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

Question

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 · Answer

The actual question is attached

anonymous · Answer

i attached the actual question

amistre64 · Answer

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

anonymous · Answer

let me get them from my notes

amistre64 · Answer

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

anonymous · Answer

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

amistre64 · Answer

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

amistre64 · Answer

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

anonymous · Answer

here is the formulas

amistre64 · Answer

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

anonymous · Answer

k i got that far

anonymous · Answer

a=-2 b=1

amistre64 · Answer

yep, them pictures is what I said lol

amistre64 · Answer

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

anonymous · Answer

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

amistre64 · Answer

need to work out something like this right?

anonymous · Answer

yea but that one is less complicated haha

amistre64 · Answer

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

anonymous · Answer

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

anonymous · Answer

is my first step correct?

amistre64 · Answer

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 · Answer

ok

amistre64 · Answer

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

anonymous · Answer

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

amistre64 · Answer

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

anonymous · Answer

its part of the reimann sum

anonymous · Answer

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

anonymous · Answer

i^3 my mistake

amistre64 · Answer

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

anonymous · Answer

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

amistre64 · Answer

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 · Answer

yes that works

amistre64 · Answer

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

amistre64 · Answer

lets work each one on its own...ok

anonymous · Answer

plug in x^3 and x

anonymous · Answer

and thats what i showed above

anonymous · Answer

as one complete summation multiplied by 3/n

amistre64 · Answer

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 · Answer

k let me show you

anonymous · Answer

hopefully this helps

anonymous · Answer

i need to break it even further for you

amistre64 · Answer

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 · Answer

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

amistre64 · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

im so confused lol

amistre64 · Answer

:)  heres what I found that helps me out

anonymous · Answer

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

amistre64 · Answer

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

anonymous · Answer

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

amistre64 · Answer

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

anonymous · Answer

no but keep going

amistre64 · Answer

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

anonymous · Answer

im trying to make sense of where you are going

amistre64 · Answer

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

anonymous · Answer

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

anonymous · Answer

let me see...

anonymous · Answer

where are you lost?

anonymous · Answer

how i formulated the entire sum?

anonymous · Answer

A=lim_{N→∞}Δx∑_{j=1}^{N}f(a+jΔx)

anonymous · Answer

does this help

anonymous · Answer

i=j

amistre64 · Answer

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 · Answer

attachment

anonymous · Answer

heres a better look

amistre64 · Answer

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

anonymous · Answer

yes

anonymous · Answer

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

amistre64 · Answer

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

anonymous · Answer

yes

anonymous · Answer

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

anonymous · Answer

thats the problem

amistre64 · Answer

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

anonymous · Answer

lol cause my professor doesnt want to make life easy

amistre64 · Answer

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 · Answer

yes

anonymous · Answer

-2

anonymous · Answer

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

amistre64 · Answer

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

anonymous · Answer

perfect

anonymous · Answer

dont forget the cubed

amistre64 · Answer

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

anonymous · Answer

yes perfect

amistre64 · Answer

f^3(x)...yeah yeah

anonymous · Answer

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 · Answer

dumb summation sign i mean*

amistre64 · Answer

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

anonymous · Answer

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

amistre64 · Answer

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

anonymous · Answer

summation but yes that is correct

anonymous · Answer

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

anonymous · Answer

you forgot the -3x

amistre64 · Answer

$$\sum_{} (-2+3i/n) (3/n)$$

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 · Answer

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