Find the number a such that the line x = a 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.

Parabola with vertex at x =-1 and y = 0 then proceeding right above and below the x axis  passing through (0,1) and (0,-1)  ∫ left of x = 0 = ∫ right of x = 0  because of symmetry we only need to do + y  y = +/-sqrt (x+1) = +/-(x+1)^.5  ∫ y dx from -1 to 0  = ∫y dx from 0 to a  ∫ y dx = ∫(x+1)^.5 dx  = (x+1)^1.5 / 1.5  at x = -1 that is 0  at x = 0 that is 1/1.5 = 2/3  so  we need to select upper limit of x = a to get the same area from 0 to a  at x = a  ∫ is (a+1)^1.5/1.5  at x = 0 we know it is 2/3  (a+1)^1.5 / (3/2) - 2/3 = 2/3  (a+1)^1.5 / (3/2) = 4/3  (a+1)^1.5 = 2  1.5 log (a+1) = .301  log (a+1) = .2  a+1 = 1.584  a = .584