Problem
Let $f(x)$ be differentiable over $[0,1]$,$f(0)=f(1)=0$, and $|f'(x)|\leq M$ for $x\in[0,1]$. Prove $$\int_0^1 f(x){\rm d}x\leq \frac{M}{4}.$$
Proof
According to Lagrange's Mean Value Theorem, $$\forall x \in (0,1/2]:f(x)=f(x)-f(0)=xf'(\xi),0<\xi<x\leq 1/2.$$ Thus$$\forall x \in (0,1/2]:|f(x)|=x|f'(\xi)|\leq Mx,$$ which also holds for $x=0$ actually, by verification.
Likewise, $$\forall x \in [1/2,1):f(x)=f(x)-f(1)=(x-1)f'(\eta),1/2 \leq x<\eta<1.$$ Thus$$\forall x \in [1/2,1):|f(x)|=(1-x)|f'(\eta)|\leq M(1-x),$$ which also holds for $x=1$.
It follows that \begin{align*} \int_0^1f(x){\rm d}x\leq \left|\int_0^1f(x){\rm d}x\right| \\ &=\left|\int_0^{\frac{1}{2}}f(x){\rm d}x+\int_{\frac{1}{2}}^1 f(x){\rm d}x\right| \\&\leq \left|\int_0^{\frac{1}{2}}f(x){\rm d}x \right|+\left|\int_{\frac{1}{2}}^1 f(x){\rm d}x\right|\\ &\leq \int_0^{\frac{1}{2}}|f(x)|{\rm d}x+\int_{\frac{1}{2}}^1 |f(x)|{\rm d}x\\ &\leq M\int_0^{\frac{1}{2}}x{\rm d}x+M\int_{\frac{1}{2}}^1 (1-x){\rm d}x\\ &=\frac{M}{8}+\frac{M}{8}\\ &=\frac{M}{4} \end{align*}