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I was reading the book interesting integrals and this came up: $$\int_{j}^{j+1} \frac{n-j}{x} \mathrm{d}x$$ it then goes on to say that $j$ is equal to floor $x$ because the integration interval $j \leq x < j+1$. I get this but aren't the limits of integration on the interval $j\leq x\leq j+1$?

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It's irrelevant. Adding or removing a single point will not change the integral; if you think about it, a single point will belong to a single very small division in the partition of the interval, so its contribution to the value of the integral is zero.

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  • $\begingroup$ Great thanks, would that also be because the point has 0 area in that line? $\endgroup$ – jake walsh May 7 '16 at 15:39
  • $\begingroup$ Yes, that's one way to see it. But if you want to justify it formally, you should probably go for the definition of integral with the partitions. $\endgroup$ – Martin Argerami May 7 '16 at 15:41
  • $\begingroup$ Glad I could help. $\endgroup$ – Martin Argerami May 7 '16 at 17:27

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