The area of the region bounded by y^2=2x+1 and x-y = 1

The correct answer is Option (4) → $\frac{16}{3}$ sq.units

$y^2=2x+1$ and $x-y = 1$

intersection point

at $x=y+1$

so $y^2=2y+3$

$y^2-2y-3=0$

$(y-3)(y+1)=0$

$⇒y=-1,3$

so area = area I + area II

$=-\int\limits_{-1}^1\frac{y^2}{2}-\frac{1}{2}dy+\int\limits_{-1}^34+1dy-\int\limits_{1}^3\frac{y^2}{2}-\frac{1}{2}dy$

$=-\left[\frac{y^3}{6}-\frac{y}{2}\right]_{-1}^1+\left[\frac{y^2}{2}+1\right]_{-1}^3-\left[\frac{y^3}{6}-\frac{y}{2}\right]_{1}^3$

$=\frac{16}{3}$ sq. units