Riemann Sum...

- Clarence

Riemann Sum...

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- Clarence

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- Clarence

I got as far as figuring out delta x to be 2/n

- 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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## More answers

- 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

You mean the part where \[x _{i}=0+i \Delta x\]?

- 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

perhaps @IrishBoy123 can help he's a bit of a god

- 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

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

2

- anonymous

hmm that won't help

- anonymous

for right hand side,
\[\Delta x \left[ f( a + \Delta x ) + f(a + 2 \Delta x)+\cdots+f(b) \right].\]

- anonymous

where a=0 b=2

- anonymous

that is your approx function if you do the right end method

- 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

is this like a web this where you enter an answer and you can see if its right or wrong?

- 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

sweet, i have something to work with now

- Clarence

Hope that it'd be more useful to you than it was to me! :P

- IrishBoy123

https://gyazo.com/ce50525553b52e9dbd904fb7970fc13a
i'm with you on this

- IrishBoy123

if that helps :p

- anonymous

ok so x(i) is defined as \[x _{i}=i \Delta x\]

- IrishBoy123

yes

- IrishBoy123

\[x_i = i \frac{2}{n}\]

- anonymous

you know \[\Delta x= \frac{ 2 }{ n }\]

- anonymous

yep!^^

- anonymous

follow @Clarence

- 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

haha

- Clarence

I'm understanding what you two are saying so far so that's good I suppose. :p

- anonymous

now since you know {x _{i}\]

- anonymous

\[x _{i}\]

- anonymous

sub that into the x variables in the function 3x^2+20x

- IrishBoy123

@chris00 yes!

- anonymous

so you get \[3x _{i}^2+20x _{i}\]

- anonymous

can you do this long hand @Clarence

- anonymous

its bringing back memories now ahaha

- anonymous

you should get \[\frac{ 12i^2 }{ n^2 }+\frac{ 60i }{ n }\]

- Clarence

I think I got 40 rather than 60 for that bit...

- IrishBoy123

\[f(x_i) = 3 (\frac{2i}{n})^2 + 20 (\frac{2i}{n})\]

- anonymous

hahah opps

- IrishBoy123

let's agree it before we move on :-)

- anonymous

i can math

- anonymous

:P

- Clarence

Aha, we all make these mistakes :)

- IrishBoy123

so we are here, right??
\[\frac{ 12i^2 }{ n^2 }+\frac{ 40i }{ n }\]

- anonymous

yeah so \[f(x _{i})=\frac{ 12i^2 }{ n^2 }+\frac{ 40i }{ n }\]

- IrishBoy123

yes, i forgot the \(f(x_i)\)
cool
next bit

- anonymous

\[f(x _{i}) \Delta x=(\frac{ 12i^2 }{ n^2 }+\frac{ 40i }{ n }) \Delta x\]

- anonymous

and you know what delta x is right clarence?

- IrishBoy123

exellent!

- Clarence

Yup, so just input that into the equation to get 2/n at the right end

- IrishBoy123

and the answer is???!!!!

- 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

which is what you got!

- Clarence

Before I got stuck, yes ;)

- anonymous

now we approximate the area

- IrishBoy123

my go!!
\[f(x _{i}) \Delta x=\frac{ 24i^2 }{ n^3 }+\frac{ 80i }{ n^2 }\]

- IrishBoy123

do we agree that?!

- anonymous

yep!

- IrishBoy123

cool, now for the real fun bit!

- anonymous

\[A _{app}\approx \sum_{i=0}^{n}(f(x _{i}) \Delta x\]

- anonymous

i think thats by definition...

- 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

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

and now we use these
https://gyazo.com/776b4188ce5e7548a6cad4c61eda1c3a

- IrishBoy123

\[=\frac{24}{n^3}\Sigma_{i = 1}^{\infty} i^2+\frac{ 80 }{ n^2 } \ \Sigma_{i = 1}^{\infty} i\]

- anonymous

beat me too it haha

- anonymous

should be hunky dory from there @Clarence

- IrishBoy123

i am latexing like a crazy man here :-))

- IrishBoy123

@Clarence
take your time
we can stop here for a while if that suits!

- 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

you are given critical information in the latter part of the question

- anonymous

\[\sum_{i}^{n}i=\frac{ n(n+1) }{ 2 }\]

- anonymous

and
\[\sum_{i}^{n}i^2=\frac{ n(n+1)(2n+1) }{ 6 }\]

- anonymous

can we not just sub these corresponding summations into our question?

- IrishBoy123

yes we can!

- 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

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

see the relation?

- 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

@Clarence
if you're doing this stuff, you *are* very bright
this is amazing maths as we shall see at the end!!

- anonymous

so fundamental to calculus.

- anonymous

can you solve the area now clarence?

- Clarence

Maybe... :p

- anonymous

its just simple substitution

- IrishBoy123

this is a pig latex-wise
shall i just do it?
it gets easier when we lose all the terms

- anonymous

go for it!

- IrishBoy123

ok
give me a few minutes to copy and paste it all together

- 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

That's what I got

- 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

\[A=\frac{24}{n^3} \frac{ 2n^3 + 3n^2 +n}{ 6 }+\frac{ 80 }{ n^2 } \ \frac{ n^2 +n }{ 2 }\]

- IrishBoy123

everyone agree?

- anonymous

yep

- Clarence

Yeah, I got up to that, so far :p

- IrishBoy123

\[A=4 (2 + \frac{3}{n} +\frac{1}{n^2})+40 (1 +\frac{1}{n} )\]

- anonymous

good so far

- anonymous

@Clarence what happens now if n approaches infinity?

- IrishBoy123

some more simplifying
\[A= 48 + \frac{52}{n} +\frac{4}{n^2} \]

- anonymous

haha more simplifying, love it

- anonymous

spelling out the answer haah

- IrishBoy123

yep!
totally unnecesary of course as we're setting n to infinity :-((

- Clarence

Aha, maths sure loves their infinities :p

- IrishBoy123

\[\int_{0}^{2} \ 3x^2 + 20x \ dx\]

- IrishBoy123

if that's not 48, we are all doomed
is it?

- IrishBoy123

please let it be so!!

- Clarence

I got 48 for that..

- anonymous

sweet as

- anonymous

what a journey

- anonymous

:P

- IrishBoy123

beautiful beautiful stuff
thank you both for involving me, really enjoyed it
@Clarence
are you good to go with this now??

- anonymous

such a pleasure

- Clarence

Aha, I think so :)

- IrishBoy123

if not keep the thread open
@chris00 and i will get a notification if there is more to do

- Clarence

Awesome, thanks! :)

- amistre64

it appears to me that you ran thru the LEFT side rundown of this

- 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

another alternative to the right side setup is simply to replace the a+ with b-

- 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

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

and we were spoon-fed because the question was so well laid out....:-))

- 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

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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