How to prove $|F(x)|\leq\frac{M(b-a)^2}{8}$ $f(x)$ is derivable in $[a,b]$, $|f^{'}(x)|\leq M$.
$\int_a^b f(x)dx=0$.
Let $F(x)=\int_a^x f(t)dt$.
Try to prove $|F(x)|\leq\frac{M(b-a)^2}{8}$

I want to use Taylor expansion at $f(\xi)=0$, but I can't continue.
 A: Assume that $|F|$ is maximum at $x$, hence $F'(x)=0$ and, for example, $F(x)=h\geqslant0$ (if $F(x)\lt0$, apply to $-f$ the reasoning below). Thus, $F(a)=F(b)=0$ and $F''=f'$ hence $F''\geqslant -M$ on the interval $(a,b)$, and the task is to show that $h\leqslant\frac18M(b-a)^2$.
First, a Taylor expansion of $F$ on the interval $(a,x)$ based at $x$ yields that there exists some $t$ in $(a,x)$ such that $F(a)=F(x)+(a-x)F'(x)+\frac12(a-x)^2F''(t)$. 
Thus, $0=h+\frac12(a-x)^2F''(t)\geqslant h-\frac12(a-x)^2M$, hence $h\leqslant\frac12(a-x)^2M$. 
Likewise, a Taylor expansion of $F$ on the interval $(x,b)$ based at $x$ yields that there exists some $s$ in $(x,b)$ such that $F(b)=F(x)+(b-x)F'(x)+\frac12(b-x)^2F''(s)$. 
Thus, $0=h+\frac12(b-x)^2F''(s)\geqslant h-\frac12(b-x)^2M$, hence $h\leqslant\frac12(b-x)^2M$.
Now, either $(x-a)^2\leqslant\frac14(b-a)^2$ or $(x-b)^2\leqslant\frac14(b-a)^2$, in both cases $h\leqslant\frac18M(b-a)^2$.
A: I can prove that $\left | F( \frac{a+b}{2})\right | \le \frac{M(b-a)^2}{8}$.
By Taylor expansion at $ x_0 = \frac{a+b}{2}$,we have
$$F(x) = F(\frac{a+b}{2}) + F'(\frac{a+b}{2})(x-\frac{a+b}{2}) + \frac{F''(\xi)}{2}(x-\frac{a+b}{2})^2$$
Substituting $x=a$ and $x=b$ into the formula above,we get
$$F(a) = F(\frac{a+b}{2}) + F'(\frac{a+b}{2})(a-\frac{a+b}{2}) + \frac{F''(\xi_1)}{2}(\frac{a-b}{2})^2$$
$$F(b) = F(\frac{a+b}{2}) + F'(\frac{a+b}{2})(b-\frac{a+b}{2}) + \frac{F''(\xi_2)}{2}(\frac{a-b}{2})^2$$
To the sum of the two formulas above,with $F(a)=F(b)=0$,we have
$$\left |F(\frac{a+b}{2}) \right | \le \left | \frac{F''(\xi_1)+F''(\xi_2)}{4} \right |(\frac{a-b}{2})^2 \le \frac{M(b-a)^2}{8}$$
But I cannot get better ideas to prove that for any $x \in [a,b]$,$\left |F(x) \right | \le \frac{M(b-a)^2}{8}$
