Question

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

Answer 1

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,
$$\begin{gathered} \mathrm{Required}\mathrm{area}={\int }_0^4\sqrt{x}dx-{\int }_0^4\frac{x}{2}dx \\ ={\left(\frac{x^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}\right)}_0^4-{\left(\frac{x^2}{4}\right)}_0^4 \\ =\frac{2}{3}{\left(x^{\frac{3}{2}}\right)}_0^4-\frac{1}{4}{\left(x^2\right)}_0^4 \\ =\frac{2}{3}\left(4^{\frac{3}{2}}-0^{\frac{3}{2}}\right)-\frac{1}{4}\left(4^2-0^2\right) \\ =\frac{2}{3}\left(2^3\right)-\frac{1}{4}\left(16\right) \\ =\frac{16}{3}-4 \\ =\frac{16-12}{3} \\ =\frac{4}{3} \\ \mathrm{Thus},\mathrm{Area}=\frac{4}{3}\mathrm{sq}.\mathrm{units} \end{gathered}$$

Hence, the correct option is (a).