anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
The actual question is attached
anonymous
  • anonymous
i attached the actual question
amistre64
  • amistre64
i see it... so what are the 3 summation formulas?

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anonymous
  • anonymous
let me get them from my notes
amistre64
  • 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
  • anonymous
yes and i she didnt give us the formulas even though she said she did, let me check the book.
amistre64
  • amistre64
(f(x+dx) - f(x)) dx; we could also take the "average height" between f(x+dx) and f(x)...
amistre64
  • amistre64
dx just stands for delta x, any arbitrary number... where dx = (b-a)/n... any of this make sense?
amistre64
  • amistre64
dx = (1--2)/n = (1+2)/n = 3/n
anonymous
  • anonymous
k i got that far
anonymous
  • anonymous
a=-2 b=1
amistre64
  • amistre64
yep, them pictures is what I said lol
amistre64
  • amistre64
the limit as n-> inf......
anonymous
  • 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
  • amistre64
need to work out something like this right?
1 Attachment
anonymous
  • anonymous
yea but that one is less complicated haha
amistre64
  • amistre64
lol .... use the Esummation for the cubic function
anonymous
  • anonymous
the answer we are going for is 3/4 for reference
anonymous
  • anonymous
is my first step correct?
amistre64
  • 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
  • anonymous
ok
amistre64
  • amistre64
[E] x^3 dx - [E] 3x dx the 3 can move out [E] x^3 (3/n) - 3 [E] x (3/n)
anonymous
  • anonymous
\[\lim_{n \rightarrow \infty}\sum_{i=1}^{n}[(-2+3i/n)^3-3(-2+3i/n)](3/n)\]
amistre64
  • amistre64
im lost on what that (-2+3i/n) stuff means, help me out :)
anonymous
  • anonymous
its part of the reimann sum
anonymous
  • anonymous
the formulas we have to use, that is the n^3
anonymous
  • anonymous
i^3 my mistake
amistre64
  • amistre64
\[\sum_{} x^{3}(3/n) -3 \sum_{}x(3/n)\] is whats in my head
anonymous
  • anonymous
well the integral is \[\int\limits_{-2}^{1}(x^3-3x)dx\]
amistre64
  • 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
  • anonymous
yes that works
amistre64
  • 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
  • amistre64
lets work each one on its own...ok
anonymous
  • anonymous
plug in x^3 and x
anonymous
  • anonymous
and thats what i showed above
anonymous
  • anonymous
as one complete summation multiplied by 3/n
amistre64
  • 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
  • anonymous
k let me show you
anonymous
  • anonymous
hopefully this helps
anonymous
  • anonymous
i need to break it even further for you
amistre64
  • 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
  • anonymous
its just pulling things out and taking individual summations, should i show you where i am so far?
amistre64
  • amistre64
yes please ∑ i^3 = (1/4)n4 + (1/2)n3 + (1/4)n2 is what I found online
anonymous
  • 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
  • anonymous
wow i guess my thing was too long, thats only half of it
anonymous
  • anonymous
im so confused lol
amistre64
  • amistre64
:) heres what I found that helps me out
1 Attachment
anonymous
  • anonymous
ok that works, lets start over then from 3/n being delta x
amistre64
  • amistre64
since 3/n = delta x, we simply multiply that to x^3? to get: 3(x^3)/n ?? is that right?
anonymous
  • anonymous
hmm sure, keep going because i wont know where you are going till you go further
amistre64
  • amistre64
\[3\sum_{n=1}^{\infty} (x^3)/n\] Does this look right?
anonymous
  • anonymous
no but keep going
amistre64
  • amistre64
lol ....if it dont look right, then why should I keep going :)
anonymous
  • anonymous
im trying to make sense of where you are going
amistre64
  • amistre64
....... you gotta drive some of the way, im lost here for the moment
anonymous
  • anonymous
oh k , im trying to think...problem is i cant find the equation that would help you
anonymous
  • anonymous
let me see...
anonymous
  • anonymous
where are you lost?
anonymous
  • anonymous
how i formulated the entire sum?
anonymous
  • anonymous
A=lim_{N→∞}Δx∑_{j=1}^{N}f(a+jΔx)
anonymous
  • anonymous
does this help
anonymous
  • anonymous
i=j
amistre64
  • 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
  • anonymous
anonymous
  • anonymous
heres a better look
amistre64
  • amistre64
ok....so delta x is a constant that gets pulled out to the left ... am I seeing that right?
anonymous
  • anonymous
yes
anonymous
  • 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
  • amistre64
and the "i" is simply each iteration of the sums....right?
anonymous
  • anonymous
yes
anonymous
  • anonymous
the integral should =3/4 after the long process, im trying to get through all the steps
anonymous
  • anonymous
thats the problem
amistre64
  • 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
  • anonymous
lol cause my professor doesnt want to make life easy
amistre64
  • 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
  • anonymous
yes
anonymous
  • anonymous
-2
anonymous
  • anonymous
limn→∞∑i=1n[(−2+3i/n)3−3(−2+3i/n)](3/n)
amistre64
  • amistre64
f(-2 + 3i/n) is a standard interation then, is that correct?
anonymous
  • anonymous
perfect
anonymous
  • anonymous
dont forget the cubed
amistre64
  • amistre64
is the "n" from 3/n is the one we are limiting to infinity right?
anonymous
  • anonymous
yes perfect
amistre64
  • amistre64
f^3(x)...yeah yeah
anonymous
  • 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
  • anonymous
dumb summation sign i mean*
amistre64
  • amistre64
n[(−2+3i/n)3−3(−2+3i/n)](3/n) this is you combining the terms together right??
anonymous
  • anonymous
the n in front shouldnt be there that was part of the format for the summation
amistre64
  • amistre64
ok...... (sigma) (-2 +3i/n) (3/n) is the right term then correct?
anonymous
  • anonymous
summation but yes that is correct
anonymous
  • anonymous
isnt it -3(-2 +3i/n) (3/n)
anonymous
  • anonymous
you forgot the -3x
amistre64
  • 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
  • anonymous
im pretty sure you do because its part of the integral function
anonymous
  • anonymous
but if you are confident, you understand this better then me so ill take your word for it
amistre64
  • amistre64
\[\int\limits_{?}^{?} -3x = -3\int\limits_{?}^{?}x \rightarrow-3 x^2/2\]
anonymous
  • 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
  • anonymous
i guess it doesnt matter go ahead, you sound excited because i think you understand it now
anonymous
  • anonymous
do your thing, ill just learn from you
amistre64
  • 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
  • anonymous
it still will work out taking as pieces?
amistre64
  • 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
  • amistre64
think of it as the granny way of doing integrals; its actually proof that we can split integerals up and work them
amistre64
  • amistre64
forgot to pu an i in there :)
anonymous
  • anonymous
ok let me write step by step following you on actual paper
amistre64
  • amistre64
we can split derivatives up and work them, integrals are just eh opposite of derivatives.....
amistre64
  • amistre64
this computer is having a hard time keeping up with formating the "equation"editor stuff, can we start a new question post?
anonymous
  • anonymous
yea
anonymous
  • anonymous
let me make a new one
anonymous
  • anonymous
k go to the top of the question list

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