# A partition of an interval to reduce the difference between the upper and lower Darboux sums

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 $q$.

• And... what did you try? – Did Apr 2 '13 at 19:16
• How can we retitle it? – rschwieb Apr 2 '13 at 19:22
• I corrected it and ı used upper and lower sums but I can not solve – e.t Apr 2 '13 at 20:11

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.