Problem:
Evaluate the line integral of the vector field $$f(x,y)=(x^2-2xy)i+(y^2-2xy)j$$ from $(-1,1)$ to $(1,1)$ along the parabola $y=x^2$.
I can't understand how to set limits on the Line Integral. While I identified $f(x,y)$ as the vector field defined and bounded on the graph of $\alpha$, I can't understand how to get $\alpha$ (the piecewise smooth path in n-space defined on the interval $[a,b]$-as it says in my textbook), nor how to work out $f.d\alpha$.
By definition, if you have a line parametrised as $\gamma: [a,b]\to \mathbb R^n$ and a vector field $f: \mathbb R^n\to\mathbb R^n$, then the interval of the vector field along the line is defined as
$$\int f d\gamma = \int_a^b f(\gamma(t))\cdot \gamma'(t) dt$$
In your case:
$$f(x,y) = (x^2 - 2xy, y^2-2xy)\\ \gamma(t) = (t, t^2) \text{ on } [-1,1]$$
Now, just plug in the numbers and see what comes out.