# Integral with limit of integration whose output is undefined

If I have the function f(x)=3 and I take the integral with respect to x from 0 to 2, the answer is 6. Now what if I have the function f(x)=3 BUT x is undefined at 0. Well, this means that a line has been subtracted from our original area of 6. But a line has zero area so we still have an area of 6 from 0 to 2 even if 0 is undefined. So why is it that f(x) must be continuous from 0 to 2 if we want to take the integral from 0 to 2.

• To talk about the integrability of a function we only need to want that it be bounded, not continuous, in that interval. – azif00 Sep 8 '19 at 1:50
• So what you're saying is if I took the integral of $f(x)=3$ from $0$ to $2$ and $x=0.1, 0.3, 0.4,$ and $0.8$ were undefined I could still take the integral? – user532874 Sep 8 '19 at 1:58
• Yes. More surprising that a function like this can be integrated :) – azif00 Sep 8 '19 at 2:10
• I have taken the liberty to suppress tags "geometry" and "paradoxes" which are not meaningful here. – Jean Marie Sep 8 '19 at 2:54
• @Azif00 I have one last question. What is the definition of bounded in this case? – user532874 Sep 8 '19 at 3:58

I leave you another, but similar, example. Take $$f: [0,2] \to \Bbb R$$ defined by f(x) = \left\{ \begin{align} 1 & \textrm{ if } x=1 \\ 0 & \textrm{ if } x\neq 1 \end{align} \right. then $$\int_0^2 f(x)dx=0$$ To prove this rigorously, we turn to the definition of integral. Suppose $$P=\{ t_0,t_1,\dots,t_n\}$$ is a partition of the interval $$[0,2]$$ (this means $$t_0=0$$, $$t_n=2$$ and $$t_{i-1} for all $$i=1,\dots,n$$) with $$t_{j-1}<1 for some $$j$$ . Let's call $$m_i=\inf \, \{ f(x): t_{i-1}\leq x\leq t_i \}$$ $$M_i=\sup \, \{ f(x): t_{i-1}\leq x\leq t_i \}$$ for $$i=1,2,\dots,n$$. Then $$m_i=M_i=0$$ if $$i\neq j$$, but $$m_j=0$$ and $$M_j=1$$. Since the lower and upper sums turns into $$L(f,P)=\sum_{i=1}^n m_i(t_i-t_{i-1})=0$$ $$U(f,P)=\sum_{i=1}^n M_i(t_i-t_{i-1})=t_j-t_{j-1}$$ it follows that, we can make the differences $$U(f,P)-L(f,P)$$ as small as possible (only choose a partition with $$t_j-t_{j-1}$$ enough small). This means that $$f$$ is integrable in that interval and since $$L(f,Q)\leq 0\leq U(f,Q) \quad \textrm{for all the partitions } Q$$ it follows $$\int_0^2 f(x)dx=0$$ as we want to show. (The latter is because the integral is defined as the infimum of all the upper sums and the supremum of all the lower sums.)