This question is in lieu of "Integrating $dψ=(x+y)dx+x_0dy$". Though I got good answers, none of them could explain the real question. Consider the integral,
$\int(x+y)dx$
In a book I recently read, such an integral is equated to $\frac{x^2}{2} + xy$, i.e. they treat x and y as constants wrt each other. However, there is no explicit mention of any dependence of x and y in the context, son naturally, you take them to functions of each other, right? But they treat (I am ignoring that it is a definite integral at the moment-unless the definiteness of the integral allows you to evaluate it )
$\int(x+y)dx = \int xdx + \int ydx$ = $x^2/2 + xy$,
i.e. they treat y as a constant wrt x in the 2nd integral. But why? can you do that for any integral? For example, can you say
$\int xy dx=y\int x dx = x^2y/2 +C$ ????
I am utterly confused, as this( integral of x+y) is what they have done in the book (pg 333 mathematical methods for physicists) for similar integrals. And again, I DON'T want to SOLVE the ODE, I merely used it as an example to see the applications of aforementioned integrals which confuse me. Thanks in advance!!
$\endgroup$1 Answer
$\begingroup$When you are calculating the integral of $f(x, y)$ with respect to $x$, the intuition is that you are summing the infinitely many $f(x, y)\cdot dx$ infinitesimals as $x$ goes through all of the values in a range but $y$ is static. Thus $x$ is the "variable" and $y$, as far as the integral is concerned, is a "constant".
This doesn't mean that $y$ is an unchanging value, $y$ may be as variable as $x$ is. It just means that $y$ is unchanging throughout the infinite sum.
If $x$ and $y$ are dependent however, the integral is taken as over $f(x, y(x))$ as $x$ changes.
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