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I am trying to show that $\log{m}\le{\int_{m}^{m+1}{\log{t}}}dt$, with $m\ge{1}$.

I tried simplifying the problem to $0\le{\int_{m}^{m+1}{\log{\big(\frac{t}{m}\big)}}}dt$, but can't seem to get any further.

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Hint: What is the minimum value of the function you are integrating ($\log t$) over the interval of integration?

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  • $\begingroup$ that would be $\log{m+1}$, so is it enough to say that $\log{m}\le{\log{m+1}}$ and that since $m\ge{1}$ then $\log$ is non-negative on the region considered ($[m,m+1]$)? $\endgroup$ – Sam.S Feb 16 at 11:18
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The idea is that log(t) is a non-decreasing function, so you may trivially write $log(m)=\int_{m}^{m+1} log(m) dt \leq \int_{m}^{m+1} log(t) dt $.

Maybe this picture is helpful:

log(t)

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