Line integral of a differential form

Parametrization. $\gamma(t) = (t,t)$, $\gamma'(t) = (1,1)$, $t \in [0,1]$.

Substitution.
$$\int_\gamma \omega = \int_0^1 \left[ (t+t)\cdot 1 + (t-t)\cdot 1 \right] dt = \int_0^1 2t \, dt = [t^2]_0^1 = 1.$$

Remark. If $\omega$ were exact, the integral would depend only on the endpoints. Here, $\omega$ is closed since $\frac{\partial}{\partial y}(x+y) = 1 = \frac{\partial}{\partial x}(x-y)$, and the domain is simply connected, so $\omega$ is exact. A primitive is $F(x,y) = \frac{x^2}{2} + xy - \frac{y^2}{2}$, giving $F(1,1)-F(0,0) = \frac12 + 1 - \frac12 = 1$.