Find the number b such that the line y = b divides the region bounded by the curves x = y^2 - 1 and the y-axis into 2 regions with equal area. Give your answer correct to 3 decimal places.

Because of the symmetry, we can just go from x=0 to x=2 to find the area between

y = x^2 and y = 4

that area = ∫4-x^2 dx from 0 to 2

= [4x - (1/3) x^3] from 0 to 2

= 8 - 8/3 - 0

= 16/3

so when y = b

x = √b

and we have the area as

∫ (b - x^2) dx from 0 to √b

= [b x - (1/3) x^3] from 0 to √b

= b√b - (1/3) b√b - 0

(2/3) b√b = 8/3

b√b = 4

square both sides

b^3 = 16

b = 16^ (1/3) = 2 cuberoot (2)

or appr 2.52