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  • Calculate the area under the curve y=2\sqrtx included between the lines x = 0 and x = 1.

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

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\\$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}

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