The area (in sq. units) of the region A = {(x | vedclass

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(D) Given the region $A = \{(x, y) / \frac{y^2}{2} \leq x \leq y+4\}$.To find the points of intersection,we set $x = \frac{y^2}{2}$ and $x = y+4$.Equa

Vedclass · Accepted Answer

$18$

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(D) Given the region $A = \{(x, y) / \frac{y^2}{2} \leq x \leq y+4\}$.To find the points of intersection,we set $x = \frac{y^2}{2}$ and $x = y+4$.Equating the two expressions for $x$:$\frac{y^2}{2} = y+4$$y^2 = 2y + 8$$y^2 - 2y - 8 = 0$$(y - 4)(y + 2) = 0$Thus,$y = 4$ or $y = -2$.For $y = 4$,$x = 4+4 = 8$. For $y = -2$,$x = -2+4 = 2$.The points of intersection are $(8, 4)$ and $(2, -2)$.The area $A$ is given by the integral with respect to $y$:$A = \int_{-2}^{4} (y + 4 - \frac{y^2}{2}) dy$$A = [\frac{y^2}{2} + 4y - \frac{y^3}{6}]_{-2}^{4}$$A = (\frac{16}{2} + 4(4) - \frac{64}{6}) - (\frac{4}{2} + 4(-2) - \frac{-8}{6})$$A = (8 + 16 - \frac{32}{3}) - (2 - 8 + \frac{4}{3})$$A = (24 - \frac{32}{3}) - (-6 + \frac{4}{3})$$A = \frac{72 - 32}{3} - \frac{-18 + 4}{3}$$A = \frac{40}{3} - (-\frac{14}{3})$$A = \frac{40 + 14}{3} = \frac{54}{3} = 18$ sq. units.

The area (in sq. units) of the region $A = \{(x, y) / \frac{y^2}{2} \leq x \leq y+4\}$ is:

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