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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    The actual question is attached

  2. anonymous
    • 5 years ago
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    i attached the actual question

  3. amistre64
    • 5 years ago
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    i see it... so what are the 3 summation formulas?

  4. anonymous
    • 5 years ago
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    let me get them from my notes

  5. amistre64
    • 5 years ago
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    if I recall correctly, the Reiman limits are just taking the area of slices as the thickness of dx goes to zero right?

  6. anonymous
    • 5 years ago
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    yes and i she didnt give us the formulas even though she said she did, let me check the book.

  7. amistre64
    • 5 years ago
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    (f(x+dx) - f(x)) dx; we could also take the "average height" between f(x+dx) and f(x)...

  8. amistre64
    • 5 years ago
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    dx just stands for delta x, any arbitrary number... where dx = (b-a)/n... any of this make sense?

  9. anonymous
    • 5 years ago
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    here is the formulas

  10. amistre64
    • 5 years ago
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    dx = (1--2)/n = (1+2)/n = 3/n

  11. anonymous
    • 5 years ago
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    k i got that far

  12. anonymous
    • 5 years ago
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    a=-2 b=1

  13. amistre64
    • 5 years ago
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    yep, them pictures is what I said lol

  14. amistre64
    • 5 years ago
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    the limit as n-> inf......

  15. anonymous
    • 5 years ago
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    so then I got lim as n goes to infinity of the summation [(-2+3i/n)^3-3(-2+3i/n)](3/n)

  16. amistre64
    • 5 years ago
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    need to work out something like this right?

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  17. anonymous
    • 5 years ago
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    yea but that one is less complicated haha

  18. amistre64
    • 5 years ago
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    lol .... use the Esummation for the cubic function

  19. anonymous
    • 5 years ago
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    the answer we are going for is 3/4 for reference

  20. anonymous
    • 5 years ago
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    is my first step correct?

  21. amistre64
    • 5 years ago
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    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

  22. anonymous
    • 5 years ago
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    ok

  23. amistre64
    • 5 years ago
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    [E] x^3 dx - [E] 3x dx the 3 can move out [E] x^3 (3/n) - 3 [E] x (3/n)

  24. anonymous
    • 5 years ago
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    \[\lim_{n \rightarrow \infty}\sum_{i=1}^{n}[(-2+3i/n)^3-3(-2+3i/n)](3/n)\]

  25. amistre64
    • 5 years ago
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    im lost on what that (-2+3i/n) stuff means, help me out :)

  26. anonymous
    • 5 years ago
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    its part of the reimann sum

  27. anonymous
    • 5 years ago
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    the formulas we have to use, that is the n^3

  28. anonymous
    • 5 years ago
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    i^3 my mistake

  29. amistre64
    • 5 years ago
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    \[\sum_{} x^{3}(3/n) -3 \sum_{}x(3/n)\] is whats in my head

  30. anonymous
    • 5 years ago
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    well the integral is \[\int\limits_{-2}^{1}(x^3-3x)dx\]

  31. amistre64
    • 5 years ago
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    yep, and the reiman is proof that we can split that into 2 parts right? \[\int\limits_{} x^3 dx -3\int\limits_{}x dx\]

  32. anonymous
    • 5 years ago
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    yes that works

  33. amistre64
    • 5 years ago
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    ∑ x^3 (3/n)−3 ∑ x (3/n) is its equivalent, tell me if im wrong... but whats our next step after this?

  34. amistre64
    • 5 years ago
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    lets work each one on its own...ok

  35. anonymous
    • 5 years ago
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    plug in x^3 and x

  36. anonymous
    • 5 years ago
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    and thats what i showed above

  37. anonymous
    • 5 years ago
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    as one complete summation multiplied by 3/n

  38. amistre64
    • 5 years ago
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    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)

  39. anonymous
    • 5 years ago
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    k let me show you

  40. anonymous
    • 5 years ago
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    hopefully this helps

  41. anonymous
    • 5 years ago
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    i need to break it even further for you

  42. amistre64
    • 5 years ago
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    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 :)

  43. anonymous
    • 5 years ago
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    its just pulling things out and taking individual summations, should i show you where i am so far?

  44. amistre64
    • 5 years ago
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    yes please ∑ i^3 = (1/4)n4 + (1/2)n3 + (1/4)n2 is what I found online

  45. anonymous
    • 5 years ago
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    \[\lim_{n \rightarrow \infty}\sum_{i=1}^{n}3/n[-8+12i/n+12i/n-18i^2/n^2+12i/n-18i^2/n^2-18i^2/n^2-27i^3/n^3-3(-2+3i/n)\]

  46. anonymous
    • 5 years ago
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    wow i guess my thing was too long, thats only half of it

  47. anonymous
    • 5 years ago
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    im so confused lol

  48. amistre64
    • 5 years ago
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    :) heres what I found that helps me out

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  49. anonymous
    • 5 years ago
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    ok that works, lets start over then from 3/n being delta x

  50. amistre64
    • 5 years ago
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    since 3/n = delta x, we simply multiply that to x^3? to get: 3(x^3)/n ?? is that right?

  51. anonymous
    • 5 years ago
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    hmm sure, keep going because i wont know where you are going till you go further

  52. amistre64
    • 5 years ago
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    \[3\sum_{n=1}^{\infty} (x^3)/n\] Does this look right?

