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$$\lim_{\Delta x\rightarrow0}\sum_{k=1}^{n-1}\int_{k+\Delta x}^{k+1-\Delta x}x^m dx$$

How to integrate the above integral?

Edit1:

$$\lim_{\Delta x\rightarrow0}\int_{2-\Delta x}^{2+\Delta x}x^m dx$$

Does this intergral give $\space\space\space\space$ $2^m\space\space$ as the output?

Edit2:

Are my following steps correct?

$\lim_{\Delta x\rightarrow0}\sum_{k=1}^{n-1}\int_{k+\Delta x}^{k+1-\Delta x}x^m dx$ =

$\lim_{\Delta x\rightarrow0}\sum_{k=1}^{n-1}\int_{k+\Delta x}^{k+1-\Delta x}x^m dx$ $+$ $\lim_{\Delta x\rightarrow0}\sum_{k=1}^{n-1}\int_{k+1-\Delta x}^{k+1+\Delta x}x^m dx$ $-$ $\lim_{\Delta x\rightarrow0}\sum_{k=1}^{n-1}\int_{k+1-\Delta x}^{k+1 +\Delta x}x^m dx$ =

$\lim_{\Delta x\rightarrow0}\int_{1+\Delta x}^{n+\Delta x}x^m dx$ $-$ $\lim_{\Delta x\rightarrow0}\sum_{k=1}^{n-1}\int_{k+1-\Delta x}^{k+1 +\Delta x}x^m dx$ =

$\lim_{\Delta x\rightarrow0}\int_{1}^{n}x^m dx$ $+$$\lim_{\Delta x\rightarrow0}\int_{n}^{n+\Delta x}x^m dx$ $-$ $\lim_{\Delta x\rightarrow0}\int_{1}^{1+\Delta x}x^m dx$ $-$

$\lim_{\Delta x\rightarrow0}\sum_{k=1}^{n-1}\int_{k+1-\Delta x}^{k+1 +\Delta x}x^m dx$ = $\lim_{\Delta x\rightarrow0}\int_{1}^{n}x^m dx$ + 0 - 0 - 0 = $\lim_{\Delta x\rightarrow0}\int_{1}^{n}x^m dx$

= $\int_{1}^{n}x^m dx$

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This version of Edit2 looks good. –  Brian M. Scott Jan 8 '13 at 21:08

3 Answers 3

up vote 3 down vote accepted

For the first question: $$ \begin{align} \lim_{\Delta x\to0}\,\left|\,\int_k^{k+1}x^m\,\mathrm{d}x-\int_{k+\Delta x}^{k+1-\Delta x}x^m\,\mathrm{d}x\,\right| &=\lim_{\Delta x\to0}\,\left|\,\int_k^{k+\Delta x}x^m\,\mathrm{d}x+\int_{k+1-\Delta x}^{k+1}x^m\,\mathrm{d}x\,\right|\\ &\le\lim_{\Delta x\to0}2\Delta x(k+1)^m\\ &=0 \end{align} $$ we get $$ \begin{align} \lim_{\Delta x\to0}\sum_{k=1}^{n-1}\int_{k+\Delta x}^{k+1-\Delta x}x^m\,\mathrm{d}x &=\sum_{k=1}^{n-1}\int_k^{k+1}x^m\,\mathrm{d}x\\ &=\int_1^nx^m\,\mathrm{d}x \end{align} $$ For the Edit:

For $\Delta x<1$, $$ \left|\,\int_{2-\Delta x}^{2+\Delta x}x^m\,\mathrm{d}x\,\right|\le2\cdot3^m\Delta x $$ Therefore, $$ \lim_{\Delta x\to0}\,\int_{2-\Delta x}^{2+\Delta x}x^m\,\mathrm{d}x=0 $$ For your steps: If you would give the justification for each step, it would help us in commenting on what is correct and what might be wrong and help you in seeing what is right.

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Ok, I didn't notice this answer. –  Inquisitive Jan 8 '13 at 21:15

HINT:

$$\sum_{k=1}^{n-1}\int_{k+\Delta x}^{k+1-\Delta x}x^m dx=\int_1^nx^mdx-\left(\int_1^{1+\Delta x}x^mdx+\sum_{k=2}^{n-1}\int_{k-\Delta x}^{k+\Delta x}x^mdx+\int_{n-\Delta x}^nx^mdx\right)$$

Let $M$ be any upper bound for $x^m$ on $[1,n]$; then

$$\begin{align*} \int_1^{1+\Delta x}x^mdx+\sum_{k=2}^{n-1}\int_{k-\Delta x}^{k+\Delta x}x^mdx+\int_{n-\Delta x}^nx^mdx&\le M\left(\Delta x+2(n-2)\Delta x+\Delta x\right)\\ &=2M\Delta x(n-1)\;. \end{align*}$$

Added: The same reasoning shows that

$$\lim_{\Delta x\rightarrow0}\int_{2-\Delta x}^{2+\Delta x}x^m dx=0\;:$$

if $M$ is any bound on $x^m$ over the interval $[1,3]$, say, the integral is bounded by $2M\Delta x$.

By the way, you really shouldn’t use $x$ both in $\Delta x$ and as the variable of integration. The new limit, for instance, ought to be

$$\lim_{\Delta x\rightarrow0}\int_{2-\Delta x}^{2+\Delta x}t^m dt$$

or the like, with similar changes in the original problem.

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$$\lim_{s\rightarrow0}\frac1s\int_{2-s}^{2+s}x^m\mathrm dx=2\cdot2^m$$

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