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Pts] (Extra Credit) Suppose for eneh e N have &n open interval In On+1 bn+ [ bn for all n € N Prove that 0Az1 In # 2.(a1,5 bn) and suppose...

Question

Pts] (Extra Credit) Suppose for eneh e N have &n open interval In On+1 bn+ [ bn for all n € N Prove that 0Az1 In # 2.(a1,5 bn) and suppose

pts] (Extra Credit) Suppose for eneh e N have &n open interval In On+1 bn+ [ bn for all n € N Prove that 0Az1 In # 2. (a1,5 bn) and suppose



Answers

Let $P=\left\{x_{0}, x_{1}, x_{2}, \ldots, x_{n-1}, x_{n}\right\}$ be a regular partition of the interval $[a, b] .$ (Sec Exercise $31 .$ ) Show that if $f$ is continuous and decreasing on $[a, b],$ then $$U_{f}(P)-L_{f}(P)=[f(a)-f(b)] \Delta x$$.

The problem, we are told to let p be a regular partition of the interval from 1-4. And to let XK star be the right endpoint of each interval. Were then asked to write the given some the limit as N approaches infinity of the sum from K equals one to infinity. Or to end rather of one plus three K over n cubed times three over N as a definite integral. One thing to note is that the general form of a sum like this would be the sum from K equals one up to n of f of X K. Star times delta X K or delta X K, I should say. So we can see based on the form that we have that delta X K appears to be three over N. This is consistent with what we have regarding the interval then considering what we have as the interior of the function Well, we have one plus three K over N. If we consider when K equals zero, That would become just a value of one. Then when we have k equals n, we should celebrate that, that has a value of one there. Then when K equals end, that would have a value of four. So we note that our interval is from X equals one up to x equals four. So it would seem then that one plus three K over n would correspond to just X. So we can write that our function F of X. That we would be integrating would be X cubed. So having all of that, we can write our definite integral form as the integral from 1 to 4 of x cubed dx.

For this problem, we are told to let p be a regular partition of the indicated interval from 0 to 2 and X K star be the right endpoint of each sub interval were then asked to write the given some the some from Kegels one up to N of one plus two K over n times two over N as a definite integral. So I'll note that the general form of a sum like this should be from Kegels one up to end of F of X K star times delta X K. So we note that it looks like our F of X K star. Well, we'll have to get we'll have to get to that in a second. We can definitely see that it seems our delta X k would equal to over end. Well, we also have that our interval is from 0-2. So if we're partitioning this regularly then our delta X would have to be too over end. So that does confirm that Yes, our delta x K here is the two over. And which means that we'd have to have that our F of X K star equals one plus two K over N. Now, does this check out? Well, we would have that starting, if we put in an equal zero, we would start with a value of one. And then as we or excuse me, if we put in K equals zero, we would start at a value of just one. And then as we bring K up to end, we would have one plus uh two times or we could have just one plus two giving us a value of three. So it does seem that in this case our XK star particularly because we are looking at the right endpoint, if we represent this as um if we represent XK star as being the two K over N, then that does mean that when we have X of zero or actually excuse me, that should be X of one, then we would have, we were told that we are looking at the right end point of each sub interval, So X one, we would expect to be two over N and then we'd have that X N is going to equal to, which does mean that yes, we're touching the right end, right endpoint of each sub interval, which then in turn means that if we want to write this as F of X, it would just be one plus X. So to translate this into a definite integral, it would be the integral from 0 to 2 of one plus XDX. That is the definite integral form

Hello. So here we have a given sequence. A sub N. Equals the quantity hoops. Uh See to the end plus the to the end. And then I'll raise to the one over and power. So first to check that were bounded. Well we have here that zero is going to be less than see to the end plus D. To the end which is going to be less than two. The to the end to the one over end. Which is going to be equal to to to the uh one over end time's D. Which is going to be less than two D. So therefore the sequence is founded. Okay then to check for um Monitor city. Well we have that um a Sub and plus one um To the n. Plus one is going to be equal to see to the end plus one plus D. To the N. Plus one. Um Which here is going to be equal to just a C. To the end. Um Yeah plus D. To the end. Uh To the um well to the one plus one over end that's to the N plus one over end. Um Which is just gonna be equal to a sub N. To the n. Plus one. So then taking the N plus one route on both sides. We then get that a sub N plus one is going to be equal to a sub N. To the one over N. So then we have that A sub N plus one to the end is going to be equal to a sub N. Which implies that a sub N plus one is less than a seven. Therefore the sequence is going to be mon atomically decreasing.

Let's legs A. Be less than B. And mhm. Equal to X. Notes. Excellent. Two X. N. Mhm. Read the partition the the partition of the interval A. B. Such. That's sledge. That's the sub intervals Sub intervals XI -1 X. I. Or how same length Its be minus age divided by mm. So since since F is continues and decreased sale on the interview uh baby then it implies that I will upper some of F would be equal to you have F. X nuts Plus FH. one last two. The F. C. We have FXN -1 then allure. So would be 58 1 less if eggs too less is it's in. So thanks we have the upper some for f minus the lua. Some for F would be equal. Soon we have this minus that. So don't give us we have F. Ex nuts. Mm X. My name is F H. M. Less. This would give us FX. one -F. x. one less. F. X. In my next one minus F eight a.m. And this is equal to if X nuts my nets F. S. X. M. Thanks to be equal to you have F A minus F. B. Thanks as the final. And so


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