Question

The area of the region bounded by the curves ${\mathrm{y}}^2=2\mathrm{x}+1$and$\mathrm{x}-\mathrm{y}=1$ is

• A. $\frac{2}{3}$
• B. $\frac{4}{3}$
• C. $\frac{8}{3}$
• D. $\frac{11}{3}$
• E. None of these

Answer 1

The correct option is E

None of these

Determining the area bounded by the curves:

Given, $y^2=2x+1,x-y=1$

Area of bounded region is the area of shaded region as shown:

For intersection points, we need to solve both the curves,

$$\begin{gathered} \Rightarrow y^2=2x+1 \\ \Rightarrow y^2=2(1+y)+1 \\ \Rightarrow y^2-2y-3=0 \\ \Rightarrow (y-3)(y+1)=0 \\ \therefore y=3,-1 \end{gathered}$$

On putting value of $y$ in the equation, we get $x=4,0$

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

Area enclosed is equal to upper curve-lower curve, $A={\int }_{-1}^3\left\{(y+1)-\frac{1}{2}\left(y^2-1\right)\right\}dy$
$$\begin{gathered} ={\int }_{-1}^3\left\{y+1-\frac{y^2}{2}+\frac{1}{2}\right\}dy \\ ={\int }_{-1}^3\left(y-\frac{y^2}{2}+\frac{3}{2}\right)dy \\ ={\left[\frac{y^2}{2}-\frac{y^3}{6}+\frac{3y}{2}\right]}_{-1}^3 \\ =\left(\frac{9}{2}-\frac{27}{6}+\frac{9}{2}\right)-\left[\frac{1}{2}+\frac{1}{6}-\frac{3}{2}\right] \\ =\frac{16}{3}\text{sq.units} \end{gathered}$$
Hence, the correct answer is option (E).