设f'(x)在[a,b]上连续,且f(a)=0,│∫(a~b)f(x)dx│≤((b-a)^2)/2)max(a≤x≤b)│f'(x)│
问题描述:
设f'(x)在[a,b]上连续,且f(a)=0,│∫(a~b)f(x)dx│≤((b-a)^2)/2)max(a≤x≤b)│f'(x)│
答
f(x)-f(a)=f'(c)(x-a) |∫f(x)dx|=|∫f'(c)(x-a)dx=(b-a)^2/2*|f'(c)|