Question

Find the area of the region bounded by $y=\sqrt{x}$ and y = x. [NCERT EXEMPLAR]

Answer 1

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

The curve y = x represents a line passing through the origin.

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

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

Thus, the given curves intersect at O(0, 0) and A(1, 1).



∴ Required area = Area of the shaded region OAO

$$\begin{gathered} ={\int }_0^1{y}_{\mathrm{parabola}}dx-{\int }_0^1{y}_{\mathrm{line}}dx \\ ={\int }_0^1\sqrt{x}dx-{\int }_0^{1}xdx \\ ={\overline{)\frac{x^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}}}_0^1-{\overline{)\frac{x^{{\displaystyle 2}}}{{\displaystyle 2}}}}_0^1 \\ =\frac{2}{3}\left(1-0\right)-\frac{1}{2}\left(1-0\right) \\ =\frac{2}{3}-\frac{1}{2} \\ =\frac{1}{6}\mathrm{square}\mathrm{units} \end{gathered}$$