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?   

Logarithm is a concave function, by Jensen inequality,
$$\ln E\left( U\right) \ge E\left( \ln (U)\right)$$
where $U \sim Uni\left( k-\frac12, k+\frac12\right)$.
$$\ln k \ge \int_{k-\frac12}^{k+\frac12} \ln (x)\, dx$$
