Question

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

Answer 1

Given the curve (represent and upward parabola with vertex (0, 0))

$y=x^2$ ...(i)

and equation of the line y = x + 2 ...(ii)

The line and parabola meet where $x^2=x+2$ (Eliminating t)

$\therefore x^2-x-2=0\Rightarrow (x-2)(x+1)=0\Rightarrow x=2or-1$

When x = 2, y = 2 + 2 = 4

When x = - 1, y = (-1) + 2 = 1

Thus, the point of intersection of the parabola and line are (-1, 1) and (2, 4). Required area = (Area under the line y = x + 2, between x = - 1, x = 2) - (Area under the curve $x^2=y$, between x =-1, x = 2)

$={\int }_{-1}^2(x+2)dx-{\int }_{-1}^2{x}^{2}dx=[\frac{x^2}{2}2+2x{]}_{-1}^2-[\frac{x^3}{3}{]}_{-1}^2=\frac{2^2}{2}+2\times 2-[\frac{(-1{)}^2}{2}+2(-1\left)\right]-\frac{1}{3}(2^3-(-1{)}^3)=6-\frac{1}{2}+2-3=\frac{9}{2}squnit$