The area (in sq. unit) bounded by the curve 4 y ^ 2 = x ^ 2 ( 4 - x...

4{y}^{2} = {x}^{2} ( 4 - x ) ( x - 2 )$</p><p>\left|y\right| = \dfrac{| x |}{2}\sqrt{( 4 - x ) ( x - 2 )}$

$⇒ {y}_{1} = \dfrac{x}{2}\sqrt{( 4 - x ) ( x - 2 )}$

and ${y}_{2} = \dfrac{- x}{2}\sqrt{( 4 - x ) ( x - 2 )}$

$D: x \in [ 2, 4 ]$

Required Area

$= {∫}_{2}^{4}$\left({y}_{1} - {y}_{2}\right)$dx = {∫}_{2}^{4}x\sqrt{( 4 - x ) ( x - 2 )}dx . . . \left(1\right)$

Applying ${∫}_{a}^{b}f \left(x\right)dx = {∫}_{a}^{b}f \left(a + b - x\right)dx$

Area $= {∫}_{2}^{4}\left(6 - x\right)\sqrt{( 4 - x ) ( x - 2 )}dx . . . \left(2\right)</p><p>$( 1 ) + ( 2 )$</p><p>$2A = 6{∫}_{2}^{4}\sqrt{( 4 - x ) ( x - 2 )}dx$</p><p>A = 3{∫}_{2}^{4}\sqrt{1 - ( x - 3{)}^{2}}dx$

A = 3 \cdot $\dfrac{\pi}{2}$ \cdot {1}^{2} = $\dfrac{3\pi}{2} sq. unit</p>$