解:因为(x+y)f(x)在R×R上连续,所以 F'(y)=Jo f(a)dar+(y+y)f(y) ∫f(x)dr+2yf(y F(y)=Jo odr+f(9)+2f(y)+2yf'(y) 3f(y)+2yf(y) 3.F(y)=bmv2+y2dx,直接计算积分求F(y,F(,用定理4计 算F(0),并验证结果正确性 解:当y≠0时, T2+y212=0-S √1+y2-(1 )d ln+y2-1+00(c+)d ln√1+y2-1+y F(0)=J alo-Jo da 0-limr-0rInr-1=-1 1 ( lim r In c= lim 所以 0 F y1+y2-1+ y arctan,y≠0 这里强调指出:利用左、右导数的定义,我们可计算 + arctan - F(y)-F(0 n(1+y2 2 故F()不存在设f(x,y)=ln(x2+y2),则f(x,y)=2y,f和f二 者均在[0,1-;-∞,+∞]上不连续,不满足定理4的条件。如果死套公式计算 F(0)=f(x,y)y=dr=Jd=0就会出现错误。即使是左右导数也不 能套用定理4中求导公式计算。因此在计算前必须先验证定理4的条件!"&# (x + y)f(x) Æ R × R  $' F
(y) =
y 0 f(x)dx + (y + y)f(y) =
y 0 f(x)dx + 2yf(y). F

(y) =
y 0 0dx + f(y)+2f(y)+2yf
(y) = 3f(y)+2yf(y). 3. F(y) =
1 0 ln x2 + y2dx, )&'( F(y), F
(0)*)* 4 ' ( F
(0), ++,,--./ !"0 y
= 0  F(y) = xlnx2 + y2| x=1 x=0 −
1 0 x2 x2+y2 dx = ln 1 + y2 −
1 0 (1 − 1 x2+y2 )dx = ln 1 + y2 − 1 + y
1 0 ( 1 ( x y )2+1 )d x y = ln 1 + y2 − 1 + y arctan 1 y . F(0) =
1 0 ln xdx = x ln x| 1 0 −
1 0 dx = 0 − limx→0 x ln x − 1 = −1. (∵ limx→0 x ln x = limx→0 ln x 1 x = limx→0 1 x −1 x2 = 0) $' F(y) = ⎧ ⎨ ⎩ −1, y = 0, ln1 + y2 − 1 + y arctan 1 y , y
= 0. .123/4"5*0 167)289'( F
+(0) = lim y→+0 F(y) − F(0) y = lim y→+0[ ln(1 + y2) 2y + arctan 1 y ] = π 2 ; F
−(0) = lim y→−0 F(y) − F(0) y = lim y→−0 [ ln(1 + y2) 2y + arctan 1 y ] = −π 2 . : F
(0) ;Æ< f(x, y) = ln(x2 + y2), 3 fy(x, y) = 2y x2+y2 , f = fy > 4?Æ [0, 1; −∞, +∞] ; ;@5)* 4 A-BCDE'( F
(0) =
1 0 fy(x, y)|y=0dx =
1 0 dx = 0 FG4HIJKL01676; MC*)* 4 ( 6DE'(&NÆ'(OPQR+,)* 4  2