The area bounded by the curve $y^2=2 x+1$ and the straight line $x-y-1=0$ is given by

• A

  $\frac{9}{2}$ sq units
• B

  $\frac{43}{6}$ sq units
• C

  $\frac{35}{6}$ sq units
• Correct answer
  D

  None of these

Solution

The point of intersection of curve $y^2=2 x+1$ and line $x-y-1=0$ is $(14,3)$ and $(0,-1)$

$\therefore$ Required area $=\int_{-1}^3\left(x_2-x_1\right) d y$ $\begin{aligned} & =\int_{-1}^3\left\{(y+1)-\frac{\left(y^2-1\right)}{2}\right\} d y \\ & =\frac{1}{2} \int_{-1}^3\left(2 y+3-y^2\right) d y \\ & =\frac{1}{2}\left[y^2+3 y-\frac{y^3}{3}\right]_{-1}^3 \\ & =\frac{1}{2}\left[(9+9-9)-\left(1-3+\frac{1}{3}\right)\right] \\ & =\frac{1}{2}\left(9+\frac{5}{3}\right) \\ & =\frac{16}{3} \mathrm{sq} \text { units }\end{aligned}$

Hence, the correct answer is option: 'D'.