If \int_0^2(\sqrt{2 x}-\sqrt{2 x-x^2}) d x=\int_0^1(1-\sqrt{1-y^2}-\frac{y^2}{2}) d y+\int_1^2(2-\frac{y^2}{2}) d y+I, then I equals [29-Jun-2022-Shift-2]

Correct Answer is : ∫01(1−√1−y2)dy

\int_0^1(\frac{y^2}{2}-\sqrt{1-y^2}+1) d y

\int_0^1(\frac{y^2}{2}+\sqrt{1-y^2}+1) d y

JEE Main 29 June 2022 Shift 2 Solved Paper

Section: Mathematics

Question : 8 of 90

Marks: +1, -0

∫

2(√2x−√2x−x2)dx=

∫

1(1−√1−y2−‌

)dy+

∫

2(2−‌

)dy+I, then I equals

[29-Jun-2022-Shift-2]

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