Find the area encolsed by the curve IyI =sin2x where x belongs to [0,2pie] a)2,b)3,c)4,d)8

Dear Student,

The graph for the given equation will be same as graph of y = sin2x but the portion below x-axis will be shifted above it (becuase y should always be positive).
And because of 2x, there will be 4 half cycles between 0 to 2pi.
Therefore, The the area enclosed will be 4 * area under sin2x from 0 to pi/2.
Simply it and get the answer.

Thanks.