@Eugene: Since $f$ is concave, for $x\in[k,k+1]$, we have
$$
f(x)\ge f(k)+(x-k)(f(k+1)-f(k))\tag{1}
$$
Integrating yields
$$
\int_k^{k+1}f(x)\,\mathrm{d}x\ge\frac{f(k)+f(k+1)}{2}\tag{2}
$$
Integrating
$$
f(x)\le f(k)+(x-k)f'(k)
$$
yields
$$
\int_k^{k+1}f(x)\,\mathrm{d}x\le f(k)+\frac12f'(k)\tag{3}
$$
Integrating
$$
f(x)\le f(k+1)+(x-k-1)f'(k+1)
$$
yields
$$
\int_k^{k+1}f(x)\,\mathrm{d}x\le f(k+1)-\frac12f'(k+1)\tag{4}
$$
Averaging $(3)$ and $(4)$ gives
$$
\int_k^{k+1}f(x)\,\mathrm{d}x\le\frac{f(k)+f(k+1)}{2}+\frac{f'(k)-f'(k+1)}{4}\tag{5}