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(A) The given equation of the ellipse is $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$.This can be written as $\frac{y^{2}}{9}=1-\frac{x^{2}}{4} \Rightarrow y^{

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) The given equation of the ellipse is $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$.This can be written as $\frac{y^{2}}{9}=1-\frac{x^{2}}{4} \Rightarrow y^{2}=9\left(1-\frac{x^{2}}{4}ight) \Rightarrow y=3 \sqrt{1-\frac{x^{2}}{4}}$ (taking the positive value for the first quadrant).Since the ellipse is symmetrical about both the $x$-axis and $y$-axis,the total area is $4$ times the area in the first quadrant.Area $= 4 \int_{0}^{2} y \, dx = 4 \int_{0}^{2} 3 \sqrt{1-\frac{x^{2}}{4}} \, dx = 4 	imes \frac{3}{2} \int_{0}^{2} \sqrt{4-x^{2}} \, dx = 6 \int_{0}^{2} \sqrt{2^{2}-x^{2}} \, dx$.Using the formula $\int \sqrt{a^{2}-x^{2}} \, dx = \frac{x}{2} \sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2} \sin^{-1} \left(\frac{x}{a}ight)$:Area $= 6 \left[ \frac{x}{2} \sqrt{4-x^{2}} + \frac{4}{2} \sin^{-1} \left(\frac{x}{2}ight) ight]_{0}^{2}$$= 6 \left[ (0 + 2 \sin^{-1}(1)) - (0 + 2 \sin^{-1}(0)) ight] = 6 \left[ 2 	imes \frac{\pi}{2} ight] = 6 \pi$ square units.

$x$	$0$	$1$	$2$	$3$	$4$	$5$
$f(x)$	$2$	$3$	$6$	$11$	$18$	$27$

Find the area of the region bounded by the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$. (in $\pi$)

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