Can you really integrate functions? Statements such as "you can't really integrate functions, the only thing you can integrate are differential forms" and "if you really believe you know how to integrate functions tell me the integral of the function 1 over this blackboard" (I read that one once on a stackexchange answer which I've been unable to track down) feel scary to me, since I would like to believe I know how to integrate functions. For instance, the Riemann integral of the function $1$ over the interval $(a,b)$ of the real numbers is $b-a$ ... do I really need a differential form for that? Is integration "coordinate dependent" in some way, even in $\mathbb{R}^N$? I am used to the nice change of variables formula:
$$\int_{U} f = \int_{\phi^{-1}(U)}f\circ \phi ~|\det{D\phi}|$$
I don't really see how this is chart dependant. From what I could gather from questions such as this one, the issue seems to be that people 
 would rather have a function multiplied with another object that spits out the jacobian automatically than place it there by hand, however, I don't really see the problem with just shoving it there when one wishes to change coordinates. I am definitely missing some essential insight and I would love to have this clarified for me. 
 A: 
I don't really see how this is chart dependant.

It is chart dependent because the original chart you start with is privileged, but privileged by you and not the manifold structure. Let me try to illustrate this. Assume that a domain $\mathcal D$ is given in a manifold $M$. The domain is completely covered by two charts, the first one's coordinates are $x^1,...,x^n$, the second one's coordinates are $y^1,...,y^n$.
You have a function $f$, whose coordinate representation in the $x$ chart is just $f(x)$, and whose rep in the $y$ chart is just $f(y)$ (I am abusive towards notation).
Remember that on a general manifold, all charts are equivalent!
Try to define the integral of $f$ over the domain $\mathcal D$ as $$ I=\int_\mathcal D f(x)\ d^nx. $$ Now imagine you want to calculate this with the chart $y$ instead. You know from multivariable calculus how integrals behave under the change of variables, so you instead calculate $$I= \int_\mathcal Df(y)\left|\det\frac{\partial x}{\partial y}\right|\ d^ny $$ (the domain's the same because I am abusing notation - I refer to the domain of abstract points as $\mathcal D$, just to simplify things. Obviously when you do integrate, you need to change the coordinate domains).
You are overjoyed that the two integrals do agree and you take the conclusion that integration is chart-independent.
But, then you realize that since all charts on a manifold are equivalently good, you could have started with the $y$ chart instead of $x$.
Then your integral will be $$ \tilde I=\int_\mathcal D f(y)\ d^ny $$ and if you transform to the chart $x$, your integral will be $$ \tilde I=\int_\mathcal D f(x)\left|\det\frac{\partial y}{\partial x}\right|\ d^nx. $$
It is fairly obvious that in general $$ \int f(y)d^ny \neq \int f(y)\left|\det\frac{\partial x}{\partial y}\right|\ d^ny $$ and $$ \int f(x)\ d^nx\neq \int f(x)\left|\det\frac{\partial y}{\partial x}\right|\ d^nx, $$ so we have $$ I\neq \tilde I. $$
But there is absolutely nothing that distinguishes the chart $x$ over the chart $y$ or vice versa. Both of them are equally fine charts. So how do you define the integral of $f$?
The integral of $f$ if not well defined, because its value depends on which chart you take as your "initial chart", and you have no mathematical reason to declare one over the other.

The situation in $\mathbb R^n$ (which includes the case of $\mathbb R$) is different, because $\mathbb R^n$ comes with coordinates. So there, you do have a preferred coordinate chart, it's standard coordinates.
Also note that while it is not without exceptions, when one speaks of $\mathbb R^n$, one also considers it to have a standard euclidean structure, where the inner product of $u=(u^1,...,u^n)$ and $v=(v^1,...,v^n)$ is given by $$ \left\langle u,v \right\rangle=\sum_i u^i v^i. $$
This structure is preserved by orthogonal transformations, which also happen to have unit determinant, so all charts related to the standard chart by orthogonal transformations are also "valid" charts.
Another thing to note is that you don't actually need to go to manifold theory to see the coordinate-dependence of integrals. Usually when one speaks of $\mathbb R^n$, one can actually replace it with an abstract real vector space $V$ of dimension $n$ and have one's results still be correct.
That's not the case here. See, $\mathbb R^n$ comes pre-coordinatized, but an abstract $V$ space doesn't. You can coordinatize it with linear coordinates by choosing a basis $E=\{e_1,...,e_n\}$, but the possible bases are related to one another by $\text{GL}(n,\mathbb R)$ transformations, which usually do not have unit determinant, so it is also impossible to integrate on a general finite dimensional vector space (!!!). Of course, if you put an inner product on your space, it will select a set of preferred coordinates (cartesian ones), which are related to one another by orthogonal transformations, so an inner product space $(V,\langle\cdot,\cdot\rangle)$ does admit integration.
A: Three comments (which maybe form an answer):


*

*In order to define the integral of differential forms you have to know what the integral of functions (on $\mathbb{R}^n$ with respect to the Lebesgue measure) is. "Integration of differential forms" is not an independent theory.

*The things you have learned about Lebesgue and Riemann integrals, the transformation formula and so on are not wrong and you can work with them without thinking about differential forms. 

*Why care about differential forms? There is a way to think about geometric objects without referring to a certain incarnation of them as a subset of $\mathbb{R}^n$, e.g. in terms of abstract manifolds. If you do that, there is no preferred way to integrate functions, while often there still is a unique way to define the integral of so called $n$-differential forms. For example you may define $X$ to be an open interval. As an abstract manifold, $X$ does not have a preferred length, you are allowed to picture it as any interval $(a,b)\subset \mathbb{R}$ (they are all diffeomorphic). A reference to such an actual interval $(a,b)$ is called a choice of coordinates and you see that it depends on this choice what say the integral of $1$ (=length) should be.   

