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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.
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@amistre64 @welshfella any ideas?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0not really sure how to go about this one.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Preetha @nincompoop @Hero @Luigi0210 @Nnesha

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1what is the limit of 5/n as n to infinity?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1the limit of a product is the product of the limits, if i recall correctly .. but you want the summation limit

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1yes, now if there is some useful way to rewrite the argument ...

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1(a+b)^3 = 1a^3b^0 +3a^2b^1 +3a^1b^2 +1a^0b^3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im sorry i dont follow

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1im wondering if we expand the ^3 part, if we cant see this thing easier, or make it funner to play with.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i dont think that would make a difference

amistre64
 one year ago
Best ResponseYou've already chosen the best response.140/n + 300k/(n^2) +750k^2/(n^3) + 625k^3/(n^4) youre prolly right

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1well, 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0don't really need to find the limit though just need to know how to write it as an integral

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do you know anyone that could solve this , im sort of new to openstudy don't know anyone :(

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1oh, i gave up trying to read the whole question when the numbers were written out in words ...

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1so this is a reimann sum thing

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes can you open this picture ?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1\[\Large\sum_{k=a}^{b}f(a+i\frac{ba}{n})\frac{ba}{n}\color{red}\implies \int_{a}^{b}f(x)dx\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1ach ... im used to i instead of k so theres some bleed over there

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1the picture is fine ...

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1k=0 to n, not a to b

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1does that seem familiar?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes sort of let me try to write it out

amistre64
 one year ago
Best ResponseYou've already chosen the best response.11 to n is a right hand rule, 0 to n1 is a left had rule ... im working from memory is all

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1either way... a=2, and ba = 5 soo b=7 seems right

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1i would venture to say f(x) = x^3, but my brain is telling me to be ware of that assumption

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how would you find k?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1k is just the kth iteration of the partition

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay so what about n?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1dw:1439506802530:dw

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1n 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i see , what makes ou think that f(x) = x^3 how would you usually go about finding f(x)?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1http://www.wolframalpha.com/input/?i=integrate+x^3+dx%2C+from+2+to+7 the good news is that my assumption is good lol

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1well the (...)^3 is a good indicator that the function is akin to x^3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh that value that i thought would be useless XD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay great thank you for your help

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how do i do the b=medal thingy?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1youre welcome :) thnx for the recollections lol

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1there should be a "best response" button in everything i post ... just click it to give a medal to someone

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you again ! :D

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1yep, and good luck ;)
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