Question

The area ( in $\text{sq units}$ ) of the region bounded by the curves $2x=y^2-1$ and $x=0$ is:

• A. $\frac{1}{3}\text{sq units}$
• B. $\frac{2}{3}\text{sq units}$
• C. $1\text{sq units}$
• D. $2\text{sq units}$

Answer 1

The correct option is B

$\frac{2}{3}\text{sq units}$

Explanation for the correct option:

Step 1: Finding the value of $x$ and $y$

The given equations are:

$$\begin{gathered} 2x=y^2-1......\left(1\right) \\ x=\frac{y^2-1}{2} \end{gathered}$$

and,

$x=0.......\left(2\right)$

Substituting $x=0$ into the equation $\left(1\right)$ we get:

$$\begin{gathered} \Rightarrow 0=y^2-1 \\ \Rightarrow y^2=1 \\ \Rightarrow y=\pm 1 \end{gathered}$$

Hence, the two intersecting points are $\left(0,1\right)$ and $\left(0,-1\right)$.

The graph is shown as,

Step 2: Finding the area of the region bounded by the curves

The area of the shaded region is,

$$\begin{gathered} A={\int }_{-1}^{1}xdy \\ A={\int }_{-1}^1\left(\frac{y^2-1}{2}\right)dy \end{gathered}$$

We know that,

${\int }_{-a}^{a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dx$

Therefore, the area will be as follows:

$$\begin{gathered} A=2{\int }_0^1\left(\frac{y^2-1}{2}\right)dy \\ A=\frac{2}{2}\times {\left[\frac{y^3}{3}-y\right]}_0^1 \\ A=\left(\frac{1}{3}-1\right)-\left(0-0\right) \\ A=\frac{1}{3}-1 \\ A=\frac{-2}{3} \end{gathered}$$

The area can never be a negative value. Hence the area of the bounded region is equal to $\frac{2}{3}\text{sq units.}$

Therefore, option (B) is the correct answer.