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Given Curves are

y^{2} = x — (i)

x = 0 — (ii)

x = 2 — (iii)

Clearly (i) is a parabola that has a verticex at (0, 0) and the line of symmetry is y = 0 with x \(\geq\) 0.

Also, (ii) and (iii) are the straight lines that are parallel to y-axis. With these information, the sketch in figure, in which the bounded region by (i), (ii), (iii) is the shaded portion.

Clearly the region has two symmetrical parts I and II. From figure, part – I, x moves from 0 to 2.

Now, the area of the bounded region is

\(A = 2\left | \int_{x=0}^{2} y dx \right | \)

= \(2\left | \int_{0}^{2} \sqrt{x} dx \right | \)

= \(2\left [ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right ]^2_0\)

= \(\frac{4}{3} (2\sqrt{2})\)

= \(\frac{8\sqrt{2}}{3}\)

Thus, the area of the region is \(\frac{8\sqrt{2}}{3}\) sq. units

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