Indefinite integral with discontinuous integrand

[PFuser1232]

Suppose $f$ is defined as follows:
$f(x) = 1$ for all $x ≠ 1$, $f(1) = 10$.
Is the indefinite integral (or the most general antiderivative) of $f$ defined at $x = 1$?
I'm asking this question because I already know how to deal with, say, $\int_0^2 f$; $f$ has only one removable discontinuity in $[a,b]$, so it's integrable. When I do the following:
$$\int_0^2 f(x) dx = \int_0^1 dx + \int_1^2 dx$$
and then apply the second fundamental theorem of calculus on each of the integrals, I am assuming that the antiderivative of $f$ exists at $x = 1$ and is equal to $1$.
Is this assumption wrong?

 

[RUber]

That is one way to say it. However, if F is the antiderivative of f, the F' = f. Perhaps, you are defining a continuous function g such that g = f almost everywhere and finding its integral.

 

[HallsofIvy]

Integration is a "smoothing" operator. If there is a singularity at a single point the integral will simply ignore it. The integral of "f(x)= 1 if x is not 1, f(1)= 10" is exactly the same as the integral of f(x)= 1.

 

[hunt_mat]

Let \varepsilon >0 and split up the integrand in the following way:
\int_{0}^{2}f(x)dx=\int_{0}^{1-\varepsilon}f(x)dx+\int_{1-\varepsilon}^{1+\varepsilon}f(x)dx+\int_{1+\varepsilon}^{2}f(x)dx
Now think about the consequences as \varepsilon\rightarrow 0

 

[PFuser1232]

Integration is a "smoothing" operator. If there is a singularity at a single point the integral will simply ignore it. The integral of "f(x)= 1 if x is not 1, f(1)= 10" is exactly the same as the integral of f(x)= 1.

But if we apply the first fundamental theorem to the antiderivative $F(x) = x + C$ (i.e. $F' = f$) we get $1 = f(x)$, which is not true for $x = 1$.

 

[Svein]

But if we apply the first fundamental theorem to the antiderivative $F(x) = x + C$ (i.e. $F' = f$) we get $1 = f(x)$, which is not true for $x = 1$.

Well, in this case you must apply the extended fundamental theorem which says F'=falmost everywhere (which means "except possibly on a set with measure 0).

 

[RUber]

The antiderivative F is defined such that F' = [f] where [f] = f almost everywhere. There is not going to be an antiderivative that captures point discontinuities, since the goal of the antiderivative is to have the property that $F(b)-F(a) = \int_a^b f(x) dx$. And the point discontinuities do not contribute to the integral.