Question

Area lying between the curves y² = 4x and y = 2x is
(a) $\frac{2}{3}$

(b) $\frac{1}{3}$

(c) $\frac{1}{4}$

(d) $\frac{3}{4}$

Answer 1

$\left(\mathrm{b}\right)\frac{1}{3}$



The points of intersection of the straight line and the parabola is obtained by solving the simultaneous equations
$$\begin{gathered} y^2=4x\mathrm{and}y=2x \\ \Rightarrow {\left(2x\right)}^2=4x \\ \Rightarrow 4x^2=4x \\ \Rightarrow x\left(x-1\right)=0 \\ \Rightarrow x=0\mathrm{or}x=1 \\ \Rightarrow y=0\mathrm{or}y=2 \\ \mathrm{Thus}O\left(0,0\right)\mathrm{and}A\left(1,2\right)\mathrm{are}\mathrm{the}\mathrm{points}\mathrm{of}\mathrm{intersection}\mathrm{of}\mathrm{the}\mathrm{parabola}\mathrm{and}\mathrm{straight}\mathrm{line} \\ \mathrm{Shaded}\mathrm{area}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{area}. \\ \mathrm{Using}\mathrm{the}\mathrm{horizontal}\mathrm{strip}\mathrm{method}, \\ \mathrm{Shaded}\mathrm{area}={\int }_0^2\left|x_2-x_1\right|dy \\ ={\int }_0^2\left[\left(\frac{y}{2}\right)-\left(\frac{y^2}{4}\right)\right]dy \\ ={\left[\frac{1}{2}\left(\frac{y^2}{2}\right)-\frac{1}{4}\left(\frac{y^3}{3}\right)\right]}_0^2 \\ =\frac{1}{4}\left(2^2\right)-\frac{1}{12}\left(2^3\right)-0 \\ =1-\frac{8}{12} \\ =\frac{12-8}{12} \\ =\frac{1}{3}\mathrm{sq}.\mathrm{units} \end{gathered}$$