Question

The area (in sq units) bounded by the curves $y=\sqrt{x}$, $2y-x+3=0$, $x$-axis and lying in the first quadrant is

• A. $9$
• B. $6$
• C. $18$
• D. $\frac{27}{4}$

Answer 1

The correct option is A

$9$

Explanation for Correct Answer:

Determining the area bounded by the curves.

We have $y=\sqrt{x}$ , given

therefore on squaring we get $y^2=x$

Similarly, We have

$$\begin{gathered} 2y-x+3=0...\left(1\right)\left[\text{Given}\right] \\ \Rightarrow x=2y+3 \end{gathered}$$

Then the point of intersection of the two curves can be obtained by, substituting equation $y=\sqrt{x}$ in equation $\left(1\right)$.

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

$$\begin{gathered} \Rightarrow \sqrt{x}=3\text{or}\sqrt{x}=-1\text{(rejected as it lies in first quadrant only)} \\ \therefore \sqrt{x}=3 \end{gathered}$$

$$\begin{gathered} y=\sqrt{x}\text{[Given]} \\ =3\left[\because \sqrt{x}=3\right] \end{gathered}$$

Now the graph of the given curves is mentioned here.

the area of the bounded region is:

$\begin{array}{l}A={\int }_0^3\left((2y+3)-y^2\right)dy\\ ={\int }_0^{3}2ydy+{\int }_0^{3}3dy-{\int }_0^3{y}^{2}dy\\ =2{\left[\frac{y^2}{2}\right]}_0^3+3[y{]}_0^3-{\left[\frac{y^3}{3}\right]}_0^3\\ =2\left[\frac{3^2}{2}-0\right]+3[3-0]-\left[\frac{3^3}{3}-0\right]\\ =9+9-\frac{27}{3}\\ =9+9-9\\ =9\end{array}$

Therefore the area bounded by the given curve is equal to $9\text{square units}$

Hence, the correct answer is option (A).