Parameterize the line segment by
[tex]\mathbf r(t)=(3\,\mathbf i+5\,\mathbf j)(1-t)+(-3\,\mathbf i+1\,\mathbf j)t=(3-6t)\,\mathbf i+(5-4t)\,\mathbf j[/tex]
with [tex]0\le t\le1[/tex]. Then the line integral is given by
[tex]\displaystyle\int_C\mathbf F\cdot\mathrm d\mathbf r=\int_{t=0}^{t=1}(-5\,\mathbf i+\mathbf j)\cdot(-6\,\mathbf i-4\,\mathbf j)\,\mathrm dt=26[/tex].
We can also use the fact that
[tex]\mathbf F(x,y)=-5\,\mathbf j+\mathbf j=\dfrac\partial{\partial x}[-5x]\,\mathbf i+\dfrac\partial{\partial y}[y]\,\mathbf j=\nabla(-5x+y)=:\nabla f(x,y)[/tex]
so that the gradient theorem applies, which means the integral is path independent and the value of the line integral can be found by evaluating
[tex]\displaystyle\int_{(3,5)}^{(-3,1)}\mathbf F\cdot\mathrm d\mathbf r=f(-3,1)-f(3,5)=16-(-10)=26[/tex]
