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

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

Answers (1)

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

So we have to integrate y=2 \sqrt{x} from 0 to 1 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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