Question

Find the area of the region bounded by $y=\sqrt{x},x=2y+3$ in the first quadrant and x-axis. [NCERT EXEMPLAR]

Answer 1

The curve $y=\sqrt{x}$ or $y^2=x$ represents the parabola opening towards the positive x-axis.

The curve x = 2y + 3 represents a line passing through (3, 0) and $\left(0,-\frac{3}{2}\right)$.

Solving $y^2=x$ and x = 2y + 3, we get

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



∴ Required area = Area of the shaded region

$$\begin{gathered} ={\int }_0^3{x}_{\mathrm{line}}dy-{\int }_0^3{x}_{\mathrm{parabola}}dy \\ ={\int }_0^3\left(2y+3\right)dy-{\int }_0^3{y}^{2}dy \\ ={\overline{)\frac{{\left(2y+3\right)}^2}{2\times 2}}}_0^3-{\overline{)\frac{y^3}{3}}}_0^3 \\ =\frac{1}{4}\left[{\left(2\times 3+3\right)}^2-3^2\right]-\frac{1}{3}\left(3^3-0\right) \\ =\frac{1}{4}\left(81-9\right)-\frac{1}{3}\left(27-0\right) \\ =18-9 \\ =9\mathrm{square}\mathrm{units} \end{gathered}$$