The area included between $y^2=4x, y=x+1$ and x-axis is :

$\frac{10}{3}$ sq.units 2 sq.units $\frac{1}{2}$ sq.units $\frac{2}{3}$ sq.units

$\frac{2}{3}$ sq.units

$y^2=4x, y=x+1$ finding intersection point $(x+1)^2=4x$ $x^2+2x+1=4x$ $x^2-2x+1=0$ $(x-1)^2=0$ $x=1$ so $y=2$ area = 0 as curves are non intersecting