JEE Main 2022 — Mathematics Calculus

We have, ${y}^{2}=2x-1...(1)$y2=2x−1...(1) and ${y}^{2}=4x-3...(2)$y2=4x−3...(2)

We first find the points of intersection of the given parabolas by solving the equations (1)&(2)simultaneously.

This gives, $2x-1=4x-3$2x−1=4x−3

$\Rightarrow 2x=2$⇒2x=2 $\Rightarrow x=1$⇒x=1

Now, put $x=1$x=1 in equation $(1)$(1), we get $y=\pm 1$y=±1.

Now, the equations (1)&(2) can be written as

$${y}^{2}=2(x-\frac{1}{2})$$y2=2(x−21​)

${y}^{2}=4(x-\frac{3}{4})$y2=4(x−43​) respectively.

The graphs of these two curves are shown in figure.

We have to find the area of the shaded region.

Now, the equations (1)&(2) can be written as

$x=\frac{{y}^{2}+1}{2}$x=2y2+1​ & $x=\frac{{y}^{2}+3}{4}$x=4y2+3​ respectively.

It is clear from the figure, required area $\displaystyle A=2{\int }_{0}^{1}({x}_{2}-{x}_{1})dy$A=2∫01​(x2​−x1​)dy ($\because$∵ Area is symmetrical about $x-$x− axis)

where ${x}_{1}=\frac{{y}^{2}+1}{2}$x1​=2y2+1​ and ${x}_{2}=\frac{{y}^{2}+3}{4}$x2​=4y2+3​

$\displaystyle \Rightarrow A=2{\int }_{0}^{1}(\frac{{y}^{2}+3}{4}-\frac{{y}^{2}+1}{2})dy$⇒A=2∫01​(4y2+3​−2y2+1​)dy $\displaystyle =\frac{2}{4}{\int }_{0}^{1}(1-{y}^{2})dy$=42​∫01​(1−y2)dy

$=\frac{1}{2}{[y-\frac{{y}^{3}}{3}]}_{0}^{1}$=21​[y−3y3​]01​ $=\frac{1}{2}[1-\frac{1}{3}]$=21​[1−31​]

$=\frac{1}{2}(\frac{2}{3})=\frac{1}{3}$=21​(32​)=31​ sq.units.