∫(0,1)dy∫(0,2y)f(x,)-搜答案-智慧作业帮

原式=∫dy∫e^(-y²/2)dx(作积分顺序变换)=∫(1-y²)e^(-y²/2)dy=∫e^(-y²/2)dy-∫y²e^(-y²/

原式=∫(-1,0)dx∫(-x,1)f(x,y)dy+∫(0,1)dx∫(0,1)f(x,y)dy+∫(1,2)dx∫(√(x-1),1)f(x,y)dy.

题目应该是e^(-y^2)交换积分次序：=∫(0,1)dy∫(0,y)e^(-y^2)dx=∫(0,1)ye^(-y^2)dy=1/2*∫(0,1)e^(-y^2)dy^2=1/2*(1-1/e)

∵根据积分上下限作图分析知,此积分区域是由直线y=x,x+y=2和y=0围城的三角形.∴∫(1,0)dx∫(x,0)f(x,y)dy+∫(2,1)dx∫(2-x,0)f(x,y)dy=∫(1,0)dy

两边对x求导2xy+x^2y'-(1+y^2)^(1/2)*y'=0前面两项是对于原方程的第一项运用积法则+链式法则得来的整理可得y'=2xy/[(1+y^2)^(1/2)-x^2]

A=y*e^y-e^y-y^2/2|(1,0)=1/2

∫[2+y/27-y^(1/3)]dy=2∫dy+(1/27)∫ydy-∫y^(1/3)dy=2y+(1/27)[y²/2]-[y^(4/3)/(4/3)]=2(27)+(1/27)(27&

=∫(0,1)dx∫(x,2-x)f(x,y)dy

∫[0,1]dx∫[-x^2,1]f(x,y)dy=∫[-1,0]dy∫[(-y)^(1/2),1]f(x,y)dx+∫[0,1]dy∫[0,1]f(x,y)dx

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∫(y^2→y)siny/ydx=[siny/yx]|(y^2→y)=(y-y^2)siny/y这里是把siny/y看成常数来积分再问：为什么可以看做常数？再答：因为这里x,y是两个自变量，互不相关，

∫（1-2x）dx/x²=∫（1/x²-2/x）dx=-1/x-2lnx+c

答：∫1/(y-y^2)dy=∫{1/[y(1-y)]}dy=∫[1/y+1/(1-y)]dy=∫1/ydy+∫1/(1-y)dy=In│y│-In│1-y│+C=In│y/(1-y)│+CC为常数.

1.确定积分区域对本题而言,即{(x,y):0

∫dy∫f(x,y)dx+∫dy∫f(x,y)dx=∫dx∫f(x,y)dy(作图分析约).再问：==求图。。求更详细过程再答：

y=∫(t-1)^3(t-2)dt,dy/dx=(x-1)^3(x-2).