Question

Find the area of the region enclosed by the parabola $x^2=y$,the line $y=x+2$ and $x-axis$

Answer 1

The correct answer is:$\frac{5}{6}units$
The area of the region enclosed by the parabola,$x^2=y$,the line,$y=x+2$,and $x-axis$
is represented by the shaded region OABCO as

The point of intersection of the parabola,$x^2=y$,and the line,$y=x+2$,is $A(–1,1).$
∴Area OABCO=Area(BCA)+Area COAC
$=∫^{-1}_{-2}(x+2)dx+∫^0_{-1}x^2dx$
$=\bigg[\frac{x^2}{2}+2x\bigg]^{-1}_{-2}+\bigg[\frac{x^3}{3}\bigg]^0_{-1}$
$=[\frac{(-1)^2}{2}+2(-1)-(\frac{-2)^2}{2}-2(-2)]+[-\frac{(-1)^3}{3}]$
$=[\frac{1}{2}-2-2+4+\frac{1}{3}]$
$=\frac{5}{6}units$