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when I simplify that long formula and take it's limit, I get a -1/4 and that's not possible is it??

It's quite possible. What is the equation?

yep...

and then is became 1/n^5 n(n+1)(2n+1)(3n^2+3n-1) all /3

the the coefficient of n^5 on bottom is 30

so we have 6/30 right? :)

no...it didn't come out for me that simply!

OH...I forgot to keep bringng down the 30!!

|dw:1336266798515:dw|

now how did you simplify that so easily??? how can I just drop out the linear terms and constant?

Ok did you see the paragraph i wrote above?

That one ^

yes, I remember that now from when I tutor precal....duh....thought about that. :-)

The degree of the top was 5
The degree of the bottom was 5

But I need to calculate the upper reimann sumof all that mess....so what does it simplify down to??

I have 6n^4 + 15n^3 -10n^2 -1/ 30n^4 and can get no further...what am I doing wriong??

you don't have to multiply all of that out and nor do i recommend it

\[\lim_{n \rightarrow \infty}\frac{6+0-0-0}{30}\]
type-o above

i left out coefficient of n^4/n^4 on bottom

so was my 6n^4 etc technically correct for the evaluation of the sum??

6/30

you don't need to multiply it all out to evaluate the limit

i don't know what you mean looking for a formula for definite integral

definite integrals you only get a number

ok, so you're saying the Reimann sum LPLUS the limit IS the same as the integral!

duh...I was trying to equate the RS ONLY to the integral....left out the limit step. :)

vs the leibnitz integral...the short cut. right?

were supposed to verify the formula fo rn=2, 3, 5, 8

yes, each time....

Then, all that changes is that the interval has to be 2/n, 3/n and so on :-)

so do the reimann sum method separately
for each n, just like I did for n=1?

that makes sense....

ok....

The limit should be : \[ \frac{(\text{end of interval})^{5}}{5} \]

gotcha...thanks SO much !! Sorry I took you away from other people with questions..

that's ok....is there anyway to priint all of this out???

ok :-)