设f(x)在[a,b]上连续可导,且f(a)=0.证明: ∫abf2(x)dx≤(b-a)2/2∫ab[f′(x)]dx

admin2022-08-19  20

问题 设f(x)在[a,b]上连续可导,且f(a)=0.证明:
abf2(x)dx≤(b-a)2/2∫ab[f′(x)]dx

选项

答案由f(a)=0,得f(x)-f(a)=f(x)=∫axf′(t)dt,由柯西不等式得 f2(x)=[∫axf′(t)dt]2≤∫ax12dt∫axf′2(t)dt≤(x-a)∫abf′2(x)dx, 积分得∫abf2(x)dx≤∫ab(x-a)dx·∫abf′2(x)dx=[(b-a)2]/2∫abf′2(x)dx.

解析
转载请注明原文地址:https://jikaoti.com/ti/eLfRFFFM
0

最新回复(0)