`y^2 = 4-x , x = 0` Find a such that the line x = a divides the region bounded by the graphs of the equations into two regions of equal area.

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Given ,

`y^2=4-x , x = 0`

=>`x=4-y^2 , x=0`

first let us find the total area of the bounded by the curves.

so we shall proceed as follows

`x=4-y^2 ,x=0`

=>` 4-y^2=0`

=> `y^2 -4 =0`

=>` (y-2)(y+2)=0`

so `y=+-2`


the the area of the region is = `int _-2 ^2 ((4-y^2)-0 ) dy`

=>`int _-2 ^2 (4-y^2) dy`

=`[4y-y^3/3] _-2 ^2`

=`[ [8-8/3]-[-8 -(-8)/3]]`

=`[[8-8/3]+[8 +(-8)/3]] = (16-16/3)=(2*16)/3=32/3`

So now we have  to find the vertical line that splits the region into two regions with area `16/3` as it is half of area of region covered by two curves `y^2=4-x ` and `x=0.`

as when the line x=a intersects the curve `x=4-y^2` then the area bounded is `16/3` ,so

let us solve this as follows

first we shall find the intersecting points

as ,




so the area bound by these curves `x=a` and `x=4-y^2 ` is as follows

A= `int _-sqrt(4-a) ^sqrt(4-a) (4-y^2-a)dy = 16/3`

=> `int _-sqrt(4-a) ^sqrt(4-a)(4-y^2-a)dy=16/3`

=> `[(4-a)(y)-y^3/3]_-sqrt(4-a) ^sqrt(4-a)`


let `t= sqrt(4-a)`



=>`[t^3-t^3/3 +t^3-t^3/3]`




but we know half the area of the region between `x=4-y^2, x=0` curves =`16/3`

so now ,

`4/3 t^3=16/3`

=>`4t^3 = 16`


Substituting `t = sqrt(4-a)` , 

`(4-a)^(3/2)= 4`





so `a= 1.4801`

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