Your answer checks out OK.
You could have included some logical steps in your solution such as where you derived values for x and why r'dt=dr. Since vector r=<x,y>=<t,t²> this implies that x=t, which is then substituted into vector F to complete the parameterisation.
r=<t,t²>,
r'=<1,2t>=dr/dt, so dr=<1,2t>dt.
<cos(t),sin(t)>・<1, 2t>=cos(t)+2tsin(t).
Line integral ∫ᶜF・dr=∫₋₁²(cos(t)+2tsin(t))dt.
Integrate 2tsin(t) by parts, u=2t, dv=sin(t)dt, du=2dt, v=-cos(t), etc.