<s>[INST] 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.
 [/INST] 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 $x \in [a,b]$,$\left |F(x) \right | \le \frac{M(b-a)^2}{8}$
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