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Hi guys I am doing a question. Well, attempting a few questions about evaluating darboux sums.

I fully understand the evaluation of setting up the upper and low sum integrals what I am having some confusion with is when evaluating the following;

$$\text{sup} \{L(P,f): P \text{ is a member of the partition } [a,b]\}$$

$$\text{inf}\{U(P,f): P \text{ is a member of the partition} [a,b]\}$$ for any given question.

Example: $g:[0,1] \rightarrow \mathbb R$ $$g(x) = \begin{cases} -1, & x = 1\ \\ 1, & x \neq 1 \ \end{cases} $$ When I evaluated the Upper sum I got a value of $1$ when I evaluated the lower sum I got the expression $2X - 1$

therefore Lower sum $= \text{sup}\{2x - 1 : X \text{ is a member of } [0,1]\}$

my thinking is if I am trying to acquire the largest value for the expressions I would substitute the smallest value $0$ which will give

$\text{sup} \{2(0) - 1\} = 1$

where the sup of the lower darboux sum is $1$, but I am not sure if this is correct and I am having the same problem when analyzing the infimum for the upper darboux sum. can anyone clear this for me please?

Donlans Donlans
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John
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  • how did you get $2x - 1$, what is $x$? – Donlans Donlans Oct 09 '19 at 01:19
  • thats the final expressions when i evaluated the lower sum for the question its suppose to be 2X_n-1 where n-1 is subscripts but i just leave it as 2x for simplicity @DonlansDonlans. The question was worked as shown above for the lower sums but im trying to understand how they arrived at the value the sup = 1 – John Oct 09 '19 at 01:25

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