As this curve is #1-t, (1-t)^2# (the #x# coordinate is #1-t#, rather than #t-1# because the former is positive in #[0,1]#, the curve is the same as the segment of the parabola #y=x^2# between #(0,0)# and #(1,1)#. The length of this segment is
#L = int_0^1 sqrt{1+(dy/dx)^2} dx = int_0^1 sqrt{1+4x^2} dx = 2int_0^1 sqrt{x^2 +1/2^2} dx#
Using the standard integral
#int sqrt{x^2+a^2 }dx = 1/2 x sqrt{x^2 +a^2 }+ a^2/2 ln | x+sqrt{x^2+a^2 }|#
this becomes
#L = x sqrt{x^2+1/4}+1/4 ln| x+sqrt{x^2+1/4}||_0^1 #
# = sqrt{5/4}+1/4ln(1+sqrt{5/4})+1/4 ln 2#
# = sqrt{5/4}+1/4ln(2+sqrt{5}) approx 1.47894#
