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Area lying between the curves y^2=4xand...

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

Text Solution

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curve `y^2=4x` is right parabola with vertex(0,0)
`y=2x` is line with pts `(0,0),(1,2),(-1,-2)`
they meet at pts
`y^2=4x` but `y=2x`
`(2x)^2=4x`
`4x^2=4x`
`x=1,y=2(1,2)`

`x=0,y=0(0,0)`
Required area is shaded area
=[Area under curve `y^2=4x` from `x=0` to `x=1`]-[Area under line `y=2x` from `x =0` to `x=1`]
`int_0^1(sqrt(4x))dx-int_0^1(2x)dx`
`2[int_0^1(sqrtx)dx-int_0^1xdx]`
`[x^(3/2)/(3/2)]_0^1-[x^2/2]_0^1`
`2[2/3-1/2]=4/3-1=1/3`sq units
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