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← JEE Main 2022 · 29 Jun · Shift 2
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JEE Main 2022 · 29 Jun · Shift 2, Mathematics question 8

If ∫02(2x−2x−x2)dx=∫01(1−1−y2−y22)dy+∫12(2−y22)dy+I\displaystyle\int_0^2\left(\sqrt{2x}-\sqrt{2x-x^2}\right)dx=\int_0^1\left(1-\sqrt{1-y^2}-\frac{y^2}2\right)dy+\int_1^2\left(2-\frac{y^2}2\right)dy+I then II equals

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Answer and worked solution

Official answer: C

Region between y2=2xy^2=2x and the circle (x−1)2+y2=1(x-1)^2+y^2=1, 0≤x≤20\le x\le2. In horizontal strips: for 0≤y≤10\le y\le1 the width is (1−1−y2)−y22\left(1-\sqrt{1-y^2}\right)-\frac{y^2}2 on the left plus 2−(1+1−y2)2-\left(1+\sqrt{1-y^2}\right) on the right; for 1≤y≤21\le y\le2 it is 2−y222-\frac{y^2}2.

So I=∫01(1−1−y2)dyI=\displaystyle\int_0^1\left(1-\sqrt{1-y^2}\right)dy.

(C)

Topics:Area between curves
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