JEE Main 2019 — Area Under Curves Question with Solution

We have to find the area of the region $A=\left\{\left(x, y\right):\frac{y^2}{2}\le x\le y+4\right\}$

Now, to find the point of intersection of the curves $x=\frac{y^2}{2}$ and $x=y+4,$ we equate the value of $x,$ to get

$\frac{y^2}{2}=y+4$

$\Rightarrow y^2=2y+8$

$\Rightarrow y^2-2y-8=0$

$\Rightarrow \left(y-4\right)\left(y+2\right)=0$

$\Rightarrow y=4$ or $y=-2$

And, the graph of the given functions is as

The shaded area is the required area, and 

Area$={\int }_{-2}^4\left(x_{line}-x_{parabola}\right)dy$

$={\int }_{-2}^4\left(y+4-\frac{y^2}{2}\right)dy$

Now, using $\int x^{n}dx=\frac{x^{n+1}}{n+1},$ we get

Area$={\left.\left(\frac{y^2}{2}+4y-\frac{y^3}{6}\right)\right|}_{-2}^4$

$=\left(8+16-\frac{32}{3}\right)-\left(2-8+\frac{4}{3}\right)$

$=18$ sq units.