1、 (xc)dx AB 是沿橢圓x2 1的正向從A(a,0)到B(0,b)yI AB (xc)2 y23一、計(c0), a解：對于函數u(x, y) ,u x,y.(xc (xc)dx AB 是沿橢圓x2 1的正向從A(a,0)到B(0,b)yI AB (xc)2 y23一、計(c0), a解：對于函數u(x, y) ,u x,y.(xc)2 y21(xc)2 y23(xc)2 y23(xc)dx (xc)dx I AB(xc)2 y23所以du(x, y) du(x, y) u(0,b)u(a,(xc)2 y231(a c)=|ac二、I lim (ey2 x2 cos2xy3y)dxey2

2、 x2 sin2xyb2dy (b0)L是依次連L),O(0,0)的有向折線(已知0 dx)22 2 解(ey x cos2xy3y)dx(ey x sin2xyb L22)dxdycos2xy3y)dxDa2 2 2 2 2Dy 2x)sin2xy2ycos2xy 2ycos2xy 2xsin 2xy 3)dxdy y y y x0a2dxdydxxx00D2 2 所I cos2xy3y)dxL 2adx)dx3= lim001三、|a|p |b|p21p(|a|b|p, (0 p1)有21 證明: 1先證當0 x10 p1 xp (1 x)p 1三、|a|p |b|p21p(|a|b|p,

3、 (0 p1)有21 證明: 1先證當0 x10 p1 xp (1 x)p 1 2Fx xp 1 xpFx) pxp1 p(1xp1Fx) 0得 x F(1 )21p，F(1) F(0)2所以當0 x10 p1F1 21p為最大值，F(1)F(01221 xp (1 x)p 有|a|b2 x , 則1 x . |a|b|a|b|a |b21(|a|b|)(|a|b|)(|a|b|)p |a|p |b|p21p(|a|b|)p, (0 p1)四fx在0,2f (x0, n 2f(2) f(0)2f (x)sinnf (x)d cosnx1( f(2) f (0) 011n222f (x)sin

4、nxdx= f(x)cos證明nn000 f f1nf (x)dx 2f(2) fn22f (x)sinn00五、設在abfx0a x1 x2 b0 f(x1)(1)f(x2) fx1(1)x2f(x1(1)x2)f(x1)(1)(x2x1)f 證明f(x1(1)x2)(x2x1)f(2f(x2)(1)(2)(1)f(x1(1)x2)f(x1)(1)f(x2)(1)(x2 x1)f(1) f(2f(x1(1)x2)(1)(x2x1)f(2) f(1)f(x1)(1)f(x2f(x1(1)x2)(1)(x2 x1)f() f(x1)(1)f(x2f(0, 所以fx11fx2 fx11)x22 六、證明x x x 2 x 2 2 n 1n nnx1x2 xn nx x x 1 證明有如下結論nnnf 六、證明x x x 2 x 2 2 n 1n nnx1x2 xn nx x x 1 證明有如下結論nnnf(pixi)，其中pi f(x) 0 f(x)為凹pi f(xi) nnnf(x0 f(x)凸pif(xi f(pixi)，其中pi (2p 1, i 1,2,1.f(x) x2f (x) 2 inx x x 22n1n1x )2x n n n由iin1xp 1, i