express limit as definite integral

**calcbeg** · September 19th 2009, 11:27 AM

Hi I think this would be easier if there was a range but one was not given. Here is the entire question:

n
Express Lim sum notation (i^5) / (n^6) as a definite integral.
n -> infinity i=1

Then evaluate the integral.

my problem is there is no x in this formula and but I guess it can be f(i) instead of the usual f(x) but then what does that mean for n?

Any insight is appreciated.

this is my second course but I am just as lost.

A calculus beginner

**galactus** · September 19th 2009, 11:39 AM

Try $\int_{0}^{1}x^{5}dx$

${\Delta}x=\frac{1-0}{n}=\frac{1}{n}$

$i{\Delta}x=\frac{i}{n}$

$f(x_{k})\cdot {\Delta}x=(\frac{i}{n})^{5}\cdot \frac{1}{n}=\frac{i^{5}}{n^{6}}$

Look anything like what you have?.

$\frac{1}{n^{6}}\sum_{i=1}^{n}i^{5}=\frac{(n+1)^{2} (2n^{2}+2n-1)}{12n^{4}}=\frac{1}{2n}+\frac{5}{12n^{2}}-\frac{1}{12n^{4}}+\frac{1}{6}$

$\lim_{n\to \infty}\left[\frac{1}{2n}+\frac{5}{12n^{2}}-\frac{1}{12n^{4}}+\frac{1}{6}\right]=\frac{1}{6}$

Check against $\int_{0}^{1}x^{5}dx$

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