[例5] 设函数f(x,y)连续,则 (int )_(1)^2dx(int )_(x)^2f(x,y)dy+(int )_(1)^2dy(int )_(y)^4-yf(x,y)dx 等于 ()-|||-(A) (int )_(1)^2dx(int )_(1)^4-xf(x,y)dy (B) (int )_(1)^2dx(int )_(x)^4-xf(x,y)dy-|||-(C) (int )_(1)^2dy(int )_(1)^4-yf(x,y)dx (D) (int )_(1)^2dy(int )_(y)^2f(x,y)dx

设 iint_(D) f(x, y), dx , dy = int_(0)^1 dx int_(x)^2x f(x, y), dy，其中 f(x, y) 是连续

五、设曲线积分(int )_(L)x(y)^2dx+yf(x)dy与路径无关,其中(int )_(L)x(y)^2dx+yf(x)dy具有连续导数,且(int

已知曲线积分(int )_(t)^yf(x)dx+((x)^2+y)dy与路径无关，则(int )_(t)^yf(x)dx+((x)^2+y)dy_____.已

在下列积分中改变累次积分的顺序:-|||-(1) (int )_(0)^2dx(int )_({x)^2x}f(x,y)dy;-|||-(2) (int )_(

[例5] 积分 (int )_(0)^2dx(int )_(0)^sqrt (2x-{x^2)}sqrt ({x)^2+(y)^2}dy= __

(int )_(0)^1dx(int )_(x)^1(e)^-(y^2)dy= () .-|||-

设 iint_(D) f(x, y)dx dy = int_(0)^1 dx int_(0)^1-x f(x, y)dy，则改变其积分次序后为A. $\int_

24.证明: (int )_(0)^1dx(int )_(0)^xf(x-y)dy=(int )_(0)^1f(x)(1-x)dx-|||-"f(x-y)dy=

[题目]设连续函数f(x)满足 (x)=(x)^2-(int )_(0)^2f(x)dx ,则-|||-(int )_(0)^2f(x)dx=

计算(int )_(0)^1dx(int )_(1-x)^sqrt (1-{x^2)}dfrac (x+y)({x)^2+(y)^2}dy=-|||-dv=__