性质26(xk=k(x)d(k为常数) 证(x)=lm∑的(5△r im∑f(5Ax=klim∑f(5)x kf(xdx 性质1+性质2得:性质2 = b a b a kf (x)dx k f (x)dx (k为常数). 证 b a kf (x)dx i i n i = kf x = → lim ( ) 1 0 i i n i = k f x = → lim ( ) 1 0 i i n i = k f x = → lim ( ) 1 0 ( ) . = b a k f x dx 性质1+性质2 得: