Let us consider some function $f(x,y):\,\mathbb R^2\to\mathbb R$, such that $$\frac{\partial f(x,y)}{\partial x}=a\,x+b\,y\qquad\qquad\frac{\partial f(x,y)}{\partial y}=b\,x+c\,y\qquad\qquad f(0,0)=0$$ where $a,b,c\in\mathbb R$.
- From this knowledge, i would like to find $f(x,y)$
I proceeded as follows $$df=\frac{\partial f}{\partial x}\,dx+\frac{\partial f}{\partial y}\,dy=(a\,x+b\,y)\,dx+(b\,x+c\,y)\,dy$$ Then $$f(x,y)=\int_{f(0,0)}^{f(x,y)}df'=\int_{0}^x(a\,x'+b\,y)\,dx'+\int_{0}^y(b\,x+c\,y')\,dy'=\frac{1}{2}\,a\,x^2+2\,b\,x\,y+\frac{1}{2}\,c\,y^2$$ where $\square'$ indicates the variable of integration
But then, if i differentiate the expression for $f$ found, partial derivatives are NOT the ones i started with $$\frac{\partial f(x,y)}{\partial x}=a\,x+2\,b\,y\neq a\,x+b\,y\qquad\qquad\frac{\partial f(x,y)}{\partial y}=2\,b\,x+c\,y\neq b\,x+c\,y$$ Where am i wrong? How to reach the correct expression for $f$, which by 'guessing' should be $f=\frac{1}{2}\,a\,x^2+\,b\,x\,y+\frac{1}{2}\,c\,y^2$?
You've gotten the formula for computing $f(x,y)$ wrong; you need to set $y$ to be $0$ in your integral using $dx'$.
The reason for this is that you're finding $f(x,y)$ by integrating along two paths; first from $(0,0)$ to $(x,0)$, and then from $(x,0)$ to $(x,y)$. The first path has $y=0$ at all times.