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Question

# The area of the region bounded by parabola y2 = x and θ the straight line 2y = x is $\left(\mathrm{a}\right)\frac{4}{3}\mathrm{sq}.\mathrm{units}\phantom{\rule{0ex}{0ex}}\left(\mathrm{b}\right)1\mathrm{sq}.\mathrm{units}\phantom{\rule{0ex}{0ex}}\left(\mathrm{c}\right)\frac{2}{3}\mathrm{sq}.\mathrm{units}\phantom{\rule{0ex}{0ex}}\left(\mathrm{d}\right)\frac{1}{3}\mathrm{sq}.\mathrm{units}$

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Solution

## To find: area of the region bounded by parabola y2 = x and the straight line 2y = x y2 = x ..(1) 2y = x ..(2) Solving (1) and (2), we find the coordinates of the point of intersection A. i.e., A(4, 2) The area of the shaded region OAO = (Area of the upper curve − Area of the lower curve) Thus, $\mathrm{Required}\mathrm{area}={\int }_{0}^{4}\sqrt{x}dx-{\int }_{0}^{4}\frac{x}{2}dx\phantom{\rule{0ex}{0ex}}={\left(\frac{{x}^{\frac{3}{2}}}{\frac{3}{2}}\right)}_{0}^{4}-{\left(\frac{{x}^{2}}{4}\right)}_{0}^{4}\phantom{\rule{0ex}{0ex}}=\frac{2}{3}{\left({x}^{\frac{3}{2}}\right)}_{0}^{4}-\frac{1}{4}{\left({x}^{2}\right)}_{0}^{4}\phantom{\rule{0ex}{0ex}}=\frac{2}{3}\left({4}^{\frac{3}{2}}-{0}^{\frac{3}{2}}\right)-\frac{1}{4}\left({4}^{2}-{0}^{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{2}{3}\left({2}^{3}\right)-\frac{1}{4}\left(16\right)\phantom{\rule{0ex}{0ex}}=\frac{16}{3}-4\phantom{\rule{0ex}{0ex}}=\frac{16-12}{3}\phantom{\rule{0ex}{0ex}}=\frac{4}{3}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{Area}=\frac{4}{3}\mathrm{sq}.\mathrm{units}$ Hence, the correct option is (a).

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