设 $f(x)$ 在 $[a,b]$ 上一阶连续可导, $f(a)=0$. 证明: $$ex int_a^b f^2(x) d xleq cfrac{(b-a)^2}{2}int_a^b [f'(x)]^2 d x -cfrac{1}{2}int_a^b [f'(x)]^2 (x-a)^2 d x. eex$$
证明: $$eex ea int_a^b f^2(x) d x &=int_a^b sez{int_a^xf'(t) d t}^2 d x\ &leq int_a^b sez{ int_a^x f'^2(t) d t cdot int_a^x 1^2 d t } d x\ &=int_a^b int_a^x (x-a)f'^2(t) d t d x\ &=int_a^b int_t^b (x-a)f'^2(t) d x d t\ &=int_a^b f'^2(t)int_t^b (x-a) d x d t\ &=int_a^b f'^2(t)cfrac{(b-a)^2-(t-a)^2}{2} d t. eea eeex$$