Question

Area bounded by parabola y² = x and straight line 2y = x is
(a) 43
(b) 1
(c) 23
(d) 13

Answer 1

Disclaimer : Non of the given option is correct



Point of intersection is obtained by solving the equation of parabola y² = x and equation of line 2y = x, we have
$$\begin{gathered} y^2=x\mathrm{and}2y=x \\ \Rightarrow y^2=2y \\ \Rightarrow y^2-2y=0 \\ \Rightarrow y=0\mathrm{or}y=2 \\ \Rightarrow x=0\mathrm{or}x=4 \\ \mathrm{Thus}O\left(0,0\right)\mathrm{and}A\left(4,2\right)\mathrm{are}\mathrm{the}\mathrm{points}\mathrm{of}\mathrm{intersection}\mathrm{of}\mathrm{the}\mathrm{curve}\mathrm{and}\mathrm{straight}\mathrm{line}. \\ \mathrm{Area}\mathrm{bound}\mathrm{by}\mathrm{them} \\ A={\int }_0^4\left(y_1-y_2\right)dx\left[\mathrm{Where},y_1=\sqrt{x}\mathrm{and}{y}_2=\frac{x}{2}\right] \\ ={\int }_0^4\left(\sqrt{x}-\frac{x}{2}\right)dx \\ ={\left[\frac{x^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}-\frac{1}{2}\times \frac{x^2}{2}\right]}_0^4 \\ ={\left[\frac{2}{3}{x}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{x^2}{4}\right]}_0^4 \\ =\frac{2}{3}{4}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{1}{4}\times 4^2-0 \\ =\frac{2}{3}\times 2^3-\frac{16}{4} \\ =\frac{16}{3}-4 \\ =\frac{16-12}{3} \\ =\frac{4}{3}\mathrm{sq}\mathrm{units} \end{gathered}$$