JEE Main 2021 — Area Under Curves Question with Solution

The given curves are $x^2+2y-1=0, y^2+4x-4=0$ and $y^2-4x-4=0.$

Drawing their graphs, we get 

The required area is shown by the shaded part and since the curves $y^2+4x-4=0$ and $y^2-4x-4=0$ are symmetric about the $y$-axis, hence they have the same area.

Hence, the required area$=2\left[{\int }_0^2\left(\frac{4-y^2}{4}\right)dy-{\int }_0^1\left(\frac{1-x^2}{2}\right)dx\right]$

$=2\left[{\int }_0^2\frac{1}{4}\left(4-y^2\right)dy-{\int }_0^1\frac{1}{2}\left(1-x^2\right)dx\right]$

$=2\left[\frac{1}{4}{\left(4y-\frac{y^3}{3}\right)}_0^2-\frac{1}{2}{\left(x-\frac{x^3}{3}\right)}_0^1\right]$

$=2\left[\frac{1}{4}\left(8-\frac{8}{3}-0\right)-\frac{1}{2}\left(1-\frac{1}{3}-0\right)\right]$

$=2\left[\frac{4}{3}-\frac{1}{3}\right]=2\mathrm{sq}.\mathrm{units}$ .