JEE Main 2021 — Mathematics Calculus

$$4{y}^{2}={x}^{2}(4-x)(x-2)$$4y2=x2(4−x)(x−2)

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

$$\Rightarrow {y}_{1}=\frac{x}{2}\sqrt{(4-x)(x-2)}$$⇒y1​=2x​(4−x)(x−2)​

and ${y}_{2}=\frac{-x}{2}\sqrt{(4-x)(x-2)}$y2​=2−x​(4−x)(x−2)​

$$D:x\in [2,4]$$D:x∈[2,4]

Required Area

$$={\int }_{2}^{4}({y}_{1}-{y}_{2})\mathrm{dx}={\int }_{2}^{4}x\sqrt{(4-x)(x-2)}\mathrm{dx}...(1)$$=∫24​(y1​−y2​)dx=∫24​x(4−x)(x−2)​dx...(1)

Applying $\displaystyle {\int }_{a}^{b}f(x)dx={\int }_{a}^{b}f(a+b-x)dx$∫ab​f(x)dx=∫ab​f(a+b−x)dx

Area $\displaystyle ={\int }_{2}^{4}(6-x)\sqrt{(4-x)(x-2)}dx...(2)$=∫24​(6−x)(4−x)(x−2)​dx...(2)

$$(1)+(2)$$(1)+(2)

$$2A=6{\int }_{2}^{4}\sqrt{(4-x)(x-2)}dx$$2A=6∫24​(4−x)(x−2)​dx

$$A=3{\int }_{2}^{4}\sqrt{1-(x-3{)}^{2}}dx$$A=3∫24​1−(x−3)2​dx

$A=3\cdot \frac{\pi }{2}\cdot {1}^{2}=\frac{3\pi }{2}$A=3⋅2π​⋅12=23π​ sq. unit