The area bounded by the curve 4y^{2} = x^{2}( | vedclass

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(C) Given the curve $4y^{2} = x^{2}(4-x)(x-2)$.For $y$ to be real,$(4-x)(x-2) \geq 0$,which implies $x \in [2, 4]$.We can write $|y| = \frac{|x|}{2} \

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$\frac{3\pi}{2}$

Vedclass · Answer

(C) Given the curve $4y^{2} = x^{2}(4-x)(x-2)$.For $y$ to be real,$(4-x)(x-2) \geq 0$,which implies $x \in [2, 4]$.We can write $|y| = \frac{|x|}{2} \sqrt{(4-x)(x-2)}$.Since $x \in [2, 4]$,$x$ is positive,so $y = \pm \frac{x}{2} \sqrt{-x^{2} + 6x - 8}$.The area $A$ is given by $\int_{2}^{4} 2 \cdot \frac{x}{2} \sqrt{-x^{2} + 6x - 8} \, dx = \int_{2}^{4} x \sqrt{-(x^{2} - 6x + 9 - 1)} \, dx = \int_{2}^{4} x \sqrt{1 - (x-3)^{2}} \, dx$.Let $x-3 = t$,then $dx = dt$. When $x=2, t=-1$; when $x=4, t=1$.$A = \int_{-1}^{1} (t+3) \sqrt{1-t^{2}} \, dt = \int_{-1}^{1} t \sqrt{1-t^{2}} \, dt + \int_{-1}^{1} 3 \sqrt{1-t^{2}} \, dt$.The first integral is $0$ because the integrand is an odd function.The second integral is $3 	imes (	ext{area of a semicircle of radius } 1) = 3 	imes \frac{\pi(1)^{2}}{2} = \frac{3\pi}{2}$.

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

The area bounded by the curve $4y^{2} = x^{2}(4-x)(x-2)$ is equal to ...... .

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