The length #(s)# along a curve #(y)# from #a# to #b# is given by:
#s = int_a^b sqrt(1+(dy/dx)^2) dx#
In this example #y = sqrt(4-x^2)#
#dy/dx = 1/2(4-x^2)^(-1/2) * (-2x)#
#= (-x)/sqrt(4-x^2)#
#:. s = int_0^1 sqrt((1+x^2/(4-x^2))) dx #
#= int_0^1 sqrt((4-x^2+x^2)/(4-x^2)) dx #
#= int_0^1 2/sqrt(4-x^2) dx#
Let #u = x/2 -> dx =2du#
#s= int_0^(1/2) (2xx2)/sqrt(4-4u^2) du#
#= 4/2*int_0^(1/2) 1/sqrt(1-u^2) du#
Apply standard integral
#s = 2* [arcsinu]_0^(1/2)#
#= 2 [arcsin(1/2) - arcsin (0)]#
#= 2 * [pi/6 -0] = pi/3#
#approx 1.0472#
Hence the length of #y# from #0# to #1# is #pi/3# units.
