r f(Mdt-f(dt ∫nf(M+」r(xy-mf(Mm x+△r f(tdt 由积分中值定理得 (x) △Φ=f4)x5∈[x,x+△x xa+△xbx →A=(5)→趣A=m 又Δx→>0→2x∴.d(x)=f(x)f t dt f t dt f t dt x a x x x x a = + − + ( ) ( ) ( ) f t dt x x x + = ( ) 由积分中值定理,得 =f( )x [x, x+x] f ( ) x = lim lim ( ) 0 0 f x→ x x→ = 又x→0→x ∴(x)=f(x) y o a x x+xbx (x) f t dt f t dt x a x x a = − + ( ) ( )