推论2 ∫rd)d(a<b) 证：-f(x)≤f(x)≤f(x) -f)dk≤fx)d≤∫fx)d 即 d)d 性质6设M=maxf(x),m=minf(x),则 [a,b] [a,b] mb-a)≤f(x)dr≤Mb-a)(a<b) BEIJING UNIVERSITY OF POSTS AND TELECOMMUNICATIONS PRESS 目录上页 下页 返回结束目录 上页 下页 返回 结束 推论2 证:   f (x)  f (x)  f (x) (a  b) f x x f x x f x x b a b a b a ( ) d ( )d ( ) d        即 f x x f x x b a b a ( )d ( ) d    性质6 设 max ( ), min ( ) , [ , ] [ , ] M f x m f x a b a b   则 (a  b)