Question

Choose the correct answer in the following questions

Area lying between the curves $y^2=4x$ and y = 2x is

(a) $\frac{2}{3}$ (b) $\frac{1}{3}$

(c) $\frac{1}{4}$ (d) $\frac{3}{4}$

Answer 1

The given curve is $y^2=4x$ ...(i)

and the given line is y = 2x ....(ii)

The two curve meet where $(2x{)}^2=4x(x-1)=0\Rightarrow x=0,1$

When $x=0,yy=2\times 0=0$ and when $x=1,y=2\times 1=2$

$\therefore$ Eqs. (i) and (ii) meet at the points (0, 0) and (1, 2).

Required area $={\int }_0^1(2\sqrt{x}-2x)dx$

$(\because y^2=4x\Rightarrow |y|=2\sqrt{x})$

$=[2\frac{x^{\frac{3}{2}}}{\frac{3}{2}}{]}_0^1=\frac{4}{3}.1^{\frac{3}{2}}-1^2-0=\frac{1}{3}squnit$

Hence, the correct option is (b).