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Definite Integration question

2022 · 29 Jun · Shift 2 · Q28
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  5. /2022 · 29 Jun · Shift 2 · Q28

Definite Integration question

2022 · 29 Jun · Shift 2 · Q28

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
If ∫02(2x−2x−x2)dx=∫01(1−1−y2−y22)dy+∫12(2−y22)dy+I\int\limits_0^2 {\left( {\sqrt {2x} - \sqrt {2x - {x^2}} } \right)dx = \int\limits_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy + \int\limits_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy + I} } }0∫2​(2x​−2x−x2​)dx=0∫1​(1−1−y2​−2y2​)dy+1∫2​(2−2y2​)dy+I, then I equals
  1. A
    ∫01(1+1−y2)dy\int\limits_0^1 {\left( {1 + \sqrt {1 - {y^2}} } \right)dy}0∫1​(1+1−y2​)dy
  2. B
    ∫01(y22−1−y2+1)dy\int\limits_0^1 {\left( {{{{y^2}} \over 2} - \sqrt {1 - {y^2}} + 1} \right)dy}0∫1​(2y2​−1−y2​+1)dy
  3. C
    ∫01(1−1−y2)dy\int\limits_0^1 {\left( {1 - \sqrt {1 - {y^2}} } \right)dy}0∫1​(1−1−y2​)dy
  4. D
    ∫01(y22+1−y2+1)dy\int\limits_0^1 {\left( {{{{y^2}} \over 2} + \sqrt {1 - {y^2}} + 1} \right)dy}0∫1​(2y2​+1−y2​+1)dy
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