Question

The area of the figure bounded by $y^2=2x+1$ and $x-y-1=0$ is

• A. $\frac{16}{3}$
• B. $\frac{8}{3}$
• C. $\frac{4}{3}$
• D. None of these

Answer 1

The correct option is A $\frac{16}{3}$

Solving $y^2=2x+1$ and $x-y-1=0$

we get $\frac{y^2-1}{2}=y+1\Rightarrow y^2-2y-3=0$

$\Rightarrow (y-3)(y+1)=0\Rightarrow y=3,-1$

Area of strip is $={\int }_{-1}^3(y+1-\frac{y^2-1}{2})dy={[\frac{y^2}{2}-\frac{y^3}{6}+\frac{3}{2}y]}_{-1}^3$

$=(\frac{9}{2}-\frac{1}{2})-(\frac{27}{6}-\frac{-1}{6})+\frac{3}{2}[3-(-1)]=\frac{16}{3}$