Clarence
  • Clarence
Riemann Sum...
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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Clarence
  • Clarence
1 Attachment
Clarence
  • Clarence
I got as far as figuring out delta x to be 2/n
Clarence
  • Clarence
And what I preceived the second one to be \[(\frac{ 12i ^{2} }{ n ^{2} } + \frac{ 40i }{ n }) (\frac{ 2 }{ n })\]

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anonymous
  • anonymous
ah its been so long since ive done this. can you jog my memory when it says the general formulae of \[x _{i}\]
Clarence
  • Clarence
You mean the part where \[x _{i}=0+i \Delta x\]?
anonymous
  • anonymous
yeah 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
  • anonymous
perhaps @IrishBoy123 can help he's a bit of a god
Clarence
  • Clarence
Aha, 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
  • anonymous
perhaps that could help but since we have n divisions, then our partitions are 0, \[\Delta x, 2\Delta x, ... , (n-2) \Delta x, (n-1) \Delta x, \]
anonymous
  • anonymous
2
anonymous
  • anonymous
hmm that won't help
anonymous
  • anonymous
for right hand side, \[\Delta x \left[ f( a + \Delta x ) + f(a + 2 \Delta x)+\cdots+f(b) \right].\]
anonymous
  • anonymous
where a=0 b=2
anonymous
  • anonymous
that is your approx function if you do the right end method
anonymous
  • anonymous
it 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
  • anonymous
is this like a web this where you enter an answer and you can see if its right or wrong?
Clarence
  • Clarence
It 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
  • anonymous
sweet, i have something to work with now
Clarence
  • Clarence
Hope that it'd be more useful to you than it was to me! :P
IrishBoy123
  • IrishBoy123
https://gyazo.com/ce50525553b52e9dbd904fb7970fc13a i'm with you on this
IrishBoy123
  • IrishBoy123
if that helps :p
anonymous
  • anonymous
ok so x(i) is defined as \[x _{i}=i \Delta x\]
IrishBoy123
  • IrishBoy123
yes
IrishBoy123
  • IrishBoy123
\[x_i = i \frac{2}{n}\]
anonymous
  • anonymous
you know \[\Delta x= \frac{ 2 }{ n }\]
anonymous
  • anonymous
yep!^^
anonymous
  • anonymous
follow @Clarence
IrishBoy123
  • IrishBoy123
i'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
  • anonymous
haha
Clarence
  • Clarence
I'm understanding what you two are saying so far so that's good I suppose. :p
anonymous
  • anonymous
now since you know {x _{i}\]
anonymous
  • anonymous
\[x _{i}\]
anonymous
  • anonymous
sub that into the x variables in the function 3x^2+20x
IrishBoy123
  • IrishBoy123
@chris00 yes!
anonymous
  • anonymous
so you get \[3x _{i}^2+20x _{i}\]
anonymous
  • anonymous
can you do this long hand @Clarence
anonymous
  • anonymous
its bringing back memories now ahaha
anonymous
  • anonymous
you should get \[\frac{ 12i^2 }{ n^2 }+\frac{ 60i }{ n }\]
Clarence
  • Clarence
I think I got 40 rather than 60 for that bit...
IrishBoy123
  • IrishBoy123
\[f(x_i) = 3 (\frac{2i}{n})^2 + 20 (\frac{2i}{n})\]
anonymous
  • anonymous
hahah opps
IrishBoy123
  • IrishBoy123
let's agree it before we move on :-)
anonymous
  • anonymous
i can math
anonymous
  • anonymous
:P
Clarence
  • Clarence
Aha, we all make these mistakes :)
IrishBoy123
  • IrishBoy123
so we are here, right?? \[\frac{ 12i^2 }{ n^2 }+\frac{ 40i }{ n }\]
anonymous
  • anonymous
yeah so \[f(x _{i})=\frac{ 12i^2 }{ n^2 }+\frac{ 40i }{ n }\]
IrishBoy123
  • IrishBoy123
yes, i forgot the \(f(x_i)\) cool next bit
anonymous
  • anonymous
\[f(x _{i}) \Delta x=(\frac{ 12i^2 }{ n^2 }+\frac{ 40i }{ n }) \Delta x\]
anonymous
  • anonymous
and you know what delta x is right clarence?
IrishBoy123
  • IrishBoy123
exellent!
Clarence
  • Clarence
Yup, so just input that into the equation to get 2/n at the right end
IrishBoy123
  • IrishBoy123
and the answer is???!!!!
anonymous
  • anonymous
simply 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
  • anonymous
which is what you got!
Clarence
  • Clarence
Before I got stuck, yes ;)
anonymous
  • anonymous
now we approximate the area
IrishBoy123
  • IrishBoy123
my go!! \[f(x _{i}) \Delta x=\frac{ 24i^2 }{ n^3 }+\frac{ 80i }{ n^2 }\]
IrishBoy123
  • IrishBoy123
do we agree that?!
anonymous
  • anonymous
yep!
