For any n and p≥0 give an argument that the following is true: ∫n0xpdx≤1+2p+3p+⋯+np≤∫n+10xpdx

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.

**Answer**

Write the integral as the sum

∫n0xpdx=∫10xpdx+∫21xpdx+∫32xpdx+⋯+∫nn−1xpdx

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

∫n0xpdx≤1p(1−0)+2p(2−1)+⋯np(n−(n−1))=1p+2p+3p+⋯+np

We also have

∫n+10xpdx=∫10xpdx+∫21xpdx+∫32xpdx+⋯+∫n+1nxpdx

Using similar reasoning, we see that

∫n+10xpdx≥0p(1−0)+1p(2−1)+⋯(n+1)p((n+1)−(n))=1p+2p+3p+⋯+np+(n+1)p

Finally, putting it all together reveals

∫n0xpdx≤1p+2p+3p+⋯+np≤1p+2p+3p+⋯+np+(n+1)p≤∫n+10xpdx

**Attribution***Source : Link , Question Author : PTiger17 , Answer Author : Mark Viola*