Here is what I've done.
We know that to solve this we must evaluate $\int_c = f(x(t),y(t)\sqrt{(\frac{dx}{dt})^2}+(\frac{dy}{dt})^2dt$
So we parametrize our x and y values.
$r_0 = (-2,-1)$ and $r_1 = (1,2)$
$r(t) = ((1-t) \times (-2,-1)) + (t(-1,2)$
$= x = -2+2t-t$ and $y = -1+t+2t$
So $\int_c = f(x(t),y(t)\sqrt{(\frac{dx}{dt})^2}+(\frac{dx}{dt})^2 = $$\int_{0}{1} = f(4(-2+2t-t)\sqrt{(12x^2)^2}dt$
$=[4(-2+2t-t) \times (\sqrt{12x^2})^2]_01$
$=4[(-2+2(1)-1) \times (\sqrt{12(1)^2})^2 ] - 4[(-2+2(0)-0) \times (\sqrt{12(0)^2})^2 ]$
I'm not sure over which interval is the integral evaluated? Would it just simply be between 0 and 1? As this is what I've done, but not sure whether thats right.
Is my parametrization right? I know the formula is $(1-t) \times$ the starting point $+t \times$ the end point. But I get numbers as well and I'm not sure whether I've classified them right for x and y.
Segment from $(-2,-1)$ to $(1,2)$ has parametric equations
$C:(x=3 t-2,y=3 t-1), \;0\le t\le 1$
$ds=\sqrt{x'^2+y'^2}=\sqrt{9+9}\,dt=3\sqrt 2\,dt$
$$\int_C 4x^2\,ds=\int_0^1 4(3t-2)^2 \cdot 3\sqrt 2\,dt=12 \sqrt{2}$$