How does the indefinite integral relate to area; My understanding of dx in the indefinite integral notation makes me think it needs to releate. I am a beginner in calculus; taking AP calc and AP Physics C
I believe I understand what f(x) * dx means in the definite integral notation: the area of an infinitely thin box. The $\int_a^b$ in the definite integral means the sum of these boxes, which is also the displacement of the anti derivative in the range ab
What I do not understand is why the indefinite integral uses the notation f(x) * dx if, based on my current understanding, it does not directly relate to area.
The reason I believe it does not relate to the area is that area is represented by displacement of the antiderivative, but the answer to the indefinite integral is the anti-derivative with + C included (which means that it is not only total displacement of the function, I think).
I have read that the dx simply represents the variable that is being integrated when used in the indefinite integral, but I have strong feelings against that answer because of how clear its purpose is in the notation of the definite integral, which is not shown in the indefinite integral.
How does the dx in the indefinite integral notation relate to area? Or does it?
 A: Hint: The notation
\begin{align*}
\color{blue}{F(x) = \int f(x)\,dx}
\end{align*}
is a short-hand notation for a function $F(x)$ of the form
\begin{align*}
F(x) = c + \int_a^xf(u)\,du
\end{align*}
for suitable constants $c$ and $a$. This notation is explained by R. Courant cited in this post .
A: I'm not totally sure this answers your question, because you seem to understand how it relates in the beginning of your question.  In any case, I will answer what I think your question is.
Since we are splitting the line into indefinitely thing boxes, each box will have a width and a height.  The height of each box is $y$, and the width of each box is this indefinitely thin amount, which we will simply call $\mathrm{d}x$.  Therefore, the area of each individual box is:
$$ \text{area}_\text{ single box} = y\,\mathrm{d}x $$
So, here, the $\mathrm{d}x$ is serving as the width.
The integral is essentially a sum of infinitely small things. Therefore, if we want to sum up all of the infinitely small boxes, we would use a definite integral:
$$ \text{area}_\text{ total} = \int_{x = a}^{x = b} y\,\mathrm{d}x $$
However, we can't solve this directly, unless we have a formula for $y$ in terms of $x$.  So, if $y = f(x)$, then we can rewrite this as:
$$ \text{area}_\text{ total} = \int_{x = a}^{x = b} f(x)\,\mathrm{d}x $$
Is that helpful or is your trouble elsewhere?
A: Feelix, the best answer to your question is that integrals are not the same thing as antiderivatives. The integral is a very general operation that involves adding up the values of a function on a set. The functions and sets can be quite exotic. It is only, say, a coincidence, but a rather nice and useful one, that this notion of adding up the values of a function coincides with the problem of finding an anti-derivative in the special case of integrable functions on compact subsets of the real line. The notation $\int f(x)\mathrm dx$ is just that - a notation. The $\mathrm dx$ in this instance does not have any special meaning. In fact it would be equally valid to just write $\int f$, if one wished.
When dealing with true integration, (not finding antiderivatives) the notation $\mathrm dx$ has to do with the idea of a measure. If we have a set $E$, and a function $f:E\to \Bbb R$, and we want to, in some sense, "add up" all of the values that $f$ takes on $E$, we need to formulate how we are going to add up these values.
The regular measure $\mathrm dx=1\mathrm dx$ on the real line essentially means that all of the points of the integration domain have equal "weight". See here for more details.
