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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.

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  • $\begingroup$ To talk about the integrability of a function we only need to want that it be bounded, not continuous, in that interval. $\endgroup$ – azif00 Sep 8 '19 at 1:50
  • $\begingroup$ 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? $\endgroup$ – user532874 Sep 8 '19 at 1:58
  • $\begingroup$ Yes. More surprising that a function like this can be integrated :) $\endgroup$ – azif00 Sep 8 '19 at 2:10
  • $\begingroup$ I have taken the liberty to suppress tags "geometry" and "paradoxes" which are not meaningful here. $\endgroup$ – Jean Marie Sep 8 '19 at 2:54
  • $\begingroup$ @Azif00 I have one last question. What is the definition of bounded in this case? $\endgroup$ – user532874 Sep 8 '19 at 3:58
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Your reasoning is right!

In principle we do not need to talk about the continuity of a function if we want to know whether or not it can be integrable in a certain closed interval. To talk about the integrability of a function we only need to want that it be bounded in that interval.

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 $t_{i-1}<t_i$ for all $i=1,\dots,n$) with $t_{j-1}<1<t_j$ 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.)

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