Flaw in calculation of $\int x \, dx=x^2$

Question

$$\int x\ dx=\int \underbrace{(1 + 1 + \cdots + 1)}_{x\text{ times}}\ dx=x^2$$ Is the algebra Ok? The professor said that the function looses continuity; could anybody explain that?

Answer 1

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.

Answer 2

$$f(x)=\underbrace{(1 + 1 + \cdots + 1)}_{x\text{ times}}$$

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 $\int_0^1 f(x)\,dx$?

Answer 3

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:

$$\int x\ dx=(\int \sum_{n=1}^x 1 dx=\int x\ dx)=\frac{x^2}{2}+C$$