Give an argument for $\int_{0}^{n} x^p dx \leq 1 +2^{p} + 3^{p} + \cdots+ n^{p}\leq \int_{0}^{n+1} x^p dx$ 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. 
 A: 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 $
A: 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
$$\int_0^n x^pdx\le1^p+2^p+3^p+\dotsb +n^p\le1^p+2^p+3^p+\dotsb +n^p+(n+1)^p\le \int_0^{n+1} x^pdx$$
A: Hint:
$$
m^p \leq \int_m^{m+1} x^p \, dp \leq (m+1)^p.
$$
