First of all we need to find a representation of C, so this is shown in the figure below.
So the integral we need to compute is this:
[tex]I=\int_c 4dy-4dx[/tex]
So, as shown in the figure, C = C1 + C2, so:
[tex]I=\int_{c_{1}} (4dy-4dx)+\int_{c_{2}} (4dy-4dx)=I_{1}+I_{2}[/tex]
Computing first integral:
[tex]c_{1}: y-y_{0}=m(x-x_{0}) \rightarrow y=x[/tex]
Applying derivative:
[tex]dy=dx[/tex]
Substituting this value into [tex]I_{1}[/tex]
[tex]I_{1}=\int_{c_{1}} (4dx-4dx)=\int_{c_{1}} 0 \rightarrow \boxed{I_{1}=0}[/tex]
Computing second integral:
[tex]c_{2}: y-y_{0}=m(x-x_{0}) \rightarrow y-0=-(x-8) \rightarrow y=-x+8[/tex]
Applying derivative:
[tex]dy=-dx[/tex]
Substituting this differential into [tex]I_{2}[/tex]
[tex]I_{2}=\int_{c_{2}} 4(-dx)-4dx=\int_{c_{2}} -8dx=-8\int_{c_{2}}dx[/tex]
We need to know the limits of our integral, so given that the variable we are using in this integral is x, then the limits are the x coordinates of the extreme points of the straight line C2, so:
[tex]I_{2}= -8\int_{4}^{8}}dx=-8[x]\right|_4 ^{8}=-8(8-4) \rightarrow \boxed{I_{2}=-32}[/tex]
Finally:
[tex]I=\int_c 4dy-4dx=0-32 \rightarrow \boxed{I=-32}[/tex]
