微积分(A)随缘一题[26]

设 \(f(x) \in C^{1}[a,b]\)，且 \(f(a)=0\)

求证：\(\int_{a}^{b}f^2(x)dx \le \frac{(b-a)^2}{2}\int_a^b[f'(x)]^2dx\)

\[\begin{aligned} &\because f(x)=f(x)-f(a)=\int_{a}^{x}f'(t)dt \\ &\therefore f^2(x)=\left( \int_{a}^{x}f'(t)dt \right)^2 \le \int_{a}^{x}[f'(t)]^2dt\int_{a}^{x}dt=(x-a)\int _a^x[f'(t)]^2dt \le (x-a)\int_a^b[f'(t)]^2dt \\ & \therefore \int _{a}^{b} f^2(x)dx \le \int _{a}^{b}(x-a) \left(\int _a^b[f'(t)]^2dt \right)dx = \frac{(b-a)^2}{2} \int _a^b[f'(t)]^2dt \end{aligned} \]