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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 1st and 2nd 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
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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