Practicing Success
The area of the region bounded by $y^2=2x+1$ and $x-y = 1 $ |
3 sq.units 16 sq.units $\frac{14}{3}$ sq.units $\frac{16}{3}$ sq.units |
$\frac{16}{3}$ sq.units |
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 |