# Thread: Fundamental Theorem of Line Integrals

1. ## Fundamental Theorem of Line Integrals

How do I use the fundamental theorem of line integrals to evaluate the integral
(3x-y+1)dx - (x+4y+2)dy from (-1,2) to (0,1)?

2. Originally Posted by noles2188
How do I use the fundamental theorem of line integrals to evaluate the integral
(3x-y+1)dx - (x+4y+2)dy from (-1,2) to (0,1)?
first find the potential function (do you know how to do this?), call it $f(x,y)$. then your line integral is equal to $f(-1,2) - f(0,1)$

3. that's the part i am having trouble with, finding f.

4. Originally Posted by noles2188
that's the part i am having trouble with, finding f.
*sigh* i know i responded to similar questions here, but i can only find the ones with 3 variables, not two, so they might confuse you. i can't find a way to give you hints for this other than do the problem...or most of it anyway. other than that, i will tell you to go read your textbook. i know they gave an example of how to do this

Originally Posted by noles2188
How do I use the fundamental theorem of line integrals to evaluate the integral
(3x-y+1)dx - (x+4y+2)dy from (-1,2) to (0,1)?
we want to find a function $f(x,y)$ so that $f_x = 3x - y + 1$ and $f_y = -x - 4y - 2$

(we know we can do this since the "equations" are exact: $\frac {\partial }{\partial y}(3x - y + 1) = \frac {\partial }{\partial x}(-x - 4y - 2)$)

so take $f_x = 3x - y + 1$, then

$f(x,y) = \int f_x~dx = \frac 32x^2 - xy + x + g(y)$

(our constant of integration is a function of y, since this will die when we differentiate with respect to x)

$\Rightarrow f_y = \frac {\partial }{\partial y}f(x,y) = -x + g'(y)$

But $f_y = -x - 4y - 2$, so we have $-x -4y - 2 = -x + g'(y)$. hence

$g'(y) = -4y - 2$

$\Rightarrow g(y) = -2y^2 - 2y + K$

Therefore, $f(x,y) = \frac 32x^2 - xy + x -2y^2 - 2y + K$ is the (potential) function we seek

5. they do show an example of this in my book but they leave out some steps so i was a little bit confused, but now i think i have decent understanding of it. thanks.