I have been losing my mind over this for a good bit, so bear with me. I've been looking into generalizing the ideas of calculus into more dimensions, and for differentiation, I've found it to be relatively simple, where you essentially "linearize" a function, you associate some linear transformation that describes the mapping of a, not necessarily linear, function between the input and output space, and if you take a "general" or undefined derivative, it gives you an, again, not necessarily linear function, but the output of said function describes the linear transformation around that same input with respect to the original function.
To maybe give an example that explains this better, let's say you have the function $f(x) = \sin(x)$, non-linear. The derivative is $f'(x) = cos(x)$, again non-linear, however, if I input, let's say, $\pi$ into the derivative, it returns $-1$, what this means is that the linear transformation that describes the original function $f(x)$ around the point $(\pi, 0)$ is $-x$. You can further generalize this to functions from $\Bbb{R}^n \to \Bbb{R}^m$ via partial derivatives and such.
So, that's the derivative out of the way, as for the integral, is it possible to get a similar view of the integral as some property or linearization or anything else about the function? I understand that it's a generalization of summation and the anti-derivative is slightly different, etc. However, it does hold some relation to the derivative as an inverse, and much in the same way that derivation with an input returns another input, but leaving it "unbounded" returns just another function, the integral seems to behave similarly, and my question is, is there any way to interpret this in the language of the integral itself? Not just as an "anti-derivative"?