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Original question

$$\int_1^2 f(x)\,\mathrm{d}x$$ where $$f(x)=\begin{cases} 1 & \text{if $x$ is rational,} \\ 0 & \text{if $x$ is irrational} \end{cases}$$

How does one interpret this to find the upper and lower sums on a regular partition

$$\Delta x={1\over n}$$

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  • $\begingroup$ The values of x are 0, 1/n, 2/n, 3/n, ..., (n-1)/n, 1. Each pair of successive x values, 0 to 1/n, 1/n to 2/n, etc. gives an interval. To find the "upper sum" take the largest value of f in each interval. To find the "lower sum", take the lowest value of f in each interval. $\endgroup$ – user247327 Apr 18 '16 at 12:39
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The values of x are 0, 1/n, 2/n, 3/n, ..., (n-1)/n, 1. Each pair of successive x values, 0 to 1/n, 1/n to 2/n, etc. gives an interval. To find the "upper sum" take the largest value of f in each interval. To find the "lower sum", take the lowest value of f in each interval. For this particular interval that is very easy- in ANY interval there exist both rational and irrational numbers so the largest value of f in any interval is 1 and the smallest is 0, 1 times 1/n is 1/n and 0 times 1/n is 0. Adding 1/n n times gives 1 and adding 0 n times gives 0.

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