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anonymous
 one year ago
Riemann Sum...
anonymous
 one year ago
Riemann Sum...

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got as far as figuring out delta x to be 2/n

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And what I preceived the second one to be \[(\frac{ 12i ^{2} }{ n ^{2} } + \frac{ 40i }{ n }) (\frac{ 2 }{ n })\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ah its been so long since ive done this. can you jog my memory when it says the general formulae of \[x _{i}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You mean the part where \[x _{i}=0+i \Delta x\]?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah is that what the formulae is for that bit? ah i should know this cause its so fundemental to calc but i only vaguely remember haha

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0perhaps @IrishBoy123 can help he's a bit of a god

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Aha, I think so. There was a previous question regarding Riemann Sum but tackling it from the left and the question is a bit different. I could potentially upload that and tell you what I got for if you'd prefer it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0perhaps that could help but since we have n divisions, then our partitions are 0, \[\Delta x, 2\Delta x, ... , (n2) \Delta x, (n1) \Delta x, \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for right hand side, \[\Delta x \left[ f( a + \Delta x ) + f(a + 2 \Delta x)+\cdots+f(b) \right].\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that is your approx function if you do the right end method

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it looks like they are leading you the way step by step, i'm just unsure what it means by the general formulae of x(i)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is this like a web this where you enter an answer and you can see if its right or wrong?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It isn't a web thing, more just like an extended question that the professor gave at the end of lecture. I've been trying to follow a link that my friend shared with me on Facebook. It seems pretty legit despite being a yahoo answer source, but even then as I say that it was hard for me to get my head around the answer: https://answers.yahoo.com/question/index?qid=20110910015823AARrDF4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sweet, i have something to work with now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hope that it'd be more useful to you than it was to me! :P

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2https://gyazo.com/ce50525553b52e9dbd904fb7970fc13a i'm with you on this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok so x(i) is defined as \[x _{i}=i \Delta x\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2\[x_i = i \frac{2}{n}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you know \[\Delta x= \frac{ 2 }{ n }\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2i'm pretty rusty on this too so careful with the symbology but i worked it through and got 48 both ways for the integral at least that means we are on track

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm understanding what you two are saying so far so that's good I suppose. :p

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now since you know {x _{i}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sub that into the x variables in the function 3x^2+20x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so you get \[3x _{i}^2+20x _{i}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can you do this long hand @Clarence

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its bringing back memories now ahaha

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you should get \[\frac{ 12i^2 }{ n^2 }+\frac{ 60i }{ n }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think I got 40 rather than 60 for that bit...

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2\[f(x_i) = 3 (\frac{2i}{n})^2 + 20 (\frac{2i}{n})\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2let's agree it before we move on :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Aha, we all make these mistakes :)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2so we are here, right?? \[\frac{ 12i^2 }{ n^2 }+\frac{ 40i }{ n }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah so \[f(x _{i})=\frac{ 12i^2 }{ n^2 }+\frac{ 40i }{ n }\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2yes, i forgot the \(f(x_i)\) cool next bit

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[f(x _{i}) \Delta x=(\frac{ 12i^2 }{ n^2 }+\frac{ 40i }{ n }) \Delta x\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and you know what delta x is right clarence?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yup, so just input that into the equation to get 2/n at the right end

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2and the answer is???!!!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0simply sub 2/n into delta x to get: \[f(x _{i}) \Delta x=(\frac{ 12i^2 }{ n^2 }+\frac{ 40i }{ n })(\frac{ 2 }{ n })\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which is what you got!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Before I got stuck, yes ;)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now we approximate the area

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2my go!! \[f(x _{i}) \Delta x=\frac{ 24i^2 }{ n^3 }+\frac{ 80i }{ n^2 }\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2do we agree that?!

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2cool, now for the real fun bit!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[A _{app}\approx \sum_{i=0}^{n}(f(x _{i}) \Delta x\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think thats by definition...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441541710476:dw somthing like that. this is not your graph but it just shows that the area approximation of one rectangle is Delta x multiplied by the function governing the rectangle

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2yes, we approximate the area by adding up all the little elements \[\Sigma_{i = 1}^{\infty} f(x _{i}) \Delta x=\Sigma_{i = 1}^{\infty} \left[ \frac{ 24i^2 }{ n^3 }+ \frac{ 80i }{ n^2 } \right]\] \[=\Sigma_{i = 1}^{\infty} \frac{ 24i^2 }{ n^3 }+\Sigma_{i = 1}^{\infty} \frac{ 80i }{ n^2 }\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2and now we use these https://gyazo.com/776b4188ce5e7548a6cad4c61eda1c3a

