Hi I'm trying to understand Second fundamental theorem of calculus when it is used for function of two variables $ f(x,y) $.
As we know from Second Fundamental Theorem, when we have a continuous function $f(x)$ and fix constant a, then
From $$ F(x) = \int_{a}^{x} f(t) dt $$ it follows that $ F'(x) = f(x) $
It's shown on the picture below:
I'm reading now a proof of theorem where is continuous function of two variables $f(x,y)$ with equation:
$$ \frac{ \partial f(x,y)}{ \mbox{d} x } = P(x,y) $$
It is written in book that from Second Fundamental Theorem it follows that:
$$ f(x,y) = \int_{x_0}^{x} P(x,y) dx + R(y) $$
where $x_0$ i constant and $R(y)$ stands for the arbitrary constant of integration.
I don't why we have here constant $R(y)$. How can this be explained? The propoal here follows from derivative to integral but in theorem it follows from integral to derivative. So is it correct proposal?
I would be greateful for explanation of my doubts. Best regards 😉
Best Answer
The main idea in the R(y) term is that the book is basically thinking that for each fixed y, there is a function $g_y(x) = f(x,y)$, so that the partial derivative of $f$ is the (ordinary) derivative of $g_y.$ Then the fundamental theorem can be applied to $g$ giving $$ g_y(x) = \int_{x_0}^x g_y'(x) dx + c.$$ This is true for any fixed $y$, although the $c$ may be different for each $y$--i.e. $c$ is a function of $y$. Changing notation from $g$ to $f$ gives the formula from your book, where $R(y)$ gives the $c$ associated to each $y$.
Also, I think you are just mixing up the first and second theorem.