设 $f(x)$ 的一阶导数在 $[0,1]$ 上连续, 且 $f(0)=f(1)=0$, 求证: $\dps{\sev{\int_0^1 f(x)\rd x}\leq \frac{1}{4}\max_{0\leq x\leq 1}|f'(x)|}$. (清华大学)

证明: 设 $\dps{M=\max_{[0,1]}|f'|}$, 则 $$\beex \bea \sev{\int_0^1 f(x)\rd x} &=\sev{\int_0^\frac{1}{2}\int_0^x f'(t)\rd t\rd x +\int_\frac{1}{2}^1 \int_1^x f'(t)\rd t\rd x}\\ &\leq \int_0^\frac{1}{2} \int_0^x |f'(t)|\rd t\rd x +\int_\frac{1}{2}^1 \int_x^1 |f'(t)|\rd t\rd x\\ &\leq M \int_0^\frac{1}{2}x\rd x +M\int_\frac{1}{2}^1 (1-x)\rd x\\ &=\frac{M}{4}. \eea \eeex$$