  53. anonymous
    • 5 years ago
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    no but keep going

  54. amistre64
    • 5 years ago
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    lol ....if it dont look right, then why should I keep going :)

  55. anonymous
    • 5 years ago
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    im trying to make sense of where you are going

  56. amistre64
    • 5 years ago
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    ....... you gotta drive some of the way, im lost here for the moment

  57. anonymous
    • 5 years ago
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    oh k , im trying to think...problem is i cant find the equation that would help you

  58. anonymous
    • 5 years ago
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    let me see...

  59. anonymous
    • 5 years ago
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    where are you lost?

  60. anonymous
    • 5 years ago
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    how i formulated the entire sum?

  61. anonymous
    • 5 years ago
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    A=lim_{N→∞}Δx∑_{j=1}^{N}f(a+jΔx)

  62. anonymous
    • 5 years ago
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    does this help

  63. anonymous
    • 5 years ago
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    i=j

  64. amistre64
    • 5 years ago
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    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....

  65. anonymous
    • 5 years ago
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  66. anonymous
    • 5 years ago
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    heres a better look

  67. amistre64
    • 5 years ago
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    ok....so delta x is a constant that gets pulled out to the left ... am I seeing that right?

  68. anonymous
    • 5 years ago
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    yes

  69. anonymous
    • 5 years ago
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    but you see how the i is appearing now. In the equation, it names it j, but the common known term is i

  70. amistre64
    • 5 years ago
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    and the "i" is simply each iteration of the sums....right?

  71. anonymous
    • 5 years ago
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    yes

  72. anonymous
    • 5 years ago
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    the integral should =3/4 after the long process, im trying to get through all the steps

  73. anonymous
    • 5 years ago
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    thats the problem

  74. amistre64
    • 5 years ago
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    yeah....when i read over that reaimann stuff I gloosed at it and figured, its true, so why do it the long way :)

  75. anonymous
    • 5 years ago
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    lol cause my professor doesnt want to make life easy

  76. amistre64
    • 5 years ago
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    f(a + i /x\) the i is each iterations from 1 to infinity. is the a the same as in our b-a interval?

  77. anonymous
    • 5 years ago
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    yes

  78. anonymous
    • 5 years ago
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    -2

  79. anonymous
    • 5 years ago
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    limn→∞∑i=1n[(−2+3i/n)3−3(−2+3i/n)](3/n)

  80. amistre64
    • 5 years ago
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    f(-2 + 3i/n) is a standard interation then, is that correct?

  81. anonymous
    • 5 years ago
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    perfect

  82. anonymous
    • 5 years ago
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    dont forget the cubed

  83. amistre64
    • 5 years ago
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    is the "n" from 3/n is the one we are limiting to infinity right?

  84. anonymous
    • 5 years ago
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    yes perfect

  85. amistre64
    • 5 years ago
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    f^3(x)...yeah yeah

  86. anonymous
    • 5 years ago
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    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

  87. anonymous
    • 5 years ago
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    dumb summation sign i mean*

  88. amistre64
    • 5 years ago
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    n[(−2+3i/n)3−3(−2+3i/n)](3/n) this is you combining the terms together right??

  89. anonymous
    • 5 years ago
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    the n in front shouldnt be there that was part of the format for the summation

  90. amistre64
    • 5 years ago
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    ok...... (sigma) (-2 +3i/n) (3/n) is the right term then correct?

  91. anonymous
    • 5 years ago
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    summation but yes that is correct

  92. anonymous
    • 5 years ago
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    isnt it -3(-2 +3i/n) (3/n)

  93. anonymous
    • 5 years ago
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    you forgot the -3x

  94. amistre64
    • 5 years ago
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    \[\sum_{} (-2+3i/n) (3/n)\] 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

  95. anonymous
    • 5 years ago
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    im pretty sure you do because its part of the integral function

  96. anonymous
    • 5 years ago
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    but if you are confident, you understand this better then me so ill take your word for it

  97. amistre64
    • 5 years ago
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    \[\int\limits_{?}^{?} -3x = -3\int\limits_{?}^{?}x \rightarrow-3 x^2/2\]

  98. anonymous
    • 5 years ago
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    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?

  99. anonymous
    • 5 years ago
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    i guess it doesnt matter go ahead, you sound excited because i think you understand it now

  100. anonymous
    • 5 years ago
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    do your thing, ill just learn from you

  101. amistre64
    • 5 years ago
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    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

  102. anonymous
    • 5 years ago
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    it still will work out taking as pieces?

  103. amistre64
    • 5 years ago
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    \[\sum{}(-2 +3/n)(3/n)\] fits into n(n+1)/2 somehow, and yes, it will work easier and just as good if we split it up

  104. amistre64
    • 5 years ago
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    think of it as the granny way of doing integrals; its actually proof that we can split integerals up and work them

  105. amistre64
    • 5 years ago
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    forgot to pu an i in there :)

  106. anonymous
    • 5 years ago
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    ok let me write step by step following you on actual paper

  107. amistre64
    • 5 years ago
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    we can split derivatives up and work them, integrals are just eh opposite of derivatives.....

  108. amistre64
    • 5 years ago
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    this computer is having a hard time keeping up with formating the "equation"editor stuff, can we start a new question post?

  109. anonymous
    • 5 years ago
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    yea

  110. anonymous
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    let me make a new one

  111. anonymous
    • 5 years ago
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    k go to the top of the question list

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