Riemann Sum...

- Clarence

Riemann Sum...

- schrodinger

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- Clarence

##### 1 Attachment

- Clarence

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

- Clarence

And what I preceived the second one to be \[(\frac{ 12i ^{2} }{ n ^{2} } + \frac{ 40i }{ n }) (\frac{ 2 }{ n })\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.