Mathematics - Area Under Curves Question with Solution | TestHub

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

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

$\Rightarrow y_1=\frac{x}{2}\sqrt{(4-x)(x-2)}$

and $y_2=\frac{-x}{2}\sqrt{(4-x)(x-2)}$

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

Required Area

$={\int }_2^4\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)\mathrm{dx}={\int }_2^4\mathrm{x}\sqrt{(4-\mathrm{x})(\mathrm{x}-2)}\mathrm{dx} ...\left(1\right)$

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

Area $={\int }_2^4\left(6-x\right)\sqrt{(4-x)(x-2)}dx ...\left(2\right)$

$\left(1\right)+\left(2\right)$

$2A=6{\int }_2^4\sqrt{(4-x)(x-2)}dx$

$A=3{\int }_2^4\sqrt{1-(x-3{)}^2}dx$

$A=3·\frac{\pi }{2}·1^2=\frac{3\pi }{2}$ sq. unit