第四章重积分 ∫y(x)/()2y()/(x →jy(x)f()-y2()/()≥0 of(x)/2()-yo)(x)] 20 ∫()2(x)-xf(x/() x(D≥x(2e j()2()-yf(o)/()+y()y()-xf(()ohy20 因:xf(x)2()-yf2(v)(x)=f(x)/(0)xy(U)-y/(x) 则,x/(x)f2()-yf2()/(x)+y/()f2(x)-xf2(x)/() f(x)()x-yf(x)-f()≥0 10.若x∈[]0<m≤/(x)≤M,证明: M+m)2 l≤ 4Mm 0≤x≤ Syst 证明: dxdy dxd 0≤ys1 首先有:2「(b dxdy d(/()f(x) 0≤ys1 (需 +2ad≥2 dxdy=1 重积分习题讨论第四章 重积分 重积分习题讨论  xf(x)f (y)dxdy yf (y)f (x)dxdy    1 0 1 0 2 1 0 1 0 2  ( ( ) ( ) ( ) ( )) 0 1 0 1 0 2 2 −   xf x f y yf y f x dxdy  ( ( ) ( ) ( ) ( )) 0 1 0 1 0 2 2 −   xf x f y yf y f x dxdy ( ( ) ( )− ( ) ( )) =  xf x f y yf y f x dxdy 1 0 1 0 2 2 (yf (y)f (x) xf (x)f (y))dxdy  − 1 0 1 0 2 2 ( ,0,1) ( ,0,1) 2 x f  x f  ( ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )) 0 1 0 1 0 2 2 2 2 − + −   x f x f y yf y f x yf y f x x f x f y dxdy 因： x f(x)f (y)− yf (y)f (x) = f (x)f (y)(x f(y)− yf (x)) 2 2 则， x f(x)f (y) yf (y)f (x) yf (y)f (x) x f (x)f (y) 2 2 2 2 − + − = f (x)f (y)(x − y)(f (x)− f (y))  0 10.若 x0,1, 0  m  f (x)  M , 证明： ( ) ( ) ( ) M m M m dxdy f y f x y x 4 1 2 0 1 0 1 +        . 证明： ( ) ( ) ( ) ( ) ( ) ( )     = =         1 0 1 0 0 1 0 1 0 1 0 1 1 dx f x dx f x dxdy f x f y dxdy f y f x y x y x 首先有： ( ) ( ) ( ) ( ) ( ) ( )                   = + 0 1 0 1 0 1 0 1 2 y x y x dxdy f x f y f y f x dxdy f y f x = ( ) ( ) ( ) ( ) 2 2 1 0 1 0 1 0 1 0 1 2  =         +         −           y x y x dxdy dxdy f x f y f y f x ;