Solution
= Solution
For the curved path $r(t)=(t,t^2,t^3)$, $r'(t)=(1,2t,3t^2)$. The <vector field> along this <parametric curve> is
$$
v(r(t))=(t^2e^t,\ e^t+2t^{11},\ 3t^{10}).
$$
Its <dot product> with the tangent is $(t^2+2t)e^t+13t^{12}$. Therefore the direct <line integral> is
$$
\boxed{\int_Cv\cdot dr=\int_0^1\bigl[(t^2+2t)e^t+13t^{12}\bigr]\,dt=\left[t^2e^t+t^{13}\right]_0^1=e+1.}
$$