The area (in sq. units) of the region enclose | vedclass

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(B) Given equations are $y^{2}=2x$ and $x+y=4$. From the line equation,$x=4-y$. Substituting this into the parabola equation: $y^{2}=2(4-y) \implies y

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(B) Given equations are $y^{2}=2x$ and $x+y=4$. From the line equation,$x=4-y$. Substituting this into the parabola equation: $y^{2}=2(4-y) \implies y^{2}=8-2y \implies y^{2}+2y-8=0$. Factoring the quadratic: $(y+4)(y-2)=0$,so $y=-4$ and $y=2$. The points of intersection are $(8, -4)$ and $(2, 2)$. The area $A$ is given by $\int_{-4}^{2} [(4-y) - \frac{y^{2}}{2}] dy$. Integrating term by term: $\left[ 4y - \frac{y^{2}}{2} - \frac{y^{3}}{6} \right]_{-4}^{2}$. Evaluating at the limits: $\left( 4(2) - \frac{2^{2}}{2} - \frac{2^{3}}{6} \right) - \left( 4(-4) - \frac{(-4)^{2}}{2} - \frac{(-4)^{3}}{6} \right)$. $= (8 - 2 - \frac{4}{3}) - (-16 - 8 + \frac{32}{3}) = (6 - \frac{4}{3}) - (-24 + \frac{32}{3}) = \frac{14}{3} - (\frac{-72+32}{3}) = \frac{14}{3} + \frac{40}{3} = \frac{54}{3} = 18$ sq. units.

The area (in sq. units) of the region enclosed between the parabola $y^{2}=2x$ and the line $x+y=4$ is

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