# Integration and Computing Area

1. Jul 16, 2008

### Beeorz

1. The problem statement, all variables and given/known data
Evaluate the sum:
$$\sum_{k=100}^{250} k^3$$

2. Relevant equations
$$\sum_{j=1}^{N} j^3 = \frac{n^2(n+1)^2}{4}$$

3. 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 heres what I tried:
$$\sum_{k=1}^{250} k^3 - \sum_{k=1}^{100} k^3 = \sum_{k=100}^{250} k^3$$

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!

2. Jul 16, 2008

### roam

You know in sigma notation, $$\sum_{k= m}^{n} f(k)$$

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, $$\int^{250}_{100} x^3 dx$$ and $$\int^{100}_{1} x^3 dx$$ ?

Last edited: Jul 16, 2008
3. Jul 16, 2008

### Piamedes

Beeorz, you're method will work out if you correct the bounds.
$$\sum_{k=1}^{100} k^3 = k_1^3 + k_2^3 + ... + k_99^3 + k_100^3$$

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

$$\sum_{k=101}^{250} k^3$$

You should subtract $$\sum_{k=1}^{99} k^3$$ from the sum from 1 to 250 in order to get the answer.

4. Jul 17, 2008

### HallsofIvy

Staff Emeritus
What in the world does this have to do with either "integration" or "Computing area"?