#### Find the area of the region $y=\sqrt x$ bounded by  and y = x.

\begin{aligned} &y=\sqrt{x}\\ &\text { squaring both sides }\\ &\Rightarrow y^{2}=x \end{aligned} \\

This is a parabola, no negative values of x, therefore it lies on the right of Y axis passing through origin.

Now  $y=\sqrt{x}$  means  y  and  x  both has to be positive hence both lie in  $1^{\text {st }}$  quadrant hence  $y=\sqrt{x}$  will be part of  $\mathrm{y}^{2}=\mathrm{x}$ which is lying only in  $1^{\text {st }}$  quadrant

And  y=x  is a straight line passing through origin

We have to find area between  $y=\sqrt{x}$ and  y=x  shown below

For finding the point of interaction, solve two equations simultaneously.
$\\Put y=x in y^{2}=x \\ \Rightarrow x^{2}=x \\ \Rightarrow x=1 \\$

Put  x=1  in  y=x  we get  y=1

The point of interaction is (1, 1)

Area between the parabolic curve and line = area under parabolic curve – area under line   $\ldots (1) \\$

For the area under parabolic curve

$\item Y = \sqrt x \\$

Integrating from 0 to 1

\begin{aligned} &\Rightarrow \int_{0}^{1} \mathrm{ydx}=\int_{0}^{1} \mathrm{x}^{\frac{1}{2}} \mathrm{~d} \mathrm{x}\\ &\Rightarrow \int_{0}^{1} \mathrm{ydx}=\left[\frac{\mathrm{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{1}\\ &\Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=\left[\frac{\mathrm{x}^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{1}\\ &\Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=\frac{2}{3}\left[1^{\frac{3}{2}}-0\right]\\ &\Rightarrow \int_{0}^{1} \mathrm{ydx}=\frac{2}{3} \end{aligned} \\

For area under straight line y = x

On integrating from 0 to 1
$\Rightarrow \int_{0}^{1} \mathrm{ydx}=\int_{0}^{1} \mathrm{xdx} \\ \Rightarrow \int_{0}^{1} y d x=\left[\frac{x^{2}}{2}\right]_{0}^{1} \\ \Rightarrow \int_{0}^{1} \mathrm{ydx}=\left(\frac{1^{2}}{2}-0\right) \\ \Rightarrow \int_{0}^{1} y d x=\frac{1}{2} \\$

Using (i)

$\Rightarrow \text{area between parabolic } = \frac{2}{3}-\frac{1}{2}= \frac{1}{6} unit ^{2} \\$

Hence area bounded is  $\frac{1}{6} unit ^{2} \\ \\$