1. ## Simple Line Integral

Hello! I need some help with a simple line integral. If there's someone out there who wouldn't mind explaining their work, could he/she lend a hand?

the problem is:

Let C be the curve which is the union of two line segments, the first going from (0, 0) to (-4, 2) and the second going from (-4, 2) to (-8, 0).
Computer the line integral
∫-4dy-2dx

i've tried solving for this by parameterizing the 2 line segments in terms of t, and calculated an incorrect answer of 32.

2. I tried and got 32 too

For the first integral:
∫-4dy-2dx
r= (-4,2)t
x=-4t
y=2t
dx=-4dt
dy=2dt
∫-8 +8 dt=0
For the second integral:
∫-4dy-2dx
r= (-4,2) + (-8+4, 0-2)t
x=-4 -4t
y=2 -2t
dx=-4dt
dy=-2dt
∫8dt+8dt=32
0<=t<=2

or maybe it is a conservative field, and the answer is U(-8,0)-U(0,0) where U= -2x-4y

3. It looks like zero to me: Start with the line going from (0,0) to (-4,2). The equation for this line is $\displaystyle y=-1/2(x+4)+2$. Parameterizing it: $\displaystyle C_1: \{x=t,\; y=-1/2(t+2)+2\}$. That integral is, remembering to keep straight the direction we're going:

$\displaystyle \mathop\int\limits_{C_1} -4dy-2dx=-4\int_0^{-4} (-1/2)dt-2\int_0^{-4}dt=0$

Second one going from (-4,2) to (-8,0). Equation is $\displaystyle y=1/2(x+8)$. Parameterizing: $\displaystyle C_2:\{x=t,\; y=1/2(t+8)$. Keeping direction straight:

$\displaystyle \mathop\int\limits_{C_2} -4dy-2dx=-4\int_{-4}^{-8} (1/2)dt-2\int_{-4}^{-8}dt=0$