补充如果f(t)连续，a(x)、b(x)可导则df(t)dt = f(b(x))b'(x);dxf(t)dt = -f (a(x)a'(x);Df(t)dt = f (b(x)b'(x)- f (a(x)a'(x)经济数学微积分
如 果 f (t)连续，a(x)、b(x)可导， 则 补充 ( )d ( ( )) ( ); d d ( ) f t t f b x b x x b x = ( ) ( )d ( ) ( ); ( ) a x d f t t f a x a x dx = − ( ) ( ) ( ) ( ) ( )d ( ) ( ) ( ) ( ). b x a x d f t t f b x b x f a x a x dx = −
b(x)证F(x) =f(t)diTXb(x)f(t)dt10a(x)a(x)b(x)f(t)dt,f(t)dt -0.:. F'(x) = f(b(x)b'(x)- f(a(x)a'(x)经济数学微积分
证 ( ) ( ) ( ) ( )d b x a x F x f t t = ( ) 0 ( )d b x = f t t ( ) 0 ( )d , a x − f t t F(x) = f (b(x))b(x)− f (a(x))a(x) ( ) 0 ( ) ( ) 0 ( )d b x a x = + f t t
dt0cosx例1求im2x-→0x0分析：这是型不定式，应用洛必达法则0ddcosx解dt.dtdxdxcosx' =sinxe-cos'xCOS(cos x)2-cos?xdt1sinx:ecosx= limlim2x2ex-→0x-0福经济数学微积分
例1 求 1 2 cos 2 0 d lim . t x x e t x − → 解 1 2 cos d t x d e t dx − cos 2 1 d , x d t e t dx − = − (cos ) 2 cos = − − e x x sin , 2 cos x x e − = 1 2 cos 2 0 d lim t x x e t x − → x x e x x 2 sin lim 2 cos 0 − → = . 2 1 e = 0 0 分析：这是 型不定式，应用洛必达法则
例 2 设,f(x)在(-0,+)内连续，且,f(x)>0.J, tf(t)dt证明函数F(x)在(0,+)内为单调增I* f(t)dt加函数dJ.(0d= (n),证tf(t)dt= xf(x),dxxf(x)f f(t)dt -f(x)f* tf(t)dtF'(x) =2(J. f()dr)华微积分经济数学
例 2 设 f (x)在(−,+)内连续，且 f (x) 0. 证明函数 0 0 ( )d ( ) ( )d x x tf t t F x f t t = 在(0,+)内为单调增 加函数. 证 0 ( )d d x tf t t dx = xf (x), 0 ( )d d x f t t dx = f (x), ( ) 0 0 2 0 ( ) ( )d ( ) ( )d ( ) ( )d x x x xf x f t t f x tf t t F x f t t − =
f(x) / (x -t)f(t)dtF'(x)=2(S. ()dr). ( f(t)dt >0,: f(x)>0, (x>0):(x-t)f(t)>0, : ((x-t)f(t)dt >0,:. F'(x)>0(x>0).故F(x)在(0,+o)内为单调增加函数微积分经济数学
( ) 0 2 0 ( ) ( ) ( )d ( ) , ( )d x x f x x t f t t F x f t t − = f (x) 0, (x 0) 0 ( )d 0, x f t t (x − t) f (t) 0, 0 ( ) ( )d 0, x − x t f t t F(x) 0 (x 0). 故F(x)在(0,+)内为单调增加函数