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
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.
1 answer
1 answer