# Indefinite integral with discontinuous integrand

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
Homework Helper
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.

PFuser1232 and WWGD
HallsofIvy
Homework Helper
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.

PFuser1232
hunt_mat
Homework Helper
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 $\varepsilon\rightarrow 0$

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