Give an argument for ∫n0xpdx≤1+2p+3p+⋯+np≤∫n+10xpdx\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 p0 give an argument that the following is true: n0xpdx1+2p+3p++npn+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++nn1xpdx

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

n0xpdx1p(10)+2p(21)+np(n(n1))=1p+2p+3p++np

We also have

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

Using similar reasoning, we see that

n+10xpdx0p(10)+1p(21)+(n+1)p((n+1)(n))=1p+2p+3p++np+(n+1)p

Finally, putting it all together reveals

n0xpdx1p+2p+3p++np1p+2p+3p++np+(n+1)pn+10xpdx

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

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