prove inequality with an integral over region with unit length

Question

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.

Answer 1

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:

Answer 2

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