(ii) 若函数g 在[a, bl 上单调增,且 g(x)≥ 0,则存在 n e[a, bl, 使 [, f(x)g(x)dx = g(b), f(x)dx证这里只证(i)，类似可证(ii).证明分以下五步：(1) 对任意分割 T: a=x<x,<.….<x,=b,1-J' f(x)g(x)dx-Z/" f(x)g(x)dxXi-1i-1Z[ (x) I g(x) - g(xi-) Idxi=1+Zg(x,-) (x)dx =1,+ 1.i-1(2) 因 / f(x) / <L, xE[a, b], 故前页后页返回
前页 后页 返回 (ii) 若函数 g 在 [a, b] 上单调增, 且 g(x) 0, 则存 在 使 [ , ], a b ( ) ( )d ( ) ( )d . b b a f x g x x g b f x x = 证 这里只证 (i), 类似可证 (ii). 证明分以下五步: (1) 对任意分割 T： , a = x0 x1 xn = b ( ) ( )d b a I f x g x x = 1 1 ( ) ( )d i i n x x i f x g x x − = = 1 1 1 ( ) [ ( ) ( ) ]d i i n x i x i f x g x g x x − − = = − . 1 2 = I + I 1 1 1 ( ) ( )d i i n x i x i g x f x x − − = + (2) | ( ) | , [ , ], 因 f x L x a b 故
≥" (x)I g(x)-g(xμ)]dx11. /=Xi-1≥"1 (x)1-I g(x)-g(xi-1)]dxsL2w'Ax,i=1i-1因g 可积,故 ET :a=x,<x,<...<x,=b,使ZoA-se8Li-1(3) 设 F(x)=[ f(t)dt, 则1, -Zg(xi-1)[F(x)-F(xi-1)]i=1= g(x,)[F(x) -F(x)I+...+ g(xn-)[F(xn)-F(xn-))后页返回前页
前页 后页 返回 1 1 1 1 | | ( ) [ ( ) ( )]d i i n x i x i I f x g x g x x − − = = − 1 1 1 | ( ) | | ( ) ( ) | d i i n x i x i f x g x g x x − − = − 1 Δ . n g i i i L x = 0 1 , : , n 因 可积 故 使 g T a x x x b = = 1 Δ n g i i i x L = 1 | | . I 2 1 1 1 ( )[ ( ) ( )] n i i i i I g x F x F x − − = = − 0 1 0 = − + g x F x F x ( )[ ( ) ( )] ( )[ ( ) ( )] + g xn−1 F xn − F xn−1 (3) ( ) ( )d , x a 设 F x f t t = 则
= F(x)[g(x)- g(x)I+...+ F(x.-)[g(xn-2)- g(xn-1)I+ F(xn)g(xn-1) F(x,)Ig(xi-1) - g(x,)I+ F(b)g(xn-1).i-1由对 g 的假设, g(xn-1)≥0, g(x-1)-g(x,)≥0. 记m= min ( F(x)), M= max (F(x)),xe(a,b)xe(a,b)n-1则 I, ≤ ME[g(xi-1)- g(x,)I+ Mg(xn-1) = Mg(a),i-1n-1I, ≥m [g(x-1)- g(x,)I + mg(x,-1) =mg(a),i-1返回前页后页
前页 后页 返回 1 1 , ( ) 0, ( ) ( ) 0. n i i 由对 的假设 记 g g x g x g x − − − 1 0 1 = − + F x g x g x ( )[ ( ) ( )] ( )[ ( ) ( )] ( ) ( ). 1 1 1 1 − = = − − + − n i F xi g xi g xi F b g xn ( )[ ( ) ( )] ( ) ( ) + F xn−1 g xn−2 − g xn−1 + F xn g xn−1 ( , ) min { ( ) }, x a b m F x = ( , ) max { ( ) }, x a b M F x = 1 2 1 1 1 [ ( ) ( )] ( ) ( ), n i i n i 则 I M g x g x Mg x Mg a − − − = − + = 1 2 1 1 1 [ ( ) ( )] ( ) ( ), n i i n i I m g x g x mg x mg a − − − = − + =
于是 mg(a)≤ I, ≤ Mg(a)(4) 综合 (2), (3), 得到mg(a) -≤ I + I, ≤Mg(a) +6.令→0, 便得 mg(a)≤I <Mg(a).(5) 若 g(a)=0, 则 1=[' f(x)g(x)dx=0, 此时任取I' f(x)g(x)dx = g(a)" f(x)dx.5e[a, b], 满足名I若 g(a)>0, 则 m≤≤M. 由 F(x)=["f(t)dig(a)前页后页返回
前页 后页 返回 (4) 综合 (2), (3), 得到 1 2 mg a I I Mg a ( ) ( ) . − + + 令 便得 → 0, ( ) ( ). mg a I Mg a (5) ( ) 0, ( ) ( )d 0, b a 若 则 此时任取 g a I f x g x x = = = [ , ], a b 满足 ( ) ( )d ( ) ( )d . b a a f x g x x g a f x x = ( ) ( ). 于 是 mg a I2 Mg a 若 则 g a( ) 0, . ( ) M g a I m ( ) ( )d x a 由 F x f t t =
的连续性，存在e[a,bl,使F(5)= f'f(t)dt=-g(a)[' f(x)g(x)dx= g(a) [" f(x) dx.即推论设 f(x)在[a,b]上可积,g(x)在[a,b]上单调,则存在[a,b],使' f(x)g(x)dx = g(a)" f(x)dx + g(b)f' (x)dx.后页返回前页
前页 后页 返回 ( ) ( )d , ( ) a I F f t t g a = = 则存在 使 [ , ], a b ( ) ( )d ( ) ( )d ( ) ( )d . b b a a f x g x x g a f x x g b f x x = + 推论 设 在 上可积, 在 上单调, f x a b g x a b ( ) [ , ] ( ) [ , ] 的连续性 ,存在 [a, b], 使 ( ) ( )d ( ) ( )d . b a a f x g x x g a f x x = 即