Flaw in calculation of $\int x \, dx=x^2$ $$\int x\ dx=\int \underbrace{(1 + 1 + \cdots + 1)}_{x\text{ times}}\ dx=x^2$$
Is the algebra Ok? The professor said that the function looses continuity; could anybody explain that?
 A: One big problem we have in integration notation is that we use the letter $x$ in a way that is inconsistent.  When writing the sum:
$$f(n)=1+2+3+\cdots+n$$
we would never write:
$$f(n)=\sum_{n=1}^n n$$
Rather, we write it as:
$$f(n)=\sum_{k=1}^n k$$
Note that $k$ is not $n$ - $k$ is a value that varies from $1$ to $n$.
In the integral, it is wrong to treat the $x$ "inside" the integral as a constant. It is not - it is like $k$ in the sum above.  Depending on how you are defining integrals (as anti-derivatives or as areas or whatever) the $x$ on the inside is not the same as the outside.  So we'd never say:
$$\int_0^x f(x) \, dx$$
But rather we say:
$$\int_0^x f(t) \, dt$$
with definite integrals.
Indefinite integrals are similar oddities.  The indefinite integral:
$$F(x) = \int f(x) \, dx$$
is, in a sense, lazy notation - we reuse the letter $x$ because we really mean $F$ and $f$ have the same domain, and $F'(x)=f(x)$ for all $x$ in that domain.
Add to that the obvious difficulty of defining what "$x$ times" means when $x$ is not an integer, and you'll see the error.
A: $$f(x)=\underbrace{(1 + 1 + \cdots + 1)}_{x\text{ times}}$$
You will note that $f$ is only defined for $x \in \mathbb{Z}^+$, i.e. at isolated points.
It is clear that
$$f(1)=1$$
$$f(2)=1+1=2$$
$$f(3)=1+1+1=3$$
etc.
however, a problem arises when we want to find $f(-1)$ or $f\left(\frac{1}{2}\right)$.  It is because of this we are not able to integrate this, as it is not a continuous (or differentiable) function. What happens when we want to find $\int_0^1 f(x)\,dx$?
A: The right hand side of the following is not using a mathematical notation:

$$\int x\ dx=\int \underbrace{(1 + 1 + \cdots + 1)}_{x\text{ times}}\ dx=x^2$$

Hence we are mixing a picture with very specific notations and this is leading to the confusion. Elaborating further on @Thomas Andrews answer (if I may),
$$\underbrace{(1 + 1 + \cdots + 1)}_{x\text{ times}}$$
can be repented by summation notation as follows:
$$x=\sum_{n=1}^x 1$$
Now you could represent the expression accurately by writing:
$$\int x\ dx=(\int \sum_{n=1}^x 1 dx=\int x\ dx)=\frac{x^2}{2}+C$$
