I should calculate
$$\int (-y+\sin x^2)dx + xdy$$
on the curve $c=c_{1}+c_{2}-c_{3}-c_{4}$ where it doesn't give me any parametrisation mappings; only the normal $$ \begin{cases} c_{1}:& [0,1] \rightarrow R^{2}, c_{1}(t)= (t,0) ,\\ c_{2}:& [0,1] \rightarrow R^{2}, c_{2}(t)= (1,t) ,\\ c_{3}:& [0,1] \rightarrow R^{2}, c_{3}(t)= (t,1) ,\\ c_{4}:& [0,1] \rightarrow R^{2}, c_{4}(t)= (0,t) . \end{cases} $$ should I just use them, for example, for $c_{1}$: $\int^{1}_{0} \sin(t^2)dt $ or am I missing something here?
With direct way $$\int_{c}(-y+\sin x^2)dx + xdy=\int_{c_1}\sin t^2dt + \int_{c_2}dt - \int_{c_3}(-1+\sin t^2)dt - \int_{c_4}0dt=2$$ where all integrals are on $[0,1]$.
With Green's theorem $$\oint_{\partial D}(P\ dx+Q\ dy)=\iint_D \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dx\ dy$$ where $D$ is the area of the region, that is $=1$, then $$\int_{c}(-y+\sin x^2)dx + xdy=\int_{D}\ 2 dx\ dy=2$$