Question

Find the area of the closed figure bounded by the following curve
$y=x^2$, $y=x+2$.

Answer 1

Given Curves $y=x^2,y=x+2$

$x^2-x-2=0$

$(x-2)(x+1)=0$

$x=2,-1$

The area enclosed between x=2 and x=-1 is ${\int }_{-1}^2\left(\right(x+2)-(x^2\left)\right)dx$

$=[\frac{x^2}{2}+2x-\frac{x^3}{3}{]}_{-1}^2=6+\frac{3}{2}-\frac{7}{3}=\frac{31}{6}sq.units$