Wrong interpretation of the indefinite integral This might sound very useless but I'd like to see what you think. Bear in mind that I'm just a novice student.
if $f$ is the original function, then it could be found this way 
$C+\int f'(x)\, dx=f(x)$
I understand this is equivalent to saying $\int f '(x)\, dx=f(x)+c$, but this way gives rise to the wrong interpretation!
If $\int f '(x)\, dx$ means the sum of all infinitesimally small increments, it is impossible that if you take the sum of those increments you'll get the original function! You'll only get the original function MINUS some constant inherent to that function. I guess this is a trivial matter but what do you think? Interpreting the indefinite integral is really making my head hurt, how do you interpret it? 
edit: The wrong interpretation is that the indefinite integral gives you the original function. Which is what my teachers have taught all along.
Thanks.
 A: First, $\int f'(x)dx$ does not mean "the sum of all infinitesimally small increments". That is what $\int_a^x f'(t)dt$ means. The former means "an anti-derivative of the derivative of f".
The source of the constant comes the $a$ in the following integral.
$$\int_a^x f'(t)dt = f(x)+C(a)$$
where $C = -f$
A: Nice question. The way I look at the indefinite integral's constant of integration (which seems to be the real crux of your question) is that the constant of integration is really a function that does not depend on the variable integrated with respect to. This "function" $C$ is the component of the original function $F(x)$ which does not depend on $x$ This really comes into play with the work of multivariable calculus.
Think about it this way. A multivariable function depends on more than one variable:
$$z=x+y$$
http://www.wolframalpha.com/input/?i=z%3Dx%2By
The slope of this plane depends on both $x$ and $y$. As a result, if we differentiate $z$ with respect to $x$ - $y$ is considered totally independent of $x$ -, then the resultant slope - in the x direction is:
$$\frac{\partial z}{\partial x}=1$$
If we integrate this function $\frac{\partial z}{\partial x}$ with respect to $x$, we will not get the original function - as you mentioned.
$$\frac{\partial z}{\partial x}= 1 \rightarrow \partial z = \partial x$$
$$\int \partial z = \int \partial x \rightarrow z = x$$
Notice how we are completely missing the $y$ term! This is because we have a component of the original function that is not dependent on $x$. Thus, to note that there might be something missing, we insert an arbitrary function $C_1$ with is not dependent on the variable of integration.
In this 2D example, $C_1$ is a function of $y$. Thus, the true expression from the integrals above is:
$$z = x + C_1(y)$$
In the case of 1D, which is what you are learning about, the only component of the original function $z$ that can be independent of $x$ is a constant number, so we remove the $(y)$ designation. I.E.
$$C_1(y)=C$$
Thus, our simplified expression is:
$$z=x+C$$
