∫[-1,0]dy∫[2,1-y]f(x,y)dx变换积分次序.如图第三题

变换积分次序∫(0,1)dy∫(-y,1+y^2)f(x,y)dx 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 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 ∫[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? 交换积分次序∫(1,2)dx∫(x,x^2)f(x,y)dy+∫(2,4)dx∫(x,4)dxf(x,y)dy y=f[(x-1)/(x+1)],f'(x)=arctanx^2,求dy/dx,dy 交换积分次序：∫(0,1/2)dx∫(x,1-x)f(x,y)dy= ∫(上1下0）dx∫（上x下x^2）f(x,y)dy=? ∫(0→1)dx∫(0→1-x)f(x,y)dy=? ∫[0,1] dx∫[0,x]f(x,y)dy= ? ∫（上限1,下限0）dy∫（上限y下限0）f(x,y)dx+∫（上限2,下限1）dy∫（上限2-y,下限0）f(x,y)dx交换耳机积分的次序 交换累次积分的次序∫（0>1) dy∫(0>2y) f(x,y)dx +∫(1>3) dy∫(0>3-y) f(x,y)dx 写下大概思路, 设f(x)= ∫0-x e^(-y+2y)dy 求∫0-1 [(1-x)^2]f(x)dx ∫[-1,0]dy∫[2,1-y]f(x,y)dx变换积分次序.如图第三题 更换积分∫（0,1）dx∫(1,1+x)f(x,y)dy+∫(1,2)dx∫(x,2)f(x,y)dy的积分顺序 函数f（x）在闭区间[0,1]连续,求证：∫（0-1）dx∫（x-1）f（x）f（y）dy=1∕ 2[∫（0-1）f（x）dx]2（帮忙解释一下第一个等号后面的等式是怎么得来的?）∫(0,1)dx∫(x,1)f(x)f(y)dy=∫(0,1)dy∫(0,y)f(x)f(y 更换积分次序∫(e,1)dx(lnx,0)f(x,y)dy