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

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

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 )_(

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

12.化下列二次积分为极坐标形式的二次积分:-|||-(1) (int )_(0)^1dx(int )_(0)^1f(x,y)dy;-|||-(2) (int

（请画图➕解答过程）3.交换积分次序:-|||-(1) (int )_(0)^1dx(int )_(x)^sqrt (x)f(x,y)dy;（请画图➕解答过程）

11.计算下列二重(二次)积分.-|||-(1) (int )_(0)^1dx(int )_(x)^3x(x-y)dy --|||-(2) iint x(e)^

把下列积分化为极坐标形式, 并计算积分值: (3)(int )_(0)^1dx(int )_({x)^2}(({x)^2+(y)^2)}^-df

二重积分(int )_(0)^1dx(int )_((x-1))^2f(x,y)dy,交换积分次序的结果是__________.二重积分交换积分次序的结果是__

计算下列二重积分:-|||-(int )_(0)^1(x)^5dx(int )_({x)^2}^1(e)^-(y^2)dy

[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