If ∫ 0 2 2 x - 2 x - x 2 d x = ∫ 0 1 1 - 1 - y 2 - y 2 2 d y + ∫ 1 2 2 - y 2 2 d y + I ,…

If 022x-2x-x2dx=

011-1-y2-y22dy+122-y22dy+I, then I equal to

  1. 011+1-y2dy
  2. 01y22-1-y2+1dy
  3. 011-1-y2dy
  4. 01y22+1-y2+1dy

Solution

Given,

022x-2x-x2dx=

011-1-y2-y22dy+122-y22dy+I

Solving LHS =022x-2x-x2dx=83-π2

Now, solving RHS =011-1-y2-y22dy+122-y22dy+I

=I+53-π4

Now equating LHS and RHS we get,  I=1-π4=011-1-y2dy as 011-y2dy=π4

Asked in: JEE Main 2022 (29 Jun Shift 2)

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