∫dx∫f(x,y)dy 0

∫dx∫f(x,y)dy 0 交换积分次序∫(1,0)dx∫(x,0)f(x,y)dy+∫(2,1)dx∫(2-x,0)f(x,y)dy 证明 ∫[0,a]dx∫[0,x]f(y)dy=∫[0,a](a-x)f(x)dx 变换积分次序∫(0,1)dy∫(-y,1+y^2)f(x,y)dx 交换积分次序∫(0,1)dy∫(0,y)f(x,y)dx+∫(1,2)dy∫(0,2-y)dxf(x,y)dx 交换积分次序 ∫(4,0)dx∫(x,2x^0.5)f(x,y)dy 更换积分次序∫(0,2)dx∫(x,3x)f(x,y)dy ∫(0→1)dx∫(0→1-x)f(x,y)dy=? ∫[0,1] dx∫[0,x]f(x,y)dy= ? 设F(X,Y)是连续函数,则∫(a,0)dx∫(x,0) f(x,y)dy= ∫[0,1] dx∫[-x^2,1] f(x,y)dy+∫[1,e] dx∫[lnx,1] f(x,y)dy交换积分次序∫[0,1] dy∫[0,1] f(x,y)dx=∫[0,1] x| [0,1]dy= ∫[0,1] dy=∫y| [0,1]=1? ∫[0,1] dx∫[-x^2,1] f(x,y)dy+∫[1,e] dx∫[lnx,1] f(x,y)dy交换积分次序∫[0,1] dy∫[0,1] f(x,y)dx=∫[0,1] x| [0,1]dy= ∫[0,1] dy=∫y| [0,1]=1? 更换积分次序∫(e,1)dx(lnx,0)f(x,y)dy 更换积分次序∫(e,1)dx(lnx,0)f(x,y)dy 二重积分计算：∫[0,a]dx∫[0,x] f ´(y)/√[(a-x)(x-y)] dy 交换积分次序,∫(上限2,下限0)dy∫(上限2y,下限y^2)f(x,y)dx 交换积分顺序后∫(0→1)dy∫(y→√y)f(x,y)dx=? 改变二次积分∫[0,2]dx∫[x,2x]f(x,y)dy的积分次序 改变二次积分∫[0,2]dy∫[y^2,2y]f(x,y)dx的积分次序①改变二次积分∫[0,2]dx∫[x,2x]f(x,y)dy的积分次序 ②改变二次积分∫[0,2]dy∫[y^2,2y]f(x,y)dx的积分次