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Let

$$ f(x) = \begin{cases} 2x+1, & x\in [2,4] \\ 7-x, & x\in(4,4.5) \\ 3, & x \in[4.5,6] \end{cases} $$

For $q = 1/4$, find a partition of $[2, 6]$ such that the difference between the upper sum and the lower sum of $f(x)$ is less than $q$.

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    $\begingroup$ And... what did you try? $\endgroup$ – Did Apr 2 '13 at 19:16
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    $\begingroup$ How can we retitle it? $\endgroup$ – rschwieb Apr 2 '13 at 19:22
  • $\begingroup$ I corrected it and ı used upper and lower sums but I can not solve $\endgroup$ – e.t Apr 2 '13 at 20:11
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There are two sources of the discrepancy between upper and lower sums: jump discontinuities, and continuous change of the function. Jump discontinuities should be boxed into tiny intervals. The effect of continuous change is reduced by choosing a fine partition. Or you can ignore all this finesse and just use partition into $N$ subintervals, where $N$ is sufficiently large. Each will have length $4/N$. Consider the following:

  1. At most two subintervals will be affected by the jump of size $6$ at $x=4$.
  2. At most two subintervals will be affected by the jump of size $0.5$ at $x=4.5$
  3. On each subintervals contained in $[2,4]$ the function changes by $2\cdot 4/N$.
  4. On each subintervals contained in $[4,4.5)$ the function changes by $1\cdot 4/N$.

Let's total these, not forgetting to multiply everything by the length $4/N$: $$\frac{4}{N}(2\cdot 6+2\cdot 0.5+ N\cdot 8/N+ N\cdot 4/N) \tag1$$ where instead of counting how many intervals fit in 3 or 4 I just say "at most $N$". To hell with finesse. Now you can choose $N$ to make (1) as small as you wish.

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