You need to solve the inequalitites such that:
`x - 2y^2gt= 0 =gt x gt= 2y^2`
`1-x-|y| gt= 0 =gt -x gt= |y| - 1=gt x =lt 1 - |y|`
Hence, the region is bounded by 3 curves `x = 1+y, x=1-y, x=2y^2` .
You need to determine the points of intersection between curves `x = 1+y` and `x=2y^2` such that:
`1 + y = 2y^2 =gt 2y^2 - y - 1 = 0`
You need to solve quadratic equation such that:
`y_(1,2) = (1+-sqrt(1 + 8))/4`
`y_(1,2) = (1+-3)/4 =gt y_1 = 1`
`y_2 = -1/2`
You need to determine the points of intersection between curves `x= 1-y` and `x=2y^2` such that:
`1 - y = 2y^2 =gt 2y^2+ y - 1 = 0`
You need to solve quadratic equation such that:
`y_(1,2) = (-1+-sqrt(1 + 8))/4`
`y_(1,2) = (-1+-3)/4 =gt y_1 = -1`
`y_2 = 1/2`
Hence, the limits of integration are `-1/2` and `1/2` and you need to evaluate the area of the region modelled by function `1 - y - 2y^2` such that:
`int_(-1/2)^(1/2) (1 - y - 2y^2 ) dy`
Since the limits of integration ar opposed numbers, hence you need to evaluate the definite integral such that:
`2int_0^(1/2) (1 - y - 2y^2 ) dy = 2(y - y^2/2 - 2y^3/3)|_0^(1/2)`
`2int_0^(1/2) (1 - y - 2y^2 ) dy = 2(1/2 -1/8- 1/12)`
`2int_0^(1/2) (1 - y - 2y^2 ) dy = (12 - 3- 2)/12`
`2int_0^(1/2) (1 - y - 2y^2 ) dy = 7/12`
Hence, evaluating the area of the region under given conditions yields`int_(-1/2)^(1/2) (1 - y - 2y^2 ) dy = 7/12.`