Question

求解这几题详细步骤！第一二三题！谢谢！

Answer 1

(1)原式=∫<1,2>dx∫<1/x,x>(x^2/y^2)dy
=∫<1,2>dx*[-x^2/y]|<1/x,x>
=∫<1,2>(x^3-x)dx
=[(1/4)x^4-(1/2)x^2]|<1,2>
=4-1-1/4+1/2
=13/4.

(2)∫<0,π>xsinxdx=(-xcosx+sinx)|<0,π>=π，
∫<0,π>(sinx)^3dx=-∫<0,π>[1-(cosx)^2]d(cosx)
=-[cosx-(1/3)(cosx)^3]|<0,π>
=2-2/3=4/3,
原式=∫<0,π>dx∫<0,sinx>(x-y^2)dy
=∫<0,π>dx[xy-(1/3)y^3]|<0,sinx>
=∫<0,π>[xsinx-(1/3)(sinx)^3]dx
=π-4/9.

（3）原式=∫<0,1/2>dx∫<-√(2x),√(2x)>xy^2dy
=(1/3)∫<0,1/2>dx[xy^3]|<-√(2x),√(2x)>
=(4√2/3)∫<0,1/2>x^(5/2)dx
=(4√2/3）(2/7)x^(7/2)|<0,1/2>
=(8√2/21）（1/2）^(7/2)
=1/21.
解2 原式=∫<-1,1>dy∫<y^2/2,1/2>xy^2dx
=∫<-1,1>y^2(1/8-y^4/8>dy
=1/12-1/28
=1/21.