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anonymous

  • one year ago

:( this question is killing me Write the limit as n goes to infinity of the summation from k equals 1 of the product of the 3rd power of the quantity 2 plus 5 times k over n and 5 over n as a definite integral.

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    @amistre64 @welshfella any ideas?

  3. anonymous
    • one year ago
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    not really sure how to go about this one.

  4. anonymous
    • one year ago
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    @Preetha @nincompoop @Hero @Luigi0210 @Nnesha

  5. amistre64
    • one year ago
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    what is the limit of 5/n as n to infinity?

  6. amistre64
    • one year ago
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    the limit of a product is the product of the limits, if i recall correctly .. but you want the summation limit

  7. anonymous
    • one year ago
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    0?

  8. amistre64
    • one year ago
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    yes, now if there is some useful way to rewrite the argument ...

  9. amistre64
    • one year ago
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    (a+b)^3 = 1a^3b^0 +3a^2b^1 +3a^1b^2 +1a^0b^3

  10. anonymous
    • one year ago
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    im sorry i dont follow

  11. amistre64
    • one year ago
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    im wondering if we expand the ^3 part, if we cant see this thing easier, or make it funner to play with.

  12. anonymous
    • one year ago
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    i dont think that would make a difference

  13. amistre64
    • one year ago
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    40/n + 300k/(n^2) +750k^2/(n^3) + 625k^3/(n^4) youre prolly right

  14. amistre64
    • one year ago
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    well, by simply putting in a large number for n http://www.wolframalpha.com/input/?i=sum%28n%3D1+to+100000%29+of+5%282%2Bn%285%2F100000%29%29^3%2F100000

  15. anonymous
    • one year ago
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    don't really need to find the limit though just need to know how to write it as an integral

  16. anonymous
    • one year ago
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    do you know anyone that could solve this , im sort of new to openstudy don't know anyone :(

  17. amistre64
    • one year ago
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    oh, i gave up trying to read the whole question when the numbers were written out in words ...

  18. amistre64
    • one year ago
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    so this is a reimann sum thing

  19. anonymous
    • one year ago
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    yes can you open this picture ?

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  20. amistre64
    • one year ago
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    \[\Large\sum_{k=a}^{b}f(a+i\frac{b-a}{n})\frac{b-a}{n}\color{red}\implies \int_{a}^{b}f(x)dx\]

  21. amistre64
    • one year ago
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    ach ... im used to i instead of k so theres some bleed over there

  22. amistre64
    • one year ago
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    the picture is fine ...

  23. amistre64
    • one year ago
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    k=0 to n, not a to b

  24. amistre64
    • one year ago
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    does that seem familiar?

  25. anonymous
    • one year ago
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    yes sort of let me try to write it out

  26. amistre64
    • one year ago
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    1 to n is a right hand rule, 0 to n-1 is a left had rule ... im working from memory is all

  27. amistre64
    • one year ago
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    either way... a=2, and b-a = 5 soo b=7 seems right

  28. amistre64
    • one year ago
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    i would venture to say f(x) = x^3, but my brain is telling me to be ware of that assumption

  29. anonymous
    • one year ago
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    how would you find k?

  30. amistre64
    • one year ago
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    k is just the kth iteration of the partition

  31. anonymous
    • one year ago
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    okay so what about n?

  32. amistre64
    • one year ago
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    |dw:1439506802530:dw|

  33. amistre64
    • one year ago
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    n is the number of partitions overall, as n to infinity the discrete reimann sum becomes a definite integral ... the area beneath the curve from a to b

  34. anonymous
    • one year ago
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    i see , what makes ou think that f(x) = x^3 how would you usually go about finding f(x)?

  35. amistre64
    • one year ago
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    http://www.wolframalpha.com/input/?i=integrate+x^3+dx%2C+from+2+to+7 the good news is that my assumption is good lol

  36. amistre64
    • one year ago
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    well the (...)^3 is a good indicator that the function is akin to x^3

  37. anonymous
    • one year ago
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    oh that value that i thought would be useless XD

  38. anonymous
    • one year ago
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    okay great thank you for your help

  39. anonymous
    • one year ago
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    how do i do the b=medal thingy?

  40. amistre64
    • one year ago
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    youre welcome :) thnx for the recollections lol

  41. amistre64
    • one year ago
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    there should be a "best response" button in everything i post ... just click it to give a medal to someone

  42. anonymous
    • one year ago
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    thank you again ! :D

  43. amistre64
    • one year ago
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    yep, and good luck ;)

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