- #1

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##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?