Question

高分求解题，如下，谢谢：

Answer 1

23、
z'x=2xe^y+y^3cosx
z''x=-2e^y-y^3sinx

24、
∫[0,π/2]x(sinx+cosx)dx
=∫[0,π/2]xsinxdx+∫[0,π/2]xcosxdx
=-∫[0,π/2]xdcosx+∫[0,π/2]xdsinx
=(xsinx-xcosx)[0,π/2]+∫[0,π/2]cosxdx-∫[0,π/2]sinxdx
=π/2+(sinx+cosx)[0,π/2]
=π/2
∫[1,e]lnx/xdx

=∫[1,e]lnxdlnx
=1/2(lnx)^2[1,e]
=1/2

Answer 2

23、dz/dx=2xe^y+y³cosx
d²z/dx²=2e^y-y³sinx
24、（1）原式=∫(0,π/2) xd(sinx-cosx)=π/2-∫(0,π/2) (sinx-cosx)dx=π/2-(0)=π/2
（2）、原式=∫(1,e) lnxdlnx=(1/2)(lnx)²|(1,e)=1/2

追问

能解一下吗？