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I posted something similar earlier, but the MathJax code in the comments got really messed up, so I'm trying to re-post it so people actually know what I am talking about!

The Problem:

In the definition of the integral, we considered partitions of intervals to be arbitrary, but whenever we actually computed one, we used equispaced partitions. The following is a famous example of how to integrate $f(x) = x^t$ for any $t > 0$ directly using non-equispaced partitions. Let $a < b$ be positive numbers, and for any positive integer n, let $r = (b/a)^{1 \over n} > 1$. Then,

$a< ar^0< ar^1< ar^2 < ... < ar^{n-1}< ar^{n}=b$

so if we set $x_i = ar^i$ for $i = 0 ... n$, then $ P_n = \lbrace x_0,...,x_n \rbrace $ is a is a partition of $[a, b]$.

Part A:

Super easy, no need for help here :)

I just had to show that $x^t$ was an increasing function.

Part B:

Using the formula for summing geometric series, show that $L(f, P_n) = a^{t+1}(r-1)(r^{n(t+1)}-1)=(b^{t+1}-a^{t+1})(r-1)/(r^{t+1}-1)$, and $U(f, P_n)=r^t(f,P_n)$

This part is killing me, so far I have that:

$$x_i=ar^i;\, m_i(f)= inf \{f(x):x \in [x_{i-1},x_i] \}$$


$$L(f,P_n)=\sum_{i=1}^k m_i \Delta x_i=\sum_{i=1}^k m_i(x_i-x_{i-1})=a* \left( {1-r^{k+1} \over {1-r}}-{1-r^k \over 1-r} \right) \sum_{i=1}^km_i$$

I'm not even sure if this much is right, but, if it is, I have no clue where to go from here!

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It looks like you've summed the $x_i-x_{i-1}$ without summing the $m_i$. That's not allowed, because in general $\sum_i a_ib_i$ is not the product of $\sum_i a_i$ and $\sum_i b_i$. I think you want to figure out the value of the $m_i$'s, and then sum $m_i(x_i-x_{i-1})$. – user108903 Dec 17 '12 at 9:07
How do I find the value of the $m_i$? – Jonathon Dec 17 '12 at 12:45
You have $m_i(f)=\inf \{f(x): x\in [x_{i-1},x_i]\}$, and $f$ is increasing as you've already observed. This should help... – user108903 Dec 17 '12 at 19:26

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