设f(x)在[a,b]上可导,且f'(x)≤M,f(a)=0,求证∫(a,b)f(x)dx≤M/2(b-a)^2

来源：学生作业帮助网编辑：六六作业网时间：2024/11/26 18:51:44

设f(x)在[a,b]上可导,且f''(x)≤M,f(a)=0,求证∫(a,b)f(x)dx≤M/2(b-a)^2设f(x)在[a,b]上可导,且f''(x)≤M,f(a)=0,求证∫(a,b)f(x)d

设f(x)在[a,b]上可导,且f'(x)≤M,f(a)=0,求证∫(a,b)f(x)dx≤M/2(b-a)^2
设f(x)在[a,b]上可导,且f'(x)≤M,f(a)=0,求证∫(a,b)f(x)dx≤M/2(b-a)^2

设f(x)在[a,b]上可导,且f'(x)≤M,f(a)=0,求证∫(a,b)f(x)dx≤M/2(b-a)^2
证明：用中值定理
∵f(a)=0
∴∫(a,b)f(x)dx
=∫(a,b)f(x)dx-f(a)(b-a)
=∫(a,b)[f(x)-f(a)]dx
=∫(a,b)f'(ξ)(x-a)dx
=f'(ξ)∫(a,b)(x-a)dx,其中ξ∈(a,b)
≤M∫(a,b)(x-a)dx
=M/2[(b-a)²]
证毕.

由题意知:f(x)在[a,b]上可导,
所以,f(x)=f(x)-f(a)=f'(ξ)(x-a)≤M(x-a).
两边积分:
∫(a,b)f(x)dx≤∫(a,b)M(x-a)dx=M/2(x-a)^2|(a,b)=M/2(b-a)^2

设f(x)在[a,b]二阶可导,且f''(x) 设函数f(x)在[a,b]上连续,在(a,b)上可导且f'(x) 设函数f x,gx在[a,b]上可导,且f'x 设函数f(x),g(x)在[a,b]上可导,且f'(x)>g'(x),则当a 设f(x)在[a,b]上连续,且a 设f(x)在[a,b]上连续,且a 设f(x)在[a,b]上连续,且a 设函数f(x)在[a,b]上连续,在(a,b)内可导且f'(x) 设f(x)在[a,b]上二阶可导,且f''(x)>0,证明:函数F(x)=(f(x)-f(a))/(x-a)在(a,b]上单调增加设函数f(x),g(x)在[a,b] 上均可导,且f'(x) 设f在[a,b]上可导,|f'(x)| 设函数f(x),g(x)在区间[a,b]上连续,且f(a) 设函数f(x)在R上的导函数为f'(x),且f(x)>f'(x).若a>b,则（）A.e^b*f(b) 设f(x),g(x)在〔a,b]上可导,且F的导数大于G的导数,当a 设函数f(x)在[a,b]可导且f'(x) 设f(x) 在[a,b] 上连续,且f(x)>0.求证:∫(a,b)f(x)dx*∫(a,bdx/f(x)≥(b-a)^2. 设f(x),g(x)在[a,b]上可导,且f’(x)＞g’(x),则当a＜x＜b时,有f(x)+g(a)＞g(x)+f(a) 设函数f(x)在[a,b]上连续,在(a,b)可导,且f(a)*f(b)>0,f(a)*f((a+b)/2)