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The parameter $t$ will run from 0 to 1 throughout.

We can parameterize the segment from (0,0) to (1,0) by $x=t,y=0$ , so $dy=0$ and the line integral is 0.

Next, from (1,0) to (1,1), we can let $x=1,y=t$ and $\int_{0}^{1}dt=1$ .

Finally, we can parameterize the opposite of the path from (1,1) to (0,0) by $x=y=t$ so $\int_{0}^{1}2t^2dt=\frac{2}{3}$ , for a grand total of $0+1-\frac{2}{3}=\frac{1}{3}\approx \boxed{0.333}$