If ∫02√2x-√(2x-x^2)dx= ∫011-√(1-y^2)-y^22dy+∫122-y^22dy+I, then I equal to

Correct answer: (C) ∫011-√(1-y^2)dy. Given, ∫02√2x-√(2x-x^2)dx= ∫011-√(1-y^2)-y^22dy+∫122-y^22dy+I Solving LHS =∫02√2x-√(2x-x^2)dx=8/3-(π)/2 Now, solving RHS =∫011-√(1-y^2)-y^22dy+∫122-y^22dy+I =I+5/3-(π)/4 Now equating LHS and RHS we get, I=1-(π)/4=∫011-√(1-y^2)dy as ∫01√(1-y^2)dy=(π)/4

Q60 / 90 · JEE Main 2022 (29 Jun Shift 2) · Mathematics

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 $I$ equal to

A
$\int_{0}^{1} (1 + \sqrt{1 - y^{2}}) d y$
B
$\int_{0}^{1} (\frac{y^{2}}{2} - \sqrt{1 - y^{2}} + 1) d y$
Correct answer
C
$\int_{0}^{1} (1 - \sqrt{1 - y^{2}}) d y$
D
$\int_{0}^{1} (\frac{y^{2}}{2} + \sqrt{1 - y^{2}} + 1) d y$

Solution

Given,

$\int_{0}^{2} (\sqrt{2 x} - \sqrt{2 x - x^{2}}) d x =$

$\int_{0}^{1} (1 - \sqrt{1 - y^{2}} - \frac{y^{2}}{2}) d y + \int_{1}^{2} (2 - \frac{y^{2}}{2}) d y + I$

Solving LHS $= \int_{0}^{2} (\sqrt{2 x} - \sqrt{2 x - x^{2}}) d x = \frac{8}{3} - \frac{π}{2}$

Now, solving RHS $= \int_{0}^{1} (1 - \sqrt{1 - y^{2}} - \frac{y^{2}}{2}) d y + \int_{1}^{2} (2 - \frac{y^{2}}{2}) d y + I$

$= I + \frac{5}{3} - \frac{π}{4}$

Now equating LHS and RHS we get, $I = 1 - \frac{π}{4} = \int_{0}^{1} (1 - \sqrt{1 - y^{2}}) d y$ as $\int_{0}^{1} (\sqrt{1 - y^{2}}) d y = \frac{π}{4}$

Hence, the correct answer is option: 'C'.

If ∫022x-2x-x2dx=∫011-1-y2-y22dy+∫122-y22dy+I, then I equal to

Solution

Relevant Questions

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 $I$ equal to