# Thread: independence of path and evaluating integral

1. ## independence of path and evaluating integral

Hi,
I have to test for independence of path and evaluate the integral.
I could prove that it's independence of path but I couldnt evaluate the integral.

(3,4)
∫ (2x+5y-7)dx + (5x-8y+3)dy
(1,1)

where C: 3x - 2y-1=0

2. Originally Posted by panda
Hi,
I have to test for independence of path and evaluate the integral.
I could prove that it's independence of path but I couldnt evaluate the integral.

(3,4)
∫ (2x+5y-7)dx + (5x-8y+3)dy
(1,1)

where C: 3x - 2y-1=0
Say that $\displaystyle \bold{f}(x,y) = (2x+5y-7,5x-8y+3)$.
We want to find $\displaystyle F(x,y)$ so that $\displaystyle \nabla F = \bold{f}$.
This mean $\displaystyle \partial_1 F = 2x+5y - 7 \text{ and }\partial_2 F = 5x - 8y+3$.
The first equation becomes $\displaystyle F(x,y) = x^2 + 5xy - 7x + g(y)$.
But then $\displaystyle \partial_2 F (x,y) = 5x + g'(y)$.
Thus we want $\displaystyle g'(y) = -8y + 3$ the function $\displaystyle g(y) = -4y^2 + 3y$ works.
Thus, an example of a function is $\displaystyle F(x,y) = x^2 + 5xy -7x - 4y^2 + 3y$.

Then the integral is $\displaystyle F(3,4) - F(1,1)$ - compute that for the answer.