Prove that $\int\limits_0^1 f(x) dx \le - \bigg |f\left(\frac{1}{2}\right)\bigg | + \frac{5}{24} \sup\limits_{x\in [0,1]}|f''(x)|$ Let $f:[0,1] \to \mathbb{R}$ be a twice differentiable function with $f''$ Riemann integrable on $[0,1]$ and $f(1)+f(0)=0$.
Prove that $$\int\limits_0^1 f(x) dx \le  - \bigg |f\left(\frac{1}{2}\right)\bigg | + \frac{5}{24} \sup\limits_{x\in [0,1]}|f''(x)|.$$
I could only observe two things and I don't know if either of them is of any use :


*

*$\int\limits_0^1 f'(x) dx=f(1)-f(0)=2f(1)$

*If we consider $g:[0,1]\to \mathbb{R}$, $g(x)=f(x)+f(1-x)$, then $g$ is twice differentiable and $g(0)=g(1)=0$. We may apply Rolle's theorem to conclude that $\exists c \in (0,1)$ such that $g'(c)=0$. But this is useless, since $g'(x)=f'(x)-f'(1-x),\forall x\in [0,1]$ and from here we have that $g'\left(\frac{1}{2}\right)=0.$
 A: The idea is to approximate $f$ by the linear interpolation polynomial which interpolates $f$ at $x=0$ and $x=1$, and that is
$$
 l(x) = f(0) + (f(1)-f(0) x =  f(0) (1 - 2x) \, .
$$
since $f(1) = -f(0)$. Note that $\int_0^1 l(x) \, dx = 0$ and $l(\frac 12) = 0$.
With $M = \sup\limits_{x\in [0,1]}|f''(x)|$ the following estimate holds:
$$ \tag{*}
 |f(x) - l(x)| \le \frac M2 x(1-x) \text{ for } 0 \le x \le 1 \, .
$$
It follows that
$$
 \int_0^1 f(x) \, dx = \int_0^1 (f(x)-l(x)) \, dx \le \frac M2 \int_0^1 x(1-x) \, dx = \frac 1{12}M
$$
and
$$
 |f(\frac 12)| = |f(\frac 12) - l(\frac 12)| \le \frac 18 M \, .
$$
Adding these inequalities  gives the desired conclusion.
Remarks: 


*

*Equality holds for $f(x) = x(1-x)$, so that the inequality is sharp.

*It is only needed that $f$ is twice differentiable, but not that $f''$ is Riemann integrable.



It remains to prove $(*)$: That is a  well-known error estimate for polynomial interpolation, I'll include a short proof for the sake of completeness.
For fixed $0 < a < 1$ consider the function
$$
 g(x) = a(1-a) (f(x)-l(x)) - x(1-x) (f(a) - l(a)) \, .
$$
Then $g(0) = g(a) = g(1) = 0$ and repeated application of Rolle's theorem gives $g''(c) = 0$ for some $c \in (0, 1)$:
$$
0 = g''(c) = a(1-a) f''(c) + 2 (f(a) -l(a)) \\
\implies |f(a) - l(a) | = \frac 12 a(1-a) |f''(c)| \le \frac M2 a(1-a) \, .
$$
