r'(t) = \langle 1, 2t \rangle. \mathbf{F}(r(t)) = \langle 3t^2, 2t^2 \rangle.

\displaystyle\int_C \mathbf{F} \cdot d\mathbf{r}= \int_0^1 \mathbf{F}(r(t)) \cdot r'(t)\, dt= \int_0^1 (3t^2 \cdot 1 + 2t^2 \cdot 2t)\, dt

\displaystyle\int_0^1 (3t^2 + 4t^3)\, dt= [t^3 + t^4]_0^1= 1 + 1 = 2