Given,
∫02(2x−2x−x2)dx=
∫01(1−1−y2−2y2)dy+∫12(2−2y2)dy+I
Solving LHS =∫02(2x−2x−x2)dx=38−2π
Now, solving RHS =∫01(1−1−y2−2y2)dy+∫12(2−2y2)dy+I
=I+35−4π
Now equating LHS and RHS we get, I=1−4π=∫01(1−1−y2)dy as ∫01(1−y2)dy=4π
If ∫02(2x−2x−x2)dx=
∫01(1−1−y2−2y2)dy+∫12(2−2y2)dy+I, then I equal to
Held on 29 Jun 2022 · Verified 6 Jul 2026.
∫01(1+1−y2)dy
∫01(2y2−1−y2+1)dy
∫01(1−1−y2)dy
∫01(2y2+1−y2+1)dy
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