$\displaystyle \lim_{n\to\infty}\sum_{i=1}^{n}\frac{3}{n} \displaystyle{[(}1+\frac{3i}{n}\displaystyle{)}^3-2(1+\frac{3i}{n}\displaystyle{)]}$

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- Nov 10th 2009, 03:46 PMEm Yeu AnhLimit
$\displaystyle \lim_{n\to\infty}\sum_{i=1}^{n}\frac{3}{n} \displaystyle{[(}1+\frac{3i}{n}\displaystyle{)}^3-2(1+\frac{3i}{n}\displaystyle{)]}$

- Nov 10th 2009, 04:31 PMadkinsjr
- Nov 10th 2009, 04:32 PMredsoxfan325
I would assume you haven't done the fundamental theorem of calculus yet, because this Riemann sum appears to be $\displaystyle \int_1^4x^3-2x\,dx=\frac{195}{4}$

Pull the $\displaystyle \frac{3}{n}$ out and expand the summand. You get

$\displaystyle \lim_{n\to\infty}\frac{3}{n}\sum_{i=1}^n\left[(3i/n)^3+3(3i/n)^2+(3i/n)-1\right]$ $\displaystyle =\lim_{n\to\infty}\frac{3}{n}\left[\frac{27}{n^3}\sum_{i=1}^n i^3+\frac{27}{n^2}\sum_{i=1}^n i^2+\frac{3}{n}\sum_{i=1}^n i-\sum_{i=1}^n 1\right]$

Evaluate that using summation formulas and then take the limit. - Nov 10th 2009, 05:22 PMEm Yeu Anh
I have studied the fundamental thereom although it was quite some time ago, would you mind explaining how you obtained 4 on the upper limit of integration for $\displaystyle \int_1^4x^3-2xdx?$

**Sorry, think I may have just found it:**

$\displaystyle \frac{3}{n} = \frac{b-1}{n} $ is that the formula? - Nov 10th 2009, 06:01 PMredsoxfan325
Yup. The reason I thought you hadn't done the FTC yet was that I assumed you wanted to evaluate this limit

*in order to*evaluate the integral. I guess not. - Nov 10th 2009, 06:09 PMEm Yeu Anh
Alright I got 195/4! :D Thank you very much redsoxfan you have been most helpful

- Nov 10th 2009, 06:15 PMredsoxfan325
You're welcome.