Integration and Computing Area

In summary, the conversation discusses evaluating the sum of k^3 from 100 to 250 and the correct method to do so using sigma notation and definite integrals. The correct solution is to subtract the sum of k^3 from 1 to 99 from the sum of k^3 from 1 to 250.
  • #1
Beeorz
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0

Homework Statement


Evaluate the sum:
[tex] \sum_{k=100}^{250} k^3 [/tex]


Homework Equations


[tex] \sum_{j=1}^{N} j^3 = \frac{n^2(n+1)^2}{4} [/tex]



The Attempt at a Solution


Little confused as how to integrate from 100 to 250. I have no examples to go off of and what I try is incorrect. But here's what I tried:
[tex] \sum_{k=1}^{250} k^3 - \sum_{k=1}^{100} k^3 = \sum_{k=100}^{250} k^3 [/tex]

984390625-25502500=958888135 (incorrect)

Any suggestions? I'm sure this is relatively easy but I just haven't seen it before and thus don't know how to go about solving it. Thanks in advance!
 
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  • #2
You know in sigma notation, [tex] \sum_{k= m}^{n} f(k) [/tex]

n is the ending value of k & m is the starting value of k. So;

Why can't you subtract the solutions of the definite integrals, [tex]\int^{250}_{100} x^3 dx[/tex] and [tex]\int^{100}_{1} x^3 dx[/tex] ?
 
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  • #3
Beeorz, you're method will work out if you correct the bounds.
[tex]
\sum_{k=1}^{100} k^3 = k_1^3 + k_2^3 + ... + k_99^3 + k_100^3
[/tex]

If you subtract that from [tex] \sum_{k=1}^{250} k^3 [/tex] , then you are left with

[tex]
\sum_{k=101}^{250} k^3
[/tex]

You should subtract [tex] \sum_{k=1}^{99} k^3 [/tex] from the sum from 1 to 250 in order to get the answer.
 
  • #4
What in the world does this have to do with either "integration" or "Computing area"?
 

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