Square both sides

⇒ y² = 4 – x²

⇒ x² + y² = 4

⇒ x² + y² = 2²

This is equation of circle with center origin and radius 2

Now in -2 ≤ x ≤ 2 and y ≥ 0 which means x and y both positive or x negative and y positive hence the curve has to be above X-axis in 1^st and 2^nd quadrant

Hence the graph of will be graph of circle x² + y² = 2² lying only above X-axis

Now equation of X-axis is y = 0

To find point of intersection of circle with X-axis put y = 0 in circle equation

⇒ x² = 4

⇒ x = ±2

Hence the intersection points with X-axis are (-2, 0) and (2, 0)

Hence the area is shown as below

Now let us find the area

Integrate from -2 to 2

Using uv rule of integration where u and v are functions of x

Here and v = 1

Hence

But

We know that

Hence area is 2π unit²