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?   
 A: Logarithm is a concave function, by Jensen inequality,
$$\ln E\left( U\right) \ge E\left( \ln (U)\right)$$
where $U \sim Uni\left( k-\frac12, k+\frac12\right)$.
$$\ln k \ge \int_{k-\frac12}^{k+\frac12} \ln (x)\, dx$$
A: 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.)
A: Too long for a comment but written for your curiosity.
You received good answers so I should use integration for illustration. Since, using one integration by parts,
$$\int \log(x)=(x-1)\log(x)$$ using the given bounds, the rhs is
$$\text{rhs}=\left(\frac{1}{2} (2 k+1) \left(\log \left(k+\frac{1}{2}\right)-1\right)\right)-\left(\frac{1}{2} (2 k-1) \left(\log \left(k-\frac{1}{2}\right)-1\right)\right)$$ Considering at least that $k$ can be large, using Taylor expansions,
$$\text{rhs}=\log(k)-\sum_{n=1}^\infty \frac {c_n}{k^{2n}}$$ and all coefficients $c_n$ are negative. Their reciprocals are
$$\{24,320,2688,18432,112640,638976,3440640,17825792,89653248,440401920\}$$ and they are related to the coefficients of Chebyshev polynomials
$$c_n=\frac {2^{-(2 n+1)} } { n (2 n+1) }$$
