Can someone recheck this answer as well?

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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'=<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.


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