IrishBoy123
  • IrishBoy123
cool, now for the real fun bit!
anonymous
  • anonymous
\[A _{app}\approx \sum_{i=0}^{n}(f(x _{i}) \Delta x\]
anonymous
  • anonymous
i think thats by definition...
anonymous
  • anonymous
|dw: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
  • IrishBoy123
yes, 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
  • IrishBoy123
and now we use these https://gyazo.com/776b4188ce5e7548a6cad4c61eda1c3a
IrishBoy123
  • IrishBoy123
\[=\frac{24}{n^3}\Sigma_{i = 1}^{\infty} i^2+\frac{ 80 }{ n^2 } \ \Sigma_{i = 1}^{\infty} i\]
anonymous
  • anonymous
beat me too it haha
anonymous
  • anonymous
should be hunky dory from there @Clarence
IrishBoy123
  • IrishBoy123
i am latexing like a crazy man here :-))
IrishBoy123
  • IrishBoy123
@Clarence take your time we can stop here for a while if that suits!
Clarence
  • Clarence
Never thought that maths could look so pretty... But now I don't even know what to do with the information that I have!
anonymous
  • anonymous
you are given critical information in the latter part of the question
anonymous
  • anonymous
\[\sum_{i}^{n}i=\frac{ n(n+1) }{ 2 }\]
anonymous
  • anonymous
and \[\sum_{i}^{n}i^2=\frac{ n(n+1)(2n+1) }{ 6 }\]
anonymous
  • anonymous
can we not just sub these corresponding summations into our question?
IrishBoy123
  • IrishBoy123
yes we can!
anonymous
  • anonymous
where @IrishBoy123 mentioned that \[A=\frac{24}{n^3}\Sigma_{i = 1}^{\infty} i^2+\frac{ 80 }{ n^2 } \ \Sigma_{i = 1}^{\infty} i\]
anonymous
  • anonymous
or 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
  • anonymous
see the relation?
Clarence
  • Clarence
Aha, 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
  • IrishBoy123
@Clarence if you're doing this stuff, you *are* very bright this is amazing maths as we shall see at the end!!
anonymous
  • anonymous
so fundamental to calculus.
anonymous
  • anonymous
can you solve the area now clarence?
Clarence
  • Clarence
Maybe... :p
anonymous
  • anonymous
its just simple substitution
IrishBoy123
  • IrishBoy123
this is a pig latex-wise shall i just do it? it gets easier when we lose all the terms
anonymous
  • anonymous
go for it!
IrishBoy123
  • IrishBoy123
ok give me a few minutes to copy and paste it all together
IrishBoy123
  • IrishBoy123
\[A=\frac{24}{n^3} \frac{ n(n+1)(2n+1) }{ 6 }+\frac{ 80 }{ n^2 } \ \frac{ n(n+1) }{ 2 }\] this good?
Clarence
  • Clarence
That's what I got
IrishBoy123
  • IrishBoy123
excellent 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
  • IrishBoy123
\[A=\frac{24}{n^3} \frac{ 2n^3 + 3n^2 +n}{ 6 }+\frac{ 80 }{ n^2 } \ \frac{ n^2 +n }{ 2 }\]
IrishBoy123
  • IrishBoy123
everyone agree?
anonymous
  • anonymous
yep
Clarence
  • Clarence
Yeah, I got up to that, so far :p
IrishBoy123
  • IrishBoy123
\[A=4 (2 + \frac{3}{n} +\frac{1}{n^2})+40 (1 +\frac{1}{n} )\]
anonymous
  • anonymous
good so far
anonymous
  • anonymous
@Clarence what happens now if n approaches infinity?
IrishBoy123
  • IrishBoy123
some more simplifying \[A= 48 + \frac{52}{n} +\frac{4}{n^2} \]
anonymous
  • anonymous
haha more simplifying, love it
anonymous
  • anonymous
spelling out the answer haah
IrishBoy123
  • IrishBoy123
yep! totally unnecesary of course as we're setting n to infinity :-((
Clarence
  • Clarence
Aha, maths sure loves their infinities :p
IrishBoy123
  • IrishBoy123
\[\int_{0}^{2} \ 3x^2 + 20x \ dx\]
IrishBoy123
  • IrishBoy123
if that's not 48, we are all doomed is it?
IrishBoy123
  • IrishBoy123
please let it be so!!
Clarence
  • Clarence
I got 48 for that..
anonymous
  • anonymous
sweet as
anonymous
  • anonymous
what a journey
anonymous
  • anonymous
:P
IrishBoy123
  • IrishBoy123
beautiful beautiful stuff thank you both for involving me, really enjoyed it @Clarence are you good to go with this now??
anonymous
  • anonymous
such a pleasure
Clarence
  • Clarence
Aha, I think so :)
IrishBoy123
  • IrishBoy123
if not keep the thread open @chris00 and i will get a notification if there is more to do
Clarence
  • Clarence
Awesome, thanks! :)
amistre64
  • amistre64
it appears to me that you ran thru the LEFT side rundown of this
amistre64
  • amistre64
on an interval from a to b, the left side setup is: \[x_i=a+i\frac{b-a}{n}~:~0\le i
amistre64
  • amistre64
another alternative to the right side setup is simply to replace the a+ with b-
amistre64
  • amistre64
hmmm, 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
  • IrishBoy123
its 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
  • IrishBoy123
and we were spoon-fed because the question was so well laid out....:-))
amistre64
  • amistre64
if a had been say 3 (not 0),and b was 5 (not 2), our interval width is still 5-3=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
  • IrishBoy123
gotcha. 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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