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(D) To find the area enclosed by the curves $y^2=4(x+7)$ and $y^2=5(2-x)$, we first find their points of intersection.Equating the expressions for $y^

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$24 \sqrt{5}$

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(D) To find the area enclosed by the curves $y^2=4(x+7)$ and $y^2=5(2-x)$, we first find their points of intersection.Equating the expressions for $y^2$:$4(x+7) = 5(2-x)$$4x + 28 = 10 - 5x$$9x = -18$$x = -2$Substituting $x = -2$ into $y^2 = 4(x+7)$, we get $y^2 = 4(-2+7) = 4(5) = 20$, so $y = \pm \sqrt{20} = \pm 2\sqrt{5}$.The points of intersection are $(-2, 2\sqrt{5})$ and $(-2, -2\sqrt{5})$.We express $x$ in terms of $y$ for both curves:For $y^2 = 4(x+7)$, $x = \frac{y^2}{4} - 7$.For $y^2 = 5(2-x)$, $x = 2 - \frac{y^2}{5}$.The area is given by the integral with respect to $y$ from $-2\sqrt{5}$ to $2\sqrt{5}$ of the right curve minus the left curve:Area $= \int_{-2\sqrt{5}}^{2\sqrt{5}} \left[ (2 - \frac{y^2}{5}) - (\frac{y^2}{4} - 7) ight] dy$$= \int_{-2\sqrt{5}}^{2\sqrt{5}} (9 - \frac{9y^2}{20}) dy$$= \left[ 9y - \frac{9y^3}{60} ight]_{-2\sqrt{5}}^{2\sqrt{5}} = \left[ 9y - \frac{3y^3}{20} ight]_{-2\sqrt{5}}^{2\sqrt{5}}$$= \left( 9(2\sqrt{5}) - \frac{3(2\sqrt{5})^3}{20} ight) - \left( 9(-2\sqrt{5}) - \frac{3(-2\sqrt{5})^3}{20} ight)$$= 2 \left( 18\sqrt{5} - \frac{3(8 	imes 5\sqrt{5})}{20} ight)$$= 2 \left( 18\sqrt{5} - 6\sqrt{5} ight) = 2(12\sqrt{5}) = 24\sqrt{5}$.

Area of the region enclosed between the curves $y^2=4(x+7)$ and $y^2=5(2-x)$ is

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