JEE Main 2021 — Area Under Curves Question with Solution

Given, the parabola ${\left(y-2\right)}^2=\left(x-1\right)$ and $y=3$, $x=2$

So, point is $(2,3)$ Differentiate given equation w.r.t. $x$, we get

$2(y-2)y^{\text{'}}=1$

$y^{\mathrm{\prime }}=\frac{1}{2(y-2)}$

${\left.y^{\mathrm{\prime }}\right|}_{(2,3)}=\frac{1}{2}$

So, the equation of tangent at $\left(2,3\right)$

$y-3=\frac{1}{2}\left(x-2\right)$

$x-2y+4=0$

Required area  $={\int }_0^3\left[\right(y-2{)}^2+1]dy-{\int }_0^3(2y-4)dy$
Required area $={\int }_0^3\left(y^2-4y+5\right)dy+{\int }_0^3\left(2y-4\right)dy$
Required area $={\left[\frac{y^3}{3}-2y^2+5y\right]}_0^3-{\left[y^2-4y\right]}_0^3$ 

Required area$=9$sq units