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Please, I wish to help with this issue. I tried with the principle of mathematical induction but could not.

Consider the following:

Let the function $$f:[a,b]\subset\mathbb{R}\to \mathbb{R},$$ The partition of the interval $ [a, b] $: $$a={x_0} < {x_1} < {x_2} < ... < {x_{n - 1}} < {x_n} = b,$$ subinterval length: $$\Delta x = \frac{{b - a}}{n},$$ the sum below: $$A({R_n}) = \sum\limits_{i = 0}^{n - 1} {f({x_i})\Delta x} $$ and the upper sum $$A({S_n}) = \sum\limits_{i = 1}^n {f({x_i})\Delta x}.$$

Well, now I want to resolve this problem:

Let $A_a^b(x^m)$ be the area under the curve $f(x) = x^m$ over the closed interval $[a, b]$. Prove that $$A_a^b(x^m)=\frac{b^{m+1}}{m+1}-\frac{a^{m+1}}{m+1}$$ whenever $m \geqslant 0$.

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Please do not copy assignment questions and post in the imperative mode. Instead, please indicate what you have tried and where you are stuck. –  Ross Millikan Aug 23 '11 at 23:02
    
@Ross Millikan. Ok Ross, I edited my post. –  mathsalomon Aug 23 '11 at 23:17
    
Have you tried substituting in the given $f(x)$ and summing the two series? –  Ross Millikan Aug 23 '11 at 23:42
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To me, everything in this post before Let $A_a^b(x^m)$ be the area... is not needed to the question nor useful to its solution. –  Did Aug 23 '11 at 23:45
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It may be easier to first find $A^b_0(x^m)$ and then note $A^b_a = A^b_0 - A^a_0 $. @Ross, summing the series is not all that easy (Bernoulli was first to solve the general case, but Faulhaber's name has stuck to it). What it all boils down to is finding $\lim_{n\to\infty} (1^p+2^p+\cdots +n^p)/(n^{p+1})$. –  Ragib Zaman Aug 24 '11 at 0:45

2 Answers 2

up vote 2 down vote accepted

The wording of the problem seems to me that he has been given the partition to work with, he can not use another one. To OP: Below I give a lemma that is crucial for solving the problem with the particular choice of partition of the interval given in the question. I leave setting up the Riemann sums, learning Big-Oh notation and filling in any details for steps you do not see immediately to you.

Proof by Induction that $ \sum_{k=1}^n k^{p-1} = \frac{n^p}{p} + O\left( n^{p-1} \right) $ where $ p \in \mathbb{N} $:

The base case $p=1$ is easily verified to be true. We make an induction hypothesis that there exists $ m \in \mathbb{N} $ such that the above equality holds for $ p = 1,2,3, \cdots m $. From the binomial theorem we have $$ (k+1)^{m+1} - k^{m+1} = \sum_{t=1}^{m+1} \binom{m+1}{t} k^{m+1-t}. $$ Letting $ k = 1,2,3, \cdots n $ and summing gives $ (n+1)^{m+1} - 1 = (m+1) \sum_{k=1}^n k^m + O(n^m) $, where the lower order terms were $ O(n^m) $ by the induction hypothesis. Thus, $ \sum_{k=1}^n k^m = \frac{n^{m+1}}{m+1} + O(n^m) $ which completes the inductive step.

From the above result, $$ \frac{ \sum_{k=1}^n k^{p} }{n^{p+1}} = \frac{1}{p+1} + O\left( \frac{1}{n}\right) \to \frac{1}{p+1} $$ as $ n\to \infty .$

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HINT:

Instead of setting up the successive $x_i$ in arithmetic progression, set them up in geometric progression. (I will not say explicitly how - I think it will become apparent).

Then, carry out the sums. Just do them. They will be geometric series, easily calculated, and the answer will appear.

If you are still stuck, then post your progress and where you are stuck.

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