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?