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2\[=\frac{24}{n^3}\Sigma_{i = 1}^{\infty} i^2+\frac{ 80 }{ n^2 } \ \Sigma_{i = 1}^{\infty} i\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0should be hunky dory from there @Clarence

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2i am latexing like a crazy man here :))

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2@Clarence take your time we can stop here for a while if that suits!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Never thought that maths could look so pretty... But now I don't even know what to do with the information that I have!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you are given critical information in the latter part of the question

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sum_{i}^{n}i=\frac{ n(n+1) }{ 2 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and \[\sum_{i}^{n}i^2=\frac{ n(n+1)(2n+1) }{ 6 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can we not just sub these corresponding summations into our question?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0where @IrishBoy123 mentioned that \[A=\frac{24}{n^3}\Sigma_{i = 1}^{\infty} i^2+\frac{ 80 }{ n^2 } \ \Sigma_{i = 1}^{\infty} i\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0or not to be confused with the infinity clarence, \[A=\frac{24}{n^3}\Sigma_{i = 1}^{n} i^2+\frac{ 80 }{ n^2 } \ \Sigma_{i = 1}^{n} i\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Aha, yes, I can see that now! Sorry, totally forgot about the original question, I'm not the brightest freshman about if you couldn't already tell :p

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2@Clarence if you're doing this stuff, you *are* very bright this is amazing maths as we shall see at the end!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so fundamental to calculus.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can you solve the area now clarence?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its just simple substitution

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2this is a pig latexwise shall i just do it? it gets easier when we lose all the terms

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2ok give me a few minutes to copy and paste it all together

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2\[A=\frac{24}{n^3} \frac{ n(n+1)(2n+1) }{ 6 }+\frac{ 80 }{ n^2 } \ \frac{ n(n+1) }{ 2 }\] this good?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2excellent so we now have a formula that approximates the area under the curve. we choose n as the number of elements and we plug it in and find the area let's simplify it remembering we want to set \(n = \infty\) to get the perfect answer

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2\[A=\frac{24}{n^3} \frac{ 2n^3 + 3n^2 +n}{ 6 }+\frac{ 80 }{ n^2 } \ \frac{ n^2 +n }{ 2 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, I got up to that, so far :p

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2\[A=4 (2 + \frac{3}{n} +\frac{1}{n^2})+40 (1 +\frac{1}{n} )\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Clarence what happens now if n approaches infinity?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2some more simplifying \[A= 48 + \frac{52}{n} +\frac{4}{n^2} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0haha more simplifying, love it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0spelling out the answer haah

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2yep! totally unnecesary of course as we're setting n to infinity :((

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Aha, maths sure loves their infinities :p

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2\[\int_{0}^{2} \ 3x^2 + 20x \ dx\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2if that's not 48, we are all doomed is it?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2please let it be so!!

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2beautiful beautiful stuff thank you both for involving me, really enjoyed it @Clarence are you good to go with this now??

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2if not keep the thread open @chris00 and i will get a notification if there is more to do

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0it appears to me that you ran thru the LEFT side rundown of this

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0on an interval from a to b, the left side setup is: \[x_i=a+i\frac{ba}{n}~:~0\le i<n\] this hits all the high points starting at the left end and moving in 'i'ncrements of the partition widths. in this case a=0 your inputs worked out. the right side setup is either an adjustment of the interval of i: 0 < i <= n this way we simply hit the right side of the partitions.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0another alternative to the right side setup is simply to replace the a+ with b

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0hmmm, ok, it was a hard read thru for these tired old eyes, but it does look like you implemented i going from 1 to n, so my observation is moot :) good job yall

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2its a really long thread but we used \(x_i = i \frac{2}{n}\) (and \(\Delta x= \frac{ 2 }{ n }\)) and ran it from i=1. so for n = i = 1 we are up at the right hand end.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2and we were spoonfed because the question was so well laid out....:))

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0if a had been say 3 (not 0),and b was 5 (not 2), our interval width is still 53=2 and we divide that into n partitions.\[\Delta x=\frac2n\] then the value of x for the ith iterations would have been: \[x_i=3+i\frac2n\] and \[f(x_i)=f(3+i\frac2n)\] just wanting to make sure that we remember that xi is not in general the same as: i /\x

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2gotcha. that particular *snag* was missing from our problem here. we had \[f(x_i)=f(0+i\frac2n) = f(i\frac2n)\] thank you @amistre.
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