Answer
$\int_C \textbf F \cdot d\textbf r = 0$
Work Step by Step
$\textbf r(t) = t \textbf i + t^2 \textbf j = x(t) \textbf i + y(t) \textbf j$
It follows that $x(t) = t , y(t) = t^2$
$\therefore \textbf r' = \textbf i +2 t \textbf j \\
\textbf F(x,y) = (x^3-2x^2)\textbf i + (x-\frac{y}{2}) \textbf j\\
\Rightarrow \textbf F (x(t),y(t))= [t^3- 2t^2] \textbf i + [t-\frac{t^2}{2}] \textbf j\\
\int_C \textbf F \cdot d\textbf r\\
= \int_C \textbf F \cdot \textbf r' dt\\
= \int ( [t^3- 2t^2] \textbf i + [t-\frac{t^2}{2}] \textbf j )\cdot ( \textbf i +2 t \textbf j ) dt\\
= \int (t^3-2t^2+2t^2-t^3) dt\\
=0$
(proved)
