Thread: Line integral

Show that the line integral is independent of path and evaluate the integral.
∫c (1-ye^(-x)dx + e^(-x)dy)
C (suppost to be a subscript to the integral) is any path from (0,1) to (1,2)

To prove it is independent of path, i did:
dP/dy = (1-ye^(-x))dy = -e^(-x)
dQ/dx = (e^(-x))dx = -e^(-x)
So, since they are the same, it is conservative, which means it is independent of path.

I'm having some trouble evaluating the integral. I think i do:
∫c (F * dr) = f(1,2)-f(0,1)

= (1-(2)e^(-1)) + e^(-1)) - (1-e^(0) + e^(0))
=1 - 2e^(-1) + e^(-1) - 1
= -2e^(-1) + e^(-1)
= -e^(-1)

I don't think that is correct. I thought is was suppost to equal zero.

The primitive function is f(x,y)=x+ye^{-x}.
That is what is used to evaluate the integral.

so,
f(x,y) = x + ye^(-x)
f(1,2)-f(0,1) = 1+2e^(-1)-1 = 2e^(-1)

Is that correct? How did you get the primitive function?

Originally Posted by justme3

so,
f(x,y) = x + ye^(-x)
f(1,2)-f(0,1) = 1+2e^(-1)-1 = 2e^(-1)

Is that correct? YES

Originally Posted by justme3

How did you get the primitive function?

I just looked at the F function and saw what primitive, f, gives F when that partials are found.