Area lying between the curves `y^2=4x`and `y = 2x`is(A) `2/3` (B) `1/3` (C) `1/4` (D) `3/4`

Options

A. `(2)/(3)`
B. `(1)/(3)`
C. `(1)/(4)`
D. `(3)/(4)`

Explanation

Given curve `y^(2)=4x " " ` … (1)
and given line `y=2x " " ` …(2)

` :. ` For their points of intersection,
`(2x)^(2)=4x`
`implies4x(x-1)=0`
`impliesx=0,1`
When `x=0, y=2xx0=0`
When `x=1, y=2xx1=2`
` :. ` Equations (1) and (2) intersect each other at points (0, 0) and (1, 2).
` :. ` Required area`=int_(0)^(1)(2sqrt(x)-2x)dx`
` " " ( :' y^(2)=4ximplies|y|=2sqrt(x))`
`=[2(x^(3//2))/(3//2)-x^(2)]_(0)^(1)`
`=(4)/(3)*1^(3//2)-1^(2)-0`
`=(1)/(3)` sq. units.