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

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