Question

Let c be the curve which is the union of two line segments, the first going from (0, 0) to (4, 4) and the second going from (4, 4) to (8, 0). computer the line integral ∫c4dy−4dx

Answer 1

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:

$I=\int_c 4dy-4dx$

So, as shown in the figure, C = C1 + C2, so:

$I=\int_{c_{1}} (4dy-4dx)+\int_{c_{2}} (4dy-4dx)=I_{1}+I_{2}$

Computing first integral:

$c_{1}: y-y_{0}=m(x-x_{0}) \rightarrow y=x$

Applying derivative:

$dy=dx$

Substituting this value into $I_{1}$

$I_{1}=\int_{c_{1}} (4dx-4dx)=\int_{c_{1}} 0 \rightarrow \boxed{I_{1}=0}$

Computing second integral:

$c_{2}: y-y_{0}=m(x-x_{0}) \rightarrow y-0=-(x-8) \rightarrow y=-x+8$

Applying derivative:

$dy=-dx$

Substituting this differential into $I_{2}$

$I_{2}=\int_{c_{2}} 4(-dx)-4dx=\int_{c_{2}} -8dx=-8\int_{c_{2}}dx$

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:

 $I_{2}= -8\int_{4}^{8}}dx=-8[x]\right|_4 ^{8}=-8(8-4) \rightarrow \boxed{I_{2}=-32}$

Finally:

$I=\int_c 4dy-4dx=0-32 \rightarrow \boxed{I=-32}$