Riemann Summ Limit Proof (strange question to me)

[Seeker555]

Recall the definition of the Riemann Sum \(\displaystyle S(f;\pi ;\xi)\) for a function \(\displaystyle f:[0,1]\rightarrow \mathbb{R}\).
Let \(\displaystyle f:[0,1]\rightarrow \mathbb{R},\ x\rightarrow \frac{5}{6}x\)[/latex] For the partition \(\displaystyle \pi_n\) of \(\displaystyle [0,1], 0=t_0<t_1<...<t_n,\ t_j\ =\ \frac{5}{n},\ j=0,...,n\) and \(\displaystyle \xi_j =\ \frac{2}{3}t_j\ +\ \frac{1}{3}t_(j+1)\in [t_j,t_(j+1)]\) find \(\displaystyle S(f;\pi_n,\xi)\).


Now Prove \(\displaystyle lim(n\rightarrow infinity)\ of\ S(f;\pi_n,\xi)=\frac{5}{12}\)


Pretty clueless on this question. A walk through on the steps would be nice for the test coming.

 

[galactus]

This notation is rough to decipher, but it would appear they are asking for the Riemann sum of

\(\displaystyle \displaystyle \int_{0}^{1}\frac{5x}{6}dx=\frac{5}{12}\)

Using the right endpoint method, \(\displaystyle x_{k}=a+k\Delta x\).

We have the interval [0,1] and break it up into n subintervals.

\(\displaystyle \frac{1}{n}\cdot k=\frac{k}{n}\)

Rectangle k has area:

\(\displaystyle \frac{5(\frac{k}{n})}{6}\cdot \frac{1}{n}=\frac{5k}{6n^{2}}\)

So, we sum up all of these rectangles by noting that \(\displaystyle \sum_{n=1}^{\infty}k=\frac{n(n+1)}{2}\).

\(\displaystyle \displaystyle\frac{5}{6n^{2}}\sum_{n=1}^{\infty}k\)

\(\displaystyle \frac{5}{6n^{2}}\cdot\frac{n(n+1)}{2}=\frac{5}{12}+\frac{5}{12n}\)

Now, take the limit as \(\displaystyle n\to \infty\) and see what you have