The answer turned out to be #2pi# units.
The arc length is essentially the usage of the distance formula with a small, independent change in x and a small change in y, where y changes according to #f(x)#. Thus:
#D = sqrt((Deltax)^2 + (Deltay)^2)#
#= sqrt((Deltax)^2 + (Deltay)^2/(Deltax)^2*(Deltax)^2)#
#= sqrt(1 + (Deltay)^2/(Deltax)^2)Deltax#
Make it a sum, and you've got the arc length over an interval #[a,b]#.
#s = sum_a^b sqrt(1 + (Deltay)^2/(Deltax)^2)Deltax#
Turn it into an integral expression to get:
#s = int_a^b sqrt(1 + ((dy)/(dx))^2)dx#
So, we could take the derivative first to get:
#(dy)/(dx) = 1/(2sqrt(4-x^2))*-2x = (-x)/(sqrt(4-x^2))#
Then, let's square it and plug it in.
#((dy)/(dx))^2 = x^2/(4-x^2)#
#-> s = int_a^b sqrt(1 + x^2/(4-x^2))dx#
Cross-multiply:
#= int_a^b sqrt((4-x^2 + x^2)/(4-x^2))dx#
#= int_a^b sqrt((4)/(4-x^2))dx#
#= 2int_a^b 1/sqrt(4-x^2)dx#
And we've got ourselves a u-substitution, if you know that #d/(dx)[arcsin(x)] = 1/sqrt(1-x^2)#.
#= 2int_a^b 1/sqrt4*1/sqrt(1-x^2/4)dx#
#= int_a^b 1/sqrt(1-(x/2)^2)dx#
Let #u = x/2# and you get:
#dx = 2du#
#= 2int_a^b 1/sqrt(1-u^2)du#
#= [2 arcsin(x/2)]|_(-2)^(2)#
#= 2 arcsin((2)/2) - 2 arcsin((-2)/2)#
#= 2 arcsin(1) - 2 arcsin(-1)#
Let:
#y = arcsin(1) -> siny = 1 -> y = pi/2#
#y = arcsin(-1) -> siny = -1 -> y = (3pi)/2#
#-> 2(pi/2) - 2((3pi)/2) -> 2(pi/2) - 2((-pi)/2)#
(so that our answer is positive)
#= pi + pi = color(blue)(2pi " u")#
