Proving $\;\ln k \geq \int_{k-\frac{1}{2}}^{k+ \frac{1}{2}}\ln x dx$ I'm trying to prove  $$\ln k \geq \int_{k-\frac{1}{2}}^{k+ \frac{1}{2}}\ln x dx$$
In other words, I'm trying to show why the area of the rectangle with height $\ln k$ and width $1$ bounds the area under the graph of $f(x)=\ln x$ in the interval $[k-\frac{1}{2},k+\frac{1}{2}].$
I tried to integrate but got stuck. Any ideas for an elegant proof for this?   

Hint: Note that $\ln$ is concave. You can generally show that for concave functions $f$, we have $$f\left(\frac{a+b}{2}\right) \ge \frac{1}{b-a}\int_a^b f(x)\, dx.$$ (This is a continuous form of Jensen's inequality.)
