Question

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

• A. 4/3
• B. 8/3
• C. 16/3
• D. None of these

Answer 1

The correct option is C 16/3

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

$y^2=2x+1$

$\Rightarrow y^2=2(1+y)+1$

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

$\Rightarrow y^2-3y+y-3=0$

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

$\therefore$ $y=3$ or $-1$

$x=4$ or $0$

$\therefore$ the curves $y^2=2x+1$ & $x-y=1$ intersect at $(4,3)$ & $(0,-1)$ and we have find the area enclosed by the curves $y^2=2x+1$ & $x-y=1$. From the graph above, it is the shaded region which we have to find

$\therefore$ area enclosed by the graph is $A={\int }_{-1}^3\{(y+1)-\frac{1}{2}(y^2-1)\}dy$

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

$={\int }_{-1}^3\{y-\frac{y^2}{2}+\frac{3}{2}\}dy={\{\frac{y^2}{2}-\frac{y^3}{6}+\frac{3y}{2}\}}_{-1}^3=\frac{16}{3}$ sq units.