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For any $n$ and $p\geq 0$ give an argument that the following is true: $$\int_{0}^{n} x^p dx \leq 1 +2^{p} + 3^{p} + \cdots+ n^{p}\leq \int_{0}^{n+1} x^p dx$$

I'm having trouble even beginning this question. My first thought it to somehow meld this with the squeeze theorem, but, again, am not sure how to begin and show any real work. Any insight is very much appreciated.

Community
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PTiger17
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3 Answers3

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I assume $n\in\mathbb{N}.$ Let $f(x)=x^p$ and consider a partiton $P=\{0,1,2,\cdots,n\}$ of $[0,n],$ then the upper sum $$U(P,f)=1^p+2^p+\cdots+n^p.$$ Also $\int\limits_{0}^{n}f(x)dx\leq U(P,f)=1^p+2^p+\cdots+n^p.$

Next, we consider the partition $\{0,1,2,\cdots,n,n+1\}$ of $[0,n+1]$ and use the fact about lower sum, i.e. $ L(P,f)=1^p+2^p+\cdots+n^p\leq\int\limits_{0}^{n+1}f(x)dx $

K_user
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Write the integral as the sum

$$\int_0^n x^pdx=\int_0^1 x^pdx+\int_1^2 x^pdx+\int_2^3 x^pdx+\dotsb +\int_{n-1}^n x^pdx$$

Now, since $p>0$, $x^p$ is an increasing function for $x>0$. Thus $m^p<(m+1)^p$ for all $m>0$. Then, we have

$$\int_0^n x^pdx\le 1^p(1-0)+2^p(2-1)+\dotsb n^p(n-(n-1))=1^p+2^p+3^p+\dotsb +n^p$$

We also have

$$\int_0^{n+1} x^pdx=\int_0^1 x^pdx+\int_1^2 x^pdx+\int_2^3 x^pdx+\dotsb +\int_{n}^{n+1} x^pdx$$

Using similar reasoning, we see that

$$\int_0^{n+1} x^pdx\ge 0^p(1-0)+1^p(2-1)+\dotsb (n+1)^p((n+1)-(n))=1^p+2^p+3^p+\dotsb +n^p+(n+1)^p$$

Finally, putting it all together reveals

$$\int_0^n x^pdx\le1^p+2^p+3^p+\dotsb +n^p\le1^p+2^p+3^p+\dotsb +n^p+(n+1)^p\le \int_0^{n+1} x^pdx$$

Mark Viola
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  • Thanks for offering to help! For the integral from 3 to 1 should it be 3 to 2 (since it's n-1) instead? Can you explain a little bit more about how for $1^{p}(1-0)$ where the $(1-0),(2-1)$, etc. come from? – PTiger17 Mar 16 '15 at 03:37
  • I figured out the second portion of the question above. Can you explain for the $n^{p+1}$ portion on the final line did you do the same as in the line above, but made it $n^{p}((n+1)-0))$ as the integrands I'm looking for are n+1 to 0? – PTiger17 Mar 16 '15 at 03:46
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    I'll add to the answer to make it more explicit. – Mark Viola Mar 16 '15 at 04:22
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    Thanks for the feedback!! Let me know if this works better for you. – Mark Viola Mar 16 '15 at 04:35
  • It's great! I much better understand how everything links. One last thing, for the notation using $m^{p}$ in the second line is that just denoting what the next line will be ($n^{p}(1)$). If so, why was it written as m not n? Is it because it's not actually part of the problem? – PTiger17 Mar 16 '15 at 04:38
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    Sure. The $m$ was used as a "dummy" variable to avoid confusion with the integration limit $n$. I hope that this was the best answer I could give you. – Mark Viola Mar 16 '15 at 04:44
  • One final thing (I promise!) for the very first integral could a $\leq$ be written instead of an equal sign? Or is the = implying that's the case ONLY if $p=0$, instead of $p \geq 0$? – PTiger17 Mar 16 '15 at 04:52
  • The first integral is an equality. The second integral is actually a strict inequality since $n^p$ is strictly increasing for $n>0$ and $p>0$. – Mark Viola Mar 16 '15 at 04:57
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Hint: $$ m^p \leq \int_m^{m+1} x^p \, dp \leq (m+1)^p. $$

Yuval Filmus
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  • Hi Yuval, could you elaborate a bit or give me any further reading on this? I am, embarrassingly, really not sure what the next steps building off what you've written would be. – PTiger17 Mar 16 '15 at 04:06