#### Calculate the area under the curve $y=2\sqrt x$  included between the lines x = 0 and x = 1.

$\\y=2 \sqrt{x}\\ \text{squaring both sides we get}\\ \Rightarrow y^{2}=4 x\\ \mathrm{y}^{2}=4 \mathrm{x} \text{ is a equation of parabola }\\\text{The equation show no negative values for x, therefore it lies to the right of Y axis passing through origin. }\\ \text{Now for} y=2 \sqrt x \text{ x and y both has to be greater than 0 that is both positive hence both lie in}\\ 1^{\text {st }} quadrant$

$\\ \text{Hence} \ y=2 \sqrt{x} \text{ will be parabolic curve of } y^{2}=4 x \ only in 1^{\text {st }} \text{quadrant}\\ \mathrm{x}=0 \text{is equation of} \mathrm{Y} -axis and \\\mathrm{x}=1\text{ is a line parallel to} \mathrm{Y} \text{-axis passing through (1,0)}\\ \text{Plot equations} y=2 \sqrt{x} and x=1$

$\\\text{So we have to integrate} y=2 \sqrt{x} from 0 to 1\\ \text{let us find area under parabola}$

$\\\Rightarrow y=2 \sqrt{x}$

$\text{Integrate from 0 to 1}$

$\Rightarrow \int_{0}^{1} \mathrm{ydx}=\int_{0}^{1} 2 \mathrm{x}^{\frac{1}{2}} \mathrm{dx}$

$\Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=2\left[\frac{\mathrm{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{1}$

\\\begin{aligned} &\Rightarrow \int_{0}^{1} \mathrm{ydx}=2\left[\frac{\mathrm{x}^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{1}\\ &\Rightarrow \int_{0}^{1} \mathrm{ydx}=2 \frac{2}{3}\left[\mathrm{x}^{\frac{3}{2}}\right]_{0}^{1}\\ &\Rightarrow \int_{0}^{1} \mathrm{ydx}=\frac{4}{3}\left(1^{\frac{3}{2}}-0\right)\\ &\Rightarrow \int_{0}^{2} \mathrm{ydx}=\frac{4}{3}\\ &\text { Therefore, the area found to be }\\ &\frac{4}{3} \text { unit }^{2} \end{aligned}