Line integral of a force field

Yes the vector field is conservative and hence its line integral is path independent. Given it is a conservative vector field, it must be gradient of a scalar function. In other words,

$ \vec F_1 = \nabla \phi(x, y)$

We can then find that $ ~ \displaystyle \phi (x, y) = \frac{x^3+y^3}{3} + xy$

Now using fundamental theorem of line integral,

$ \displaystyle \int_C \nabla \phi(x,y) \cdot d\vec r = \phi(\vec r(b)) - \phi(\vec r(a)) $

where $\vec r(a)$ and $\vec r(b)$ represent the start and end points on the curve which in this case are $(0, 0)$ and $(1, 1)$.

Alternatively you can parametrize the curve $C$ from $(0, 0)$ to $(1, 1)$ as $ ~ r(t) = (t, t), 0 \leq t \leq 1$

Then $F_1(r(t)) = (t^2 + t, t^2 + t)$

$r'(t) = (1, 1)$

So the line integral is,

$ \displaystyle \int_0^1 (t^2 + t, t^2 + t) \cdot (1, 1) ~ dt$

It is simple enough to parameterize the lines.

The line from $(0,0)$ to $(1,0)$ is $(t,0)$ with $0\le t\le 1$
And, the line from $(1,0)$ to $(1,1)$ is $(1,t)$ with $0\le t\le 1$

Your integrals become:

$\int_0^1 (t^2, t)\cdot(1,0)\ dt = \int_0^1 t^2\ dt$

and

$\int_0^1 (1+t,t^2 + 1)\cdot(0,1)\ dt = \int_0^1 t^2+1\ dt$

But as you note $f(x,y) = (x^2 + y, y^2+ x)$ is conservative.

Which means that there exists some $F(x,y)$ such that $\nabla F(x,y) = f(x,y)$

$F(x,y) = \frac 13 x^3 + \frac 13 y^3 + xy$

$\int_0^1 (t^2, t)\cdot(1,0)\ dt = F(1,0) - F(0,0)$

and

$\int_0^1 (1+t,t^2 + 1)\cdot(0,1)\ dt = F(1,1) - F(0,1)$