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#### Find the area of region bounded by the line x = 2 and the parabola $y^2 = 8x$

The equation $y\textsuperscript{2} = 8x$  is a parabola not defining negative values of x, therefore it lies to the right of the Y axis passing through origin.

And x = 2 is a straight line parallel to Y axis.

Below figure shows the area to be calculated.

We have to integrate $y\textsuperscript{2} = 8x \\$

$\\Y = 2 \sqrt 2 \sqrt x \text{ from 0 to 2} \\$

On integrating the above equation from 0 to 2 it would give area enclosed under quadrant 1 only. Therefore, for finding the area in quadrant 2 we have to multiply it by 2 since it is symmetrical.

Therefore the area ODBC = 2 x area OBC  $\ldots .(1) \\$

let us find area under parabola

$\\ \Rightarrow y^{2}=8 x\\ \Rightarrow y=2 \sqrt{2} \sqrt{x}\\ \text{Integrate from 0 to 2}\\ \Rightarrow \int_{0}^{2} y d x=\int_{0}^{2} 2 \sqrt{2} x^{\frac{1}{2}} d x\\ \Rightarrow \int_{0}^{2} \mathrm{ydx}=2 \sqrt{2}\left[\frac{\mathrm{x}^{ \frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{2}\\ \Rightarrow \int_{0}^{2} \mathrm{ydx}=2 \sqrt{2}\left[\frac{\mathrm{x}^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{2} \\$

\\\begin{aligned} &\Rightarrow \int_{0}^{2} \mathrm{ydx}=2 \sqrt{2} \frac{2}{3}\left[\mathrm{x}^{\frac{3}{2}}\right]_{0}^{2}\\ &\Rightarrow \int_{0}^{2} y d x=\frac{4}{3}\left(2^{\frac{1}{2}}\right)\left(2^{\frac{3}{2}}-0\right)\\ &\Rightarrow \int_{0}^{2} y d x=\frac{4}{3}\left(2^{\frac{1}{2}+\frac{3}{2}}\right)\\ &\Rightarrow \int_{0}^{2} y d x=\frac{4}{3}\left(2^{2}\right)\\ &\Rightarrow \int_{0}^{2} \mathrm{ydx}=\frac{16}{3}\\ &\Rightarrow \operatorname{areaOBC}=\frac{16}{3}\\ &\text { Using (i) } \end{aligned} \\