A. $\int_{0}^{1}dy\int_{\sqrt{y}}^{2-y}f(x,y)dx$
B. $\int_{0}^{2}dy\int_{x^{2}}^{2-x}f(x,y)dx$
C. $\int_{0}^{1}dy\int_{0}^{2}f(x,y)dx$
D. $\int_{0}^{1}dy\int_{x^{2}}^{2-x}f(x,y)dx$
1.交换积分次序：int_(1)^2dxint_((1)/(x))^xf(x,y)dy.1.交换积分次序：$\int_{1}^{2}dx\int_{\frac{
1.积分int_(0)^1dxint_(x)^1e^-y^(2)dy的值是____1.积分$\int_{0}^{1}dx\int_{x}^{1}e^{-y^{2
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_
267 累次积分 int_(-(pi)/(2))^(pi)/(2)dxint_(0)^sin x(x^2+ycosx)sqrt(1-y^2)dy=A. $\fr
（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）
【例7.16】已知函数f(t)=int_(1)^t^(2)dxint_(sqrt(x))^tsin(x)/(y)dy,则f((pi)/(2))=____.【例7
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__
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
24.证明: (int )_(0)^1dx(int )_(0)^xf(x-y)dy=(int )_(0)^1f(x)(1-x)dx-|||-"f(x-y)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