The area bounded by the curve

Solution:

$4 y^{2}=x^{2}(4-x)(x-2)$

$|y|=\frac{|x|}{2} \sqrt{(4-x)(x-2)}$

$\Rightarrow y_{1}=\frac{x}{2} \sqrt{(4-x)(x-2)}$

and $y_{2}=\frac{-x}{2} \sqrt{(4-x)(x-2)}$

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

Required Area

$=\int_{2}^{4}\left(y_{1}-y_{2}\right) d x=\int_{2}^{4} x \sqrt{(4-x)(x-2)} d x$  .......(1)

Applying $\int_{\mathrm{a}}^{\mathrm{b}} f(\mathrm{x}) \mathrm{d} \mathrm{x}=\int_{\mathrm{a}}^{\mathrm{b}} f(\mathrm{a}+\mathrm{b}-\mathrm{x}) \mathrm{dx}$

Area $=\int_{2}^{4}(6-x) \sqrt{(4-x)(x-2)} d x$

$(1)+(2)$

$2 \mathrm{~A}=6 \int_{2}^{4} \sqrt{(4-\mathrm{x})(\mathrm{x}-2)} \mathrm{dx}$

$A=3 \int_{2}^{4} \sqrt{1-(x-3)^{2}} d x$

$\mathrm{A}=3 \cdot \frac{\pi}{2} \cdot 1^{2}=\frac{3 \pi}{2}$