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