Question

高数打钩题

Answer 1

∫[0,2]∫[0,1]（x-y）dydx
=∫[0,2]（xy-1/2y^2）[0,1]dx
=∫[0,2](x-1/2)dx
=1
∫[0,π]∫[x,π]xcos（x+y）dydx
=∫[0,π]xsin（x+y）[x,π]dx
=-∫[0,π]x(sin2x+sinx)dx
=-[∫[0,π]xsin2xdx+∫[0,π]xsinxdx]
=-(-xcosx+sinx][0,π]+1/2[-cos2x+1/2sin2x][0,π])
=-(π-π/2)
=-π/2