Explain this inequality, related to logarithms I am trying to understand a proof of Stirling's formula.
One part of the proof states that, 'Since the log function is increasing on the interval  $(0,\infty)$, we get 
$$\int_{n-1}^{n} \log(x) dx < \log(n) < \int_{n}^{n+1} \log(x) dx$$ for $n\geq 1$.'
Please could you explain why this is true?
In particular, I am struggling to visualise this inequality graphically/geometrically.

Let $f$ be a strictly increasing continuous function in the interval $[a,b]$ (with $a<b$. Then, for $x\in[a,b]$, we have $f(x)\le f(b)$. Therefore
$$
\int_{a}^{b}f(x)\,dx\le (b-a)f(b)
$$
because $f(b)$ is the maximum of $f$ in the interval. Equality would mean
$$
\int_{a}^{b}(f(b)-f(x))\,dx=0
$$
and, as $g(x)=f(b)-f(x)$ is a continuous non negative function, this implies $f(x)=f(b)$ for all $x$, so the function is constant. This is impossible if $f$ is strictly increasing. Thus
$$
\int_{a}^{b}f(x)\,dx< (b-a)f(b)
$$
Similarly,
$$
(b-a)f(a)<\int_{a}^{b}f(x)\,dx
$$
In your case there's only to verify that
$$
\int_0^1\log x\,dx<\log 1=0
$$
because $\log$ is not defined at $0$, so this is an improper integral. You should be able to do it.
