I am trying to find the line integral of the field: $F = (3y- \frac{y}{x^2}e^{1/x},sin(y)+ e^{1/x})$ for the line: $$C: x=t,y=0,-6\leq t \leq -2$$ So I used the formula and got to find:
$$\int^{-6}_{-2} (3\cdot 0 - \frac{0}{t^2}\cdot e^{1/t})\cdot x'(t)\cdot dt = 0$$ I checked my result many times for mistakes but couldn't find any . Also checked the internet for similar situations but didn't find any explanation for that... Is my caclualtion wrong? and if it isn't, how is it possible to get a 0 result for non closed line?