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JEE Main 202229 JuneEvening Shift
Exam
JEE Main
Year
2022
Date
29 June
Shift
Evening Shift
Subject
Mathematics
Chapter
Definite Integration

JEE Main 2022 · 29 June · Evening Shift · Mathematics · Definite Integration

If ∫ _ 0 ^ 2 ( 2x - 2x - x ^ 2 )dx = ∫ _ 0 ^ 1 (1 - 1 - y ^ 2 - y ^...

If ∫02(2x−2x−x2)dx={∫}_{0}^{2}\left(\sqrt{2x} - \sqrt{2x - {x}^{2}}\right)dx =

∫01(1−1−y2−y22)dy+∫12(2−y22)dy+I{∫}_{0}^{1}\left(1 - \sqrt{1 - {y}^{2}} - \dfrac{{y}^{2}}{2}\right)dy + {∫}_{1}^{2}\left(2 - \dfrac{{y}^{2}}{2}\right)dy + I, then II equal to

Options

  1. A∫01(1+1−y2)dy{∫}_{0}^{1}\left(1 + \sqrt{1 - {y}^{2}}\right)dy
  2. B∫01(y22−1−y2+1)dy{∫}_{0}^{1}\left(\dfrac{{y}^{2}}{2} - \sqrt{1 - {y}^{2}} + 1\right)dy
  3. C∫01(1−1−y2)dy{∫}_{0}^{1}\left(1 - \sqrt{1 - {y}^{2}}\right)dy
  4. D∫01(y22+1−y2+1)dy{∫}_{0}^{1}\left(\dfrac{{y}^{2}}{2} + \sqrt{1 - {y}^{2}} + 1\right)dy

Correct answer

C. ∫01(1−1−y2)dy{∫}_{0}^{1}\left(1 - \sqrt{1 - {y}^{2}}\right)dy

Step-by-step solution

Given,

∫02(2x−2x−x2)dx={∫}_{0}^{2}\left(\sqrt{2x} - \sqrt{2x - {x}^{2}}\right)dx =

∫01(1−1−y2−y22)dy+∫12(2−y22)dy+I{∫}_{0}^{1}\left(1 - \sqrt{1 - {y}^{2}} - \dfrac{{y}^{2}}{2}\right)dy + {∫}_{1}^{2}\left(2 - \dfrac{{y}^{2}}{2}\right)dy + I

Solving LHS =∫02(2x−2x−x2)dx=83−π2= {∫}_{0}^{2}\left(\sqrt{2x} - \sqrt{2x - {x}^{2}}\right)dx = \dfrac{8}{3} - \dfrac{\pi}{2}

Now, solving RHS =∫01(1−1−y2−y22)dy+∫12(2−y22)dy+I= {∫}_{0}^{1}\left(1 - \sqrt{1 - {y}^{2}} - \dfrac{{y}^{2}}{2}\right)dy + {∫}_{1}^{2}\left(2 - \dfrac{{y}^{2}}{2}\right)dy + I

=I+53−π4= I + \dfrac{5}{3} - \dfrac{\pi}{4}

Now equating LHS and RHS we get, I=1−π4=∫01(1−1−y2)dyI = 1 - \dfrac{\pi}{4} = {∫}_{0}^{1}\left(1 - \sqrt{1 - {y}^{2}}\right)dy as ∫01(1−y2)dy=π4{∫}_{0}^{1}\left(\sqrt{1 - {y}^{2}}\right)dy = \dfrac{\pi}{4}

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