JEE Main 2019 — Area Under Curves Question with Solution

The two given curves are $y=x+2$ and $y=x^2$

To find their points of intersection, we equate the value of $y$ to get

$x^2=x+2$

$\Rightarrow x^2-x-2=0$

$\Rightarrow \left(x-2\right)\left(x+1\right)=0$

$\Rightarrow x=2$ or $x=-1.$

The graphs of the curves are given below.

Required area is the area of the shaded portion.

Area $={\int }_{-1}^2\left(y_{line}-y_{parabola}\right)dx$
$={\int }_{-1}^2\left(x+2-x^2\right)dx$

Using $\int x^{n}dx=\frac{x^{n+1}}{n+1}$

$={\left[\frac{x^2}{2}+ 2x-\frac{x^3}{3}\right]}_{-1}^2$

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

$=6-\frac{8}{3}-\frac{1}{2}+ 2-\frac{1}{3}$

$=8-3-\frac{1}{2}$

$=\frac{9}{2}$ sq units.