Find the value of the following definite integral using the limit of a Riemann sum, and express as a common fraction. You should use the formulas for ∑i, ∑i^2, and ∑i^3 given in class.
S(Integral from -2 to 1) (x^3-3x)dx

- anonymous

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- anonymous

The actual question is attached

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- anonymous

i attached the actual question

- amistre64

i see it... so what are the 3 summation formulas?

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## More answers

- anonymous

let me get them from my notes

- amistre64

if I recall correctly, the Reiman limits are just taking the area of slices as the thickness of dx goes to zero right?

- anonymous

yes and i she didnt give us the formulas even though she said she did, let me check the book.

- amistre64

(f(x+dx) - f(x)) dx; we could also take the "average height" between f(x+dx) and f(x)...

- amistre64

dx just stands for delta x, any arbitrary number... where dx = (b-a)/n... any of this make sense?

- anonymous

here is the formulas

- amistre64

dx = (1--2)/n = (1+2)/n = 3/n

- anonymous

k i got that far

- anonymous

a=-2 b=1

- amistre64

yep, them pictures is what I said lol

- amistre64

the limit as n-> inf......

- anonymous

so then I got lim as n goes to infinity of the summation [(-2+3i/n)^3-3(-2+3i/n)](3/n)

- amistre64

need to work out something like this right?

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- anonymous

yea but that one is less complicated haha

- amistre64

lol .... use the Esummation for the cubic function

- anonymous

the answer we are going for is 3/4 for reference

- anonymous

is my first step correct?

- amistre64

we can split the addition up into to parts just like regualr integration, because regular integration is the result of this reimann sum stuff, so do x^3 and x sepertaely

- anonymous

ok

- amistre64

[E] x^3 dx - [E] 3x dx the 3 can move out
[E] x^3 (3/n) - 3 [E] x (3/n)

- anonymous

\[\lim_{n \rightarrow \infty}\sum_{i=1}^{n}[(-2+3i/n)^3-3(-2+3i/n)](3/n)\]

- amistre64

im lost on what that (-2+3i/n) stuff means, help me out :)

- anonymous

its part of the reimann sum

- anonymous

the formulas we have to use, that is the n^3

- anonymous

i^3 my mistake

- amistre64

\[\sum_{} x^{3}(3/n) -3 \sum_{}x(3/n)\]
is whats in my head

- anonymous

well the integral is \[\int\limits_{-2}^{1}(x^3-3x)dx\]

- amistre64

yep, and the reiman is proof that we can split that into 2 parts right?
\[\int\limits_{} x^3 dx -3\int\limits_{}x dx\]

- anonymous

yes that works

- amistre64

∑ x^3 (3/n)−3 ∑ x (3/n) is its equivalent,
tell me if im wrong... but whats our next step after this?

- amistre64

lets work each one on its own...ok

- anonymous

plug in x^3 and x

- anonymous

and thats what i showed above

- anonymous

as one complete summation multiplied by 3/n

- amistre64

you say "plug in" but i have no idea how that happens, im pretty much an idiot when it comes to the limit stuff, show me step by step please how we "plug in" to get:
lim n→∞ ∑ i=1 n [(−2+3i/n)^3 −3(−2+3i/n)](3/n)

- anonymous

k let me show you

- anonymous

hopefully this helps

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- anonymous

i need to break it even further for you

- amistre64

lol.... well, I got that part :)
How do we move from that to the "formula for"
[E] i^3 and [E] i
I am not familiar with those formulas, yet :)

- anonymous

its just pulling things out and taking individual summations, should i show you where i am so far?

- amistre64

yes please
∑ i^3 = (1/4)n4 + (1/2)n3 + (1/4)n2 is what I found online

- anonymous

\[\lim_{n \rightarrow \infty}\sum_{i=1}^{n}3/n[-8+12i/n+12i/n-18i^2/n^2+12i/n-18i^2/n^2-18i^2/n^2-27i^3/n^3-3(-2+3i/n)\]

- anonymous

wow i guess my thing was too long, thats only half of it

- anonymous

im so confused lol

- amistre64

:) heres what I found that helps me out

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- anonymous

ok that works, lets start over then from 3/n being delta x

- amistre64

since 3/n = delta x, we simply multiply that to x^3?
to get:
3(x^3)/n ?? is that right?

