The area ( in sq units ) of the region described by x,y : y2≤2x and y≥4x−1 is “A” then 32 A =

y22=y+14⇒2y2−y−1=0⇒y=1,−12∫−121y+14−y22dy=14y22+y−2y33−121=932A=932⇒32A=9

The area ( in sq units ) of the region described by x,y : y2≤2x and y≥4x−1 is “A” then 32 A =

y22=y+14⇒2y2−y−1=0⇒y=1,−12∫−121y+14−y22dy=14y22+y−2y33−121=932A=932⇒32A=9

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The area ( in sq units ) of the region described by $\{(x, y) : y^{2} \leq 2 x a n d y \geq 4 x - 1\}$ is “A” then 32 A =

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$\frac{y^{2}}{2} = \frac{y + 1}{4} \Rightarrow 2 y^{2} - y - 1 = 0 \Rightarrow y = 1, \frac{- 1}{2}$

$\int_{\frac{- 1}{2}}^{1} (\frac{y + 1}{4} - \frac{y^{2}}{2}) d y$

$= \frac{1}{4} {[\frac{y^{2}}{2} + y - \frac{2 y^{3}}{3}]}_{\frac{- 1}{2}}^{1}$

$= \frac{9}{32}$

$A = \frac{9}{32} \Rightarrow 32 A = 9$

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The area ( in sq units ) of the region described by x,y : y2≤2x and y≥4x−1 is “A” then 32 A =

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The area ( in sq units ) of the region described by $\{(x, y) : y^{2} \leq 2 x a n d y \geq 4 x - 1\}$ is “A” then 32 A =