MHB Total Differential: Find f(3,1) from f(1,2)=0

[Yankel]

Hello again,

Following my previous post, I have a total differential:

\[df=(2xy+1)dx+(x^{2}-2y)dy\]and it is also given that f(1,2)=0. I am asked to find f(3,1). The final answer is 12. I don't get it.

When I integrate the first part by x, and the second part by y, I get two different functions. So according to what you guys told me in the previous post, this is not a df in the first place ! Where do I get it wrong ?

thanks !

 

[I like Serena]

Hello again,

Following my previous post, I have a total differential:

\[df=(2xy+1)dx+(x^{2}-2y)dy\]and it is also given that f(1,2)=0. I am asked to find f(3,1). The final answer is 12. I don't get it.

When I integrate the first part by x, and the second part by y, I get two different functions. So according to what you guys told me in the previous post, this is not a df in the first place ! Where do I get it wrong ?

thanks !

Hi! :)

You have:
$$\frac{\partial f}{\partial x} = 2xy+1$$
Integrate it to find:
$$f = x^2y+x + g(y)$$
where $g(y)$ is an arbitrary function of $y$.

What can you deduce about $g(y)$ by integrating $x^{2}-2y$ with respect to $y$?