A refinement of Hermite-Hadamard inequality

If $af(b)-bf(a)=0$ with $f"$a continuous and concave function then $\frac{f(b)-f(a)}{2}(b-a)<\int^b_af(x)dx \leq \frac{f(a)+f(b)}{2}(a+b)$ Proof:the graph of f is above the line determined by AB having the equation $ y=\frac{f(b)-f(a)}{b-a}x$ so f(x)>$ \frac{f(b)-f(a)}{b-a}x=\frac{f(b)-f(a)}{2}(a+b)$.

For the second part of inequality the area of trapeze ABCD is higher than of the surface between lines x=a,x=b and the grapf of f hence the conclusion.