# Computer Integral with Non-Equispaced Partitions

Just about to finish my analysis course, but I've having some trouble with a few problems for my final assignment. Any help would be awesome as this course is driving me mad! Thanks so much for the help. You have no clue what a relief resources like this site are!

Forward to 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 $U(f, P_n)=r^t(f,P_n)$

Part C: Use the last problem and part (b) to prove that:

$inf_{n\ge\ 1} U (f,P_n)=sup_{n\ge\ 1}=(b^{t+1}-a^{t+1})/(t+1)$

The last problem:

Let $a_n=\gamma^{s \over n}$ then for $\gamma, s > 0$ and all positive integers $n$:

$\lim_{n\to \infty} a_n = 1$

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Hint: write expressions for the lower and upper Darboux sums; express each of them as the difference of two geometric progressions. – Yury Dec 17 '12 at 4:50
Still super confused! So far I have: $S_t=\sum_i=1^k m_i \delta x_i$ with $x_i=ar^i$. So $$L(f,P_n)=\sum_i=1^k m_i \Delta x_i=a*\sum_i=1^k (x_i-x_{i-1}=a \left {1-r^{k+1}\over {1-r}-{1-r^{k}\over {1-r} \right$$ This is clearly not what I'm looking for! Any extra help? Thanks!!!!! – Jonathon Dec 17 '12 at 5:41
Fermat described this method well before Newton was born. – André Nicolas Dec 17 '12 at 5:47
@Yuri Still super confused! So far I have: $S_t=\sum_i=1^k m_i \delta x_i$ with $x_i=ar^i$. So $$L(f,P_n)=\sum_{i=1}^k m_i \Delta x_i=a*\sum_{i=1}^k x_i-x_{i-1}=a \left( {1-r^{k+1} \over {1-r}}-{1-r^k \over 1-r} \right)$$ This is clearly not what I'm looking for! Any extra help? Thanks!!!!! – Jonathon Dec 17 '12 at 5:49
Please don't blank questions. If you found the answer, you could write it up for posterity. If you don't want to invest the time, you could just delete the question. – joriki Dec 17 '12 at 23:57