As this curve is 1-t, (1-t)^21−t,(1−t)2 (the xx coordinate is 1-t1−t, rather than t-1t−1 because the former is positive in [0,1][0,1], the curve is the same as the segment of the parabola y=x^2y=x2 between (0,0)(0,0) and (1,1)(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} dxL=∫10√1+(dydx)2dx=∫10√1+4x2dx=2∫10√x2+122dx
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 }|∫√x2+a2dx=12x√x2+a2+a22ln∣∣x+√x2+a2∣∣
this becomes
L = x sqrt{x^2+1/4}+1/4 ln| x+sqrt{x^2+1/4}||_0^1 L=x√x2+14+14ln∣∣∣x+√x2+14∣∣∣∣∣∣10
= sqrt{5/4}+1/4ln(1+sqrt{5/4})+1/4 ln 2=√54+14ln(1+√54)+14ln2
= sqrt{5/4}+1/4ln(2+sqrt{5}) approx 1.47894=√54+14ln(2+√5)≈1.47894
