Finding f(x,y) given its partial derivatives 
This is from Fitzpatrick's advanced calculus chapter 13 which is a analysis text and the start of analysis of several variables. Now the solution manual uses a multivariable version of the mean value theorem and this problem is actually from the chapter on the multivariable mean value theorem. Now based on what we did in multivariable calculus, an easy way to solve this problem is to just integrate with respect to x which gives us an an arbitrary constant and an arbitrary function of y, then differentiate that. However, because this is an analysis text, I am not sure if that is correct from a rigorous standpoint. For one thing, indefinite integrals has never been defined, but I do want to ask whether that is a valid solution method.
 A: This is correct from a rigorous standpoint. We can define a function $g: \mathbb{R} \to \mathbb{R}$ by \begin{equation} 
g(t) = f( t , 0 ). \end{equation}
Since $f$ has all partial derivatives at each point in $\mathbb{R}^2$ it follows that the function $g$ we defined is differentiable (in the single variable sense). This is because $g’(t) = \frac{d}{dt} f(t,0) = \partial_x f(t,0). $ Now, also from single variable calculus we know from the fundamental theorem of calculus that \begin{equation} f(t,0) - f(0,0) = g(t) - g(0) = \int_0^t g’(s) ds = \int_0^t  \partial_x f(s,0) ds. \end{equation}
Since we assume that $\partial_x f (x,0) \equiv 2$ it follows from this equation that $$ f(t,0) - f(0,0) = 2t.$$
Moreover, if we do the same thing for $y$ and we define a function $$ h_x(t) = f(x,t)$$ then we get the similar conclusion that $$ f(x,t)  - f(x,0) = 3t .$$
Finally we can notice that $$f (x,y) - f(0,0) = f(x,y) - f(x,0) + f(x,0) - f(0,0) = 3y + 2x$$ and so $$ f(x,y) = 3y + 2x + f(0,0) = 3y + 2x + 1. $$
This should be totally rigorous since we are just leveraging facts we know from single variable calculus.
