Integration ‘with respect’ to a variable.

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I'm slightly confused by 'integrating' with respect to a given variable, and in particular why for a function $f$ we integrate $f(x)$ to yield another function applied to $x$. I understand generally integration with respect to Riemann sums, and the idea of the sum of infinitesimal (I understand this is not-rigorous) terms $dx$ and the sum of possible derivatives at points $x$ $F(x)'=f(x)$ in this case. But what if we wanted to integrate:

$\int\ f(y)\ \mathrm dx\ $ for $x$ and $y$ which do not depend on each other in this case.

I understand that for a infinitesimal $dx$ around $x$ we would need to multiply by the infinitesimal rate of change around that point $F'(x)$ so would it be possible to perform such an integral as I've defined, and is the integral of $f(x)$ the intrgral of the application $x$ to $f$ or the integral of $f$ itself?

Best Answer

In multi-variable calculus, partial differentiation is the process of differentiating one variable while keeping others constant: thus, the derivative of $3y-2x$ with respect to (wrt) $x$ is $-2$, and wrt $y$ it is $3$.

There is an analogue of integration: given $f(y)$, what are all the possible functions of $x,y$ that have gradient $f(y)$ wrt $x$?

We might naively say that if $x$ and $y$ are unrelated, then $f(y)$ is a constant wrt $x$ and so $\int f(y) \, \mathrm d x=f(y)\int 1\, \mathrm d x=xf(y)+c$ for any constant $c$.

However, the partial integral does not necessarily have a constant $c$ as its integral, but anything that is constant with respect to $x$. This means that it can be a function of every variable other than $x$: in this case,

$\int f(y) \, \mathrm d x=xf(y)+g(y)$

for any univariate function $g$.

If we were working with three variables, $x,y,z$, you'd have $\int f(y) \, \mathrm d x = xf(y)+g(y,z)$ for any bivariate function $g$. And so on.

You ask, "is the integral of $f(x)$ the integral of the application $x$ to $f$ or the integral of $f$ itself?" Well, there is no "integral of $f(x)$". There is an integral of $f$ itself, which is equivalent to an integral of $f(x)$ wrt $x$, and either is what we are finding here.

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