Without knowing exactly what [tex]f[/tex] is, this is impossible to do. So let's assume [tex]f(x,y)=1[/tex]. Then the line integral over the given rectangle will correspond to the "signed" perimeter of the region.
You don't specify that the loop is complete, so in fact the integral will only give the "signed" length of three sides.
Parameterize the region by first partitioning the contour into three sub-contours:
[tex]C_1:\mathbf r_1(t)=(3,0)(1-t)+(3,2)t=(3,2t)\implies\dfrac{\mathrm d\mathbf r_1}{\mathrm dt}=(0,2)[/tex]
[tex]C_2:\mathbf r_2(t)=(3,2)(1-t)+(-2,2)t=(3-5t,2)\implies\dfrac{\mathrm d\mathbf r_2}{\mathrm dt}=(-5,0)[/tex]
[tex]C_3:\mathbf r_3(t)=(-2,2)(1-t)+(-2,0)t=(-2,2-2t)\implies\dfrac{\mathrm d\mathbf r_3}{\mathrm dt}=(0,-2)[/tex]
where [tex]0\le t\le1[/tex] for each sub-contour. Then the line integral is given by
[tex]\displaystyle\int_Cf\,\mathrm dS=\int_{C_i}f(\mathbf r_i(t))\cdot\frac{\mathrm d\mathbf r_i}{\mathrm dt}[/tex]
with [tex]i\in\{1,2,3\}[/tex]. You have
[tex]\displaystyle\int_{C_1}f\,\mathrm dS=\int_0^1(1,1)\cdot(0,2)\,\mathrm dt=2[/tex]
[tex]\displaystyle\int_{C_2}f\,\mathrm dS=\int_0^1(1,1)\cdot(5,0)\,\mathrm dt=5[/tex]
[tex]\displaystyle\int_{C_1}f\,\mathrm dS=\int_0^1(1,1)\cdot(0,-2)\,\mathrm dt=-2[/tex]
Then the integral over the entire contour would be [tex]2+5-2=5[/tex]. Note that if the loop is complete, then the last leg of the contour would evaluate to -5, and so the total would end up as 0. This result would also follow from the fact that [tex]f(x,y)[/tex] is conservative, i.e. [tex]f(x,y)=\nabla g(x,y)[/tex] for some scalar field [tex]g[/tex], and so the line integral is path independent. Its value would depend only on the endpoints of the contour, which in the case of a closed loop would simply be 0.
