\pi/2}(1+\cos 2u)\,\mathrm du=2u+\sin2u\bigg|_{u=0}^{u=\pi/2}=\pi\)
