In what ratio does the x-axis divide the area of the region bounded by the parabolas `y=4x-x^(2) and y=x^(2)-x`?

Answer

`121:4`

Explanation

Given parabolas are ` y=4x-x^(2)`
and `y=-(x-2)^(2)+4`
or `(x-2)^(2)=-(y-4)`
Therefore, it is a vertically downward parabola with vertex at (2,4) and its axis is `x=2`
and ` y=x^(2)-x rArr y=(x-(1)/(2))^(2)-(1)/(4) `

` rArr (x-(1)/(2))^(2)=y+(1)/(4)`
This is a paraboal having its vertex at `((1)/(2),-(1)/(4)). `
Its axis is at `x=(1)/(2)` and opening upwards.
The points of intersection of given curves are
` 4x-x^(2)=x^(2)-x rArr 2x^(2)=5x `
` rArr x(2-5x)=0 rArr x=0,(5)/(2) `
Also, ` y=x^(2)-x ` meets X-axis at (0,0) and (1,0).
`therefore " Area, " A_(1)=int_(0)^(5//2)[(4x-x^(2))-(x^(2)-x)]dx `
`=int_(0)^(5//2)(5x-2x^(2)) dx `
`=[(5)/(2)x^(2)-(2)/(3) x^(3) ]_(0)^(5//2)=(5)/(2)((5)/(2))^(2)-(2)/(3)*((5)/(2))^(3)`
` =(5)/(2) *(25)/(4)-(2)/(3)*(125)/(8)`
`=(125)/(8) (1-(2)/(3))=(125)/(24)` sq units
This area is considering above and below X-axis both. Now, for area below X-axis separately, we consider
`A_(2) = - int_(0)^(1)(x^(2)-x)dx=[(x^(2))/(2)-(x^(3))/(3)]_(0)^(1) = (1)/(2)-(1)/(3)=(1)/(6)` sq units
Therefore, net area above the X-axis is
` A_(1)-A_(2)=(125-4)/(24)=(121)/(24)` sq units
Hence, ratio of area above the X-axis and area below X-axis
` =(121)/(24):(1)/(6)=121: 4`