# Thread: Having problems with another line integral

1. ## Having problems with another line integral

int[(x+2y)dx + (x-2y)dy]; C is the line segment from (1,1) to (3,-1). The answer is 0. I got an answer of -8 by just taking the integral with respect to x before dx and respect to y after that. Then f(3,-1) - f(1,1). Obviously I did something wrong, any ideas?

2. Originally Posted by pakman
int[(x+2y)dx + (x-2y)dy]; C is the line segment from (1,1) to (3,-1). The answer is 0. I got an answer of -8 by just taking the integral with respect to x before dx and respect to y after that. Then f(3,-1) - f(1,1). Obviously I did something wrong, any ideas?
I am not sure what you did, but you tought you can use the fundamental theorem of line integral you cannot.
Because the cross partials test fails.

The parametrization of the curve is,
x=t and y=2-t for 1<=t<=3.

That means,
INT(1,3)(x+2y)*(dx/dt)dt+INT(1,3)(x-2y)(dy/dt)dt
Insert the parametrization.

INT(1,3)(t+2(2-t))*(dt/dt)dt+INT(1,3)(t-2(2-t))*(dy/dt)dt

INT(1,3) 4-t dt + INT(1,3) 3t-4 dt

What ever that is should be the answer.

3. Originally Posted by ThePerfectHacker
INT(1,3)(t+2(2-t))*(dt/dt)dt+INT(1,3)(t-2(2-t))*(dy/dt)dt
INT(1,3) 4-t dt + INT(1,3) 3t-4 dt
There is a minor sign error there: (dy/dt)=(-1).