Well, the answer of Tasneem is cool, but I suppose not very helpful.
You have a vector X[ t ], its variation is d X[ t ] = D[ X[ t ], t ] dt, where D[..., t ] ist the differentiation with repect to t. The length of this variation is Sqrt[ d X[ t ] . d X[ t ] ] dt , where the . indicates the scalar product of two vectors. This expression ist to be integrated between t1 and t2 to get the length you are looking for.
In[9]:= X[t] = { t^2, (1 - t)^2, 2 t};
dx = D[X[t], t]
dx2 = dx.dx // Expand
jj = Sqrt[dx2]
arc = Integrate[jj, {t, t1, t2}]
Out[10]= {2 t, -2 (1 - t), 2}
Out[11]= 8 - 8 t + 8 t^2
Out[12]= Sqrt[8 - 8 t + 8 t^2]
Out[13]= If[((Im[t2] Re[t1] <= Im[t1] Re[t2] &&
Im[t1] >= Im[t2]) || (Im[t2] Re[t1] >= Im[t1] Re[t2] &&
Im[t1] <= Im[t2])) && ((Im[t2] Re[t1] -
Im[t1] Re[t2])^2/(Re[t1] - Re[t2])^2 ==
1 || (Im[t2] Re[t1] - Im[t1] Re[t2])^2/(Re[t1] - Re[t2])^2 <=
1), -((2 (-1 + 2 t1) Sqrt[1 - t1 + t1^2] -
3 ArcSinh[(1 - 2 t1)/Sqrt[3]])/(2 Sqrt[2])) + (
2 (-1 + 2 t2) Sqrt[1 - t2 + t2^2] - 3 ArcSinh[(1 - 2 t2)/Sqrt[3]])/(
2 Sqrt[2]),
Integrate[Sqrt[8 - 8 t + 8 t^2], {t, t1, t2},
Assumptions -> ! (((Im[t2] Re[t1] <= Im[t1] Re[t2] &&
Im[t1] >= Im[t2]) || (Im[t2] Re[t1] >= Im[t1] Re[t2] &&
Im[t1] <= Im[t2])) && ((Im[t2] Re[t1] -
Im[t1] Re[t2])^2/(Re[t1] - Re[t2])^2 ==
1 || (Im[t2] Re[t1] - Im[t1] Re[t2])^2/(Re[t1] - Re[t2])^2 <=
1))]]