- anonymous

hmm sure, keep going because i wont know where you are going till you go further

- amistre64

\[3\sum_{n=1}^{\infty} (x^3)/n\]
Does this look right?

- anonymous

no but keep going

- amistre64

lol ....if it dont look right, then why should I keep going :)

- anonymous

im trying to make sense of where you are going

- amistre64

....... you gotta drive some of the way, im lost here for the moment

- anonymous

oh k , im trying to think...problem is i cant find the equation that would help you

- anonymous

let me see...

- anonymous

where are you lost?

- anonymous

how i formulated the entire sum?

- anonymous

A=lim_{N→∞}Δx∑_{j=1}^{N}f(a+jΔx)

- anonymous

does this help

- anonymous

i=j

- amistre64

yup.... im still at the split and trying to apply the formula for [E] x^3 to the left part.
the top part of the [E] is gonna be (b-a) right? which we determined to be 3.... but its not sinking in....

- anonymous

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- anonymous

heres a better look

- amistre64

ok....so delta x is a constant that gets pulled out to the left ... am I seeing that right?

- anonymous

yes

- anonymous

but you see how the i is appearing now. In the equation, it names it j, but the common known term is i

- amistre64

and the "i" is simply each iteration of the sums....right?

- anonymous

yes

- anonymous

the integral should =3/4 after the long process, im trying to get through all the steps

- anonymous

thats the problem

- amistre64

yeah....when i read over that reaimann stuff I gloosed at it and figured, its true, so why do it the long way :)

- anonymous

lol cause my professor doesnt want to make life easy

- amistre64

f(a + i /x\) the i is each iterations from 1 to infinity. is the a the same as in our b-a interval?

- anonymous

yes

- anonymous

-2

- anonymous

limn→∞∑i=1n[(−2+3i/n)3−3(−2+3i/n)](3/n)

- amistre64

f(-2 + 3i/n) is a standard interation then, is that correct?

- anonymous

perfect

- anonymous

dont forget the cubed

- amistre64

is the "n" from 3/n is the one we are limiting to infinity right?

- anonymous

yes perfect

- amistre64

f^3(x)...yeah yeah

- anonymous

hey if you dont wanna type the dumb sigma and lim n thing, just say lim and sigma, ill know what you are saying or use the equation feature on here

- anonymous

dumb summation sign i mean*

- amistre64

n[(−2+3i/n)3−3(−2+3i/n)](3/n) this is you combining the terms together right??

- anonymous

the n in front shouldnt be there that was part of the format for the summation

- amistre64

ok......
(sigma) (-2 +3i/n) (3/n) is the right term then correct?

- anonymous

summation but yes that is correct

- anonymous

isnt it -3(-2 +3i/n) (3/n)

- anonymous

you forgot the -3x

- amistre64

\[\sum_{} (-2+3i/n) (3/n)\]
would be the \[\sum_{} x \]
formula part right?
the -3 part of the -3x doesnts need to be included in this process if i recall correctly

- anonymous

im pretty sure you do because its part of the integral function

- anonymous

but if you are confident, you understand this better then me so ill take your word for it

- amistre64

\[\int\limits_{?}^{?} -3x = -3\int\limits_{?}^{?}x \rightarrow-3 x^2/2\]

- anonymous

well you are going to have a -8 from the cube equation so its better if you keep everything together when you distribute it right?

- anonymous

i guess it doesnt matter go ahead, you sound excited because i think you understand it now

- anonymous

do your thing, ill just learn from you

- amistre64

keep this split up, dont try to solve it all as one summation, work that parts.... use the formulas for [E]x and [E]x^3 seperately

- anonymous

it still will work out taking as pieces?

- amistre64

\[\sum{}(-2 +3/n)(3/n)\] fits into
n(n+1)/2 somehow, and yes, it will work easier and just as good if we split it up

- amistre64

think of it as the granny way of doing integrals; its actually proof that we can split integerals up and work them

- amistre64

forgot to pu an i in there :)

- anonymous

ok let me write step by step following you on actual paper

- amistre64

we can split derivatives up and work them, integrals are just eh opposite of derivatives.....

- amistre64

this computer is having a hard time keeping up with formating the "equation"editor stuff, can we start a new question post?

- anonymous

yea

- anonymous

let me make a new one

- anonymous

k go to the top of the question list

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