If ∫ limits02(√2 x-√2 x-x2) d x= ∫ limits01(1-√1-y2-(y2/2)) d y+∫ limits12(2-(y2/2)) d y+I

Question

If ∫ limits02(√2 x-√2 x-x2) d x= ∫ limits01(1-√1-y2-(y2/2)) d y+∫ limits12(2-(y2/2)) d y+I (A) ∫ limits01(1+√1- y 2) d y (B) ∫ limits01((y2/2)-√1-y2+1) d y (C) ∫ limits01(1-√1-y2) d y (D) ∫ limits01((y2/2)+√1-y2+1) d y

Tardigrade Admin · Accepted Answer

LHS =∫ limits02(√2 x-√2 x-x2) d x=(8/3)-(π/2) R H S=∫ limits01(1-√1-y2-(y2/2)) d y+∫ limits12(2-(y2/2)) d y+I I+(5/3)-(π/4) So, I=1-(π/4)=∫ limits01(1-√1-y2) d y

Q. If $0 \int 2 (2 x - 2 x - x^{2}) d x =$
$0 \int 1 (1 - 1 - y^{2} - \frac{y ^{2}}{2}) d y + 1 \int 2 (2 - \frac{y ^{2}}{2}) d y + I$

$0 \int 1 (1 + 1 - y^{2}) d y$

$0 \int 1 (\frac{y ^{2}}{2} - 1 - y^{2} + 1) d y$

$0 \int 1 (1 - 1 - y^{2}) d y$

$0 \int 1 (\frac{y ^{2}}{2} + 1 - y^{2} + 1) d y$

$L H S = 0 \int 2 (2 x - 2 x - x^{2}) d x = \frac{8}{3} - \frac{π}{2}$
$R H S = 0 \int 1 (1 - 1 - y^{2} - \frac{y ^{2}}{2}) d y + 1 \int 2 (2 - \frac{y ^{2}}{2}) d y + I$
$I + \frac{5}{3} - \frac{π}{4}$
So, $I = 1 - \frac{π}{4} = 0 \int 1 (1 - 1 - y^{2}) d y$

Q. If 0∫2​(2x​−2x−x2​)dx= 0∫1​(1−1−y2​−2y2​)dy+1∫2​(2−2y2​)dy+I

0∫1​(1+1−y2​)dy

0∫1​(2y2​−1−y2​+1)dy

0∫1​(1−1−y2​)dy

0∫1​(2y2​+1−y2​+1)dy