The proof of $\frac{f(a)+f(b)}{2}\le\frac{1}{b-a}\int^b_af(x)dx$ provided that $f''(x)<0,x \in [a,b]$ and $f\in \mathcal C^2[a,b]$ This question is geometrically obvious, but I got a little trouble when proving it.
Let's consider the Taylor expansion of  $F(x)=\int^x_af(x)\mathrm dx$ at  $a$ and  $b$ and subsitute  $b,a$ into it respectively.
$$\int_a^bf(x)\mathrm dx=F_a(b)=f(a)(b-a)+f'(a)(b-a)^2/2+f''(\xi_1)(b-a)^3/6\\0=F_b(a)=\int^b_af(x)\mathrm dx+f(b)(a-b)+f'(b)(a-b)^2/2+f''(\xi_2)(a-b)^3/6$$
And subtract the second from the first
$$\int_a^bf(x)\mathrm dx=F_a(b)-F_b(a)=-\int_a^bf(x)\mathrm dx+(b-a)(f(a)-f(b))+(f'(a)-f'(b))(b-a)^2/2+(f''(\xi_1)+f''(\xi_2))(b-a)^3/6$$
And divide both side by 2(b-a) and subsitute $f'(a)-f'(b)=-f(\xi_3)(b-a)$ into it
$$\frac{1}{b-a}\int_a^bf(x)\mathrm dx=\frac{f(a)-f(b)}2-f''(\xi_3)(b-a)^2/4+(f''(\xi_1)+f''(\xi_2))(b-a)^2/12$$
since $f''(\xi_3)<0$ we have
$$\frac{1}{b-a}\int_a^bf(x)\mathrm dx\ge\frac{f(a)-f(b)}2+(f''(\xi_1)+f''(\xi_2))(b-a)^2/12$$
I guess my approximation is too rough so I get an additional term $(f''(\xi_1)+f''(\xi_2))(b-a)^2/12$.
Maybe the glitch is that I use $f''(x)<0$ just at one point i.e. $\xi_3$ but actually it holds at every point in  $[a,b]$ which is a decisive condition.
 A: Since $f''(x)<0$, the function $f$ is concave, i.e. for all $t\in [0,1]$, \begin{align*}
f(a)+t\big(f(b)-f(a)\big)&=tf(b)+(1-t)f(a)\\
&\leq f\big(tb+(1-t)a\big).
\end{align*}
Integrating both sides yields
$$\frac{f(a)+f(b)}{2}\leq \int_0^1 f\big(tb+(1-t)a\big)\,\mathrm d t=\frac{1}{b-a}\int_a^b f(x)\,\mathrm d x,$$
as wished.

Remark that the result hold for all concave functions, no need to be $\mathcal C^2$.
A: If this question appeared in a numerical analysis course, it is just stating that, for concave functions, the trapezoidal method underestimates the value of the integral.
Trapezoidal rule:
$$
\int_a^b f(x) dx \approx \frac{b-a}{2}(f(a)+f(b))
$$
Using the interpolation error by linear function and the mean value theorem for integrals, you can easily obtain the error formula:
\begin{align*} 
\int_a^b f(x) dx - &\int_a^b p_1(x) dx = \int_a^b \frac{f''(\xi_x)}{2}(x-a)(x-b) dx\\
=&\frac{f''(\xi)}{2}\int_a^b p_1(x) dx = -\frac{(b-a)^3}{12} f''(\xi).
\end{align*}
Hence,
$$
\int_a^b f(x) dx = \frac{b-a}{2}(f(a)+f(b)) -\frac{f''(\xi)}{12}(b-a)^3
$$
Since in this case $f''(\xi)<0$, you easily obtain
$$\int_a^b f(x) dx \ge \frac{b-a}{2}(f(a)+f(b)),$$
which is equivalent to the proposed inequality.
note: In the picture the inequality just states that the area of the triangle (in general trapezoid) is less than the area under the graphic of $f$.
